Hölder's Inequality

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Theorem

InfoTheory

Let $(X,\mathscr{M},\mu)$ be a measure space, and let $f,g:X\to[0,+\infty]$ be measurable functions. Let $1<p,q<\infty$ be Conjugate Exponents then $\int\limits fg \, d\mu\le \left( \int\limits f^{p} \, d\mu \right)^{1/p}\left( \int\limits g^{q} \, d\mu \right)^{1/q}$ or equivalently: Let $1\le p\le \infty$ and suppose $p$ and $q$ are Conjugate Exponents. If $f\in L^{p}(X,\mathscr{M},\mu)$ and $g\in L^{q}(X,\mathscr{M},\mu)$ then $fg\in L^{1}(X,\mathscr{M},\mu)$ and we have $\|fg\|_{1}\le \|f\|_{p}\|g\|_{q}$

Lemma (474)

Assume $p>1$ and $q>1$ are s.t. $\frac{1}{p}+\frac{1}{q}=1$ . Let $u(x)\ge 0$ and $v(x)\ge 0$ satisfy $\int\limits _{-\infty}^{\infty}u(x)^{p} \, dx \ \ \ \int\limits _{-\infty}^{\infty}v(x)^{q} \, dx$ Then $\int\limits _{-\infty}^{\infty}u(x)v(x) \, dx \le \left( \int\limits _{-\infty}^{\infty} u(x)^{p} \, dx \right)^{1/p}\left( \int\limits _{-\infty}^{\infty} v(x)^{q} \, dx \right)^{1/q}$ or $\int\limits _{-\infty}^{\infty}u(x)v(x) \, dx \le \|u\|_{p}\|v\|_{q}$ Moreover, equality holds if and only if $v(x)^{q}=Cu(x)^{p}$ for some $C>0$ .

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A Summary of MATH 891

Bucklew and Wise