Haven’t had a straight up “math” day in quite a while here. That ends now. While reading an article [PDF link], I came across the definition of a Young function: a function which is convex and where F(t) = 0 only when t = 0. Recall that for a real valued function, convex means that every secant line lies above the curve. So far, so good. Such a function might serve as a measure of how alike two functions are: rather than looking at (for example) the norm of u-v, we might first see what we can say about F(|u-v|), a more general question.
But here’s the proposition that made me pause: For 1<q<m<p, there is a differentiable Young function F and a constant C(m,p,q)>0 so that
1. , and
2. for all t > 0.
In fact, there was one more somewhat technical, but non-trivial assertion about this F (proposition 2.1 in the linked paper), but let’s focus on these two. Initially I was convinced that no such F existed, even satisfying these two conditions. Here’s how my erroneous thoughts went: suppose such an F were to exist. Property 2 gives us then that . Solving this “differential inequality” gives us that . A similar calculation will also yield that .
Now as a “back of the envelope” calculation, I tried plugging the bounds of F into property 1. Specifically, first I computed
Since q<m, the exponent is (strictly) smaller than 1, and the integral diverges (the indefinite integral looks like , where ). In particular, it does great from 0 to any finite number, but has a “fat tail”. Similarly, the integral diverges, but this time because its singularity near zero is too big (the indefinite integral is the same as the previous one, though now $\alpha < 0$. So this one does great from a very small number to infinity, but ultimately diverges.
Possibly I’ve given away the answer since I have emphasized my mis-steps, but here is an example of a Young function satisfying the first two properties. The trick is to construct the function out of two pieces:
for small t, and for large t. You can even select so that the derivative is continuous. Explicitly, suppose m = 3. Then we may set F(t) = t for t < 1, and for . Notice that F(t) is continuously differentiable, and that , so we know that .