ACIT4330-Page/content/Lectures/Lecture 17 - Lp Spaces.md

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17	2025-03-13

Let (X, \mu) be a Measure space. For p > 0 and a 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).

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 f : X \to \mathbb{C} with \| f \|_{p} \lt \infty on (X, \mu) is a 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 \| \cdot \|_{p} on V, except \| f \|_{p} = 0 \;\not\!\!\!\implies f = 0 almost everywhere.

In fact f = 0 almost everywhere, in that $\exists$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 and with 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.

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Exercises

This is the exercise part of the lecture.

Exercise 3.3.6 Question 2

(X, \mu), f : X \to [0, \infty \rangle 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?

[!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, 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 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