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math104-s21:s:martinzhai

Martin Zhai's Review Note

Content Summary

Week 1

Lecture 1 (Jan 19) - Covered Ross Section 1.1, 1.2, 1.3

  • Natural Numbers(N\natnums) {1,2,3,}\{ 1, 2, 3, … \}
  • Integers(Z\Z) {,2,1,0,1,2,}\{…, -2, -1, 0, 1, 2, … \} (an example of Ring structure)
  • Rational Numbers(Q\mathbb{Q}) {nm,n,mZ,m0}\{\frac{n}{m}, n,m\isin \Z, m\neq0\}
    • Proposition: if rQr\isin\mathbb{Q} with gcd(c,d)=1, and is a root of Cnxn+Cn1xn1++C0C\scriptstyle{n} \normalsize {x^n}+ C\scriptstyle{n-1} \normalsize {x^{n-1}}+ … +C\scriptstyle{0}, C0/=0,Cn/=0C\scriptstyle{0} \normalsize \, \mathrlap{\,/}{=} \, 0, C\scriptstyle{n} \normalsize \, \mathrlap{\,/}{=}\,0, then d divides CnC\scriptstyle{n}, c divides C0C\scriptstyle{0}
    • Corollary: if r=cd/=0r\,=\,\frac{c}{d}\,\mathrlap{\,/}{=}\,0 is a root of a “monic polynomial”, i.e. leading term has coefficient 1, then r is an integer.

Lecture 2 (Jan 21) - Covered Ross Section 1.4

  • Maximum and Minimum: let SRS\, \subset\, \Reals and SøS\, \neq\, \text{\o}, we say αS\alpha\, \isin\, S is a maximum if βS,αβ\forall\beta\, \isin\, S,\, \alpha\, \geq\, \beta. Similarly, αS\alpha\, \isin\, S is a minimum if βS,αβ\forall\beta\, \isin\, S,\, \alpha\, \leq\, \beta.
    • Remark: both maximum and minimum are elements in a set, and they are not guaranteed to exist.
    • Examples: S=[2,100]RS\, =\, [\sqrt{2},100]\, \subset\, \Reals, max(S)=100,min(S)=2\max(S)\, =\, 100,\, \min(S)\, =\, \sqrt{2}
    • Non-Examples: S=[2,100]QS\, =\, [\sqrt{2},100]\, \subset\, \mathbb{Q}, max(S)=100,min(S)\max(S)\, =\, 100,\, \min(S) does not exist (since 2/Q\sqrt{2}\, \mathrlap{\,/}{\isin}\, \mathbb{Q} and Q\mathbb{Q} is dense)
  • Upper and Lower Bounds: let øSR\text{\o}\, \neq\, S\, \subset\, \Reals
    1. αR\alpha\, \isin\, \Reals, we say α\alpha is an upper bound of S, if βS,βα\forall\, \beta\, \isin\, S,\, \beta\, \leq\, \alpha
      • Examples: Define S=x210x+24,xRS={x^2-10x+24,\, x \isin \Reals}, 1010 would be an upper bound for SS, since for x100x\geq100, then x210x+240{x^2}-10x+24 \geq 0, hence x/Sx \mathrlap{\,/}{\isin}S.
      • Non-Examples: With the same set S defined as above, 5 is not an upper bound, since consider 656\, \geq\, 5 and x210x+24{x^2}-10x+24 evaluated at x=6x\, =\, 6 is 0, i.e. 6S6\, \isin\, S.
    2. αR\alpha\, \isin\, \Reals, we say α\alpha is a lower bound of S, if βS,βα\forall\, \beta\, \isin\, S,\, \beta\, \geq\, \alpha
      • Examples: Define S=[5,10]ZS\, =\, [5,10]\cup\Z, both 1 and 4 would be a lower bound for S, since for xS,x4x\, \isin\, S,\, x\, \geq\, 4 and x1x\, \geq\, 1.
      • Non-Examples: Consider S=RS\, =\, \Reals, there is no such lower bound for R\Reals, since for any number ss, we could always find an element αS\alpha\, \isin\, S and αs\alpha\, \leq\, s.
    • Remark: upper bounds and lower bounds are not unique, and may not exist.
  • Supremum and Infimum: let øSR\text{\o}\, \neq\, S\, \subset\, \Reals, i.e. SS be a non-empty subset of R\Reals
    1. if SS is bounded above (i.e. there exists an upper bound for SS), and SS has a least upper bound, then it is called supremum, denoted as sup(S)\sup(S)
      • Examples: S=[0,5]ZS\, =\, [0,\sqrt{5}]\cup\Z, then sup(S)=5\sup(S)\, =\, \sqrt{5} regardless of 5/S\sqrt{5}\, \mathrlap{\,/}{\isin}\, S.
      • Non-Examples: S=ZS\, =\, \Z, then SS does not have a supremum since SS is not bounded above, hence no such least upper bound exist.
    2. if SS is bounded below (i.e. there exists a lower bound for SS), and SS has a greatest lower bound, then it is called infimum, denoted as inf(S)\inf(S)
      • Examples: S=NS\, =\, \natnums, then inf(S)=1\inf(S)\, =\, 1.
      • Non-Examples: S=QS\, =\, \mathbb{Q}, then there is no infimum for SS.
  • Completeness Axiom: Every non-empty subset SRS\, \subset\, \Reals that is bounded above has a least upper bound, i.e. a supremum.
    • Corollary: if SS is bounded from below, then inf(S)\inf(S) exists.
  • Archimedean Property: If a>0a\, >\, 0, b>0b\, >\, 0 are real numbers, then for some nNn\, \isin\, \natnums, we have na>bn*a\, >\, b.

Homework 1

  • if STS\, \subset\, T, then inf(T)inf(S)sup(S)sup(T)\inf(T)\, \leq\, \inf(S)\, \leq\, \sup(S)\, \leq\, \sup(T)
  • sup(ST)=max(sup(S),sup(T))\sup({S}\cup{T})\, =\, \max(\sup(S),\, \sup(T))
  • sup(A+B)=sup(A)+sup(B)\sup(A+B)\, =\, \sup(A)\, +\, \sup(B)
  • inf(A+B)=inf(A)+inf(B)\inf(A+B)\, =\, \inf(A)\, +\, \inf(B)
  • if ab+1n,nNa\, \leq\, b+\frac{1}{n},\, \forall\, n\, \isin\, \natnums, then aba\, \leq\, b

Week 2

Lecture 3 (Jan 26) - Covered Ross Section 2.7, 2.9

  • Sequence: a1,a2,a3,,an,Ra\scriptstyle{1}\normalsize,\, a\scriptstyle{2}\normalsize,\, a\scriptstyle{3}\normalsize,\, …,\, a\scriptstyle{n}\normalsize,\,\isin\, \Reals, could be denoted as (an)nN(a\scriptstyle{n}\normalsize)\scriptstyle{n\isin\natnums}
    • Remark: sequence is not a set where sequence considers the order of all its elements while set only record what element is in it.
  • Limit of Sequence: we say a sequence (an)n(a\scriptstyle{n}\normalsize)\scriptstyle{n} has limit αR\alpha\, \isin\, \Reals, if ϵ>0\forall\, \epsilon>0, there exists a real number N>0N>0 such that for all integer n>Nn>N, we have anα<ϵ\lvert\, a\scriptstyle{n}\normalsize\, -\, \alpha\, \rvert\, <\, \epsilon. We denote this by limnan=α\lim_{n\to\infty}a_{n}=\alpha
    • Example: Prove limn1n=0\lim_{n\to\infty}\frac{1}{n}=0
      • Fix a ϵ>0\epsilon>0, we take N=1ϵN=\frac{1}{\epsilon}, then n>N    n>1ϵ    1n<ϵ    1n0<ϵn>N \iff n>\frac{1}{\epsilon} \iff \frac{1}{n}<\epsilon \iff \lvert \frac{1}{n}-0 \rvert\, <\epsilon, completing the proof.
    • Non-Examples: Determine if limnn=5\lim_{n\to\infty}n=5
      • Consider for all ϵ>0\epsilon>0. In order to have an5<ϵ\lvert a_{n}-5 \rvert <\epsilon, then an<5+ϵa_{n} < 5 + \epsilon, but since nn\rightarrow\infty, it is impossible to find a NN satisfying n>N,an<5+ϵ\forall n>N,\, a_{n} <5+\epsilon, hence the claim is incorrect.
  • Properties and Tools to find Limit:
    • Bounded sequence: (an)n(a_{n})_{n} is a bounded sequence if M>0\exists M>0 such that ManM,nN-M\leq a_{n} \leq M,\, \forall n \isin \natnums
      • Examples: Consider an=1n,nNa_{n} = \frac{1}{n},\, \forall n \isin \natnums, then it is a bounded sequence where we could take M=1M=1 such that anM=1nNa_{n} \leq M=1\, \forall n\isin \natnums.
      • Non-Examples: Define an=n,nNa_{n} = n,\, \forall n \isin \natnums. For this sequence, there is no such MM to bound the elements in the sequence since for arbitrary MM, it is always possible to find a ana_{n} such that an>Ma_{n} > M (ana_{n} tends to \infty as nn \rightarrow \infty).
    1. Theorem 9.1: All convergent sequences are bounded.
    2. Theorem 9.2: If liman=α\lim a_{n} = \alpha, and if kRk\isin \Reals, then lim(kan)=kα\lim (k*a_{n})= k* \alpha.
    3. Theorem 9.3, 9.4, 9.6: Let an,bna_{n},\, b_{n} be convergent sequences with liman=α\lim a_{n} =\alpha and limbn=β\lim b_{n} =\beta, then:
      1. lim(an+bn)=α+β\lim (a_{n}+b_{n}) = \alpha + \beta
      2. lim(anbn)=αβ\lim (a_{n}*b_{n}) = \alpha * \beta
      3. if an0  na_{n} \neq 0\,\, \forall n and α0\alpha \neq 0, then lim(bnan)=βα\lim (\frac{b \scriptscriptstyle n \normalsize}{a \scriptscriptstyle n \normalsize}) = \frac{\beta}{\alpha}

