Does there exist an essentially nonzero function $f:[0,1] \mapsto \mathbb{R}$ so that $$ \int_0^1 |y-x| f(x) \, dx = 0 $$ for every $y \in [0,1]$? I think I see how to show that any such $f$ can't be continuous at any $x$ for which $f(x) \neq 0$ (since, if $K_y(x) := |y-x|$, then $-2K_y + K_{y - \epsilon} + K_{y + \epsilon}$ is positive on $[y-\epsilon,y+\epsilon]$ but zero elsewhere), but does there still exist any measurable $f$ with the above property?
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No, linear combinations of functions of the form $|x-y|$ are dense in $C([0, 1])$ (because you can approximate any continuous function by the piecewise-linear), so (if $f\in L^1(0,1)$, otherwise the integral may not exist) $f$ is orthogonal to all continuous functions, hence almost everywhere zero.
answered Nov 9, 2023 at 17:22
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$\begingroup$ I didn't really read or think about the problem, but why are you assuming $f \in L^1$? You say the integral may not exist otherwise, but so? $\endgroup$ Commented Nov 9, 2023 at 17:23
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2$\begingroup$ @mathworker21 : Since $|0-x|+|1-x|=1$ for all $x\in[0,1]$, we have $\int_0^1 f=0$, so that $f\in L^1$. $\endgroup$ Commented Nov 9, 2023 at 18:03
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$\begingroup$ Excellent, that does it. Thanks! $\endgroup$ Commented Nov 9, 2023 at 19:03