Mixed Derivative bounded by Laplacian in 2D?

Question

Suppose that I have a $C^2$ odd periodic function $u$ on $[0,1]^2$. Is it true that $\| \partial_1 \partial_2 u \|_\infty \leq C \|\Delta u\|_\infty$ for some constant $C$ independent of $u$? The norm is the usual sup norm: $\|u\|_\infty = \text{sup}_{x\in [0,1]^2} |u(x)|$.

Answer 1

No. Technically speaking: Riesz transform is not bounded on $L^\infty$. A concrete example is below.

Being odd or periodic has little to do with it; there is a local obstruction. Consider the function $$u(x,y) = xy\, g(x^2+y^2)$$ where $g$ is smooth on $[0,\infty)$ and vanishes on $[1/4,\infty)$ (so it can be extended in odd periodic fashion). A computation shows $$ \Delta u(x,y) = 12xy\,g' + 4xy(x^2+y^2)\,g'' $$ $$ \partial_{1,2}^2 u = g + 2(x^2+y^2)\, g'+4x^2y^2\,g'' $$ where the argument of $g$ and its derivatives is $x^2+y^2$. To reduce this to one-variable analysis, let $t=x^2+y^2$ and use the inequality $|2xy|\le t$ to obtain $$ |\Delta u(x,y)| \le 6t|g'| + 2t^2|g''| $$ $$ |\partial_{1,2}^2 u| \ge |g| - 2t|g'| - t^2 |g''| $$ So, for this special case your question becomes: do we have $$ |g(0)| \le C\sup( t|g'| + t^2 |g''|) $$ for smooth $g$ that vanish for $t>1/4$? And the answer is no.

For example, let $g(t)=\log(\epsilon+t)$ with small $\epsilon>0$; vanishing can be arranged by multiplying this by a smooth cutoff. Then $|g(0)|= |\log \epsilon|\to\infty$ as $\epsilon\to 0$, while $|tg'(t)| \le 1$ and $|t^2g''(t)|\le 1$ for all $t\ge 0$.