Lecture 4 (Jan 28) - Covered Ross Section 2.9, 2.10

  • Properties and Tools to find Limit:
    1. Theorem 9.7:
      1. limn1np=0,p>0\lim_{n\to\infty} \frac{1}{n^p} = 0,\, \forall p > 0
      2. limnan=0,\lim_{n\to\infty} {a^n} = 0, if a<1\lvert a \rvert < 1
      3. limnn1n=1\lim_{n\to\infty} {n^\frac{1}{n}} = 1
      4. limna1n=1\lim_{n\to\infty} {a^\frac{1}{n}} = 1 for a>0a>0
    • Limit of Infinity: we say limsn=+\lim s_n = +\infty provided for each M>0M>0, N\exists N such that n>N    sn>Mn>N \implies s_n >M; similarly limsn=\lim s_n = -\infty provided for each M>0M>0, N\exists N such that n>N    sn<Mn>N \implies s_n <M
      • Example: lim(n+7)=+\lim (\sqrt{n} + 7)= +\infty
        • Fix a M>0M>0, define N=(M7)2N = (M-7)^2. Then n>N    n>(M7)2    n>M7    n+7>Mn>N \implies n>(M-7)^2 \implies \sqrt{n} > M-7 \implies \sqrt{n} +7>M, completing the proof.
      • Non-Example: Determine if lim1n=+\lim \frac{1}{n}= +\infty
        • No. Consider M=1M = 1, in order to have 1n>1\frac{1}{n} > 1, nn must be smaller than 11. Thus it is not possible to define a NN such that n>N    1n>1\forall n>N \implies \frac{1}{n} > 1.
      1. Theorem 9.9: Let (sn),(tn)(s_n),(t_n) be sequences such that limsn=+,limtn>0\lim s_n = +\infty,\, \lim t_n > 0, then lim(sntn)=+\lim (s_n*t_n) = +\infty.
      2. Theorem 9.10: Let a sequence (sn)(s_n) of positive real numbers, then limsn=+    lim1sn=0\lim s_n = +\infty \iff \lim \frac{1}{s_n} = 0.
  • Monotone Sequence and Cauchy Sequence:
    • Cauchy Sequence: (an)(a_n) is a Cauchy sequence if ϵ>0,N>0\forall \epsilon >0,\, \exists N>0 such that n1,n2>N\forall n_1,n_2>N, we have an1an2<ϵ\lvert a_{n_1} - a_{n_2} \rvert < \epsilon i.e. oscillation amplitude gets smaller and smaller.
      • Example: an=1n2nNa_n = \frac{1}{n^2}\, \forall n \isin \natnums is a Cauchy sequence.
        • Fix a ϵ>0\epsilon >0. Since we already know from Theorem 9.7a), lim1n2=0\lim \frac{1}{n^2} = 0, then we could find a N>0N>0 such that n>N    1n2<ϵn>N \implies \frac{1}{n^2} < \epsilon. Thus take the same NN, we get a,b>N\forall a,b >N, 1a21b21a2<ϵ\lvert \frac{1}{a^2} - \frac{1}{b^2} \rvert \leq \lvert \frac{1}{a^2} \rvert < \epsilon, completing the proof.
      • Non-Example: Consider an=na_n = n, then if we fix ϵ=0.5\epsilon = 0.5, there is no such N>0N>0 such that n1,n2>N    an1an2<0.5n_1, n_2>N \implies \lvert a_{n_1} - a_{n_2} \rvert < 0.5 since the smallest difference between any ana_n is 1, thus not a Cauchy sequence.
    • Monotone Sequence:
      1. A monotone increasing sequence is such that n>m,anam\forall n > m,\, a_{n} \geq a_m.
      2. A monotone decreasing sequence is such that n>m,anam\forall n > m,\, a_{n} \leq a_m.
        • Example: an=11na_n = 1 - \frac{1}{n} is a monotone increasing sequence, since n>m,1n<1m\forall n > m,\, \frac{1}{n} < \frac{1}{m}, hence 11n>11m1 - \frac{1}{n} > 1 - \frac{1}{m}.
        • Non-Example: an=(n2)2a_n = (n-2)^2 is not a monotone sequence.
          • Consider n=2,m=1,n>mn = 2,\, m = 1,\, n>m, but a2=0<a1=1a_2 = 0 < a_1 = 1, hence not monotone increasing.
          • Then consider n=2,m=3,n<mn = 2,\, m = 3,\, n<m, but a2=0<a3=1a_2 = 0 < a_3 = 1, hence not monotone decreasing.
    1. Theorem 10.2: All bounded monotone sequences are convergent.
    2. Theorem 10.11: Let (an)(a_n) be a sequence. Then (an)(a_n) is Cauchy     (an)\iff (a_n) converges.

Homework 2

  • 9.9c): If there exists N0N_0 such that sntnn>N0s_n \leq t_n\, \forall n > N_0,and limsn\lim s_n and limtn\lim t_n exists, then limsnlimtn\lim s_n \leq \lim t_n.

Week 3

Lecture 5 (Feb 2) - Covered Ross Section 2.10

  • Monotone and Cauchy Sequences:
    • lim sup\limsup: Let (an)(a_n) be a sequence in R\Reals, lim supan=limN(supn>Nan)\limsup a_n = \lim_{N\to\infty}(\sup_{n>N} {a_n})
    • lim inf\liminf: Let (an)(a_n) be a sequence in R\Reals, lim infan=limN(infn>Nan)\liminf a_n = \lim_{N\to\infty}(\inf_{n>N} {a_n})
    • Example:(an)=0(a_n) = 0 if n2=0\frac{n}{2} = 0, (an)=1n(a_n) = \frac{1}{n} if n2=1\frac{n}{2} = 1.
      • lim supan=0\limsup a_n = 0 since the supremum of the subsequence (an)n>N(a_n)_{n>N} for all NN is 1N+1\frac{1}{N+1} if NN is even and 1N\frac{1}{N} if NN is odd, which converges to 00 as NN \rightarrow \infty.
      • lim infan=0\liminf a_n = 0 since the infimum of the subsequence (an)n>N(a_n)_{n>N} for all NN is 00 (there is always an even number N\geq N)
      • Property of Monotone Sequences: If it is bounded, then its limit exists; if it is unbounded, then liman=+/\lim a_n = +/- \infty.
      • Lemma: If (an)(a_n) is a bounded sequence and α+=lim supan\alpha_+ = \limsup a_n, then for any ϵ>0,N\epsilon >0,\, \exists N such that n>N\forall n>N, we have anα++ϵa_n \leq \alpha _+ + \epsilon
      • Theorem 10.7: Let (an)(a_n) be a bounded sequence, then liman\lim a_n exists     lim supan=lim infan\iff \limsup a_n = \liminf a_n.

Lecture 6 (Feb 4) - Covered Ross Section 2.11

  • Subsequences:
    • Subsequence: Let (sn)nN(s_n)_{n\isin \natnums} be a sequence of real numbers. Given a list of indices, n1<n2<<nk<n_1 < n_2 < … <n_k< …. Take tk=snkt_k = s_{n_k}, then (tm)(t_m) is called a subsequence of (sn)(s_n), we write (snk)k(s_{n_k})_k for the subsequence.
      • Example: Let (sn)(s_n) denote some sequence with real numbers, define (tm)(t_m) as tm=s2mt_m = s_{2*m}. Then (tm)(t_m) is a subsequence of (sn)(s_n) (could be denoted as (s2n)n(s_{2n})_n)
    • Theorem 11.2: Let (sn)(s_n) be a sequence.\\ If tRt\isin \Reals, then there is a subsequence of (sn)(s_n) converging to t    t \iff the set {nN:snt<ϵ}\{n\isin \natnums :\, \lvert s_n - t\rvert < \epsilon\} is infinite for all ϵ>0\epsilon >0.
    • Theorem 11.3: If (sn)(s_n) is convergent, then any subsequence converges to the same point.
    • Theorem 11.4: Every sequence has a monotonic subsequence.

Homework 3

  • (sn)(s_n) a bounded sequence, then lim infsnlim supsn\liminf s_n \leq \limsup s_n and lim supsn=inf{supnNsn:nN}\limsup s_n = \inf \{\sup_{n\geq N} s_n:\, n\isin \natnums\}.
  • If (an),(bn)(a_n), (b_n) are two bounded sequences, then lim sup(an+bn)lim sup(an)+lim sup(bn)\limsup (a_n+b_n) \leq \limsup (a_n) + \limsup (b_n).

Week 4

Lecture 7 (Feb 9) - Covered Ross Section 2.11

  • Subsequences
    • Theorem 11.5 (Bolzano-Weiestrass Theorem): Every bounded sequence has a convergent subsequence.
    • Subsequential Limit: Let (sn)(s_n) be a sequence in R\Reals, a subsequential limit is any real number or +/+/- \infty that is the limit of a subsequence of (sn)(s_n).
      • Example: Let (rn)(r_n) be the enumeration of Q\mathbb{Q}, then rR\forall r \isin \Reals is a subsequential limit of (rn)(r_n). This is because the denseness of Q\mathbb{Q} in R\Reals, which means for any rR,ϵ>0,(rϵ,r+ϵ)r\isin \Reals,\, \forall \epsilon >0,\, (r - \epsilon, r+\epsilon) is an infinite set. Hence by Theorem 11.2, there is some subsequence of (sn)(s_n) that converges to such rr, i.e. a subsequential limit.
    • Theorem 11.7: Let (sn)(s_n) be any sequence. Then there exists a monotonic subsequence that converges to lim supsn\limsup s_n and a monotonic subsequence that converges to lim infsn\liminf s_n.
    • Theorem 11.8: Let (sn)(s_n) be a sequence, SS be the set of subsequential limits of (sn)(s_n), then
      • S is non-empty
      • supS=lim supsn\sup S = \limsup s_n, and infS=lim infsn\inf S = \liminf s_n
      • S={α}    limsn=αS=\{ \alpha \} \iff \lim s_n = \alpha
    • Theorem 11.9: Let SS be the set of subsequential limits of (sn)(s_n). Suppose (tn)(t_n) is a sequence in SRS\cap \Reals and t=limtnt = \lim t_n. Then tSt \isin S.

Lecture 8 (Feb 11) - Covered Ross Section 2.12

  • lim sup's and lim inf's:
    • Recall that in general, lim supsnlim infsn\limsup s_n \geq \liminf s_n, and if they are equivalent, then limsn\lim s_n exists.
    • Theorem 12.1: Let (sn)(s_n) be a sequence that converges to a positive real number ss, and (tn)(t_n) be any sequence. Then lim sup(sntn)=slim suptn\limsup (s_n*t_n) = s*\limsup t_n. (Here we allow notation of s(+)=+s*(+\infty) = +\infty and s()=s*(-\infty) = -\infty for s>0s>0).
    • Theorem 12.2: Let (sn)(s_n) be a sequence of positive real numbers, then we have\\ lim inf(sn+1sn)lim inf(sn)1nlim sup(sn)1nlim sup(sn+1sn)\enspace \liminf (\frac{s_{n+1}}{s_n}) \leq \liminf (s_n)^{\frac{1}{n}} \leq \limsup (s_n)^{\frac{1}{n}} \leq \limsup (\frac{s_{n+1}}{s_n}).

Homework 4

  • (sn)(s_n) bounded     lim supsn<+\iff \limsup \lvert s_n \rvert < +\infty
  • Let (sn)(s_n) be a bounded sequence in R\Reals. If A={aR:A = \{ a \isin \Reals: only finitely many sn<a}s_n <a\} and B={bR:B = \{ b \isin \Reals: only finitely many sn>b}s_n >b\}, then supA=lim infsn\sup A = \liminf s_n and infB=lim supsninf B = \limsup s_n.

