Holder Inequality

Holder's inequality

For $1\le p\le \infty$, $1\le q\le \infty$ with $\frac{1}{p}+\frac{1}{q}=1$, $f\in L^p(X,\mathcal{M},\mu )$,$g\in L^q(X,\mathcal{M},\mu )$, then $fg\in L^1(X,\mathcal{M},\mu )$ and ${\int }_X|fg(x)|d\mu (x)\le {\left({\int }_X|f(x){|}^{p}d\mu (x)\right)}^{\frac{1}{p}}{\left({\int }_X|g(x){|}^{q}d\mu (x)\right)}^{\frac{1}{q}}$

Remark

Proof of Holder’s inequality relies on generalized form of arithmetic-geometric mean inequality: For $A,B\ge 0$, $0\le \theta \le 1$, then $A^{\theta }{B}^{1-\theta }\le \theta A+(1-\theta )B$

Proof Consider $f_1=\frac{f}{\Vert f{\Vert }_{L^p}}$, $g_1=\frac{g}{\Vert g{\Vert }_{L^q}}$ $A=|f_1{|}^p$, $B=|g_1{|}^q$, by above inequality, $A^{\frac{1}{p}}{B}^{\frac{1}{q}}\le \frac{1}{p}A+\frac{1}{q}B$ so we have

$\int A^{\frac{1}{p}}{B}^{\frac{1}{q}}\le \int \frac{1}{p}A+\frac{1}{q}B,$ then $\int \frac{|f|}{\Vert f{\Vert }_{L^p}}\frac{|g|}{\Vert g{\Vert }_{L^q}}\le \frac{1}{p}\int \frac{|f{|}^p}{\Vert f{\Vert }_{L^p}^p}+\frac{1}{q}\int \frac{|g{|}^q}{\Vert g{\Vert }_{L^q}^q}=\frac{1}{p}+\frac{1}{q}=1.$

So we have

$$\int |fg| \leq |f|{L^{p}}|g|{{L^{q}}}$$$\square$

Multiple Holder Inequality

If $f_j\in L^{p_j}(X)$, where $X$ is a measure space, $j=1,\dots ,N$, and ${\sum }_{j=1}^{N}1/p_j=1$ with $p_j\ge 1$, then ${\Vert {\prod }_{j=1}^N{f}_j\Vert }_{L^1}\le {\prod }_{j=1}^N\Vert f_j{\Vert }_{L^{p_j}}$