Holder's inequality

Theorem (Holder’s inequality) Suppose $(X,\mathcal{A},\mu )$ is a measure space, $1\le p\le +\infty$, and $u,v:X\to F$ are measurable. Then

$$\Vert uv{\Vert }_1\le \Vert u{\Vert }_p\Vert v{\Vert }_{p^{\prime }}.$$

Proof The cases $p=1,\infty$ are trivial. Suppose $1<p<\infty$. Consider the special cases where $u$ and $v$ are normalized, $\Vert u{\Vert }_p=\Vert v{\Vert }_{p^{\prime }}=1$, using Young’s inequality on the absolute values

$$|uv|\le \frac{|u{|}^p}{p}+\frac{|v{|}^{p^{\prime }}}{p^{\prime }}$$

integrating both sides

$$\Vert uv{\Vert }_1\le \frac{1}{p}+\frac{1}{p^{\prime }}=1=\Vert u{\Vert }_p\Vert v{\Vert }_{p^{\prime }}$$

If one of the norms are zero or infinity the inequality works, so assume that $0<\Vert u{\Vert }_p,\Vert v{\Vert }_{p^{\prime }}<\infty$.

We can normalize such funtions! Just apply the special case. $\square$