Week 5

Lecture 9 (Feb 16) - Covered Ross Section 2.13

  • Metric Space & Topology:
    • Metric Space: A metric space is a set SS together with a distance function d:S×SRd: S \times S \rightarrow \Reals such that
      1. d(x,y)0,d(x,y) \geq 0, and d(x,y)=0    x=yd(x,y) = 0 \iff x=y
      2. d(x,y)=d(y,x)d(x,y) = d(y,x)
      3. d(x,y)+d(y,z)d(x,z)d(x,y) + d(y,z) \geq d(x,z) (Triangle Inequality)
      • Example: Let d(x,y)=xyd(x,y) = \sqrt{\lvert x-y \rvert}, then with the set R\Reals, it forms a metric space. The first two criteria is obvious. For the triangle inequality, (xy+yz)2=xy+yz+2xyyz(\sqrt{\lvert x-y \rvert} + \sqrt{\lvert y-z \rvert})^2 = \lvert x-y \rvert + \lvert y-z \rvert + 2\sqrt{\lvert x-y \rvert} \sqrt{\lvert y-z \rvert}. By Euclidean distance metric, we know xy+yzxz\lvert x-y \rvert + \lvert y-z \rvert \geq \lvert x-z \rvert. Also 2xyyz02 \sqrt{\lvert x-y \rvert} \sqrt{\lvert y-z \rvert} \geq 0. Therefore d(x,y)=xyd(x,y) = \sqrt{\lvert x-y \rvert} is a metric, hence with R\Reals is a metric space.
      • Non-Example: Let d(x,y)=(xy)2d(x,y) = (x-y)^2. It is not a metric since consider x=2,y=1,z=0x=2, y = 1, z=0, then d(x,y)=1d(x,y) = 1 and d(y,z)=1d(y,z)=1 but d(x,z)=4d(x,z) = 4 which means d(x,y)+d(y,z)<d(x,z)d(x,y) + d(y,z) < d(x,z), thus not a metric.
    • Cauchy Sequence (in metric space (S,d)): A sequence (sn)(s_n) in SS is Cauchy if ϵ>0,N>0\forall \epsilon >0,\, \exists N >0 such that n,m>N,d(sn,sm)<ϵ\forall n,m >N,\, d(s_n,s_m) < \epsilon.
    • Convergence (in metric space (S,d)): A sequence (sn)(s_n) converge to a point sSs\isin S if ϵ>0,N>0\forall \epsilon >0,\exists N>0 such that d(sn,s)<ϵn>Nd(s_n,s)<\epsilon \enspace \forall n>N.
    • Completeness: A metric space (S,d)(S,d) is complete if every Cauchy sequence is convergent.
      • Example: The Euclidean k-space Rk\Reals^k is complete by Theorem 13.4.
      • Non-Example: Q\mathbb{Q} with the Euclidean distance function is not complete. Consider a sequence of rational numbers converging to 2\sqrt{2} (possible by example in Lecture 7). This sequence is clearly Cauchy but not convergent since 2/Q\sqrt{2} \mathrlap{\,/}{\isin} \mathbb{Q}. Thus Q\mathbb{Q} with Euclidean distance formula is not a complete metric space.
    • Induced Distance Function: If (S,d)(S,d) is a metric space and ASA \subset S is any subset, then (A,dA×A)(A, d|_{A\times A}) is a metric space.
    • Theorem 13.5 (Bolzano-Weiestrass Theorem for Rn\Reals^n): Every bounded sequence (sm)mRn(s_m)_m \isin \Reals^n has a convergent subsequence.
    • Topology: Let SS be a set. A topological structure on SS is the data of a collection of subsets in S. This collection needs to satisfy:
      1. SS and ø\text{\o} are open
      2. arbitrary union of open subsets is still open
      3. finite intersections of open sets are open

Lecture 10 (Feb 18) - Midterm 1

Homework 5

  • Every open subset of R\Reals is the disjoint union of finite or countably infinite sequence of open intervals.

Week 6

Lecture 11 (Feb 23) - Covered Ross Section 2.13, Rudin Chapter 2

  • Topology of Metric Space:
    • Basic Notions: Let ESE \subset S
      • Interior point: pEp\isin E is an interior point of E if δ>0\exist \delta>0 such that Bδ(p)={qSd(p,q)<δ}EB_{\delta}(p) = \{q \isin S| d(p,q)< \delta\} \subset E.
        • Example: Consider E=[0,1]RRE = [0,1] \cap \Reals \subset \Reals, then p=0.5p=0.5 is an interior point of EE. If we take δ=0.3\delta = 0.3, then B0.3(0.5)=[0.2,0.8]REB_{0.3}(0.5) = [0.2, 0.8]\cup \Reals \subset E.
        • Non-Example: Take the same E as above, consider p=0p=0. P is not an interior point since for any δ0\delta \neq 0, the points in R\Reals on the left side of 00 is not in EE, hence not an interior point.
      • Interior: The set of all interior points of E, denote as EoE^o.
    • Open: ESE\subset S is an open subset of SS if E=EoE=E^o, i.e. pE,δ>0\forall p \isin E,\, \exists \delta >0 such that Bδ(p)EB_{\delta}(p) \subset E.
      • Example: Take E=(0,1)RRE = (0,1) \cap \Reals \subset \Reals, EE is a subset of R\Reals. For any point pEp \isin E, let δo=max(d(p,0),d(p,1))\delta_o = \max (d(p,0), d(p,1)). Then we could choose δ=δo2\delta = \frac{\delta_o}{2}, then Bδ(p)EB_{\delta}(p) \subset E for arbitrary point pEp \isin E, hence EE is an open subset of R\Reals.
      • Non-Example: Take E=[0,1]RRE = [0,1] \cap \Reals \subset \Reals. As the non-example for interior points shown, 00 is not an interior point in EE, hence EE is not an open subset of R\Reals.
    • Close: ESE\subset S is a closed subset of S if the complement Ec=SEE^c = S \setminus E is open.
      • Propositions:
        1. SS and ø\text{\o} are closed
        2. An arbitrary collection of closed sets is closed
        3. A finite intersection of closed sets is closed
    • Limit Point: Let ESE\subset S, pSp\isin S is a limit point of E    δ>0,Bδ(p)E \iff \forall \delta >0,\, B_{\delta}(p) intersects EE non-empty, i.e. qE,qp,d(p,q)<δ\exists q\isin E,\, q \neq p,\, d(p,q)<\delta. E=E' = set of limit points of EE.
      • Example: Let S=R,E={1n:nN}S=\Reals,\, E = \{\frac{1}{n}:n\isin \natnums\}. Then 00 is a limit point since for any δ\delta, we could always find a N>0N>0 such that 1n<δ,n>N\frac{1}{n} < \delta, \forall n>N, hence Bδ(0)EøB_{\delta}(0) \cap E \neq \text{\o}.
      • Non-Example: Consider the same scenario as the previous example. 12\frac{1}{2} is not a limit point if we take δ=110\delta = \frac{1}{10}, then B110(12)=[410,610]E=øB_{\frac{1}{10}}(\frac{1}{2}) = [\frac{4}{10}, \frac{6}{10}] \cap E = \text{\o}.
    • Closure: ESE\subset S, the closure of EE is the intersection of all closed subsets containing EE. Denote as EE^-.
      • Proposition: E=EEE^- = E \cup E'
      • Example: Let S=R,E={1n:nN}S=\Reals,\, E = \{\frac{1}{n}:n\isin \natnums\}. Then E={1n:nN}{0}E^- = \{\frac{1}{n}:n\isin \natnums\} \cup \{0\} by proposition above ({0}=E\{0\} = E')
    • Boundary: The boundary points of ESE\subset S is the set EEoE^-\setminus E^o.
    • Proposition 13.9: Let EE be a subset of a metric space (S,d)(S,d)
      1. The set EE is closed if and only if E=EE=E^-
      2. The set EE is closed if and only if it contains the limit of every convergent sequence of points in EE
      3. An element is in EE^- if and only if it is the limit of some sequence of points in EE
      4. A point is in the boundary of EE if and only if it belongs to the closure of both EE and its complement
    • Isolated Point: If pEp \isin E and pp is not a limit point of EE, then pp is called an isolated point.
    • Perfect: EE is perfect if EE is closed and every point of EE is a limit point of EE.
    • Dense: EE is dense in SS if every point of SS is a limit point of EE or a point of EE or both.
    • Rudin 2.30: Suppose YXY\subset X. A subset EE of YY is open relative to YY if and only if E=YGE = Y\cap G for some open subset GG of XX.
  • Compact Set:
    • Open Cover: Let (S,d)(S,d) be a metric space, ESE\subset S, {Gα}\{ G_{\alpha} \} is a collection of open sets. We say {Gα}\{ G_{\alpha} \} is an open cover of EE if EαGαE \subset \cup_{\alpha} G_{\alpha}.
    • Compact Set: KSK\subset S is a compact subset, if for any open cover of KK, there exists a finite subcover, i.e. if {Gα}\{ G_{\alpha} \} is an open cover, then α1,,αn\alpha_1, …, \alpha_n indices such that KGα1GαnK\subset G_{\alpha_1} \cup … \cup G_{\alpha_n}.
      • Example: K=[0,1]K=[0,1] is a compact subset of R\Reals.
      • Non-Example: K=(0,1]K=(0,1] is not a compact subset of R\Reals. Consider the open cover Gn=B12n(1n),nN,G_n = B_{\frac{1}{2n}}(\frac{1}{n}), n\isin \natnums, then KnNK \subset \cup_{n\isin \natnums}. In this case, there is no finite subcover. If we take indices n1<n2<<nmn_1 < n_2 < … < n_m, then x<12nmx < \frac{1}{2n_m} is not in the union.
    • Sequentially Compact: ESE\subset S is sequentially compact if any sequence in EE has a convergent subsequence in EE (the limit point is also in EE).
    • Theorem: For any metric space (S,d)(S,d), ESE\subset S, EE compact     E\iff E sequentially compact.
    • Theorem 13.12 (Heine-Borel Theorem): Consider Rn\Reals^n with Euclidean metric d(x,y)=xyd(x,y)= \lvert x-y \rvert, ERnE\subset \Reals^n is compact     E\iff E is closed and bounded.
    • Rudin 2.33: KYXK\subset Y\subset X, then KK is compact relative to YY if and only if KK is compact relative to XX.
    • Rudin 2.34: Compact subsets of metric spaces are closed.
    • Rudin 2.35: Closed subsets of compact sets are compact.

Lecture 12 (Feb 25) - Covered Ross Section 2.14, 2.15

  • Series:
    • Infinite Sum: An infinite sum of sequence (an)(a_n) is defined as a1+a2+=n=1ana_1 + a_2 + … = \sum_{n=1}^{\infty} a_n.
    • Partial Sum: Defined as a1+a2++an=i=1naia_1 + a_2 + … + a_n = \sum_{i=1}^{n} a_i.
    • Convergence: A series converge to α\alpha if the corresponding partial sum converges to α\alpha.
    • Cauchy Condition for Series Convergence: ϵ>0,N>0\forall \epsilon>0,\, \exists N>0 such that n,m>N,i=n+1mai<ϵ\forall n,m>N,\, \lvert \sum_{i=n+1}^{m} a_i \rvert < \epsilon.
    • Absolute Convergence: If an<\sum \lvert a_n \rvert < \infty, we say an\sum a_n converges absolutely.
    • Recall Geometric Series: n=0arn\sum_{n=0}^{\infty} ar^n converges to a11ra\frac{1}{1-r} if r<1\lvert r \rvert <1.
  • Tests for Series Convergence:
    • Comparison Test:
      • Suppose nan<,an>0\sum_{n} a_n <\infty,\, a_n >0, and bnR<anb_n \isin \Reals < a_n, then nbn<\sum_{n} b_n < \infty.
      • Suppose nan=,an>0\sum_{n} a_n =\infty,\, a_n >0, and bnanb_n \geq a_n, then nbn=\sum_{n} b_n = \infty.
    • Ratio Test (Test for Absolute Convergence):
      • If lim supan+1an<1\limsup \lvert \frac{a_{n+1}}{a_n} \rvert <1, then nan\sum_{n} \lvert a_n \rvert converges.
      • If lim infan+1an>1\liminf \lvert \frac{a_{n+1}}{a_n} \rvert >1, then nan\sum_{n} \lvert a_n \rvert diverges.
      • Otherwise the test does not give any information on its convergence.
    • Root Test: Let nan\sum_{n} a_n be a series, α=lim sup(an)1n\alpha = \limsup (\lvert a_n \rvert)^{\frac{1}{n}}, then nan\sum_{n} a_n:
      • converges absolutely if α<1\alpha <1.
      • diverges if α>1\alpha >1.
      • gives no information if α=1\alpha = 1.
    • Alternating Series Test: Let a1a2a_1 \geq a_2 \geq … be a monotone decreasing series, an0a_n \geq 0. And assuming liman=0\lim a_n = 0. Then n=1(1)n+1an=a1a2+a3\sum_{n=1}^{\infty} (-1)^{n+1}a_n = a_1 - a_2 + a_3 - … converges. Moreover, the partial sums sn=k=1n(1)k+1aks_n = \sum_{k=1}^{n} (-1)^{k+1}a_k satisfy ssnan\lvert s- s_n \rvert \leq a_n for all nn.
    • Integral Test: If the terms ana_n in nan\sum_{n} a_n are non-negative and f(n)=anf(n) = a_n is a decreasing function on [1,)[1, \infty), then let α=limn1nf(x)dx\alpha = \lim_{n\to\infty} \int_{1}^{n} f(x)dx
      • If α=\alpha = \infty, then the series diverge
      • If α<\alpha < \infty, then the series converge

Homework 6

  • Rudin 2.36: If {Kα}\{K_{\alpha}\} is a collection of compact subsets of a metric space such that the intersection of every finite subcollection of {Kα}\{K_{\alpha}\} is non-empty, then Kα\cap K_{\alpha} is non-empty. (This statement is incorrect if compact is replaced by closed or bounded).
  • If an>0a_n >0 and an\sum a_n diverges, then an1+an\sum \frac{a_n}{1+a_n} also diverges.
  • If an>0a_n >0 and an\sum a_n converges, then ann\sum \frac{\sqrt{a_n}}{n} also converges.

