# More Banach Spaces (Thursday, January 20)

:::{.remark}
Last time: complete iff absolutely convergent implies convergent.
Today: wrapping up some results on Banach and Hilbert spaces, skipping a review of $L^p$ spaces, and starting on operators next week.
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:::{.remark}
Note that $S \da (0, 1]$ is open and not complete, but $\cl_\RR(S) = [0, 1]$ is both closed and complete.
Generalizing:
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:::{.lemma title="?"}
A subset $S \subseteq B$ of a Banach space is complete iff $S$ is closed in $B$.
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:::{.proof title="?"}
$\impliedby$:
If $S$ is closed, a Cauchy sequence $(f_n)$ in $S$ converges to some $f\in B$.
Since $S$ is closed in $B$, in fact $f\in S$.

$\implies$:
Suppose that for $f\in \cl_B(S)$, there is a Cauchy sequence $f_n\to f$ with $f_n \in S$ and $f\in B$.
Since $S$ is complete, $f\in S$, so $\cl_B(S) \subseteq S$ making $S$ closed.
:::

:::{.theorem title="?"}
For $\Omega \subseteq \RR^n$, the space $X = (C(\Omega; \CC), \norm{\wait}_\infty)$ is a Banach space. 
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:::{.warnings}
This is *not* complete with respect to any other $L^p$ norm with $p<\infty$!
:::

:::{.proof title="of theorem"}
Use the lemma -- write $B$ for the space of all bounded (not necessarily continuous) functions on $\Omega$, which is clearly a normed vector space, so it suffice to show

- $X \subseteq B$ is closed,
- $B$ is a Banach space (i.e. complete).

**Step 1**:
we'll show $\cl_B(X) \subseteq X$.
Take $f$ to be a limit point of $X$, then for every $\eps>0$ there is a $g\in X$ with $\norm{f-g} < \eps$.
Apply the triangle inequality:
\[
\norm{f} \leq \norm{f-g+g}\leq \norm{f-g} + \norm{g} = \eps + C < \infty
,\]
so $f\in \cl_B(X) \subseteq B$ since it is bounded.
It remains to show $f$ is continuous.
Use that $\norm{f-g}\infty <\eps$ and continuity of $g$ to get $\abs{g(x) - g(x_0)} < \eps$ for $\abs{x-x_0}<\eps$.
Now
\[
\abs{f(x) - f(x_0)} 
&= \abs{f(x) - g(x) + g(x) -g(x_0) + g(x_0) - f(x_0)} \\
&\leq \abs{f(x) - g(x)} + \abs{g(x) - g(x_0)} + \abs{f(x_0) - g(x_0)} \\
&\leq 3\eps
.\]
So $X$ is closed in $B$.

**Step 2**:
Let $f_n$ be Cauchy in $B$, and note that we have a pointwise bound $\abs{f_n(x) - f_n(x_0)} \leq \norm{f_n - f_m} \to 0$.
So pointwise, $f_n(x)$ is a Cauchy sequence in $\CC$ which is complete, so $f_n(x) \to f(x)$ for some $f: \Omega\to \CC$.
We now want to show $f_n\to f$ in $X$.
Using that $f_n$ is Cauchy in $X$, produce an $N_0$ such that $n, m\geq N_0 \implies \norm{f_n - f_m}< \eps$.
Now
\[
\norm{f - f_n} 
&= \sup_{x\in \Omega} \abs{f(x) - f_n(x)} \\
&\leq \sup_{x\in \Omega} \sup_{m\geq N_0} \abs{f_m(x) - f_n(x)} \\
&= \sup_{m\geq N_0} \sup_{x\in \Omega} \abs{f_m(x) - f_n(x)} \\
&= \sup_{m\geq N_0} \norm{f_m - f_n} \\
&\leq \eps
.\]
Now use the reverse triangle inequality to show $f_n$ is bounded 
\[
\norm{f} - \norm{f_n} \leq \norm{f-f_n} < \eps \implies \norm{f} < \infty
.\]

Now by problem 1.13, every Cauchy sequence is bounded, so $f_n \to f\in B$. 
:::

