Sketch 6

A Quick Result Not Shown In Zhu

Huh...

Kehe Zhu, in his book Operator Theory in Function Spaces, leaves many results without proofs for the sake of brevity. I'm rereading his book, and I realize I still don't have command over the geometry and ideas of Banach spaces. This will be one of a sequence of sketches proving results that Zhu leaves out as well-known.

Date Started: May 16, 2024
Date Finished: May 16, 2024

Banach Spaces

I decided to first brainstorm all the properties that I think of when I think of a linear function:

1. $F(ax+by)=aF(x)+bF(y)$ (lines map points proportionately).
2. Lines are continuous, and differentiable, and $C^{\mathrm{\infty }}$.
3. The "slope" of a line determines the proportions on the line to which points are mapped.
4. The derivative of a line is its slope.
5. The second derivative of a line is zero.

anyway, let's get started. Let $X$ be a Banach space.

Proposition. Let $F:X\to \mathbb{C}$ be a linear functional on a Banach space $X$. $F$ is bounded if and only if $F$ is continuous in the norm topology of $X$.

Proof.

Suppose $F$ is bounded, so that $|F(x)|\le C||x||$ for all $x\in x$. Then for any $x$ and $y$ satisfying $||x-y||<\frac{ϵ}{C}$, $$|F(x)-F(y)|=|F(x-y)|\le C||x-y||<C\cdot \frac{ϵ}{C}=ϵ$$ so indeed $F$ is continuous.

Suppose $F$ is continouous. Notice that $S:=\{x:||x||=1\}$ is a compact set, so $F(S)$ is compact. In particular, it is bounded. Let $C=\underset{S}{sup}|F|$. Then $$|F(x)|=||x|||F(\frac{x}{||x||})|\le ||x||\cdot C$$ as desired.

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Proposition. The space of all bounded linear functionals on $X$ is a Banach space with the norm: $$||F||=\underset{S}{sup}|F|.$$ This is called the dual space of $X$.

Proof.

Firstly, bounded linear functionals form an abelian group (they are closed under addition and inverses, and zero is a bounded linear functional), and are compatible with scalar multiplication, so they form a vector space. Our norm is clearly nonnegative, and satisfies:

1. $||F+G||=\underset{S}{sup}|F+G|\le \underset{S}{sup}(|F|+|G|)\le \underset{S}{sup}|F|+\underset{S}{sup}|G|=||F||+||G||$
2. $||aF||=\underset{S}{sup}|aF|=\underset{S}{sup}|a||F|=|a|\underset{S}{sup}|F|=|a|||F||$
3. $||F||=0\iff \underset{S}{sup}|F|=0\iff F=0$
4. If $\{F_n\}$ is a Cauchy sequence, then for each $x\in S$, $\{F_n(x)\}$ is Cauchy, so there is a limit $F(x)$ for each $x$. I believe the uniform continuity of the $F_n(x)$ convergence guarantees continuity of $F$ (although I'm kind of sleepy, it's really late). This is now linear because we define it by $F(y)=F(\frac{y}{||y||})$, and it must be the limit everywhere, and each $F_n$ is linear.

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