Hölder’s Inequality

Analysis,

Let {(\Omega,\mu)} be a {\sigma}–finite measure space and {f,g : \Omega \rightarrow \mathbb{C}} be measurable functions. Hölder’s inequality asserts for {p,q \in [1, \infty]}, {p^{-1} + q^{-1} = 1} one has

\displaystyle \int |fg| \leq ||f||_{p} ||g||_{q}.

One standard proof is to integrate the AM-GM inequality

\displaystyle \frac{|f(x)||g(x)|}{||f||_{p} ||g||_q} \leq \frac{|f(x)|^p}{p||f||_p^p} + \frac{|g(x)|}{q ||g||_q^q}.

Here, I provide an alternative proof. First consider the case where {g = \chi_E} is the characteristic function of a set {E} of finite measure. Let

\displaystyle f_{<M} = f(x) 1_{f(x) < M} , \ \ \ \ f_{\geq M}(x) = f(x) 1_{f(x) \geq M} .

Then

\displaystyle \int |f| \chi_E = \int |f_{<M} |\chi_E + \int |f_{\geq M} |\chi_E \leq M \mu(E) + \frac{1}{M^{p-1}} \int |f|^p.

By choosing the optimal value for {M= \frac{||f||_p}{\mu(E)^{\frac{1}{p}}}}, we obtain

\displaystyle \int |f| \chi_E \leq 2 \mu(E)^{1-\frac{1}{p}} ||f||_p.\ \ \ \ \ (1)

Remark: Note that if {|f|} is bounded away from {\lambda}, we may improve the above argument. For instance, suppose {f(x) \notin [M^{-1} \lambda , M^{\frac{1}{p-1}} \lambda]} for some {M\geq 1} and for all {x \in \Omega}. Then

\displaystyle \int |f| \chi_E \leq 2 M^{-1} \mu(E)^{1-\frac{1}{p}} ||f||_p.

♠

We can get rid of the constant 2 in (1) by using the tensor product trick. That is, apply (1) to

\displaystyle F: \Omega^n \rightarrow \mathbb{C} , \ \ \ \ \ \ \ F(x_1 , \ldots , x_n) = f(x_1) \cdots f(x_n).

It follows that

\displaystyle \left(\int |f| \chi_E \right)^n = \int |F|\chi_{E^n} \leq \mu(E^n)^{1-\frac{1}{p}} ||F||_p = 2 \left(\mu(E)^{1-\frac{1}{p}} ||f||_p\right)^n.

Taking {n^{\rm th}} roots and letting {n \rightarrow \infty} establishes Hölder’s inequality in the case {g = \chi_E}.

Remark: In many applications, (1) is enough. Nevertheless, extending to simple functions is an easy matter and Hölder’s inequality follows by the density of simple functions in {L^p}. ♠

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