Hölder's Inequality

Definition (Conjugate Exponents)

Let 1<p<∞1<p<\infty. If 1p+1q=1\frac{1}{p}+\frac{1}{q}=1then we call pp and qq conjugate exponents.

Theorem (Hölder)

Let (X,M,μ)(X,\mathscr{M},\mu) be a Measure Space, and let f,g:X→[0,+∞]f,g:X\to[0,+\infty] be Measurable Functions. Let 1<p,q<∞1<p,q<\infty be Conjugate Exponents then ∫fg dμ≤(∫fp dμ)1/p(∫gq dμ)1/q\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≤p≤∞1\le p\le \infty and suppose pp and qq are Conjugate Exponents. If f∈Lp(X,M,μ)f\in L^{p}(X,\mathscr{M},\mu) and g∈Lq(X,M,μ)g\in L^{q}(X,\mathscr{M},\mu) then fg∈L1(X,M,μ)fg\in L^{1}(X,\mathscr{M},\mu) and we have ∥fg∥1≤∥f∥p∥g∥q\|fg\|_{1}\le \|f\|_{p}\|g\|_{q}

Lemma ((474))

Assume p>1p>1 and q>1q>1 are s.t. 1p+1q=1\frac{1}{p}+\frac{1}{q}=1. Let u(x)≥0u(x)\ge 0 and v(x)≥0v(x)\ge 0 satisfy ∫−∞∞u(x)p dx   ∫−∞∞v(x)q dx\int\limits _{-\infty}^{\infty}u(x)^{p} \, dx \ \ \ \int\limits _{-\infty}^{\infty}v(x)^{q} \, dx Then ∫−∞∞u(x)v(x) dx≤(∫−∞∞u(x)p dx)1/p(∫−∞∞v(x)q dx)1/q\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 ∫−∞∞u(x)v(x) dx≤∥u∥p∥v∥q\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)pv(x)^{q}=Cu(x)^{p} for some C>0C>0.

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