# Tuesday November 19

## Lp Spaces

Given $f:\RR^n \to \CC$ and $0 < p < \infty$, we define
$$
\norm{f}_p = \left( \int \abs{f}^p \right)^{1/p}.
$$

and $L^p(\RR^n) = \theset{f \suchthat \norm{f}_p \infty }$.

We also define 
$$
\norm{f}_\infty = \inf_{a \geq 0} \theset{ m(\theset{x \suchthat f(x) > a}) = 0 }
$$

which is morally the "best upper bound almost everywhere".

> **Qual problem alert**: If $X \subseteq \RR^n$ with $\mu(X) < \infty$ then $\norm{f}_p \to \norm{f}_\infty$.

Note that $\abs{f(x)} \leq \norm{f}_\infty$ almost everywhere, and if $\abs{f(x)} \leq M$ almost everywhere, then $\norm{f}_\infty \leq M$.

For $1 \leq p \leq \infty$, $( L^p, \norm{\wait}_p)$ yields a complete normed vector space.
Scaling and non-degeneracy are fairly clear, it just remains to show the triangle inequality (sometimes referred to as Minkowski's inequality), i.e. 
$$
f, g\in L^p \implies \norm{f+g}_p \leq \norm{f}_p + \norm{g}_p.
$$
For $p=2$, this boiled down to Cauchy-Schwarz, here we'll need a souped-up version.

**Definition:**
If $1 \leq p \leq \infty$, we define the *conjugate exponent* of $p$ as the $q$ satisfying $\frac 1 p + \frac 1 q  = 1$.

An immediate consequence is that 
$$
q = \frac{p}{p-1}
.$$

**Holder's Inequality:**
If $f, g$ are measurable functions then 
$$
\norm{fg}_1 \leq \norm{f}_p \norm{g}_q.
$$


*Proof of Minkowski:*

\begin{align*}
\abs{f + g}^p 
&= \abs{f+g} \abs{f+g}^{p-1} \\
&\leq ( \abs{f} + \abs{g} ) \abs{f+g}^{p-1} \\
&\implies \int \abs{f+g}^p \leq \int \abs{f} \abs{f+g}^{p-1} + \int \abs{g} \abs{f + g}^{p-1} \\
&\leq \norm{f}_p ( \int \abs{f+g}^{(p-1)q}  )^{1/q}
+ \norm{g}_p ( \int \abs{f+g}^{(p-1)q}  )^{1/q} \\
&= (\norm{f}_p + \norm{g}_p) + ( \int\abs{f+g}^p )^{1 - 1/p} \\
&= (\norm{f}_p + \norm{g}_p) + ( \int\abs{f+g}^p )^{1/q}
,\end{align*}

and taking $p$th roots yields the result. (?? Revisit)
$\qed$

> Note: For $1\leq p \leq \infty$, $L^p$ is a Banach space.

> AM-GM: $\sqrt{ab} \leq \frac{a+b}{2}$.

*Proof of Holder:*

We'll use the following key fact:
$$
a^\lambda b^{1-\lambda} \leq \lambda a + (1-\lambda)b
$$

with equality iff $a=b$. This can be verified by the first derivative test.

**Important simplfication:** 
we can assume that $\norm{f}_p = \norm{g}_q = 1$, since 
$$
\norm{fg}_1 \leq \norm{f}_p \norm{f}_q \iff \int \frac{\abs{f}}{\norm{f}_p} \frac{\abs{g}}{\norm{g}_q}
.$$

Applying the key fact, we can choose $\lambda = \frac 1 p, a =\abs{f}^p, b = \abs{g}^q$.

We then obtain
$$
\int \abs{f} \abs{g} \leq \int \frac{\abs{f}^p}{p} \frac{\abs{g}^q}{q} = \frac 1 p + \frac 1 q = 1.
$$
$\qed$

## Dual of $L^p$

Given $g\in L^q$, define an operation

\begin{align*}
\Lambda_g(f): L^p \to \CC \\
f \mapsto \int f g
.\end{align*}

Note that this makes sense by Holder's inequality, i.e. $fg$ is integrable.
This defines a linear functional on $L^p$, which is continuous by Holder, since we have 
$$
\abs{\Lambda_g(f)} \leq \norm{g}_q \norm{f}_p
$$ 
where $\norm{g}_q$ is a constant that works for all $f \in L^p$.

> Recall: linear functionals are continuous iff bounded.

We have 
$$
\norm{\Lambda_g}_{(L^p)\dual} \definedas \sup_{\norm{f}_p = 1} \abs{\int fg} \leq \norm{g}_q
.$$

> In fact, we have equality here for every $g\in L^q$. This is sometimes referred to as the converse of Holder.

