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.