---
lecture: 17
date: 2025-03-13
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Let $(X, \mu)$ be a [[Measure|measure]] space. For $p > 0$ and a [[Measurable|measurable]] function $f : X \to \mathbb{C}$, then we define $\| f \|_{p} = (\int \mid f \mid^{p} \, d\mu)^{\frac{1}{p}}$ .
> [!note]
> $\mid f + g \mid \leq \mid f \mid + \mid g \mid$
> $\| f + g \|_{p} \leq \| f \|_{p} + \| g \|_{p}$

A pair of **conjugate exponents** is $(1, \infty)$, $(\infty, 1)$, or $(p, q)$ with $p, q \gt 0$ and $\frac{1}{p} + \frac{1}{q} = 1$. Note that $(2, 2)$ is okay (it also gives you a [[Hilbert Spaces|Hilbert space]]).
# Proposition
Let $f, g : X \to [0, \infty]$ be measurable on $(X, \mu)$. Then $\| f \times g \|_{1} \leq \| f \|_{p} \times \| g \|_{q}$ ([[Hölder's Inequality]]) and $\| f + g \|_{p} \leq \| f \|_{p} + \| g \|_{p}$ ([[Minkowski's Inequality]]) for any pair of [[Conjugate Exponents]] $(p, q)$ with $p, q \gt 0$.
## Proof
### Hölder's Inequality
To show [[Hölder's Inequality]], we may assume that $\| f \|_{p}$, $\| g \|_{q}$ are positive real numbers

> [!note] Need:
> $\Pi_{i=1}^{n} y_{i}^{a_{i}} \leq^{(*)} \sum_{i=1}^{n} a_{i} \times y_{i}$ when $a_{i}, y_{i} \gt 0$ and $\sum_{i = 1}^{n} a_{i} = 1$.

Use (* from the note above) with $a_{1} = \frac{1}{p}$, $a_{2} = \frac{1}{q}$, $y_{1} = \left(\frac{\mid f(x) \mid}{\| f \|_{p}}\right)^{p}$, $y_{2} = \left(\frac{\mid g(x) \mid}{\| g \|_{q}} \right)^{q}$ for $x$ such that $y_{i} \gt 0$.

Then
$$\frac{\mid f(x) \times g(x) \mid}{\| f \|_{p} \times \| g \|_{q}} \leq \frac{\mid f(x) \mid^{p}}{p \times \| f \|_{p}^{p}} + \frac{\mid g(x) \mid^{q}}{q \times \| g \|_{q}^{q}}$$
Integration gives
$$\frac{\| f \times g \|_{1}}{\| f \|_{p} \times \| g \|_{q}} \leq \frac{1}{p} + \frac{1}{q} = 1 \implies \text{Hölder's Inequality}$$
### Minkowski's Inequality
To get [[Minkowski's Inequality]], note that $\| f \times (f + g)^{p-1} \|_{1} \leq \| f \|_{p} \times \| (f + g)^{p - 1} \|_{q}$ by [[Hölder's Inequality]].

And similarly
$$\| g \times (f + g)^{p - 1} \|_{1} \leq \| g \|_{p} \times \| (f + g)^{p - 1} \|_{q}$$
Then combine the two
$$\| f + g \|_{p}^{p} = \| (f + g)^{p} \|_{1} = \| (f + g) \times (f + g)^{p-1} \|_{1} \leq \| f \times (f + g)^{p-1} \|_{1} + \| g \times (f + g)^{p-1} \|_{1}$$
$$\leq \| f \|_{p} \times \| (f + g)^{p-1} \|_{q} + \| g \|_{p} \times \| (f + g)^{p-1} \|_{q}$$
$$= \| (f + g)^{p-1} \|_{q} \times \left( \| f \|_{p} + \| g \|_{p} \right)$$
Then we have
$$\frac{\| f + g \|_{p}^{p}}{\| (f + g)^{p-1} \|_{q}} = \| f + g \|_{p}$$
(This is something that you can check)

And then we get the [[Minkowski's Inequality]].
Provided the denominator is $\neq 0, \infty$, this can be assumed: [[Minkowski's Inequality]] holds if $\text{LHS} = 0$. If $\text{RHS} \lt \infty$, then $\| f + g \|_{p} \lt \infty$ since $\left| \frac{(f + g)}{2} \right|^{p} \leq \frac{1}{2} \mid f \mid^{p} + \frac{1}{2} \mid g \mid^{p}$.

QED.
# $L^{p}(\mu)$-spaces
Let $p \in [1, \infty \rangle$. By [[Minkowski's Inequality]] the set of [[Measure|measure]] $f : X \to \mathbb{C}$ with $\| f \|_{p} \lt \infty$ on $(X, \mu)$ is a [[Complex Vector Space|complex vector space]] $V$. Note $\| a \times f \|_{p} = ( \int \underbrace{\mid a f \mid^{p}}_{\mid a \mid^{p} \times \mid f \mid^p} \, d\mu )^{\frac{1}{p}} = \mid a \mid \times \left( \int  \mid f \mid^{p} \, d\mu \right)^{\frac{1}{p}} = \mid a \mid \times \| f \|_{p}$.
> [!note]-
> 
> $$\| f \|_{p} = \left( \int \mid f \mid^{p} \, d\mu  \right)^{\frac{1}{p}}$$
> $$\| f + g \|_{p} \leq \| f \|_{p} + \| g \|_{p}$$

We have almost a [[Norm|norm]] $\| \cdot \|_{p}$ on $V$, except $\| f \|_{p} = 0 \centernot\implies f = 0$ almost everywhere.

