In order to test the latex2wp software, I’ve dug out an old file and will post it here. There might be something interesting here. But maybe not.
1. Kashin’s Inequality
Kashin’s Inequality: Let be a Lipschitz function, i.e., there is a constant such that
Note that the left hand-side is also known as the Besov semi-norm . The constant is probably not sharp. However, the function shows that it’s at least .
Proof: Use the Lipschitz hypothesis on and integrate in first to get
Likewise, on we integrate in first.
Now we need to consider . Majorize by and split into two integrals. Then, considering first, change variables with , and integrate in for to get
Likewise, for use the lower bound instead and integrate in first.
2. Complex variables approach
Here we follow the lead of Theorem 2-5 page 32 of Ahlfors’s book Conformal Invariants (hat tip to Don Marshall for this reference).
Given , Lipschitz with constant , let its harmonic extension to the upper half-plane be:
Then there is analytic on , which is given by the following formula
To see this one checks that
Now , where and .
Note first that
for fixed, as .
On the other hand, by the residue theorem,
So, for ,
Moreover, Cauchy’s theorem (on similar contours) shows that, for fixed ,
Expanding the square we therefore see that
By Lebesgue dominated convergence we can let tend to zero and get that
The middle integral needs justification, e.g., it is ok if .
In particular, we get
- The Besov semi-norm above can be interpreted as a Dirichlet energy by changing variables:
and, at least for smooth functions,
where the kernel is quite similar to the Poisson kernel.
- By Plancherel for the Hilbert transform, we have that the Dirichlet integral of
- Green’s theorem applied to the function and , and the fact that for harmonic functions, gives:
where we have skipped many technical details and is a harmonic conjugate of . So we also have
However, the Hilbert transform sends to so it doesn’t seem possible to replace by this way.
- Finally, remark that a simple computation shows