Holder's inequality

Given two n-tuples $(x_1,x_2,\dots ,x_n)$ and $(y_1,y_2,\dots ,y_n)$, and $p,q\ge 1$ such that $\frac{1}{p}+\frac{1}{q}=1$, we have Holder’s inequality:

$$\sum _{j=1}^n|x_j{y}_j|\le {\left(\sum _{j=1}^n|x_j{|}^p\right)}^{1/p}{\left(\sum _{j=1}^n|y_j{|}^q\right)}^{1/q}$$

If we denote the two tuples as $\mathbf{x}$ and $\mathbf{y}$, the inequality can be expressed as

$$\sum _{j=1}^n|x_j{y}_j|\le \Vert \mathbf{x}{\Vert }_p\Vert \mathbf{y}{\Vert }_q$$

Theorem 1(Lemma).

If $A,B>0$, then $AB\le \frac{1}{p}{A}^p+\frac{1}{q}{B}^q$.

Proof
We know that $f(x)=\ln (x)$ is concave. Thus,

$$\begin{array}{rl}\ln (\lambda x+(1-\lambda )y)\ge & \lambda \ln (x)+(1-\lambda )\ln (y)\end{array}$$

Plug in $x=A^p$, $y=B^q$, $\lambda =\frac{1}{p}$.

$$\begin{array}{r}\ln \left(\frac{1}{p}{A}^p+\frac{1}{q}{B}^q\right)\ge \ln A+\ln B=\ln AB\end{array}$$

Now, let

$$\begin{array}{rl}{A}_i& =\frac{|x_i|}{\Vert \mathbf{x}{\Vert }_p}\\ B_i& =\frac{|y_i|}{\Vert \mathbf{y}{\Vert }_q}\end{array}$$

So, we have the inequalities

$$\begin{array}{r}\frac{|x_i{y}_i|}{\Vert \mathbf{x}{\Vert }_p\Vert \mathbf{y}{\Vert }_q}\le \frac{1}{p}\frac{|x_i{|}^p}{\Vert \mathbf{x}{\Vert }_p^p}+\frac{1}{q}\frac{|y_i{|}^q}{\Vert \mathbf{y}{\Vert }_q^q}\end{array}$$

If we add these inequalities over $i$, we get

$$\begin{array}{r}\frac{{\sum }_{i=1}^n|x_i{y}_i|}{\Vert \mathbf{x}{\Vert }_p\Vert \mathbf{y}{\Vert }_q}\le \frac{1}{p}+\frac{1}{q}=1.\end{array}$$

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Note how Holder’s generalizes the Cauchy Schwarz inequality: for $p=q=\frac{1}{2}$, we have

$$\begin{array}{rl}\left|\sum _{j=1}^n{x}_j{y}_j\right|& \le \sum _{j=1}^n|x_j{y}_j|\le {\left(\sum _{j=1}^n|x_j{|}^2\right)}^{1/2}{\left(\sum _{j=1}^n|y_j{|}^2\right)}^{1/2}\\ |⟨\mathbf{x},\mathbf{y}⟩|& \le \Vert \mathbf{x}{\Vert }_2\Vert \mathbf{y}{\Vert }_2\end{array}$$