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Denote $I=(0, 1)$, and let $\mu$ be the Lebesgue measure on $I$. Does there exist a function $f$ on $I\times I$ viewed as an element of the space $L^\infty(\mu\times\mu)$ such that $$ f^2=f $$ (that is, $f$ takes values 0 and 1); $$ f(x, y)+f(y, x)=1 $$ (that is, if $f(x, y)=0$ then $f(y, x)=1$, and vice versa); and $f$ admits a representation $$ f(x, y)=\int_I g_t(x)\cdot h_t(y)\,dt $$ (which in particular contains representations as sums $\sum_k g_k(x)\cdot h_k(y)$) with $$ \int_I \|g_t\|_\infty\cdot \|h_t\|_\infty\,dt<\infty, $$ where $\|\cdot\|_\infty$ stands for the norm in the space $L^\infty(\mu)$?

Remark (update): If we replace the $L^\infty$-norms by the $L^2$-norms in the last formula, we obtain the condition for the integral operator on $L^2(\mu)$ with kernel $f$ to belong to the trace class. Observe that this operator belongs to the Hilbert-Schmidt class for any kernel $f$ satisfying the above properties because $f$ is square-summable.

Update No.2. The integral operator whose kernel is the indicator of the triangle $\{x<y\}$ is not of trace class. (Proof: If it is of trace class, then so is the integral operator whose kernel is the indicator of the rhombus, but this is not true.) Therefore, the set $\{f=1\}$ must be very complicated near the diagonal: no subset of $I$ can give us a 'triangular' structure of $f$ even after an arbitrary rearrangement.

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  • $\begingroup$ Doesn't a "triangle" function defined by $f(x,y) = 1$ iff $x \leq y$ work ? It satisfies obviously the two first conditions and regarding the third, you may write $f$ as a series of "rectangular" functions, simply tiling the support of $f$ by rectangles. $\endgroup$ – Hachino May 12 '15 at 6:51
  • $\begingroup$ I think it does not work, the norm will be infinite. $\endgroup$ – limanac May 12 '15 at 7:36
  • $\begingroup$ Hm, indeed, you're right. $\endgroup$ – Hachino May 12 '15 at 7:36
  • $\begingroup$ That is why I believe that the answer is negative; nevertheless, I have no proof, and there is a chance that the function may exist. $\endgroup$ – limanac May 12 '15 at 7:43
  • $\begingroup$ So that's where your motivation comes from ? I suggest you add it somewhere in the description, it never hurts to provide some context, even if it eventually turns out to be irrelevant. $\endgroup$ – Hachino May 12 '15 at 8:02
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Well, I am not sure, but let me try to prove that the answer is negative.

The idea is to prove that any function $f(x,y)$ given by $\int g_t(x)h_t(y)dt$ is continuous with respect to appropriate admissible metric, where admissible means ``separable on the set of full measure.'' Namely, for any $x$ define $F_x(t)=g_t(x)\|h_t(y)\|_{\infty}$. For any $x$ we get a function $F_x(t)$ in (the unit ball of) $L^1(I)$. Thus the pushforward of the metric in $L^1(I)$: $\rho(x_1,x_2):=\|F_{x_1}-F_{x_2}\|_{L^1(I)}$ defines an admissible metric on $I$. Define analogous metric for $y$'s. Next, sum they up and we get still admissible metric on $I$ which we denote $\rho$ too. Note that $f(x,y)$ is continuous and even 1-Lipschitz in variable $x$ in our metric $\rho$. Analogously for $y$. Note that what we actually use is that $\int |g_t(x)|\cdot \|h_t(y)\|_{\infty} dt<\infty$ for any $x$ and viceversa, this is bit weaker than your condition.

Now consider the metric measure space $(I,\mu,\rho)$. Since $(I,\rho)$ is separable, there exists a ball $B(x_0,1/10)$ of radius $1/10$ which has positive measure. Thus $|f(x,y)-f(y,x)|\leq 4/10$ for almost all $x,y$ in this ball. This is impossible by the very definition of $f$.

The above argument is essentially borrowed from http://lanl.arxiv.org/abs/1410.0898 and I am glad if our theory may be helpful.

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  • $\begingroup$ Fedor, many thanks! I will get into all details carefully and I hope I will accept your answer: for the first look at it, everything seems to be nice. Do you think this proof can be strengthened to cover the $L^2$-norms in place of $L^\infty$-norms (see my 'update' no.1)? $\endgroup$ – limanac May 14 '15 at 10:01
  • $\begingroup$ @limanac Kernels of nuclear (trace class $S^1$) operators are also always virtually continuous. The proof may be found in the reference above. $\endgroup$ – Fedor Petrov May 14 '15 at 11:57

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