note

note .

“How to measure the distance between two functions?”

Just as you can have a sequence

• of number in $\R$
• of vectors in $\R^n$
• of points in a general metric space $X$.

You can have a sequence of functions. $f_n(x)$

Let $V = \{f: f: [0,1] \to \R\}$ be the space of bounded functions from $[0,1]$ to $\R$, we will consider various ways to measure size of a function

• sup norm: $\| f \|_\infty = \sup \{ f(x) : x \in [0,1]\}$
• $L^1$ norm $\| f\|_1 = \int |f(x)| dx$
• $L^2$ norm $\| f\|_2 = (\int |f(x)|^2 dx)^{1/2}$
• in general, for $1 \leq p < \infty$, we have $L^p$ norm.

The different norms equip the vector space $V$ with different topologies. This is different from the finite dimensional vector space case.

In sport, sometimes you measure the score using the best score (like freestyle skiing in the winter Olympics); sometimes you use the average score of several try (like Tour de France, you add up the points in each leg); this is like sup or $L^1$ norm.

In this class, we will consider sup norm, and we define distance as $d_\infty(f,g) = \| f-g\|_\infty$

We say a sequence of functions $f_n$ converge to $f$ uniformly, if $\lim_n d_\infty(f_\infty, f) = 0$.

Q: can you find a metric on the space of functions $V$ so that metric convergence means pointwise convergence? (No, you but you can still define a topology on $V$ so that convergence in that topology means pointwise convergence)

In general, if $M = \{f| f: X \to Y, f(X) \text{bounded}\}$ is the space of maps with bounded images, then for any $f, g \in M$, we can define $d_\infty(f,g) = \sup_{x} d_\infty(f(x), g(x))$.

• The running bump $f_n = 1_{[n,n+1]}(x)$, which is 1 on $[n,n+1]$ and $0$ elsewhere. They converge to $0$ pointwise, but not uniformly.
• The shrinking and rising bump $f_n = n 1_{(0,1/n)}$ converge to $0$ pointwise, but not uniformly.

Thm: If $f_n: \R \to \R$ are continuous and bounded, and $f_n \to f$ uniformly, then $f$ is continuous.

Proof: we need to show that, for any $x \in \R$, for any $\epsilon>0$, there is $\delta > 0$, such that for any $x'$ with $|x' - x| < \delta$, we have $|f(x') - f(x)|<\epsilon$.

First, we choose $n$ large enough, such that $d_\infty(f_n, f) < \epsilon/3$. Then, for $f_n$ and $x$ and $\epsilon/3$, we find $\delta$, such that any $|x'-x|<\delta$ implies $|f_n(x) - f_n(x')|<\epsilon/3$. Finally, if any $|x'-x|<\delta$, then $|f(x) - f(x')| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(x')| + |f_n(x') - f_n(x)| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon.$ Done

• Devil's staircase
• power series
• Weierstrass M-test

Prop: If $f: X \to Y$ is continuous, and $K \In X$ is compact, then $f(K)$ is compact.
Proof: any open cover of $f(K)$ can be pulled back to be an open cover of $K$, then we can pick a finite subcover in the domain, and the corresponding cover in the target forms a cover of $f(K)$.

We can also prove using sequential compactness. To see any sequence in $f(K)$ subconverge, we can lift that sequence back to $K$, find a convergent subsequence, and push them back to $f(K)$.

Lemma: if $f: X \to Y$ is continuous, then for any $E \In X$, $f|_E: E \to Y$ is continuous.

Lemma: if $f: X \to Y$ is continuous, then $f: X \to f(X)$ is continuous.

Prop: If $f: X \to Y$ is continuous, and $K \In X$ is connected, then $f(K)$ is connected.
Pf: first, note that map $f: K \to f(K)$ is also continuous. If $f(K)$ is the disjoint union of two non-empty open subsets, $f(K) = U \cap V$, then $K = f^{-1}(U) \cap f^{-1}(V)$ the disjoint union of two non-empty open subsets of $K$.

Intermediate value theorem: if $[a,b] \In \R$, and $f: \R \to \R$ is continuous, then $f([a,b])$ is also a closed interval.
Proof: since $[a,b]$ is compact, hence $f([a,b])$ is compact, hence closed. Since $[a,b]$ is connected, hence $f([a,b])$ is connected, hence an interval, a closed interval.

We say a function $f: X \to Y$ is uniformly continuous, if for any $\epsilon >0$, there exists $\delta > 0$, such that for any pair $x_1, x_2 \in X$ with $d(x_1, x_2)<\delta$, we have $d(f(x_1), f(x_2))< \epsilon$.

For example, the function $f: (0, 1) \to \R$ $f(x) =1 /x$ is continuous but not uniformly continuous.

Theorem: if $f: X \to Y$ is continuous, and $X$ is compact, then $f$ is uniformly continuous.

Proof: Fix $\epsilon>0$. For each $x \in X$, let $r_x > 0$ be small enough such that $f(B_{2r_x}(x)) \In B_{\epsilon/2}(f(x))$. Let $B_x = B_{r_x}(x)$. The, $\{B_x: x \in X\}$ forms an open cover of $X$. Pass to finite subcover, $X = \cup_{i=1}^N B_{x_i}$. Then let $\delta = \min \{r_{x_i}\}$. Then, suppose $x$ and $x'$ has distance less than $\delta$, then $x$ is contained in some $B_{r_{x_i}}(x_i)$, and $x' \in B_{2 r_{x_i}}(x_i)$, thus $f(x), f(x') \in B_{\epsilon/2}(f(x_i))$, thus $f(x)$ and $f(x')$ has distance less than $\epsilon$.

Now we will leave the safe world of continuous functions. We consider more subtle cases of maps.

