Math-ing the place up

A convex function, with one secant line drawn in red.

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 F: [0,\infty) \to [0,\infty) 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 L^2 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. \int_0^{\infty}F'(t)^{-\frac{1}{m-1}}~dt\leq C, and

2. q\leq \frac{tF'(t)}{F(t)}\leq p for all t > 0.

A non-convex function, with a secant line drawn in passing beneath the function.

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 \frac{qF(t)}{t} \leq F'(t).  Solving this “differential inequality” gives us that F(t) \geq A_1t^q.  A similar calculation will also yield that F(t) \leq A_2t^p.

Now as a “back of the envelope” calculation, I tried plugging the bounds of F into property 1.  Specifically, first I computed

\int_0^{\infty} A_1qt^{\frac{q-1}{m-1}}dt.

Since q<m, the exponent is (strictly) smaller than 1, and the integral diverges (the indefinite integral looks like ct^{\alpha}, where 0<\alpha < 1).  In particular, it does great from 0 to any finite number, but has a “fat tail”.  Similarly, the integral \int_0^{\infty} A_1qt^{\frac{p-1}{m-1}}dt 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:

The constructed Young function, near where the two "pieces" join.

F(t) = c_1 t^q for small t, and F(t) = c_2t^p for large t.  You can even select c_1, c_2 so that the derivative is continuous.  Explicitly, suppose m = 3.  Then we may set F(t) = t for t < 1, and F(t) = \frac{t^5+4}{5} for t\geq 1.  Notice that F(t) is continuously differentiable, and that \int_0^{\infty}F'(t)^{-\frac{1}{m-1}}~dt = 2+1 = 3, so we know that C(3,1,5) \geq 3.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s