I have a simple little analysis question that I'm hoping is well known.

Suppose $D=\lbrace(x,y): x^2+y^2<1\rbrace$ is the unit disk and that $u$ is a harmonic function on $D$. Suppose in addition that $u(0)=0$ and $\nabla u(0)=0$. Lets also assume $u$ has finite $L^2$ norm -- i.e. $||u||_2<\infty$

If $u_{xy}=\partial_{xy}u=0$ on $D$ then it is clear that $u=a(x^2-y^2)$ for some $a$.

Suppose instead that we only know that $|| u_{xy}||_2<<1$. I'm pretty sure a compactness argument implies that there is an $a$ so that $||u-a(x^2-y^2)||_2<<1$ (or at least this is true on any smaller disk).

What I'm most interested in is whether there is a constant $C>0$ so that

$ \inf_{a\in \mathbb{R}} ||u-a(x^2-y^2)||_2\leq C ||u_{xy}||_2$.

(I'm also happy if the control was only on a smaller disk).

On a related note, does anyone know if $u_{xy}=0$ in a distributional sense then $u=F(x)+G(y)$. I can't remember now if this is true or not...


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

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

Now pick a $F$ so that $F'(x)=f(x)$.
Set $G(t)=\frac{1}{2\pi}\int_0^{2\pi} u(s,t)-F(s) ds$ so $\int_0^{2\pi} u(s,t)-F(s)-G(t) ds=0$ 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^{2\pi} (u(s,t)-F(s)-G(t))^2 ds \leq \int_0^{2\pi} (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...


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