Quartz sync: Mar 13, 2025, 1:03 PM

content/Definitions/Measure Theory/Hölder's Inequality.md

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+$\| f \times g \|_{1} \leq \| f \|_{p} \times \| g \|_{q}$

content/Definitions/Measure Theory/Minkowski's Inequality.md

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+$\| f + g \|_{p} \leq \| f \|_{p} + \| g \|_{p}$

content/Definitions/Measure Theory/Sigma-Algebra/Borel Sets.md

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 # Definition
-The $\sigma$[[Sigma-Algebra|-algebra]] generated by the [[Open Sets|open sets]] in a [[Topological Space|topological space]] $X$ is the $\sigma$[[Sigma-Algebra|-algebra]] of **Borel Sets** of $X$.
+The $\sigma$[[Sigma-Algebra|-algebra]] generated by the [[Open Sets|open sets]] in a [[Topological Space|topological space]] $X$ is the $\sigma$[[Sigma-Algebra|-algebra]] of **Borel Sets** of $X$.

content/Definitions/Measure Theory/Sigma-Algebra/Borel Sigma-Algebra.md

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 # Definition
-**Borel $\sigma$-algebra** on a [[Topological Space|topological space]] $X$ is the one generated by the [[Open Sets|open sets]] $\tau$.
+**Borel $\sigma$-algebra** on a [[Topological Space|topological space]] $X$ is the one generated by the [[Open Sets|open sets]] $\tau$.

content/Lectures/Lecture 17 - Lp Spaces.md

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+---
+lecture: 17
+date: 2025-03-13
+---
+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:
+> $\Pi_{i=1}^{n} y_{i}^{a_{i}} \leq^{(*)} \Sigma_{i=1}^{n} a_{i} \times y_{i}$ when $a_{i}, y_{i} \gt 0$ and $\Sigma_{i = 1}^{n} a_{i} = 1$.
+
+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 $(f|g) = \int f \times \bar{g} \, d\mu$.

content/index.md

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 - [[Lecture 13 - Measure Theory]]
 - [[Lecture 14]]
 - [[Lecture 15]]
-- [[Lecture 16]]
+- [[Lecture 16]]
+- [[Lecture 17 - Lp Spaces]]