Week 7

Lecture 13 (Mar 2) - Covered Rudin Chapter 4

  • Continuous Functions:
    • Function: A function from set AA to set BB is an assignment for each element αA\alpha \isin A an element f(α)Bf(\alpha) \isin B.
      1. Injective: A function ff is injective/one-to-one if x,yA,xy\forall x,y\isin A,\, x \neq y, then f(x)f(y)f(x) \neq f(y)
      2. Surjective: A function ff is surjective/onto if βB,\beta \isin B, there exists at least one element αA\alpha \isin A such that f(α)=βf(\alpha)=\beta
      3. Bijective: A function ff is bijective if ff is both injective and surjective
      • Example: f:RR,f(x)=xf: \Reals \rightarrow \Reals,\, f(x)=x is a bijective function since if f(x)=f(y)f(x) = f(y), then x=yx = y. And for any yRy \isin \Reals, we take x=yRx=y \isin \Reals as the domain will prove its surjectivity. Hence a bijection.
      • Non-Example:
        1. f:RR,f(x)=x2f: \Reals \rightarrow \Reals,\, f(x)=x^2 is not injective since for x=1x=1 and y=1y=-1, f(x)=f(y)f(x) = f(y) even if xyx\neq y.
        2. f:RR,f(x)=1xf: \Reals \rightarrow \Reals,\, f(x)=\frac{1}{x} is not surjective since for y=0Ry = 0 \isin \Reals, there is no element in domain that get mapped to 00 under such ff.
    • Pre-image: If f:ABf: A\rightarrow B. Given a subset EBE\subset B, f1(E)={αAf(α)E}f^{-1}(E) = \{\alpha \isin A|\, f(\alpha) \isin E\} is called the pre-image of EE under ff, which is a subset of AA.
      • Example: f:RR,f(x)=lnxf: \Reals \rightarrow \Reals,\, f(x)=\ln x and let E=[0,)E=[0,\infty). Then pre-image of EE under ff is [1,)[1,\infty).
    • Limit of a Function: Suppose pEp\isin E'(set of limit points of EE), we write f(x)q(Y)f(x) \rightarrow q(\isin Y) as xpx \rightarrow p or limxpf(x)=q\lim_{x\to p} f(x) = q if ϵ>0,δ>0\forall \epsilon >0,\, \exists \delta >0 such that xE,0<dX(x,p)<δ    dY(f(x),q)<ϵ\forall x \isin E,\, 0<d_X(x,p)<\delta \implies d_Y(f(x),q)<\epsilon.
      • Example: Consider f:RRf: \Reals \rightarrow \Reals f(x)=x21x1f(x)=\frac{x^2-1}{x-1}. Claim $\lim_{x\to 1} f(x) = 2.
        • Fix a ϵ>0\epsilon >0, consider δ=ϵ1\delta = \lvert \epsilon - 1\rvert. Using the Euclidean distance formula, x1<δ=ϵ1    ϵ+1<x1<ϵ1    ϵ+3<x+1<ϵ+1    ϵ+3<(x+1)(x1)x1<ϵ+1    ϵ+2<(x+1)(x1)x11<ϵ    (x+1)(x1)x11<ϵ\lvert x - 1\rvert < \delta = \lvert \epsilon - 1\rvert \implies -\epsilon + 1 < x - 1< \epsilon -1 \implies -\epsilon + 3 < x + 1< \epsilon + 1 \implies -\epsilon + 3 < \frac{(x + 1)(x-1)}{x-1}< \epsilon + 1 \implies -\epsilon + 2 < \frac{(x + 1)(x-1)}{x-1} - 1< \epsilon \implies \lvert \frac{(x + 1)(x-1)}{x-1} - 1 \rvert < \epsilon, hence proving the claim.
    • Rudin 4.2: With the same notation as above, limxpf(x)=q\lim_{x\to p} f(x) = q if and only iff limnf(pn)=q\lim_{n\to\infty} f(p_n) = q for every sequence (pn)(p_n) in EE such that pnp,limnpn=pp_n \neq p, \lim_{n\to\infty} p_n = p.
      • Example: f:RRf:\Reals \rightarrow \Reals and f(x)=x2f(x)=x^2. limx0f(x)=0\lim_{x\to 0} f(x) = 0 and then consider sequence pn=1np_n = \frac{1}{n} which converges to 00, and f(pn)=1n2f(p_n) = \frac{1}{n^2} which also converges to 00 when nn \rightarrow \infty
    • Corollary: If ff has a limit at point pp, then it is unique.
    • Rudin 4.4: Suppose f,g:ERf,g: E \rightarrow \Reals, suppose pE;p\isin E; and limxpf(x)=A,limxpg(x)=B\lim_{x\to p} f(x) = A, \lim_{x\to p} g(x) = B, then
      • limxpf(x)+g(x)=A+B\lim_{x\to p} f(x) +g(x)= A+B
      • limxpf(x)g(x)=AB\lim_{x\to p} f(x)g(x) = AB
      • limxpf(x)g(x)=AB\lim_{x\to p} \frac{f(x)}{g(x)} = \frac{A}{B} if B0B\neq 0 and g(x)0xEg(x)\neq 0 \, \forall x\isin E
      • cR\forall c\isin \Reals, limxpcf(x)=cA\lim_{x\to p} c*f(x)=cA
  • Continuity of Functions:
    • Continuity at a Point: Let (X,dX),(Y,dY)(X,d_X), (Y,d_Y) be metric spaces, EXE\subset X, f:EYf:E\rightarrow Y, pEp \isin E, q=f(p)q=f(p). We say ff is continuous at pp, if ϵ>0,δ>0\forall \epsilon>0, \exists \delta>0 such taht xE\forall x\isin E with dX(x,p)<δ    dY(f(x),q)<ϵd_X(x,p) <\delta \implies d_Y(f(x),q)<\epsilon.
    • Rudin 4.6: If pEp\isin E is also a limit point of EE, then ff is continuous at p    limxpf(x)=f(p)p \iff \lim_{x\to p} f(x) = f(p).
    • Continuity: We say ff is continuous on EE if ff is continuous at every point in EE.
    • Rudin 4.7: (X,dX),(Y,dY)(X,d_X), (Y,d_Y), f:XYf:X\rightarrow Y as above. Then ff is continuous     \iff for every open set VYV\subset Y, f1(V)f^{-1}(V) is open in XX.
    • Lemma: If f:ABf: A\rightarrow B is a function and EA,FBE\subset A, F\subset B. The f(E)=F    Ef1(F)f(E)=F \iff E\subset f^{-1}(F).
    • Rudin 4.7: Let X,Y,ZX,Y,Z be metric spaces and f:XYf:X\rightarrow Y and g:YZg:Y\rightarrow Z continuous functions. We define h:XZh:X\rightarrow Z by h(x)=g(f(x)h(x)=g(f(x). Then hh is also continuous (hh is called the composition of ff and gg).
    • Rudin 4.9_: If f,g:XRf,g:X\rightarrow \Reals continuous, then f+g,fg,fgf+g,\, f-g,\, fg are continuous functions, and if g(x)0g(x)\neq 0 for any xXx\isin X, then fg\frac{f}{g} is also continuous.
      • Result: All polynomials are continuous since f(x)=xf(x)=x is continuous.
    • Rudin 4.10: Let f:XRnf:X\rightarrow \Reals^n with f(x)=(f1(x),f2(x),,fn(x))f(x) = (f_1(x), f_2(x), …, f_n(x)). Then ff is continuous     \iff each fif_i is continuous.

Lecture 14 (Mar 4) - Covered Rudin Chapter 4

  • Review of Compact Sets:
    • Compactness: KXK\subset X is compact if \forall open cover of KK, \exists a finite subcover.
    • Propositions:
      • KK compact     K\implies K bounded
      • KK compact     K\implies K closed
      • EXE\subset X is closed, KK is compact, EK    EE\subset K \implies E is compact.
    • Theorems:
      • Compactness     \iff Sequential Compactness
      • Heine-Borel (Rudin 2.41): in Rn\Reals^n, KK compact     K\iff K closed and bounded.
    • Remark: The notion of “compact” is intrinsic, while open and closed depends on the ambient space.
  • Continuous Maps and Compactness:
    • Three Definitions of Continuous Maps:
      • ff is continuous if and only if pX,ϵ>0,δ>0\forall p\isin X, \forall \epsilon >0, \exists \delta >0 such that f(Bδ(p))Bϵ(f(p))f(B_{\delta}(p)) \subset B_{\epsilon}(f(p))
      • ff is continuous if and only if VY\forall V\subset Y open, f1(V)f^{-1}(V) is open
      • ff is continuous if and only if xnx\forall x_n \rightarrow x in XX, we have f(xn)f(x)f(x_n) \rightarrow f(x) in YY
    • Rudin 4.14: Suppose ff is a continuous map from a compact metric space XX to another compact metric space YY, then f(X)Yf(X) \subset Y is compact.
    • Rudin 4.16: Suppose ff is a continuous real function on a compact metric space XX, and M=suppXf(p)M = \sup_{p\isin X} f(p), m=infpXf(p)m=\inf_{p\isin X} f(p). Then there exists point p,qXp,q \isin X such that f(p)=Mf(p)=M and f(q)=mf(q)=m.
      • Recall: if KRK\subset \Reals, KK is compact, then sup(K)K\sup(K) \isin K and inf(K)K\inf(K) \isin K.
      • Remark: If f:XYf:X\rightarrow Y is continuous, ff sends compact set XX to compact set YY, but given EYE\subset Y compact, f1(E)f^{-1}(E) is not guaranteed to be compact.

Homework 7

  • If f:XYf:X\rightarrow Y is a continuous function from a metric space XX to a metric space YY, f(E)f(E)f(E^-)\subset {f(E)}^- for any EXE\subset X.
  • Let f,gf,g be continuous maps from XX to YY. Suppose there is a dense set EXE\subset X such that fE=gEf|_E=g|_E, then f=gf=g.

