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Start with the Hilbert space $L^2([0, 1])$ with Lebesgue measure. Fix some Borel-to-Borel measurable function $f: [0, 1] \times [0, 1] \rightarrow \mathbb{R}$. I present 4 scenarios, each more restrictive (and hence more likely to give a positive answer) than the previous one.

  1. $f(\cdot, y)$ belongs to $L^2([0, 1])$ for all $y \in [0, 1]$.

  2. $f$ is bounded.

  3. $f$ is continuous.

  4. $f$ is continuous and for all non-zero $g \in L^2([0, 1])$ the integral $\int f(x, y) g(x) dx$ attains value $0$ only finitely many times.

In each of these scenarios I ask the following question: does there necessarily exist a non-zero element $g \in L^2([0, 1])$ such that $\int f(x, y) g(x) dx \ge 0$ for all $y \in [0, 1]$?

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    $\begingroup$ At least for 3. the answer is negative with a fairly simply counterexample: just arrange all the cosines and sines from the basis for $L^2$ with positive and negative signs as $f(x, y_n)$ for different $y_n\to 1$, connect them continuously and multiply by something positive converging to $0$ to get continuity at $1$. Then the function $g$ would have to be orthogonal to all of them (since we include them with both positive and negative sign), hence zero. $\endgroup$ Commented Dec 8, 2023 at 20:37
  • $\begingroup$ That's brilliant! Thanks for solving fist three. $\endgroup$ Commented Dec 8, 2023 at 20:53
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    $\begingroup$ And, well, the answer for 4. is also obviously negative (if such functions $f$ exist at all): give such an $f$, just consider $yf(x, |y|)$ on $[-1, 1]$ and then rescale $[-1, 1]$ to $[0, 1]$. For any $g$ the number of zeroes is at most doubled plus $1$, so still finite, and again all the integrals have to be zero, hence $g$ is identically zero $\endgroup$ Commented Dec 8, 2023 at 20:59
  • $\begingroup$ Right, that solves it. $\endgroup$ Commented Dec 8, 2023 at 21:08
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    $\begingroup$ What if $\int_0^1 f(x, y) dy=0$ for all $x$? $\endgroup$ Commented Dec 8, 2023 at 21:16

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