# A problem concerning $L^2([0,1]\times[0,1])$

Trying to solve a conjecture in differential geometry, I am leaded to the following problem (which may seem weird to a analyst). I wonder if anyone know some techniques that happen to solve it.

Let $f$ be $\mathbb{Z}^2$-periodic positive-real-valued function on $\mathbb{R}^2$, which you may assume to be as soomth as you like. Let $h$ be another such function without the positivity assumption. We suppose that both $f$ and $h$ are unit vectors in $L^2([0,1]\times[0,1])$ (defined with the usual Euclidean measure), and they are orthogonal.

Problem: Is it always possible to write $h$ as a sum of two functions $a,b\in L^2([0,1]\times[0,1])$ with the following conditions:

(1) $\int_0^1 f(x,y)a(x,y)dx=0$ for any $y$

(2) $\int_0^1 f(x,y)b(x,y)dy=0$ for any $x$

(3) $\int_0^1\int_0^1a(x,y)b(x,y)dxdy\geq 0$

Remark: Using Fourier expansion, the proof is easy if $f(x,y)$ is a product such as $\cos(2\pi mx)\sin(2\pi ny)$ (although this $f$ is not positive). In fact in this case we may take $a$ and $b$ to be orthogonal, instead of having inner product $\geq 0$. But for general $f$, the analysis of Fourier coefficients seems hard.

Addendum: As fedja mentioned below, the above assertion is not always true. So I feel like to mention here another assertion which a consequence of the above problem. You can also try to prove it directed. This would provide the last ingredient that I need to solve some conjecture.

For unit vectors $f, h_1, h_2, h_3, h_4\in L^2([0,1]\times[0,1])$ such that each $h_i$ is orthogonal to $f$, we define $$I(f,h_1,h_2,h_3,h_4)=\int_0^1[\sum_{i=1}^4(\int_0^1fh_idy)^2]^{1/2}dx\times \int_0^1[\sum_{i=1}^4(\int_0^1fh_idx)^2]^{1/2}dy$$ we consider $I$ as a functional depending on a point $f$ in the unit sphere of $L^2([0,1]\times[0,1])$ and four unit tangent vectors of the unit sphere at $f$. Then the assertion is that the maximal value of $I$ is $2$.

(The value $2$ is achieved when, for instance, $f=1$ and $h_1=\sqrt{2}\cos(2\pi x)$, $h_2=\sqrt{2}\sin(2\pi x)$, $h_3=\sqrt{2}\cos(2\pi y)$, $h_4=\sqrt{2}\sin(2\pi y)$.)

A first attempt to prove this is to apply Cauchy-Schwarz two times, one can then get $I\leq 4$, which is not enough. Then I came up with the idea: If the $h_i$'s can be written as $a_i+b_i$ satisfying above conditions (1), (2) and (3), then the proof goes through.

-
It is possible to achieve (1) and (2), but not (3) unless $f(x,y)=u(x)v(y)$ . All you can do instead of (3) in general is $(a,b)_{L^2}\ge -c(f)\|a\|\cdot\|b\|$ with $c(f)<1$. If that is good enough for your purposes, I'll post the details. –  fedja Nov 20 '11 at 16:17
Unfortunately this is not good enough... I've edited the text to include my exact purpose. But can you still discribe your method briefly? –  Xin Nie Nov 20 '11 at 21:20
In your new version of the problem, are $h_j$ always pairwise orthogonal like in your example, or they are unrestricted (say, can all be the same)? –  fedja Nov 20 '11 at 21:27
I only need the bound $2$ for pairwise orthogonal $f$ and $h_j$'s. But for some geometric reason, I had the impression that generally the maximum is achieve only in this case. –  Xin Nie Nov 21 '11 at 7:47

Alas, Cauchy-Schwarz is the best you can do. Indeed, imagine $n$ squares with side $1/n$ arranged along the diagonal of the unit square. Now let $f$ be $\sqrt n$ on each such square and $0$ outside (well, you said "positive", but since it is not quantitative, it is just as good as "non-negative"). Let $h_i$ be $\pm\sqrt n$ on each such square and $0$ outside with $\pm$ chosen to satisfy orthogonality relations (which is possible with $n=16$, say). Now, all inner integrals are $1$ in absolute value, so you get exactly $4$ instead of the $2$ you were looking for.