Week 8

Lecture 15 (Mar 9) - Covered Ross Section 3.19 Rudin Chapter 2 and 4

  • Uniform Continuity:
    • Uniform Continuous Function: f:XYf:X\rightarrow Y. Suppose for all ϵ>0\epsilon >0, δ>0\exists \delta >0 such that p,qX\forall p,q\isin X with dX(p,q)<δd_X(p,q)<\delta, we have dY(f(p),f(q))<ϵd_Y(f(p),f(q)) <\epsilon. Then we say ff is a uniform continuous function.
      • Example: f:[0,1]Rf:[0,1] \rightarrow \Reals and f(x)=x2f(x)=x^2. Then ff is uniformly continuous function. ϵ>0\forall \epsilon >0, we can take δ=ϵ2\delta=\frac{\epsilon}{2}, then p,q[0,1]\forall p,q \isin [0,1], pq<δ\lvert p-q \rvert < \delta we have p2q2=(pq)(p+q)=(pq)(p+q)<δ2=ϵ\lvert p^2-q^2 \rvert = \lvert (p-q)(p+q) \rvert = \lvert (p-q) \rvert \lvert (p+q) \rvert < \delta * 2 = \epsilon.
      • Non-Example 1: f:RRf:\Reals \rightarrow \Reals and f(x)=x2f(x)=x^2. Then ff is not uniformly continuous. δ>0\delta > 0 , we could always find p,qRp,q\isin \Reals and pq<δ\lvert p-q \rvert < \delta such that f(p)f(q)=1\lvert f(p) - f(q) \rvert = 1. If we take pq=δ2p-q = \frac{\delta}{2} and p+q=2δp+q = \frac{2}{\delta}, then f(p)f(q)=p2q2=(pq)(p+q)=(pq)(p+q)=δ22δ=1\lvert f(p) - f(q)\rvert = \lvert p^2-q^2 \rvert = \lvert (p-q)(p+q) \rvert = \lvert (p-q) \rvert \lvert (p+q) \rvert = \frac{\delta}{2} * \frac{2}{\delta} = 1.
      • Non-Example 2: f:(0,)Rf:(0, \infty) \rightarrow \Reals, f(x)=1xf(x)=\frac{1}{x} is not uniformly continuous. Intuition: when x0x\rightarrow 0, then distance between two points p,qp,q may be close enough but 1p1q\lvert \frac{1}{p} - \frac{1}{q} \rvert may be large.
    • Theorem: Suppose f:XYf:X\rightarrow Y is a continuous function between metric spaces. If XX is compact, then ff is uniformly continuous.
      • Theorem 19.2: If ff is continuous on a closed interval [a,b][a,b], then ff is uniformly continuous on [a,b][a,b].
    • Proposition: If f:XYf:X\rightarrow Y is uniformly continuous and SXS\subset X subset with induced metric, then the restriction fS:SYf|_S:S\rightarrow Y is uniformly continuous.
  • Connectedness:
    • Connected: Let XX be a set. We say XX is connected if SX\forall S\subset X and SS is both open and closed, then SS has to be either XX or ø\text{\o}.
      • Non-Example (From Midterm 2): (0,2)Q(0,2)\cap \mathbb{Q} is not connected. Consider the set (0,2)Q(0,\sqrt{2}) \cap \mathbb{Q} and (2,2)Q(\sqrt{2}, 2) \cap \mathbb{Q}. Both open and closed but none of those two are ø\text{\o}.
    • Proposition: XX is connected     \iff if X=UVX = U\sqcup V and U&VU\&V are both open, then one of U,VU,V is empty set.
    • Rudin 4.22: If f:XYf:X\rightarrow Y is continuous, if EXE\subset X is connected, then f(E)f(E) is connected. (Continuity preserves connectedness)
    • Proposition: [0,1]R[0,1] \subset \Reals is a connected subset.

Lecture 16 (Mar 11) - Covered Rudin Chapter 2 and 4

  • Review:
    • Induced Topology: Given a topological space XX, and SXS\subset X, we endow SS with the induced toplogy: USU\subset S is open in SS if and only if VX\exists V \subset X open in XX such that U=VSU = V\cap S.
      • Example: X=RX= \Reals and S=ZRS= \Z \subset \Reals, with induced topology on SS, nZ\forall n \isin \Z, {n}\{ n \} is open in SS since {n}=(n12,n+12)S\{ n \} = (n-\frac{1}{2}, n+\frac{1}{2}) \cap S
    • Corollary:
      • If SXS\subset X is open in XX, then USU\subset S is open in SS if and only if UU is open in XX.
      • If SXS\subset X is closed in XX, then ESE\subset S is closed in SS if and only if EE is closed in XX.
  • Connectedness:
    • Lemma: EE is connected if and only if EE cannot be written as ABA\cup B when AB=øA^- \cap B = \text{\o} and AB=øA\cap B^- = \text{\o} (closure taken with respect to ambient space XX).
    • Rudin 4.27: ERE\subset \Reals is connected    x,yE,x<y\iff \forall x,y\isin E, x<y, we have [x,y]E[x,y]\subset E
    • Rudin 4.23: Let ff be a continuous real function on the interval [a,b][a,b]. If f(a)<f(b)f(a)<f(b) and if cc is a number such that f(a)<c<f(b)f(a)<c<f(b), then there exists a point x[a,b]x\isin [a,b] such that f(x)=cf(x)=c.
  • Discontinuities
    • Discontinuous: f:XYf:X\rightarrow Y is discontinuous at xXx\isin X if and only if xx is a limit point of XX and limxpf(q)\lim_{x\to p} f(q) either does not exist or f(x)\neq f(x).
    • Right and Left Limit: Let f:(a,b)Rf:(a,b)\rightarrow \Reals (not necessarily continuous)
      • x[a,b)\forall x\isin [a,b), we say f(x+)=qf(x+) = q if for all sequence (tn)(t_n) in (x,b)(x,b) that converge to xx, we have limit limnf(tn)=q\lim_n f(t_n) = q.
        • Example: f(x)=1f(x) = 1 if x0x\geq 0, f(x)=0f(x)=0 if x<0x < 0. Then f(x+)=1f(x+) =1 .
      • x(a,b]\forall x\isin (a,b], we say f(x)=qf(x-) = q if for all sequence (tn)(t_n) in (a,x)(a,x) that converge to xx, we have limit limnf(tn)=q\lim_n f(t_n) = q.
        • Example: f(x)=1f(x) = 1 if x0x\geq 0, f(x)=0f(x)=0 if x<0x < 0. Then f(x+)=1f(x+) =1.
    • Discontinuity of First and Second Kind: f:(a,b)Rf:(a,b)\rightarrow \Reals, x(a,b)x\isin (a,b). Suppose ff is discontinuous at xx
      • We say ff has a simple discontinuity or discontinuity of the first kind at xox_o if both f(xo+)f(x_o+) and f(xo)f(x_o-) exists.
        • Example: f(x)=1nf(x) = \frac{1}{n} if xQ,x=mnx\isin \mathbb{Q}, x= \frac{m}{n} with m,nm,n coprime, f(x)=0f(x) =0 otherwise. xQ\forall x\isin \mathbb{Q}, we have a simple discontinuity.
      • We say ff has a discontinuity of second kind, if it is not a simple discontinuity.
        • Example: f(x)=sin1x,x>0f(x) = \sin \frac{1}{x}, x>0, f(x)=x,x0f(x) = x, x\leq 0. Since f(0+)f(0+) does not exist, thus ff has a discontinuity of second kind at x=0x=0.

Homework 8

  • If KRnK\subset \Reals^n is compact and CRnC\subset \Reals^n is closed, then K+CK+C is closed.

Week 9

Lecture 17 (Mar 16) - Covered Rudin Chapter 4 and 7

  • Monotonic Functions:
    • Monotonic Functions: A function f:(a,b)Rf:(a,b)\rightarrow \Reals is monotone increasing if x>y\forall x>y, we have f(x)f(y)f(x) \geq f(y). Similarly one can define monotone decreasing functions.
      • Example: f(x)=log(x)f(x) = \log(x) is a monotone increasing function, g(x)=1g(x) = 1 is both a monotone increasing function and a monotone decreasing function.
      • Non-Example: f(x)=x2+2x+1f(x) = x^2+2x+1.
    • Rudin 4.29: Suppose f:(a,b)Rf:(a,b)\rightarrow \Reals is a monotone increasing function, then x(a,b)\forall x\isin (a,b), the left limit f(x)f(x-) and the right limit f(x+)f(x+) exists, satisfying sup{f(t)t<x}=f(x)f(x+)=inf{f(t)t>x}\sup\{f(t)|\, t<x\} = f(x-) \leq f(x+) = \inf\{f(t)|\, t>x\}; and given x<yx<y in (a,b)(a,b), then f(x+)f(y)f(x+) \leq f(y-).
    • Corollary: If ff is monotone, then f(x)f(x) only has discontinuity of the first kind/simple discontinuity.
    • Rudin 4.30: If ff is monotone, then there are at most countably many discontinuities.
  • Sequence and Convergence of Functions:
    • Pointwise Convergence of Sequence of Sequences: Let (xn)n(x_n)_n be a sequence of sequences, xnRNx_n\isin \Reals^{\natnums}, we say (xn)n(x_n)_n converges to xRNx\isin \Reals^{\natnums} pointwise if iN\forall i\isin \natnums, we have limnxni=xi\lim_{n\to\infty} x_{ni} = x_i.
      • Example: xni=in+ix_{ni} = \frac{i}{n+i}, then this seq to 00 pointwise, since for arbitrary fixed ii, we have limnxni=limnin+i=0\lim_{n\to\infty} x_{ni} = \lim_{n\to\infty} \frac{i}{n+i} = 0.
    • Uniform Convergence of Sequence of Sequences: Let (xn)n(x_n)_n be a sequence of sequences, xnRNx_n\isin \Reals^{\natnums}, we say xnxx_n \rightarrow x uniformly if ϵ>0\forall \epsilon >0, N>0\exists N>0 such that n>N\forall n>N, sup{xnixi:iN}<ϵ\sup\{\lvert x_{ni} - x_i \rvert :\, i\isin\natnums\} <\epsilon (also known as d(xn,x)d_{\infty}(x_n, x)).
      • Non-Example: xni=in+ix_{ni} = \frac{i}{n+i} failed to converge uniformly to 00, since d(xn,0)=sup{xni:iN}=sup{in+i:iN}=1d_{\infty}(x_n,0) = \sup \{\lvert x_{ni} \rvert :\, i\isin\natnums\} = \sup \{\frac{i}{n+i}:\, i\isin\natnums\} = 1 (n is fixed).
    • Pointwise Convergence of Sequence of Functions: Given a sequence of functions fnf_n \isin Map(R,R)(\Reals, \Reals), we say fnf_n converge to ff pointwise if xR\forall x\isin\Reals, limnfn(x)=f(x)    limnfn(x)f(x)=0\lim_{n\to\infty} f_n(x) = f(x) \iff \lim_{n\to\infty} \lvert f_n(x) - f(x) \rvert = 0.
      • Examples:Running and Shrinking Bumps.

