# Tuesday, February 15

:::{.remark}
Last time:

- Continuous operators are bounded:
  - If $\norm{Lf_n} = 1$ and $\norm{f_n} \to 0$, check $\lim (Lf_n) = L(\lim f_n) = L0 = 0$.
  - Take norms to contract $\norm{Lf_n} = 1$.
:::

:::{.theorem title="3.4.4"}
If $L: B\to C$ with dense image (so $\cl_B(D(L)) = B$), if $L$ is continuous on $D(L)$ then it has a unique extension $\tilde L$ to all of $B$, so $D(\tilde L) = B$, with $\norm{L} = \norm{\tilde L}$.
:::

:::{.proof title="of theorem"}
In steps:

- Defining the extension:
  - For $f\in B$, pick $f_n \to f$ with $f_n \in D(L)$ using density.
  - Convergent implies Cauchy, so estimate:
  \[
  \norm{Lf_n - Lf_m} = \norm{L(f_n - f_m)} \leq \norm{L} \norm{f_n - f_m} \to 0
  .\]
  - Thus $Lf_n$ is Cauchy, by completeness $Lf_n \to g$ for some $g$.

- Preservation of norm:
  - Define the extension as $\tilde L f \da g$, by continuity it is independent of the sequence $\ts{f_k}\to f$.
  - Check that $\tilde L$ is a bounded operator:
  \[
  \norm{\tilde L f} \da \norm{g} = \norm{\lim L f_n} = \lim \norm{Lf_n} \leq \lim \norm{L} \norm{f_n} = \norm{L} \norm{f} \\
  \implies \norm{\tilde L} \leq \norm{L}
  .\]
  - Since $\norm{A}_{\op}$ is defined in terms of sups over test functions in $D(A)$ for any operator $A$ and here $D(\tilde L) \contains D(L)$ is a larger set, we have $\norm{\tilde L} \geq \norm{L}$ by definition, yielding $\norm{\tilde L} = \norm{L}$.

- Uniqueness of the extension:
  - Take $\tilde L_1, \tilde L_2$ extending $L$, then
  \[
  \tilde L_1 f = \lim \tilde L_1 f_n = \lim L f_n = \lim \tilde L_2 f_n = \tilde L_2 f
  .\]
  - Use linearity:
  \[
  \tilde L_1 f - \tilde L_2 f = (\tilde L_1 - \tilde L_2)f = 0 \implies \tilde L_1 - \tilde L_2 = 0
  .\]





:::

:::{.example title="?"}
Let $\mcl \in L(\CC^n, \CC^n)$ be defined in coordinates by $(\mcl f)_i \da \sum_{1\leq j\leq n} \alpha_{ij} f_j$ for $1\leq i\leq n$.
Take $\norm{\wait}_{\ell^\infty}$ and check
\[
\norm{Lf}_\infty 
&\da \sup_i \abs{\sum \alpha_{ij} f_j} \\
&\leq \qty{ \sup_i \sum \abs{\alpha_{ij} } } \sup_j \abs{f_j} \\
&\da m \norm{f}_{\ell^\infty}
.\]
So $\norm{L} \leq m$, where $m$ is the largest row sum.
Is there an $f$ for which equality holds?
In this case, we'd need
\[
\norm{Lf}_{\ell^\infty} \geq m \norm{f}_{\ell^\infty}
.\]
Identify the row $k$ so that $m = \sum_{1\leq j\leq n} \abs{\alpha_{kj}}$.
Set $f$ to be a unit vector with coefficients $(f)_j = \bar{\alpha_{kj}} / \abs{\alpha_{kj}}$.
Then
\[
\norm{Lf}_\infty
&= \sup_i \abs{ \sum_j \alpha_{ij} f_j } \\
&\geq \abs{\sum_j \alpha_{kj} f_j } \\
&= \abs{\sum_j \alpha_{kj} {\bar \alpha_{kj} / \abs{\alpha_{k j}} } } \\
&= \sum_j \abs{\alpha_{kj}} \\
&= m \norm{f}_{\ell^\infty}
.\]

So the answer is yes in this case.
Does this also work for $\norm{\wait}_{\ell^p}$ with $p\in (1, \infty)$?
Recall Holder's inequality:
\[
\abs{\sum \alpha_{ij} f_j } 
&\leq \qty{ \sum \abs{\alpha_{ij}}^q }^{1\over q} \qty{\sum \abs{f_j}^p }^{1\over p} \\
&= \qty{ \sum \abs{\alpha_{ij}}^q }^{1\over q} \norm{f}_{\ell^p}
.\]
Check that
\[
\norm{Lf}_{\ell^p}^p
&= \sum_i \abs{(Lf)_i }^p \\
&= \sum_i \abs{\sum_j \alpha_{ij} f_j }^p \\
&\leq \sum_i \qty{ \sum_j \abs{\alpha_{ij} } }^{p\over q} \norm{f}_{\ell^p}^p
,\]
where we've applied Holder in the last line.
Thus
\[
\norm{L} \leq \qty{ \sum_i \qty{ \sum_j \abs{ \alpha_{ij}} }^{p\over q} }^{1\over p}
.\]

:::{.exercise title="?"}
Is there an $f$ that attains this bound in the $\ell^p$ case?
:::

:::

:::{.remark}
For $\mcl \in L(\CC^\infty, \CC^\infty)$ defined by $(Lf)_i = \sum_{j\geq 11} \alpha_{ij} f_j$ for $j\geq 1$, one needs a notion of convergence of the coordinates $\alpha_{ij}$ in order for $\mcl$ to be bounded.
A sufficient condition is $m\da \sup_i \sum_{j\geq 1} \abs{\alpha_{ij}} < \infty$.
:::

:::{.definition title="?"}
Some notation:

\[
\normm{\alpha}_1 &\da \sup_j \sum_i \abs{ \alpha_{ij}} \\
\normm{\alpha}_p &\da \qty{ \sum_i \qty{ \sum_j \abs{\alpha_{ij} }^{q} }^{p\over q} }^{1\over p} \\
\normm{\alpha}_\infty &\da \sup_i \sum_j \abs{ \alpha_{ij}}
.\]
:::

:::{.remark}
Note that if $\mcl: \ell^p\to \ell^p$, then $\norm{L} \leq \normm{\alpha}_p$ for $p\in [1, \infty)$ and for $p=\infty$ this is an equality.
:::

:::{.example title="Kernels"}
Consider $C[a, b]$ with $\norm{\wait}_\infty$ and $k\in C^0( [a,b]\cartpower{2}, \CC)$.
Define
\[
K: C[a, b] &\to C[a, b] \\
f &\mapsto \int_a^b k(x, y) f(y) \dy
.\]
What is $\norm{K}$?
Estimate
\[
\norm{Kf} 
&\leq \sup_{y\in [a, b]}\abs{f(y)}  \sup_{x\in [a, b]} \int_a^b \abs{k(x, y)} \dy \\
&\leq \norm{f}_\infty \norm{k}_\infty
,\]
so $\norm{K} \leq \norm{k}_\infty$.

Define
\[
\normm{k}_1 &\da \sup_y \int \abs{k} \dx \\
\normm{k}_p &\da \qty{\int \qty{ \int \abs{k}^q \dy }^{p\over q} \dx }^{1\over p} \\
\normm{k}_\infty &\da \sup_x \int \abs{k} \dy
.\]

:::