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.




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**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. 



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