Lecture 18 (Mar 18) - Covered Rudin Chapter 7

  • Uniform Convergence:
    • Uniform Convergence of Sequence of Functions: Given a sequence of functions (fn):XY(f_n): X\rightarrow Y, is said to converge uniformly to f:XYf:X \rightarrow Y, if ϵ>0,N>0\forall \epsilon >0, \exists N>0 such that n>N,xX\forall n>N, \forall x\isin X, we have fn(x)f(x)<ϵ\lvert f_n(x) - f(x) \rvert <\epsilon.
      • Remark: The integer NN depends only on ϵ\epsilon in uniform convergence while NN could depend on both ϵ\epsilon and xx in pointwise convergence.
      • Example: fn(x)=sin(x)1+nx2f_n(x) = \frac{\sin(x)}{1+nx^2} converges uniformly. See Homework 9 Question 4 for the detailed proof.
    • Rudin 7.8: Suppose fn:XRf_n: X\rightarrow \Reals satisfies that ϵ>0,N>0\forall \epsilon >0,\exists N>0 such that xX,fn(x)fm(x)<ϵ\forall x\isin X, \lvert f_n(x) - f_m(x) \rvert < \epsilon, then fnf_n converges uniformly (Uniform Cauchy     \iff Uniform Convergence).
    • Rudin 7.9: Suppose fnff_n\rightarrow f pointwise, then fnff_n \rightarrow f uniformly     limn(supfn(x)f(x))=0\iff \lim_{n\to\infty} (\sup \lvert f_n(x) - f(x) \rvert) = 0.
    • A sequence of functions {fn}\{ f_n\} is uniformly convergent to f:DR    limnsup{fn(x)f(x):xD}f:D\to\Reals\iff \lim_{n\to\infty} \sup \{\lvert f_n(x) - f(x) \rvert : x\isin D\}.
    • Rudin 7.10 (Weiestrass M-Test): Suppose f(x)=n=1fn(x)xXf(x) = \sum_{n=1}^{\infty} f_n(x)\, \forall x\isin X. If Mn>0\exists M_n>0 such that supxfn(x)Mn\sup_x \lvert f_n(x) \rvert \leq M_n and nMn<\sum_{n} M_n < \infty, then the partial sum FN(x)=n=1Nfn(x)F_N(x)=\sum_{n=1}^{N} f_n(x) converges to f(x)f(x) uniformly.
      • Example: See last question on Midterm 2 version A.
  • Uniform Convergence and Continuity:
    • Rudin 7.11: Suppose fnff_n \rightarrow f uniformly on set EE in a metric space. of EE, and suppose that limtxfn(t)=An\lim_{t\to x} f_n(t) = A_n. Then {An}\{ A_n\} converges and limtxf(t)=limnAn\lim_{t\to x} f(t) = \lim_{n\to\infty} A_n. In conclusion, limtxlimnfn(t)=limnlimtxfn(t)\lim_{t\to x} \lim_{n\to\infty} f_n(t) = \lim_{n\to\infty} \lim_{t\to x} f_n(t).
    • Rudin 7.12: If {fn}\{f_n\} is a sequence of continuous functions on EE, and if fnff_n\rightarrow f uniformly on EE, then ff is continuous on EE.
    • Rudin 7.13: Suppose KK compact and
      1. {fn}\{f_n\} is a sequence of continuous functions on KK
      2. {fn}\{f_n\} converges pointwise to a continuous function f(x)f(x) on KK
      3. fn(x)fn+1(x)xK,n=1,2,f_n(x) \geq f_{n+1}(x) \forall x\isin K,\, \forall n = 1,2,…
      • Then fnff_n\rightarrow f uniformly on KK.

Homework 9

  • Let f:XRf:X\rightarrow \Reals be a function on a metric space. We say ff is Lipschitz Continuous if there exists a K>0K>0 such that for any x,yXx,y\isin X, we have f(x)f(y)Kd(x,y)\lvert f(x)-f(y) \rvert \leq K*d(x,y). Such KK is called a Lipschitz constant for ff.

Week 10

Spring Break (Mar 23, Mar 25)

Week 11

Lecture 19 (Mar 30) - Review for Midterm 2

Lecture 20 (Apr 1) - Midterm 2

Week 12

Lecture 21 (Apr 6) - Covered Rudin Chapter 5

  • Derivative:
    • Differentiability: Let f:[a,b]Rf:[a,b]\rightarrow \Reals be a real valued function. Define x[a.b]\forall x\isin [a.b], f(x)=limtxf(t)f(x)txf'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x}. If f(x)f'(x) exists, we say ff is differentiable at this point xx.
      • Example: f(x)=x2f(x)=x^2. For x=3x=3, define g(x)=f(x)f(3)x3=x29x3=x+3g(x) = \frac{f(x)-f(3)}{x-3} = \frac{x^2-9}{x-3} = x+3. Then f(3)=limx3g(x)=limx3x+3=6f'(3) = \lim_{x\to 3} g(x) = \lim_{x\to 3} x+3 = 6.
      • Non-Example: f(x)=xsin1xf(x) = x\sin \frac{1}{x} if x>0x>0 and f(x)=0f(x)=0 if x0x\leq 0, then f is not differentiable at x=0x=0. For x>0x>0, g(x)=f(x)f(0)x0=xsin1xx=sin1xg(x) = \frac{f(x) - f(0)}{x-0} = \frac{x\sin \frac{1}{x}}{x} = \sin \frac{1}{x}, and limx0+g(x)\lim_{x\to 0^+} g(x) does not exist, thus f(0)f'(0) does not exist as well.
    • Proposition: If f:[a,b]Rf:[a,b]\rightarrow \Reals is differentiable at xo[a,b]x_o\isin [a,b], then ff is continuous at xox_o, i.e. limxxof(x)=f(xo)\lim_{x\to x_o} f(x) = f(x_o).
    • Rudin 5.3: Let f,g:[a,b]Rf,g: [a,b]\rightarrow \Reals. Assume f,gf,g are differentiable at point xo[a,b]x_o\isin [a,b], then
      1. cR\forall c\isin \Reals, (cf)(xo)=cf(xo)(cf)'(x_o)=cf'(x_o)
      2. (f+g)(xo)=f(xo)+g(xo)(f+g)'(x_o) = f'(x_o)+g'(x_o)
      3. (fg)(xo)=f(xo)g(xo)+f(xo)g(xo)(fg)'(x_o)=f'(x_o)g(x_o)+f(x_o)g'(x_o) (Leibniz's Rule)
      4. if g(xo)0g(x_o)\neq 0, then (fg)(xo)=f(xo)g(xo)f(xo)g(xo)(g(xo))2(\frac{f}{g})'(x_o) = \frac{f'(x_o)g(x_o)-f(x_o)g'(x_o)}{(g(x_o))^2}
    • Rudin 5.5 (Chain Rule): Suppose f:[a,b]IRf:[a,b]\rightarrow I\subset \Reals and g:IRg: I\rightarrow \Reals. Suppose for some xo[a,b]x_o\isin [a,b], f(xo)=yof(x_o) = y_o, yoRy_o\isin \Reals, f(xo)f'(x_o) and g(yo)g'(y_o) exists. Then, the composition h=gf:[a,b]Rh=g \circ f:[a,b]\rightarrow \Reals. h(x)=g(f(x))h(x)=g(f(x)) is differentiable at xox_o, h(xo)=g(yo)f(xo)h'(x_o) = g'(y_o)f'(x_o).
      • Example: h(x)=sin(x2)h(x) = \sin(x^2), then by Chain rule, h(x)=2xcos(x2)h'(x) = 2x\cos(x^2)
  • Mean Value Theorem:
    • Local Maximum and Minimum: Let f:[a,b]Rf:[a,b]\rightarrow \Reals. We say ff has a local maximum at point p[a,b]p\isin [a,b], if δ>0\exists \delta >0 and x[a,b]Bδ(p),f(x)f(p)\forall x\isin[a,b]\cap B_{\delta}(p),\, f(x) \leq f(p). Local minimum is defined in the similar way.
      • Example: Say f:[π,π]Rf:[-\pi,\pi] \rightarrow \Reals and f(x)=sin(x)f(x)=\sin(x). Then we can say p=π2p= \frac{\pi}{2} is a local maximum with δ=π4\delta = \frac{\pi}{4}. And similarly, we can say q=0q=0 is a local minimum with δ=π2\delta = \frac{\pi}{2}.
    • Rudin 5.8: Let f:[a,b]Rf:[a,b]\rightarrow \Reals. If ff has a local maximum or minimum at p(a,b)p\isin (a,b), and if ff is differentiable at pp, then f(p)=0f'(p)=0.
      • Remark: The point pp cannot be taken at the endpoints of the domain.
    • Rolle's Theorem: Suppose f:[a,b]Rf:[a,b]\rightarrow \Reals is a continuous function and ff is differentiable in (a,b)(a,b). If f(a)=f(b)f(a)=f(b), then there is some d(a,b)d\isin (a,b) such that f(d)=0f'(d) = 0.

Lecture 22 (Apr 8) - Covered Rudin Chapter 5

  • Mean Value Theorem:
    • Rudin 5.9 (Mean Value Theorem): Let f,g:[a,b]Rf,g: [a,b]\rightarrow \Reals be continuous function differentiable on (a,b)(a,b). Then d(a,b)\exists d\isin (a,b) such that [f(a)f(b)]g(d)=[g(a)g(b)]f(d)[f(a)-f(b)]g'(d) = [g(a)-g(b)]f'(d).
    • Rudin 5.10: Let f:[a,b]Rf: [a,b]\rightarrow \Reals be continuous function differentiable on (a,b)(a,b). Then d(a,b)\exists d\isin (a,b) such that
      [f(b)f(a)]=[ba]f(d)[f(b)-f(a)] = [b-a]f'(d).
      • Remark : Mean Value Theorem relates slope at a point to the difference of values of the functions.
    • Corollary: Suppose f:[a,b]Rf: [a,b]\rightarrow \Reals be continuous function, f(x)f'(x) exists for all x(a,b)x\isin (a,b), and f(x)M\lvert f'(x) \rvert \leq M for some constant MM. Then ff is uniformly continuous.
    • Rudin 5.11: Suppose ff is differentiable in (a,b)(a,b), then
      1. If f(x)0f'(x) \geq 0 for all x(a,b)x\isin (a,b), then ff is monotonically increasing.
      2. If f(x)=0f'(x) = 0 for all x(a,b)x\isin (a,b), then ff is constant.
      3. If f(x)0f'(x) \leq 0 for all x(a,b)x\isin (a,b), then ff is monotonically decreasing.
  • Intermediate Value Theorem:
    • Rudin 5.12: Assume ff is differentiable over [a,b][a,b] with f(a)<f(b)f'(a)<f'(b). Then from each λ(f(a),f(b))\lambda \isin (f'(a),f'(b)), there exists a d(a,b)d\isin (a,b) such that f(d)=λf'(d)=\lambda.
  • L'Hospital's Rule:
    • Rudin 5.13: Suppose ff and gg are real and differentiable in (a,b)(a,b), and g(x)0g'(x)\neq 0 for all x(a,b)x\isin (a,b), where a<b-\infty \leq a<b\leq \infty. Suppose f(x)g(x)A\frac{f'(x)}{g'(x)} \to A as xax\to a. Then if f(x)0f(x) \to 0 and g(x)0g(x) \to 0 as xax\to a, or if g(x)+g(x) \to +\infty as xax\to a, then f(x)g(x)A\frac{f(x)}{g(x)}\to A as xax\to a.
      • Example: f(x)=sin(x)xf(x)=\frac{\sin(x)}{x}. limx0sin(x)=0\lim_{x\to 0} \sin(x) = 0 and limx0x=0\lim_{x\to 0} x= 0, then by L'Hospital's Rule, limx0f(x)=limx0cos(x)1=1\lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{\cos(x)}{1} = 1.
      • Another Example: f(x)=log(x)xf(x)=\frac{\log(x)}{x}. limxx=+\lim_{x\to\infty} x = +\infty. Thus by L'Hospital's Rule, limxf(x)=limx1x1=0\lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{\frac{1}{x}}{1} = 0.

Homework 10

  • Let f:[a,b]Rf:[a,b]\rightarrow \Reals be differentiable, then f(x)f'(x) cannot have any simple discontinuities.
  • Even if a sequence of differentiable functions converges uniformly to ff, ff is not guaranteed to be differentiable.