In fact $f = 0$ **almost everywhere**, in that $\exists$[[Measure|measure]] $A \subset X$ such that $f\restriction_{A} = 0$ and $\mu(A^{\complement})= 0$.

> [!example]
> $\int f \, d\mu = \int_{A \cup A^{\complement}} f \, d\mu$
> $= \int_{A} f \, d\mu + \int_{A^{\complement}} f \, d\mu$

> [!example]
> $f \restriction_{A} = f$ on $A$.
> $g(x) = \begin{cases} f(x), & \forall x \in A\\ 0, & \forall x \in A^{\complement}, A = f^{-1}(\mathbb{C} \setminus \{ 0 \})\end{cases}$

> [!example]
> $\int g \, d\mu = \int_{A} g \, d\mu + \int_{A^{\complement}} g \, d\mu = \int_{A} g \, d\mu = \int f \, d\mu$

Define equivalence relation on $V$ by $f \sim g$ if $f - g = 0$ almost everywhere.
Then $V \setminus \sim$ will be a [[Vector Space|vector space]] and with [[Norm|norm]] $\| \, [ \, f \, ] \, \|_{p} \equiv \| \, f \, \|_{p}$. Then $\| \, [ \, f \, ] \, \|_{p} = 0 \implies f = 0$ almost everywhere, so $[\, f \, ] = [\, 0 \,] = 0$.

# Theorem
$L^{p}(\mu) = V \setminus \sim$ is a [[Banach Space]].
$L^{2}(\mu)$ is a [[Hilbert space]] with $(f \mid g) = \int f \times \bar{g} \, d\mu$.

---
# Exercises
This is the exercise part of the lecture.
# Exercise 3.3.6 Question 2
$(X, \mu)$, $f : X \to [0, \infty \rangle$ [[Measurable|measurable]] such that $\int f \, d\mu = 0$.
Show that $f = 0$ almost everywhere.
**Hint:** $A \equiv \{ x \mid f(x) \gt 0 \} = \cup_{n = 1}^{\infty} A_{n}$, where $A_{n} = \underbrace{\left\{  x \mid f(x) \gt \frac{1}{n}  \right\}}_{f^{-1} \left( \langle \frac{1}{n}, \infty \rangle \right)} \in \sigma$.
## Proof
If $\mu(A) \gt 0$, then $0 \lt \mu(A) \lt \sum_{n=1}^{\infty} \mu(A_{n})$, so $\mu(A_{m}) \gt 0$ for some $m$.

Then $\int f \, d\mu \geq \int_{A} f \, d\mu \geq \int_{A_{m}} f \, d\mu \geq \int_{A_{m}} \frac{1}{m} \, d\mu = \frac{1}{m} \times \mu(A_{m}) \gt 0$, which is a contradiction.

So $\mu(A) = 0$ and $f = 0$ almost everywhere.

QED.
# Exercise 3.4.5 Question 3
What is $L^{p}(\mu)$ for $p \in [1, \infty \rangle$ when $\mu$ is the [[Measure|counting measure]]?
> [!note] On "counting measure"
> I thing "counting measure" is the same way that has been said in short hand with "measure" for all the previous lectures.

## Proof
$f : X \to \mathbb{C}$ all functions. $\sigma = \wp(X)$.

> [!note] What is $L^{p}(\mu)?$
> $\| f \|_{p} = (\int \mid f \mid^{p} \, d\mu)^{\frac{1}{p}}$
> $V = \left\{  f : X \to \mathbb{C} \mid \int \mid f \mid^{p} \, d\mu \lt 0  \right\}$
> $L^{p}(\mu) = V \setminus \sim$

Say $g : X \to [0, \infty \rangle$.
Then (define: $s = \sum_{i = 1}^{n} a_{i} X_{A_{i}}$) $\int g \, d\mu = \sup_{0 \leq s \leq g} \int s \, d\mu = \sup_{0\leq s\leq g} \sum_{i=1}^{n} \overbrace{a_{i}}^{\neq 0} \times \mu(A_{i}) = \infty$ if any $A_{i}$ is infinite.

For any finite subset $F$ of $X$, define
$$S_{F}(x) = \begin{cases}
g(x), & x\in F \\
0, & x \not\in F
\end{cases}$$
Then $S_{F}$ is a [[Simple Function|simple function]], and $0\leq S_{F} \leq g$. Have $\int S_{F} \, d\mu = \int_{F} g \, d\mu = \sum_{x \in F} g(x)$

$S_{F} = \sum_{x \in F} g(x) X_{\{ x \}}$ since $\int S_{F} \, d\mu = \sum_{x \in F} g(x) \mu \underbrace{(\{ x \})}_{1}$
$\sum_{x \in F} g(x) \times \underbrace{X_{\{ x \}} (y)}_{\delta x, y} = g(y) = S_{F}(y)$

$X = \mathbb{N}$
$F = \{ 1, \dots, n \}$

Then $\int g \, d\mu = \sup_{0 \leq s \leq g} \int s \, d\mu \geq \sup_{\text{F finite}} \sum_{x \in F} g(x) \overbrace{=}^{x =\mathbb{N}} \sum_{n=1}^{\infty} g(n)$
($\mathbb{N}$, $\mu$ [[Continuous|continuous]] [[Measure|measure]]) $\implies l^{p}(\mathbb{N}) = L^{p}(\mu) = \left\{  \{ x_{n} \} \mid \sum_{n=1}^{\infty} |x_{n}|^{p} \lt \infty  \right\}$

$\| f \|_{p} = 0 \implies f = 0$ almost everywhere $\iff f = 0$ since $\mu(A) \gt 0 \; \forall A \neq \emptyset$
$V \setminus \sim = V$