ap.analysis of pdes - Partial $L^2$ control on (part of) the Hessian of a harmonic function.

Okay so I thought about this some more and I believe it is just a (really) straightforward application of the Poincare/Wirtinger inequality and the obvious way you would solve $u_{xy}=0$ classically.

Lets work on the square $S=(0,2pi)_xtimes(0,2pi)_y$ as it is simpler to work there and doesn't change much.
We assume $u$ is $C^infty_0(S)$ (i.e. smooth and of compact support) for the time being but do not need it to be harmonic.

Set $f(s)=frac{1}{2pi}int_0^{2pi} u_x(s, t)dt$ so $int_0^{2pi} u_{x}(s, t)-f(s) dt=0$ for all $s$.

By Wirtinger's inequality we have
$int_0^{2pi} (u_x(s,t)-f(s))^2 dt leq int_0^{2pi} u_{xy}^2(s,t) dt$.

Now pick a $F$ so that $F'(x)=f(x)$.
Set $G(t)=frac{1}{2pi}int_0^{2pi} u(s,t)-F(s) ds$ so $int_0^{2pi} u(s,t)-F(s)-G(t) ds=0$ for all $t$. Since $frac{d}{ds}left( u(s,t)-F(s)-G(t)right)=u_x(s,t)-f(s)$ we have by Wirtinger's inequality that:

$int_0^{2pi} (u(s,t)-F(s)-G(t))^2 ds leq int_0^{2pi} (u_x(s,t)-f(s))^2 $ for all $t$.
Then by Fubini we have that

$int_{S} (u(x,y)-F(x)-G(y))^2 leq int_S u^2_{xy}$

Everything is valid if $u$ is in $H^2(S)$. Notice the $F,G$ are given explicitly in terms of $u$.
If $u$ is harmonic with the normalizing properties at $0$ then the $F=ax^2, G=-ay^2$.

At least I think this works...