Week 13

Lecture 23 (Apr 13) - Covered Rudin Chapter 5

  • Higher Order Derivatives:
    • Definition: If f(x)f'(x) is differentiable at xox_o, then we define f(xo)=(f)(xo)f“(x_o)=(f')'(x_o). Similarly, if the (n-1)th derivative f(n1)f^{(n-1)} exists and differentiable at xox_o, we define f(n)(xo)=(f(n1))(xo)f^{(n)}(x_o) = (f^{(n-1)})'(x_o)
      • Example: f(x)=sin(x)f(x)=\sin(x). Then f(x)=cos(x)f'(x)=\cos(x), f(x)=sin(x)f”(x)=-\sin(x), f(3)(x)=cos(x)f^{(3)}(x)=-\cos(x), f(4)(x)=sin(x)f^{(4)}(x)=\sin(x), …
    • Smooth Function: f(x)f(x) is a smooth function on (a,b)(a,b) if x(a,b)\forall x\isin (a,b), k{1,2,}\forall k\isin \{1,2,…\}, f(k)(x)f^{(k)}(x) exists.
      • Example: f(x)=sin(x)f(x)=\sin(x); f(x)=e1x2f(x)=e^{\frac{-1}{x^2}}; smooth step function; smooth bump function.
  • Taylor Theorem:
    • Rudin 5.15: Suppose ff is a real function on [a,b][a,b], nn is a positive integer, f(n1)f^{(n-1)} is continuous on [a,b][a,b], f(n)(t)f^{(n)}(t) exists for every t(a,b)t\isin (a,b). Let α,β\alpha, \beta be distinct points of [a,b][a,b], and define P(t)=k=0n1f(k)(α)k!(tα)kP(t) = \sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!} (t-\alpha)^k. Then there exists a point xx between α\alpha and β\beta such that f(β)=P(β)+f(n)(x)n!(βα)nf(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta - \alpha)^n.
    • Taylor Series for a Smooth Function: If ff is a smooth function on (a,b)(a,b), and α(a,b)\alpha \isin (a,b), we can form the Taylor Series:
      Pα(x)=k=0f(k)(α)k!(xα)kP_{\alpha}(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}(\alpha)}{k!} (x-\alpha)^k.
      • Remark: RHS is not guaranteed to converge and even if it converges, it may not equal to f(x)f(x).

Lecture 24 (Apr 15) - Covered Rudin Chapter 3 and 6

  • Taylor Series:
    • Remark: Taylor expansion is finite term expansion with remainder while Taylor series is infinite sum with no remainder.
    • Nth Order Taylor Expansion: Pxo,N(x)=n=0Nfn)(xo)1n!(xxo)nP_{x_o,N}(x) = \sum_{n=0}^{N} f^{n)}(x_o) * \frac{1}{n!} (x-x_o)^n.
    • Definition: Let NN\to\infty, we write Pxo(x)=n=0f(n)(xo)n!(xxo)nP_{x_o}(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_o)}{n!} (x-x_o)^n (Taylor Series at xox_o).
    • Rudin 3.39: Consider power series ncnzn\sum_{n} c_n z^n, put α=limnsupcn1n\alpha = \lim_{n\to\infty} \sup \lvert c_n \rvert^{\frac{1}{n}}. Let R=1αR=\frac{1}{\alpha} (if α=0\alpha = 0 then R=+R=+\infty; if α=+\alpha =+\infty then R=0R=0), then the series is convergent if z<R\lvert z \rvert <R and the series is divergent if z>R\lvert z \rvert >R. Such RR is called the radius of convergence.
      • Remark: If z=R\lvert z \rvert = R, it depends.
    • Example: f(x)=11+x2f(x)=\frac{1}{1+x^2}, find its Taylor series based at x=0x=0.
      • Directly manipulate, x2<1\forall \lvert x^2 \rvert < 1, 11+x2=11α=1+α+α2+=1+(x2)+(x2)2+=1x2+x4+(x2)n+\frac{1}{1+x^2} = \frac{1}{1-\alpha}= 1+\alpha +{\alpha}^2 + … = 1 + (-x^2) + (-x^2)^2+…=1-x^2+x^4-… +(-x^2)^n+…
      • Radius of convergence: cn=0\lvert c_n \rvert = 0 if odd and cn=1\lvert c_n \rvert = 1 if even. Hence limnsupcn1n=1=α\lim_{n\to\infty} \sup \lvert c_n \rvert^{\frac{1}{n}} = 1 =\alpha, then R=1α=1R=\frac{1}{\alpha}=1.
    • Remark: Taylor expansion is a way to approximate a smooth function near a given point, but the approximation is not uniform over the entire domain of ff.
  • Riemann Integral:
    • Partition: Let [a,b]R[a,b]\subset \Reals be a closed interval. A partition PP of [a,b][a,b] is finite set of number in [a,b][a,b]: a=x0x1xn=ba=x_0 \leq x_1 \leq … \leq x_n=b. Define Δxi=xixi1\Delta x_i=x_i-x_{i-1}.
      • Example: [10,20]R[10, 20]\subset \Reals, then a partition would be P={10,15,18,19,20}P = \{10, 15, 18, 19, 20\}.
    • U(P,f) and L(P,f): Given f:[a,b]Rf:[a,b]\to \Reals bounded, and partion p={x0x1xn}p = \{x_0 \leq x_1 \leq … \leq x_n\}, we define U(P,f)=i=1nΔxiMiU(P,f) = \sum_{i=1}^{n} \Delta x_i M_i where Mi=sup{f(x),x[xi1,xi]}M_i= \sup \{f(x), x\isin [x_{i-1},x_i]\}; L(P,f)=i=1nΔximiL(P,f) = \sum_{i=1}^{n} \Delta x_i m_i where mi=inf{f(x),x[xi1,xi]}m_i= \inf \{f(x), x\isin [x_{i-1},x_i]\}.

  • U(f) and L(f): Define U(f)=infPU(P,f)U(f) = \inf_{P} U(P,f) and L(f)=supPL(P,f)L(f)= \sup_{P} L(P,f).
    • Since ff is bounded, hence m,MR\exists m,M\isin \Reals such that mf(x)Mm\leq f(x)\leq M for all x[a,b]x\isin [a,b], then P\forall P partition of [a,b][a,b], U(P,f)i=1nΔxiM=M(ba)U(P,f) \leq \sum_{i=1}^{n} \Delta x_i M = M(b-a), and L(P,f)m(ba)L(P,f) \geq m(b-a), and m(ba)L(P,f)leqU(P,f)M(ba)m(b-a)\leq L(P,f) leq U(P,f) \leq M(b-a).
  • Riemann Integrable: We say a function ff is Riemann integrable if U(f)=L(f)U(f)=L(f),
    • Some sufficient conditions:
      1. If ff is continuous, then ff is Riemann integrable.
      2. If ff is monotone, then ff is Riemann integrable.

Homework 11

  • A function ff is convex if for any x,yRx,y\isin\Reals and any t[0,1]t\isin [0,1], we have tf(x)+(1t)f(y)f(tx+(1t)y)tf(x)+(1-t)f(y)\geq f(tx + (1-t)y).
  • If f:RRf: \Reals\to\Reals is a differentiable and convex function, then f(x)f'(x) is monotone increasing.
  • Real and bounded function \neq Riemann integrable.
    • Example: f(x)=1f(x) =1 if xQx\isin \mathbb{Q} and f(x)=0f(x)=0 if xRQx\isin \Reals\setminus\mathbb{Q}. Then U(f)=1U(f) = 1 and L(f)=0L(f) = 0, since U(f)L(f)U(f) \neq L(f), then ff is not Riemann integrable even if ff is real and bounded.

Week 14

Lecture 25 (Apr 20) - Covered Rudin Chapter 6

  • Stieltjes Integral:
    • Weight Function: Let α:[a.b]R\alpha: [a.b]\to\Reals be a monotone increasing function, then α\alpha could be referred to as a weight function for Stieltjes Integral. We refer to Δαi=α(xi)α(xi1)\Delta \alpha _i = \alpha(x_i) - \alpha(x_{i-1}).
    • Basic Notions: Similar to what we defined in Riemann Integral, we define U(P,f,α)=i=1nMiΔαiU(P,f,\alpha) = \sum_{i=1}^{n} M_i \Delta\alpha_i and L(P,f,α)=i=1nmiΔαiL(P,f,\alpha) = \sum_{i=1}^{n} m_i \Delta\alpha_i.
    • Stilejes Integrable: If U(f,α)=L(f,α)U(f,\alpha) = L(f,\alpha), we say ff is integrable with respect to α\alpha and write fR(α)f\isin \mathscr{R}(\alpha) on [a,b][a,b].
      • Remark: If x[a,b]\forall x\isin [a,b], mf(x)Mm\leq f(x)\leq M, then m(α(b)α(a))L(P,f,α)U(P,f,α)M(α(b)α(a))m (\alpha(b)-\alpha(a)) \leq L(P,f,\alpha) \leq U(P,f,\alpha) \leq M(\alpha(b) - \alpha(a)).
    • Refinement: Let PP and QQ be 2 partitions of [a,b][a,b], then PP and QQ can be identifies as a finite subset of [a,b][a,b]. We say QQ is a refinement of PP if PQP\subset Q as subsets of [a,b][a,b].
      • Example: [a,b]=[0,10][a,b]=[0,10]. Let P={0,1,2,3,4,5,6,7,8,9,10}P = \{ 0, 1, 2,3,4,5,6,7,8,9,10\}, and Q={0,0.5,1,1.5,2,3,4,5,6,7,8,9,9.9,10}Q=\{0, 0.5, 1, 1.5,2,3,4,5,6,7,8,9,9.9,10\}. We could claim that QQ is a refinement of PP on [0,10][0,10].
    • Common Refinement: Let P1P_1 and P2P_2 be 2 partitions of [a,b][a,b], then P1P2P_1 \cup P_2 is the common refinement of P1P_1 and P2P_2.
    • Rudin 6.4: If PP' is a refinement of PP, then L(P,f,α)L(P,f,α)L(P',f,\alpha) \leq L(P,f,\alpha) and U(P,f,α)U(P,f,α)U(P',f,\alpha) \leq U(P,f,\alpha).
    • Rudin 6.5: L(f,α)U(f,α)L(f,\alpha) \leq U(f,\alpha).
    • Rudin 6.6(Cauchy Condition): fR(α)    ϵ>0,Pf\isin \mathscr{R}(\alpha) \iff \forall \epsilon >0, \exists P partition such that U(P,f,α)L(P,f,α)<ϵU(P,f,\alpha)-L(P,f,\alpha) < \epsilon.
    • Rudin 6.7:
      • If Rudin 6.6 holds for PP, then for any refinement QQ of PP, U(Q,f,α)L(Q,f,α)<ϵU(Q,f,\alpha)-L(Q,f,\alpha) < \epsilon.
      • If Rudin 6.6 holds for PP, and let si,ti[xi1,xi]i=1,2,,ns_i, t_i\isin [x_{i-1},x_i] \forall i = 1,2,…,n, then i=1nf(si)f(ti)Δαi<ϵ\sum_{i=1}^{n} \lvert f(s_i) - f(t_i) \rvert \Delta\alpha_i < \epsilon.
      • If fR(α)f\isin\mathscr{R}(\alpha) and the above holds, then i=1nf(si)Δαifdα<ϵ\sum_{i=1}^{n} \lvert f(s_i) \Delta\alpha_i - \int fd\alpha \rvert < \epsilon.
    • Rudin 6.8: If ff is continuous on [a,b][a,b], then fR(α)f\isin\mathscr{R}(\alpha) on [a,b][a,b].
    • Rudin 6.9: If ff is monotonic on [a,b][a,b] and α\alpha is continuous, then fR(α)f\isin \mathscr{R}(\alpha).

