Elementary Properties and Examples

Complete the proof of Proposition 1.3: if $\mathcal{X}$ is a normed space, then

Complete the proof of Proposition 1.5: if $\left\| \cdot \right\|_{1}$ and $\left\| \cdot \right\|_{2}$ are two norms on $\mathcal{X}$, then these norms are equivalent if and only if there are positive constants $c$ and $C$ such that

for all $x$ in $\mathcal{X}$. (The implication ``constants exist $\Rightarrow$ equivalent'' has been proved in the text; only the converse implication remains.)

For $1 \leq p < \infty$ and $x = (x_1, \ldots , x_d)$ in $\mathbb {F}^d$, define $\left\| x\right\|_{p} \equiv \left[\sum_{j=1}^{d} \left|x_j\right|^p\right]^{1/p}$; define $\left\| x\right\|_{\infty } \equiv \sup \left\{ \left|x_j\right| : 1 \leq j \leq d\right\}$. Show that all of these norms are equivalent. For $1 \leq p, q \leq \infty$, what are the best constants $c$ and $C$ such that $c \left\| x\right\|_{p} \leq \left\| x\right\|_{q} \leq C \left\| x\right\|_{p}$ for all $x$ in $\mathbb {F}^d$?

If $1 \leq p \leq \infty$ and $\left\| \cdot \right\|_{p}$ is defined on $\mathbb {R}^2$ as in Exercise 3, graph $\left\{ x \in \mathbb {R}^2 : \left\| x\right\|_{p} = 1\right\}$. Note that if $1 < p < \infty$, $\left\| x\right\|_{p} = \left\| y\right\|_{p} = 1$, and $x \neq y$, then for $0 < t < 1$, $\left\| tx + (1-t)y\right\|_{p} < 1$. The same cannot be said for $p = 1, \infty$.

Let $c =$ the set of all sequences $\left\{ \alpha_n\right\}_1^\infty$, $\alpha_n$ in $\mathbb {F}$, such that $\lim \alpha_n$ exists. Show that $c$ is a closed subspace of $l^\infty$ and hence is a Banach space.

Let $X = \left\{ n^{-1} : n \geq 1\right\} \cup \left\{ 0\right\}$. Show that $C(X)$ and the space of $c$ of Exercise 5 are isometrically isomorphic.

If $l^\infty (I)$ and $l^\infty (J)$ are isometrically isomorphic, do $I$ and $J$ have the same cardinality?

Show that $l^\infty$ is not separable.

Complete the proof of Proposition 1.7: if $X$ is a locally compact space and $C_0(X) =$ all continuous functions $f : X \to \mathbb {F}$ such that for all $\varepsilon > 0$, $\left\{ x \in X : \left|f(x)\right| \geq \varepsilon \right\}$ is compact, then $C_0(X)$ is a closed subspace of $C_b(X)$ and hence is a Banach space. (It has been shown in the text that $C_0(X)$ is closed in $C_b(X)$; only the fact that $C_0(X)$ is a linear manifold in $C_b(X)$ remains.)

Verify the statements in Example 1.10: let $n \geq 1$ and let $C^{(n)}[0,1] =$ the collection of functions $f : [0,1] \to \mathbb {F}$ such that $f$ has $n$ continuous derivatives. Define $\left\| f\right\| = \sup_{0 \leq k \leq n} \left\{ \sup \left\{ \left|f^{(k)}(x)\right| : 0 \leq x \leq 1\right\} \right\}$. Then $C^{(n)}[0,1]$ is a Banach space.

Verify the statements in Example 1.11: let $1 \leq p < \infty$ and $n \geq 1$ and let $W_p^n[0,1] =$ the functions $f : [0,1] \to \mathbb {F}$ such that $f$ has $n - 1$ continuous derivatives, $f^{(n-1)}$ is absolutely continuous, and $f^{(n)} \in L^p[0,1]$. For $f$ in $W_p^n[0,1]$, define

Then $W_p^n[0,1]$ is a Banach space.

Let $X$ be locally compact and let $X_\infty = X \cup \left\{ \infty \right\}$ be the one-point compactification of $X$. Show that $C_0(X)$ and $\left\{ f \in C(X_\infty ) : f(\infty ) = 0\right\}$, with the norm it inherits as a subspace of $C(X_\infty )$, are isometrically isomorphic Banach spaces.

Let $X$ be locally compact and define $C_c(X)$ to be the continuous functions $f : X \to \mathbb {F}$ such that $\mathrm{spt}\, f \equiv \mathrm{cl}\left\{ x \in X : f(x) \neq 0\right\}$ is compact ($\mathrm{spt}\, f$ is the support of $f$). Show that $C_c(X)$ is dense in $C_0(X)$.

If $W_p^n[0,1]$ is defined as in Example 1.11 and $f \in W_p^n[0,1]$, let $\vertiii {f} \equiv \left[\int \left|f(x)\right|^p \, dx\right]^{1/p} + \left[\int \left|f^{(n)}(x)\right|^p \, dx\right]^{1/p}$. Show that $\vertiii {\cdot }$ is equivalent to the norm defined on $W_p^n[0,1]$.

Let $\mathcal{X}$ be a normed space and let $\hat{\mathcal{X}}$ be its completion as a metric space. Show that $\hat{\mathcal{X}}$ is a Banach space.

Show that the norm on $C([0,1]) = C_b([0,1])$ does not come from an inner product by showing that it does not satisfy the parallelogram law.