Originally published at 狗和留美者不得入内. You can comment here or there.

I had, a while ago, the great pleasure of going through the proof of the Riesz-Thorin interpolation theorem. I believe I understand the general strategy of the proof, though for sure, I glossed over some details. It is my hope that in writing this, I can fill in the holes for myself at the more microscopic level.

Let us begin with a statement of the theorem.

**Riesz-Thorin Interpolation Theorem.** *Suppose that and are measure spaces and . If , suppose also that is semifinite. For , define and by*

*.*

*If is a linear map from into such that for and for , then for , .*

We begin by noticing that in the special case where ,

,

wherein the first inequality is a consequence of Holder’s inequality. Thus we may assume that and in particular that .

Observe that the space of all simple functions on that vanish outside sets of finite measure has in its completion for and the analogous holds for . To show this, take any and any sequence of simple that converges to almost everywhere, which must be such that , from which follows that they are non-zero on a finite measure. Denote the respective spaces of such simple functions with and .

To show that for all , we use the fact that

,

where is the conjugate exponent to . We can rescale such that .

From this it suffices to show that across all with and , .

For this, we use the three lines lemma, the inequality of which has the same value on its RHS.

**Three Lines Lemma****.** *Let be a bounded continuous function on the strip that is holomorphic on the interior of the strip. If for and for , then for , .*

This is proven via application of the maximum modulus principle on for . The serves of function of as for any .

We observe that if we construct such that for some . To do this, we can express for convenience and where the ‘s and ‘s are nonzero and the ‘s and ‘s are disjoint in and and take each to power for such a fixed for some with . We let be the value corresponding to the interpolated . With this, we have

.

Needless to say, we can do similarly for , with ,

.

Together these turn the LHS of the inequality we desire to prove to a complex function that is

.

To use the three lines lemma, we must satisfy

.

It is not hard to make it such that . A sufficient condition for that would be integrands associated with norms are equal to and respectively, which equates to and . Similarly, we find that and . From this, we can solve that

.

With these functions inducing a that satisfies the hypothesis of the three lines lemma, our interpolation theorem is shown for such simple functions, from which extend our result to all .

To extend this to all of , it suffices that a.e. for some sequence of measurable simple functions with and pointwise. Why? With this, we can invoke Fatou’s lemma (and also that by dominated convergence theorem) to obtained the desired result, which is

.

Recall that convergence in measure is a means to derive a subsequence that converges a.e. So it is enough to show that for all . This can be done by upper bounding with something that goes to zero. By Chebyshev’s inequality, we have

.

However, recall that in our hypotheses we have constant upper bounds on in the and norms respectively assuming that is in and , which we can make use of. So apply Chebyshev on any one of (let’s use this) and , upper bound its upper bound with or times , which must go to zero by pointwise convergence.