Lecture 26 (Apr 22) - Covered Rudin Chapter 6

  • More on Integrations:
    • Rudin 6.10: If ff is discontinuous only at finitely many points, and α\alpha is continuous where ff is discontinuous, then fR(α)f\isin \mathscr{R}(\alpha).
    • Rudin 6.11: Let f:[a,b][m,M]f:[a,b]\to [m,M] and ϕ:[m,M]R\phi:[m,M]\to \Reals is continuous. If ff is integrable with respect to α\alpha, then h=ϕfh=\phi \circ f is integrable with respect to α\alpha.
      • If f1,f2R(α)f_1,f_2\isin\mathscr{R}(\alpha) and cRc\isin\Reals, then f1+f2,cf1R(αf_1+f_2,cf_1\isin\mathscr{R}(\alpha, and f1+f2dα=f1dα+f2dα\int f_1+f_2 d\alpha = \int f_1 d\alpha + \int f_2 d\alpha, cf1dα=cf1dα\int cf_1 d\alpha = c \int f_1 d\alpha.
      • If f,gR(α)f,g\isin\mathscr{R}(\alpha) and f(x)g(x)x[a,b]f(x)\leq g(x)\forall x\isin [a,b], then abfdαabgdα\int_{a}^{b} fd\alpha \leq \int_{a}^{b} gd\alpha.
      • If fR(α)f\isin\mathscr{R}(\alpha) on [a,c][a,c], then fR(α)f\isin\mathscr{R}(\alpha) on [a,b][a,b] and on [b,c][b,c] if a<c<ba < c< b, and acfdα=abfdα+bcfdα\int_{a}^{c} fd\alpha = \int_{a}^{b} fd\alpha + \int_{b}^{c} fd\alpha.
      • If fR(α)f\isin\mathscr{R}(\alpha) on [a,b][a,b], and f(x)M\lvert f(x) \rvert \leq M on [a,b][a,b], then abfdαM(α(b)α(a))\lvert \int_{a}^{b} fd\alpha \rvert \leq M(\alpha(b)-\alpha(a)).
      • If fR(α1)f\isin\mathscr{R}(\alpha_1) and fR(α2)f\isin\mathscr{R}(\alpha_2) and let cc be a positive constant, then fR(α1+α2)f\isin\mathscr{R}(\alpha_1 +\alpha_2) and fR(cα1)f\isin\mathscr{R}(c \alpha_1) with fd(α1+α2)=fdα1+fdα2\int fd(\alpha_1 + \alpha_2) = \int fd\alpha_1 + \int fd\alpha_2 and fd(cα1)=cfdα1\int fd(c \alpha_1) = c \int fd\alpha_1.
    • Rudin 6.13:
      • If f,gR(α)f,g\isin\mathscr{R}(\alpha), then fgR(α)fg\isin\mathscr{R}(\alpha).
      • If fR(α)f\isin\mathscr{R}(\alpha), then fR(α)\lvert f\rvert \isin\mathscr{R}(\alpha) and abfdαabfdα\lvert \int_{a}^{b} fd\alpha \rvert \leq \int_{a}^{b} \lvert f\rvert d\alpha.
    • Unit Step function: The unit step function II is defined by I(x)=0I(x) = 0 if x0x\leq 0 and I(x)=1I(x) = 1 if x>0x > 0.
    • Rudin 6.15: If f:[a,b]Rf:[a,b]\to\Reals and is continuous at s[a,b]s\isin [a,b] and α(x)=I(xs)\alpha(x) = I(x-s), then fdα=f(s)\int fd\alpha = f(s).
    • Rudin 6.16: Suppose cn0c_n\geq 0 for n=1,2,3,n=1,2,3,…, cn<\sum c_n < \infty, {sn}\{ s_n \} is a sequence of distinct points in (a,b)(a,b), and α(x)=n=1cnI(xsn)\alpha(x) = \sum_{n=1}^{\infty} c_n I(x-s_n). Let ff be continuous on [a,b][a,b], then fdα=n=1cnf(sn)\int fd\alpha = \sum_{n=1}^{\infty} c_n f(s_n).

Homework 12

  • If ff is a continuous non-negative function on [a,b][a,b] and abfdx=0\int_{a}^{b} fdx=0, then f(x)=0f(x) = 0 for all x[a,b]x\isin [a,b].
  • If f,gRf,g\isin\mathscr{R} are real and bounded functions and p,q>0p,q>0 such that 1p+1q=1\frac{1}{p} + \frac{1}{q} = 1, then fgdx[fpdx]1p[gqdx]1q\int fgdx \leq [\int \lvert f \rvert^p dx]^{\frac{1}{p}} [\int \lvert g \rvert^q dx]^{\frac{1}{q}}.
    • If u,v>0u,v>0, then uvupp+vqquv \leq \frac{u^p}{p} + \frac{v^q}{q}.
    • If f,gf,g are non-negative Riemann integrable functions on [a,b][a,b], and fpdx=gqdx=1\int f^p dx = \int g^q dx = 1 then fgdx1\int fgdx \leq 1.

Week 15

Lecture 27 (Apr 27) - Covered Rudin Chapter 6

  • Integration:
    • Rudin 6.17: Assume α\alpha is increasing and αR\alpha' \isin \mathscr{R} on [a,b][a,b]. ff is a bounded real function on [a,b][a,b], then fR(α)    fαRf\isin\mathscr{R}(\alpha) \iff f\alpha' \isin \mathscr{R} and if so, abfdα=abfαdx\int_{a}^{b} fd\alpha = \int_{a}^{b} f\alpha' dx.
    • Rudin 6.19 (Change of Variable): Suppose α\alpha is increasing on [a,b][a,b] and fR(α)f\isin\mathscr{R}(\alpha). Suppose ϕ:[A,B][a,b]\phi :[A,B]\to [a,b] is a strictly increasing continuous and surjective function. Define β(y)=α(ϕ(y))\beta(y) = \alpha(\phi(y)) and g(y)=f(ϕ(y))g(y) = f(\phi(y)). Then gR(β)g\isin\mathscr{R}(\beta) and ABgdβ=abfdα\int_{A}^{B} gd\beta = \int_{a}^{b} fd\alpha.
  • Relation Between Integration and Differentiation:
    • Rudin 6.20: Let fRf\isin\mathscr{R} on [a,b][a,b]. For axba\leq x\leq b, let F(x)=axf(t)dtF(x) = \int_{a}^{x} f(t)dt. Then F is continuous on [a,b][a,b]; furthermore if f(x)f(x) is continuous at xo[a,b]x_o\isin [a,b], then F(x)F(x) is differentiable at xox_o, F(xo)=f(xo)F'(x_o) = f(x_o).
    • Rudin 6.21 (Fundamental Theorem of Calculus): If fRf\isin\mathscr{R} on [a,b][a,b] and if there is a differentiable function FF on [a,b][a,b] such that F=fF'=f, then abf(x)dx=F(b)F(a)\int_{a}^{b} f(x)dx = F(b) - F(a).
    • Rudin 6.22 (Integration by Parts): Suppose F,GF,G are differentiable, F,GF',G' are integrable, f=Ff=F' and g=Gg=G'. Then abF(x)g(x)dx=F(b)G(b)F(a)G(a)abf(x)G(x)dx\int_{a}^{b} F(x)g(x)dx = F(b)G(b) - F(a)G(a) - \int_{a}^{b} f(x)G(x)dx.

Lecture 28 (Apr 29) - Covered Rudin Chapter 7

  • Uniform Convergence :
    • Rudin 7.16: Let α\alpha be monotone increasing on [a,b][a,b]. Suppose fnR(α)f_n\isin\mathscr{R}(\alpha), and fnff_n\to f uniformly on [a,b][a,b]. Then ff is integrable and abfdα=limnabfndα\int_{a}^{b} fd\alpha = \lim_{n\to\infty} \int_{a}^{b} f_n d\alpha.
    • Corollary: Suppose fnR(α)f_n\isin\mathscr{R}(\alpha) and F(x)=n=1fn(x)F(x) = \sum_{n=1}^{\infty} f_n(x), the series converges uniformly, then FR(α)F\isin\mathscr{R}(\alpha) and abF(x)dα=n=1abfn(x)dα\int_{a}^{b} F(x)d\alpha = \sum_{n=1}^{\infty} \int_{a}^{b} f_n(x)d\alpha.
    • Theorem: Suppose {fn}\{ f_n \} is a sequence of differentiable functions on [a,b][a,b] such that fn(x)f_n'(x) converges uniformly to g(x)g(x) and xo[a,b]\exists x_o\isin [a,b] such that {fn(xo)}\{f_n(x_o)\} converges. Then fn(x)f_n(x) converges to some function ff uniformly and f(x)=g(x)=limnfn(x)f'(x)=g(x)=\lim_{n\to\infty} f_n'(x).

Questions

  1. I understand the visualization of this recursive sequence, but to 5\sqrt{5}?
  2. In general, how to prove a set is infinite(in order to use theorem 11.2 in Ross)?
  3. Is there a way/analogy to understand/visualize the closure of a set?
  4. Is there a way to actually test if a set is compact or not instead of merely finding finite subcovers for all open covers?
  5. Rudin 4.6 states that if pEp\isin E, a limit point, and ff is continuous at pp if and only if limxpf(x)=f(p)\lim_{x\to p} f(x) = f(p). Does this theorem hold for pEp\isin E but not a limit point of EE?
  6. How is the claim at the bottom proved?
  7. Could we regard the global maximum as the maximum of all local minimums?
  8. Under what circumstance would Taylor Theorem be essential to our proof, since for the question appeared in HW11 Q2, I really do not think using Taylor Theorem is a better way to prove the claim?
  9. In order for a Taylor series to converge (ncnzn\sum_{n} c_n z^n), z<R\lvert z \rvert < R where RR is the radium of convergence. But if z=R\lvert z \rvert = R, how can we tell?
  10. If we are claiming ff is continuous on [a,b][a,b], , i.e. do we just extend our interval to the left side of aa and right side of bb to do so?
  11. What information can we extract from the line “ff has a bounded first derivative (i.e. fM\lvert f' \rvert \leq M for some M>0M>0)”?
  12. How sequentially compact without proving that it is compact? (Starting from ms too complicated to take into account all sequences in the set)
  13. If an+1=cos(an)a_{n+1} = \cos (a_n) and choose a1a_1 such that 0<a1<10 < a_1 < 1, is ana_n a ?
  14. Does uniform convergence on a sequence of functions {fn}\{f_n\} in FF to ff imply ?
  15. If fn\sum f_n converges uniformly, does it imply fnf_n satisfies Weiestrass M-test?
  16. For the alternating series test, if instead of sequence of numbers we have sequence of functions and those functions {fn}\{ f_n \} satisfies f1f2f3f_1 \geq f_2 \geq f_3 … and fn0f_n \geq 0 for all xXx\isin X, limfn=0\lim f_n = 0, does that mean n(1)nfn\sum_{n} (-1)^n f_n converges uniformly?
  17. What is measure zero? (Related to Lebesgue measure and volume of open balls)
  18. Is there any covering for a compact set that satisfies total length of the finite subcovers for the set is less than ba\lvert b-a \rvert? (Answer is no)
  19. Question 16 on Prof Fan's practice exam.
  20. This is my solutions towards the practice exam: practice_solutions.pdf
math104-s21/s/martinzhai.txt · Last modified: 2022/01/11 18:31 by 24.253.46.239