Let be a –finite measure space and be measurable functions. Hölder’s inequality asserts for , one has

One standard proof is to integrate the AM-GM inequality

Here, I provide an alternative proof. First consider the case where is the characteristic function of a set of finite measure. Let

Then

By choosing the optimal value for , we obtain

**Remark:** Note that if is bounded away from , we may improve the above argument. For instance, suppose for some and for all . Then

♠

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

It follows that

Taking roots and letting establishes Hölder’s inequality in the case .

**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 . ♠