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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: $\|u\|_\infty = \text{sup}_{x\in [0,1]^2} |u(x)|$.

user44442
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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 $t\ge 0$.

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    Hi Michelle thanks for the reply. I am not fully convinced. I understand that the Riesz transform is not bounded on $L^\infty$. This is why I added the constraints of odd and periodic. However, in your example, the function you provided is in fact even. If you attempt to extend it oddly, its 2nd order derivatives becomes discontinuous. I'd like to rule this out by assuming $u$ is $C^2$ (now I added this to the question). I believe that in your example, you exploit this discontinuity feature. Namely, $g(0)$ should be zero for the 2nd derivatives to be continuous at zero. – user44442 Aug 26 '17 at 15:58
  • The example is local. It's a narrow bump. Move it to the middle of $[0,1]^2$ and extend any way you want. –  Aug 26 '17 at 16:14