Def: Let $f: X \to Y$ be any map, and let $x \in X$ be a point, we say $f$ is continuous at $x$, if for any $\epsilon>0$, there exists $\delta>0$, such that $f(B_\delta(x)) \In B_\epsilon(f(x))$.

Def: Let $E \In X$ and $f: E \to Y$. Suppose $x \in \bar E$. We say $\lim_{p \to x} f(p) = y$ if for any convergent sequence $p_n \to x$ with $p_n \in E$, we have $f(p_n) \to y$.

note that $x$ may not be in $E$.

Prop: $f: X \to Y$ is continuous at $x \in X$, if and only if $\lim_{p \to x} f(p) = f(x)$.

If $f: (a,b) \to \R$ is a function, and $f$ is not continuous at some $x \in (a,b)$, then

• if $\lim_{t \to x-} f(t)$ and $\lim_{t \to x+} f(t)$ both exists, but does not equal to $f(x)$, we say this is a simple discontinuity, or first kind discontinuity
• otherwise, it is called a second kind.

Last time we didn't introduce the notion of a connected space. Let's do it now.

Sometimes, we can tell from a glance, whether a space is connected or not: the interval $[0,1]$ is connected, the set $\{1,2,3\}$ (with induced topology from $\R$) is not (there are gaps between points). But, we also know that topological space can be weird.

Ex: let $X = \{1,2,3,\cdots \}$ be a set, we declare the following subsets in $X$ are open:

• $\emptyset$ and $X$,
• $\{1,2,\cdots, n\}$, for some $n \geq 1$.

One can easily check that this collection of open sets are closed under arbitrary union and finite intersections. Hmm, is it connected? What does it mean when we ask this question?

Here is the general definition:

We say a metric space $X$ is connected, if $X$ cannot be written as disjoint union of two non-emtpy open subset. In other word, the only subsets in $X$ that is both open and closed are $X$ and $\emptyset$.

For example, $\Q$ is not connected, since $(-\infty, \sqrt{5})$ is both open and closed in $Q$. (why?)

For example, $X=\{1,2,3\}$ (with induced metric from $\R$) is not connected, since $\{1\}$ is both open and closed in $X$. (Discussion: Equip $X$ with the induced metric, can you show that $\{1\}$ is both open and closed? Equip $X$ with the induced topology, can you show that $\{1\}$ is both open and closed? )

For example, in that weird topology example above, $X$ is connected, since $X$ cannot be written as disjoint union of two open sets.

Theorem: a subset $E \In \R$ is connected, if and only if, for any $x,y \in E$, we have $[x,y] \In E$. Proof: suppose $E$ is connected, and $x,y \in E$, we need to show that $[x,y] \In E$. If not, say $z \in (x,y)$ is not in $E$, then let $E_1 = E \cap (-\infty,z)$, $E_2 = E \cap (z,+\infty)$, they are open subsets of $E$ (since they are open subsets of $\R$ intersecting $E$), they are non-empty (containing $x$ and $y$ respectively), and $E = E_1 \sqcup E_2$, therefore $E$ is not connected. Contradiction.

Suppose for any $x,y \in E$, we have $[x,y] \In E$, and suppose $E$ is not connected. Then $E$ can be written as disjoint union of two non-empty open sets $E = A \sqcup B$, where $A, B$ open in $E$. Pick $x \in A$ and $y \in B$, w.l.o.g, assuming $x<y$. By hypothesis, $[x,y] \In E$. Let $A' = A \cap [x,y]$ and $B' = B \cap [x,y]$. Then $[x,y] = A' \cup B'$, and $A', B'$ are open in $[x,y]$, and $x \in A'$, $y \in B'$. Let $z = \sup A'$. We discuss the following cases:

• if $z=x$, then $A' = \{x\}$, and $A'$ cannot be an open subset of $[x,y]$
• If $x< z < y$, we split into two case:
  • (a) if $z \in A'$, then $A'$ is not an open subset of $[x,y]$ at point $z$.
  • (b) if $z \in B'$, then $B'$ is not an open subset of $[x,y]$ at $z$.
• if $z = y$, then $z \in B'$, we claim that $B'$ is not open in $[x,y]$.

Thus, we get a contradiction.

Remark: Rudin uses a seeming different definition of connectedness, but it is equivalent. We say two subsets $A, B$ are 'separated', if $\bar A \cap B = \emptyset$ and $A \cap \bar B = \emptyset$. It is possible for two sets to be disjoint, but not separated, like $A = (0,1), B=[1,2)$.

Prop: A subset $E \In X$ is connected $\Leftrightarrow$ if $E$ cannot be written as the union of two non-empty separated sets.

Proof: $\Rightarrow$ Suppose $E \In X$ is connected, and $E = G \sqcup H$ with $G, H$ non-empty separated subsets in $X$. Then, $\bar G \cap E = \bar G \cap (G \sqcup H) = (\bar G \cap G) \sqcup (\bar G \cap H) = G \sqcup \emptyset = G$, hence $G$ is closed in $E$. Similarly, $H$ is closed in $E$. That means the complement of $G$ in $E$, ie. $H$ is open, similarly, $G$ is open. Thus, $E = G \sqcup H$ writes $E$ as disjoint union of two non-emptyset open subsets of $E$, contradicting with $E$ being connected.

$\Leftarrow$: suppose $E$ is not connected, thus $E = G \sqcup H$ with $G, H$ open in $E$ (hence both $G, H$ are closed in $E$, since $G = E \RM H$, $H = E\RM G$). Let $G' \In X$ be a closed subset, such that $G' \cap E = G$, then $G' \cap H = \emptyset$. In particular, $\bar G \In G'$, hence $\bar G \cap H = \emptyset$. Similarly $\bar H \cap G=\emptyset$, thus $G,H$ are separated.