Reading Difficulty: 6/5
Prerequisites: Know what a Sobolev space is

When we have that continuously embeds into . For the sake of my fingers I'm going to drop the ; all spaces referenced in this post are over unless otherwise specified. Anyways, I have a hard time remembering the proof. I think the following approach is far easier to remember and more useful since it relies on a result which is useful in other areas.

The first "step" is to prove the following very cool characterization of Holder continuity. Throughout, we use the notation to mean the average integral
where is the measure of the -ball. We also write to mean that there exists a constant depending only on such that . Also, to avoid writing or et cetera, all unimportant "constants" hereafter are referred to as .

Theorem (Campanato's Criterion): Let , . Then iff there exists such that
for all and all .


Written plainly: A function is -Holder iff the -average of the deviation from the -average is controlled by .

(For a reason that we'll see in the remarks, we'll be keeping track of the constant, which is why it's special.)

Proof.

The forward direction is a simple exercise, so we'll just consider the backward direction. Assume that holds for every and .

Step 1

We begin by investigating how much the average changes when the radius changes. To wit, fix and take with . What can we say about ? Since we want to use , it makes sense to take an arbitrary and go up with the triangle inequality:
Now we average the RHS over (the smaller ball). From there, the bounding here is quite natural:
So . That's a nice estimate.

Step 2

Now we use this estimate to get information on how differs from the average around . Fix a Lebesgue point of (since they are the only points where a pointwise value of makes sense) and . What can we say about ? The natural idea is that the value is recovered by the limit of as . So we apply Step 1 for very small radii. Particularly, we apply Step 1 for , etc. An "infinite triangle inequality" gives
and the sum bounds as
So . Good.

Step 3

We're basically ready to show is -Holder now. Fix Lebesgue points. We're done if we show (because then there is clearly a -Holder extension from the Lebesgue points by density, and this is the desired Holder continuous representative). Step 2 gives
Well hey, we've never actually chosen . The obvious value to take is , so let's do that. Now it remains to prove that is controlled by for this value of . In particular we'll show that it's .

Again, we need to use , and to do that we use the same trick of taking an arbitrary point and going up by triangle inequality.
Now average the RHS over . Note that since , the measure of is times some constant depending only on . Hence
Now grow the domain of integration to get
and this is by .

A few important remarks:

• Thanks to our dilligence in keeping track of , we've actually shown that
  for a constant depending only on . That is, . So a more descriptive statement of the criterion is that the quantity
  is "equivalent" to the Holder seminorm in the sense that .
• Campanato generalizes to other open domains. I'm pretty sure that for any , we have that holds with replaced with iff is Holder continuous on . I can't guarantee that the previous remark still holds though, I'll leave that to you to figure out.

With this criterion proven, Morey's embedding falls out pretty much immediately.

Proof. Take . Then by the Poincare inequality,
So
giving us that by Campanato. In fact, the above inequality tells us that , so by the first remark we may deduce that
So we've tamed the seminorm. As for the sup norm, we write
and averaging the RHS over gives
Thus , which completes the proof that .