Kreyszig Chapter 2 Commentary

Kreyszig Functional Analysis

First, a quick exercise:

Theorem) For $p>1$, the dual space $({\ell }^p{)}^{\prime }$ is norm space-isomorphic to ${\ell }^q$ where $1/p+1/q=1$.

Proof) Note that $e_i$ is a Schauder basis of ${\ell }^p$, where $e_i$ has $1$ in the $i$th position and $0$ elsewhere. Let $f$ be a bounded linear functional on ${\ell }^p$ such that $f(e_i)={\gamma }_i$; any $x\in {\ell }^p$ has a unique representation $x={\sum }_{i=1}^{\infty }{\xi }^i{e}_i$, and $f(x)={\sum }_{i=1}^{\infty }{\xi }^i{\gamma }_i$, which must converge. We first show, given these facts, that $({\gamma }_i)\in {\ell }^q$, creating a map from $({\ell }^p{)}^{\prime }\to {\ell }^q$.

The crux of the problem is cleverly exploiting the boundedness of $f$. Define the sequence ${\xi }_n$ to be ${\xi }_n^i=0$ if $i>n$ and ${\xi }_n^i=|{\gamma }_i{|}^q/{\gamma }_i$ for $i\le n$. Then, we have $|{\sum }_{i=1}^n{\xi }_n^i{\gamma }_i|\le \Vert f\Vert ({\sum }_{i=1}^n(|{\gamma }_i{|}^{q-1}{)}^p{)}^{1/p}$, which implies ${\sum }_{i=1}^n|{\gamma }_i{|}^q\le \Vert f\Vert ({\sum }_{i=1}^n|{\gamma }_i{|}^q{)}^{1/p}\Rightarrow ({\sum }_{i=1}^n|{\gamma }_i{|}^q{)}^{1/q}\le \Vert f\Vert$. Taking the $n\to \infty$ limit gives $({\gamma }_i)\in {\ell }^q$, as desired. Furthermore, we have that $\Vert f\Vert \ge \Vert {\gamma }_i{\Vert }_{{\ell }^q}$.

In the other direction, we have to show that given $({\gamma }_i)\in {\ell }^q$, the function $f:{\ell }^p\to \mathbb{F}$ such that $f(e_i)={\gamma }_i$ is a well-defined bounded linear functional. For any $x\in {\ell }^p$ where $x={\sum }_{i=1}^{\infty }{\xi }^i{e}_i$, we have that ${\sum }_{i=1}^{\infty }|{\xi }^i{\gamma }_i|\le \Vert {\xi }^i{\Vert }_{{\ell }^p}\Vert {\gamma }_i{\Vert }_{{\ell }^q}$, and thus $\sum {\xi }^i{\gamma }_i$ is well-defined. Taking $f$ to be this limit, $f$ is clearly a linear functional. Boundedness follows from $|{\sum }_{i=1}^{\infty }{\xi }^i{\gamma }_i|\le {\sum }_{i=1}^{\infty }|{\xi }^i{\gamma }_i|\le \Vert {\xi }^i{\Vert }_{{\ell }^p}\Vert {\gamma }_i{\Vert }_{{\ell }^q}$, which implies $\Vert f\Vert \le \Vert {\gamma }_i{\Vert }_{{\ell }^q}$. Therefore, each $({\gamma }_i)\in {\ell }^q$ corresponds to a functional in the dual space of ${\ell }^q$.

We have thus found a bijective correspondence between $({\ell }^p{)}^{\prime }$ and ${\ell }^q$. As $\Vert f\Vert \le \Vert {\gamma }_i{\Vert }_{{\ell }^q}$ and $\Vert f\Vert \ge \Vert {\gamma }_i{\Vert }_{{\ell }^q}$, we have that $\Vert f\Vert =\Vert {\gamma }_i{\Vert }_{{\ell }^q}$, so this correspondence is norm-preserving. It suffices to show the correspondence is linear; this follows trivially from definition.

The completion theorems are also strangely undersold in Kreyszig. The theorems can be stated, in its full power, as follows:

(Metric Completion) Let $M$ be any metric space. There is a complete metric space $S$ and a map $\varphi :M\to S$ such that $\varphi$ is a bijective isometry between $M$ and the image $\varphi (M)$, and $\varphi (M)$ is dense in $S$. If $(S^{\prime },{\varphi }^{\prime })$ also satisfies these conditions, there exists a unique bijective isometry $\psi :S\to S^{\prime }$ such that $\psi \circ \varphi ={\varphi }^{\prime }$.

(Banach Completion) Let $V$ be any normed space, and $(S,\varphi )$ be its “unique” metric completion. There is a unique norm and vector space structure on $S$ that induces the metric of $S$ and such that $\varphi$ is linear. With this structure, $S$ is a Banach space.

Notice that not much needs to be said about the Banach completion, as the uniqueness of the metric completion suffices. The proofs are given in Kreyszig, so we omit them here.