An interesting inequality due, perhaps, to Kashin

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 $f:{\mathbb R}\rightarrow {\mathbb R}$ be a Lipschitz function, i.e., there is a constant ${L}$ such that

$\displaystyle |f (x)-f (y)|\leq L|x-y|\qquad \forall\ x,y\in{\mathbb R}.$

Then,

$\displaystyle \int_{{\mathbb R}}\int_{{\mathbb R}}\left(\frac{f (x)-f (y)}{x-y} \right)^{2}dxdy\leq 10L\int_{{\mathbb R}}|f (x)|dx.$

Note that the left hand-side is also known as the Besov semi-norm ${{\mathcal B}_{2}}$ . The constant ${10}$ is probably not sharp. However, the function ${(1-|x|)_{+}}$ shows that it’s at least ${2}$ .

Proof: Use the Lipschitz hypothesis on ${\{(x,y): |x-y|\leq |f (x)|/L \}}$ and integrate in ${y}$ first to get

$\displaystyle \int_{{\mathbb R}}\int_{|x-y|\leq \frac{|f (x)|}{L}}\left(\frac{f (x)-f (y)}{x-y} \right)^{2}dydx\leq \int_{{\mathbb R}}2L^{2}\frac{|f (x)|}{L}dx=2L\|f\|_{1}.$

Likewise, on ${\{(x,y):|x-y|\leq |f (y)|/L \}}$ we integrate in ${x}$ first.

Thus,

$\displaystyle \int\int_{|x-y|\leq \max (|f (x)|,|f (y)|)/L}\left(\frac{f (x)-f (y)}{x-y} \right)^{2}dxdy\leq 4\|f\|_{1}L.$

Now we need to consider ${\{(x,y):|x-y|>\max (|f (x)|,|f (y)|)/L \}}$ . Majorize ${(f (x)-f (y))^{2}}$ by ${2 (f (x)^{2}+f (y)^{2})}$ and split into two integrals. Then, considering ${f (x)^{2}}$ first, change variables with ${s=x-y}$ , and integrate in ${y}$ for ${|y-x|>\frac{|f (x)|}{L}}$ to get

$\displaystyle \int_{{\mathbb R}}\int_{|x-y|>\frac{|f (x)|}{L}}\frac{f (x)^{2}}{(x-y)^{2}} dydx=2\int_{R}L|f (x)|dx=2L\|f\|_{1}.$

Likewise, for ${f (y)^{2}}$ use the lower bound ${|f (y)|/L}$ instead and integrate in ${x}$ first. $\Box$

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 ${f\in L^{1} ({\mathbb R})}$ , Lipschitz with constant ${L}$ , let its harmonic extension to the upper half-plane ${{\mathbb H}=\{y>0 \}}$ be:

$\displaystyle u (x+iy)= \int_{{\mathbb R}}f (t)\frac{y}{(x-t)^{2}+y^{2}}\frac{dt}{\pi}.$

Then there is ${F(z)=u(z)+iv(z)}$ analytic on ${\mathbb H}$ , which is given by the following formula

$\displaystyle F(x+iy)=\int_{-\infty}^{+\infty} f(t)\frac{i}{z-t}\frac{dt}{\pi}.$

To see this one checks that

$\displaystyle {\mathcal Re} \left(\frac{i}{z-t}\right)=\frac{{\mathcal Im} z}{|z-t|^2}.$

Consider domains ${\Omega_{\epsilon,R}=\{{\mathcal Im} z>\epsilon, |z|<R\}}$

$\displaystyle \int_{\Gamma_{\epsilon ,R}}\overline{F(z)}F^{\prime}(z)dz=-\int\int_{\Omega_{\epsilon ,R}}|F^{\prime}|^2dz\wedge d\overline{z} =2i\int\int_{\Omega_{\epsilon ,R}}|\nabla u|^2 dxdy.$

Now ${\Gamma_{\epsilon ,R}={\mathcal C}_{\epsilon, R}\cup{\mathcal L}_{\epsilon, R}}$ , where ${{\mathcal C}_{\epsilon, R}=\partial\Omega_{\epsilon ,R}\cap \{|z|=R\}}$ and ${{\mathcal L}_{\epsilon, R}=\partial\Omega_{\epsilon ,R}\cap \{{\mathcal Im} z=\epsilon\}}$ .

Note first that

$\displaystyle \int_{\Gamma_{\epsilon,R}}\overline{F(z)}F^{\prime}(z)dz= -\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(t)f(s) \int_{\Gamma_{\epsilon, R}}\frac{1}{(\overline{z}-t) (z-s)^{2}}dzdsdt,$

and that

$\displaystyle \left|\int_{{\mathcal C}_{\epsilon, R}}\frac{1}{(\overline{z}-t) (z-s)^{2}}dz\right|\leq \pi R \frac{1}{R-t}\frac{1}{(R-s)^2}\rightarrow 0$

for ${\epsilon>0}$ fixed, as ${R\rightarrow\infty}$ .