:::{.remark}
Extending to vector-valued functions: for $\Omega \subseteq \RR^n$, take $\vector x = \tv{x_1, \cdots, x_n}$ and $F = \tv{f_1, \cdots, f_m}: \CC^n\to \CC^m$.
Then there is a Banach space
\[
X = C^(\Omega, \CC^m), \qquad \norm{f}_{C_1(\Omega)} \da \sum_{i\leq m} \sup_{x\in \Omega} \abs{f(x)} + \sum_{i\leq m, j\leq n} \sup_{x\in \Omega} \abs{\dd{f_i}{x_j}(x) }
.\]
Similarly define $L^p(\Omega)$, noting that $\norm{f}_{L^\infty(\Omega)}$ is the *essential supremum*.
:::

:::{.theorem title="?"}
For $p\in (1, \infty)$, the sequence space $X = (\ell^p, \norm{\wait}_{\ell^p})$ is a Banach space.
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:::{.definition title="Closed subspaces"}
A **closed subspace** of a Banach space is a linear subspace $M \leq B$ which is norm-closed in $B$.
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:::{.example title="?"}
\[
M \da \ker \divergence = \ts{f\in C(\Omega) \st \divergence f = 0} \leq C(\Omega)
\]
is closed, where $\divergence f$ is the divergence of a function $f$.

For any $S \subseteq B$, one can also take the corresponding closed subspace $\bar{[S]} \da \cl_B \spanof_\CC\ts{s\in S}$, i.e. all linear combinations of elements in $S$ and their limits.
This is called the **closed linear span** of $S$.
:::

:::{.exercise title="?"}
Let $B = ( C[a, b], \norm{\wait}_{L^\infty}$ and for $x_0 \in [a, b]$ define
\[
M \da \ts{f\in C[a, b] \st f(x_0) = 0}, \qquad
N \da \ts{f\in C[a, b] \st f(x_0) \leq c}
.\]
Show that these are closed subspaces with no nontrivial open subsets of $B$, since any $f\in M$ can be perturbed to be nonzero at $x_0$ with an arbitrarily small norm difference.
:::

:::{.remark}
Recall that for $S_1 \subseteq S_2 \subseteq B$, $S_1$ is **dense** in $S_2$ iff $\cl_{S_2}(S_1) = S_2$.
Recall Weierstrass' theorem: for $\Omega \subseteq \RR^n$ is closed and bounded and write $\OO \da \RR[x_1, \cdots, x_n]$ for the polynomials in $n$ variables.
Then $\OO \subseteq C(\Omega)$, and $\cl_{C(\Omega)} \OO = C(\Omega)$, i.e. $\OO$ is a dense subspace in the $L^\infty$ norms.
In fact, piecewise linear functions are dense.
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:::{.remark}
Norms are equivalent iff $c_1 \norm{f}_a \leq \norm{f}_b\leq c_2 \norm{f}_a$ for some constants $c_i$.
All norms on $\RR^n$ (resp. $\CC^n$) are equivalent.
:::

:::{.example title="?"}
For $a>0, f\in C[0, 1]$, define
\[
\norm{f} \da \sup_{x\in I} \abs{e^{-ax} f(x)}
,\]
which can be thought of as a weighting on the uniform norm which de-emphasizes the tails of functions near the endpoints.
This is equivalent to $\norm{\wait}_\infty$, since
\[
e^a \sup \abs{f} \leq \sup \abs{e^{-ax} f} \leq 1\cdot \sup \abs{f}
.\]
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:::{.remark}
Note that a basis for a norm can be used as a basis with respect to an equivalent norm in finite dimensions.
In infinite dimensions this may not hold -- e.g. for Fourier series, $\ts{e_k(x)}_{k\in \ZZ}$ is not a basis for $C[0, 2\pi]$ with the sup norm.
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:::{.definition title="Separable Banach spaces"}
An $B\in \Banach$ is **separable** iff $X$ contains a countable dense subset $S = \ts{f_k}_{k\geq 0}$ such that for each $f\in B$ and $\eps>0$, there is an $f_k\in S$ with $\norm{f-f_k} < \eps$.
:::

:::{.example title="?"}
Show that

- $\Omega \subseteq \RR^n$ a bounded subset, $C(\Omega), \norm{\wait}_{\sup}$ is separable.
- $\ell^p$ is separable for $p\in (1,\infty)$.
- $\ell^\infty$ is *not* separable.

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## Random Notes

:::{.remark}

![](figures/2022-01-20_14-48-53.png)

![](figures/2022-01-20_14-49-39.png)

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