Thus the map $g \mapsto \Lambda_g$ is an *isometric* map $L^q \injects (L^p)\dual$ for $1 \leq p,q \leq \infty$.

By Riesz representation, it turns out that this is a surjection as well for $p\neq \infty$.

**Big Fact**: 
This breaks for $p=\infty$, but for $1 \leq p < \infty$, this mapping is surjective.

## Riesz Representation
**Theorem (Riesz Representation):**
Suppose $1\leq p < \infty$ and let $q$ be its conjugate exponent, and let $X\subseteq \RR^n$ be measurable.

Given any $\Lambda \in (L^p(X))\dual$, there exists a unique $g\in L^q(X)$ such that for all $f\in L^p(X)$, we have 
$$
\Lambda(f) = \int_X fg
\quad \text{ and } \quad 
\norm{\Lambda}_{(L^p(X))\dual} = \norm{g}_{L^q(X)}
.$$

**Summary:**

- If $1\leq p < \infty$, we have $(L^p)\dual = L^q$.
- $(L^\infty)\dual \supset L^1$, since the isometric mapping is always injective, but *never* surjective, so this containment is always proper (requires Hahn-Banach Theorem).

> For qual, supposed to know this for $p=1$. $p=2$ case is easy by Riesz Representation for Hilbert spaces.

*Proof (in the special case where $1\leq p < 2$ and $m(X) < \infty$):*

We'll use the fact that we know this for $p=2$ already.

Let $\Lambda \in (L^p)\dual$, then we know 
$$
\abs{\Lambda(f)} \leq \norm{\Lambda}_{(L^p)\dual} \norm{f}_p
,$$ 
since $\norm{\Lambda}$ is the *best* upper bound.

> Note: in general, there are no inclusions between $L_p, L_q$, but restricting to a compact set changes this fact. 
Example from homework: 
$$
L^2(X) \subseteq L^1(X) \text{ for } m(X) < \infty
.$$

> This follows from 
$$\norm{f}_1 \leq \norm{f}_2 \norm{1}_2 = m(X)^{1/2} \norm{f}
.$$ 
But this works for $L^2(X) \subseteq L^p(X)$ by taking 
$$
\norm{f}_p^p = \int \abs{f}^p \leq ( \int \abs{f}^2 )^{p/2} (\int \abs{1}^{?})^{1 - \frac 2 p}
$$ 
by Holder with $\frac 2 p$.

So we can write
$$
\abs{\Lambda(f)} = \norm{\Lambda} m(X)^{\frac 1 p - \frac 1 2} \norm{f}_2 \quad \forall f\in L^2,
$$

which verifies that $\Lambda$ is a continuous linear functional on $L^2$, so $\Lambda \in (L^2)\dual$, and by Riesz Representation, $\exists g \in L^2$ such that $\Lambda(f) = \int fg$ for all $f\in L^2$.

This is almost what we want, but we need $g\in L^q$ and $f\in L^p$. 
We also want to show that $\norm{\Lambda} = \norm{g}_q$.

**Claim**:
$g\in L^q$ and $\norm{g}_q \leq \norm{\Lambda}$.

> Pause on the proof, we'll come back to it!

Note that since $L_2 \subseteq L^p$ and both have simple functions as a dense subset, $L^2$ is in fact dense in $L^p$.

So let $f \in L^p$ and pick a sequence $f_n \subset L^2$ converging to $f$ in the $L^p$ norm.

Then $\Lambda(f_n) \to \Lambda(f)$ by continuity, and since $g\in L^q$, integrating against $g$ is a linear functional $\Lambda_q (f_n)$ on $L^q$ converging to $\int f g$, so $\Lambda(f) = \int fg$. 
$\qed$

> Definitely need to know: $(L^1)\dual = L^\infty$!

*Proof of claim:*
Suppose it's not true, so $\norm{g}_\infty > \norm{\Lambda}_{(L^1)\dual}$.

Using the fact that $\norm{g}$ is the best lower bound, there must be a positive measure set such that $\abs{g(x)} \geq \norm{\Lambda}$.
So there is some set $E = \theset{x \suchthat \abs{g(x) > \norm{\Lambda}}}$ with $m(E) > 0$.

Let 
$$
h = \frac{\overline{g}}{\abs{g}} \frac{\chi_E}{m(E)}
.$$

> Note: useful technique!

Then $h\in L^2$ and $\norm{h} = 1$.

Then 
$$
\Lambda(h) = \frac{1}{m(E)} \int_E \abs{g} \geq \norm{\Lambda} O(1)
,$$ 
which is a contradiction.