On the other hand, by the residue theorem,

$\displaystyle \begin{array}{rcl} \int_{\Gamma_{\epsilon, R}}\frac{1}{\overline{z}-t}\frac{1}{(z-s)^2}dz & = & \int_{\Gamma_{\epsilon, R}}\frac{1}{z-2\epsilon i-t}\frac{1}{(z-s)^2}dz\\ & = & 2\pi i {\mathcal Res} _{t+2\epsilon i}\left(\frac{1}{z-2\epsilon i-t}\frac{1}{(z-s)^2} \right)\\ & = & 2\pi i\frac{1}{(t-s+2\epsilon i)^2}. \end{array}$

So, for ${\epsilon>0}$ ,

$\displaystyle \begin{array}{rcl} \lim_{R\rightarrow\infty}\int\int_{\Omega_{\epsilon,R}}|\nabla u|^2 \frac{dxdy}{\pi} & = & \int_{{\mathcal Im} z=\epsilon}\overline{F(z)}F^\prime(z)\frac{dz}{2\pi i}\\ & = & -\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{f(t)f(s)}{(t-s+2\epsilon i)^2}dtds. \end{array}$

Moreover, Cauchy’s theorem (on similar contours) shows that, for fixed ${a\in{\mathbb R}}$ ,

$\displaystyle \int_{-\infty}^{+\infty}\frac{dt}{(t-a+2\epsilon i)^{2}}=0,$

Hence,

$\displaystyle \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{(f(s))^2}{(t-s+2\epsilon i)^{2}}dtds=0.$

Expanding the square we therefore see that

$\displaystyle -2\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{f(t)f(s)}{(t-s+2\epsilon i)^2}dtds =\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{(f(t)-f(s))^2}{(t-s+2\epsilon i)^2}dtds.$

By Lebesgue dominated convergence we can let ${\epsilon}$ tend to zero and get that

$\displaystyle \int\int_{\mathbb H}|\nabla u|^2\frac{dxdy}{\pi }= \int_{{\mathbb R}}\overline{F(x)}F^\prime(x)\frac{dx}{2\pi i}= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{(f(t)-f(s))^2}{(t-s)^2}dtds.$

The middle integral needs justification, e.g., it is ok if ${(F^{2})^{\prime}\in H^{1} (\mathbb H)}$ .

In particular, we get

$\displaystyle \int\int_{\mathbb H}|\nabla u|^2 \frac{dxdy}{\pi}\leq \frac{1}{2\pi}\|F^{\prime}\|_{L^\infty({\mathbb R})}\|F\|_{L^1({\mathbb R})}.$

3. Remarks

The Besov semi-norm above can be interpreted as a Dirichlet energy by changing variables:
$\displaystyle \int_{{\mathbb R}}\int_{{\mathbb R}}\left(\frac{f (x)-f (y)}{x-y} \right)^{2}dxdy = 2\int_{x\in{\mathbb R}}\int_{y>0}\left(\frac{f (x+y)-f (x)}{y} \right)^{2}dydx$

and, at least for smooth functions,

$\displaystyle \frac{f (x+y)-f (x)}{y}=\int_{{\mathbb R}}f^{\prime} (t)\frac{\chi_{(x,x+y)} (t)}{y}dt$

where the kernel ${\chi_{(x,x+y)} (t)/y}$ is quite similar to the Poisson kernel.
By Plancherel for the Hilbert transform, we have that the Dirichlet integral of ${u}$

$\displaystyle \int_{\mathbb H}|\nabla u|^{2}dxdy=2\int_{\mathbb H}|\partial_{x}u|^{2}dxdy. \ \ \ \ \ (1)$
Green’s theorem applied to the function ${u^{2}}$ and ${1}$ , and the fact that ${\Delta u^{2}=|\nabla u|^{2}}$ for harmonic functions, gives:
$\displaystyle \int_{\mathbb H}|\nabla u|^{2}dxdy=\int_{\mathbb H}\Delta u^{2} dxdy=\int_{{\mathbb R}}2u\partial_{y}u dx =\int_{{\mathbb R}}2u\partial_{x}v dx$

where we have skipped many technical details and ${v}$ is a harmonic conjugate of ${u}$ . So we also have

$\displaystyle \int_{\mathbb H}|\nabla u|^{2}dxdy\leq C \|f\|_{1}\|\partial_{x}v\|_{\infty}$

However, the Hilbert transform sends ${L^{\infty}}$ to ${BMO}$ so it doesn’t seem possible to replace ${\|\partial_{x}v\|_{\infty}}$ by ${\|\partial_{x}u\|_{\infty}}$ this way.
Finally, remark that a simple computation shows
$\displaystyle \int_{{\mathbb R}}\int_{{\mathbb R}}\left(\frac{f (x)-f (y)}{x-y} \right)^{2}dxdy=C\int_{{\mathbb R}}|\xi||\hat{f} (\xi)|^{2}d\xi.$