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I don't think there exists such a function. So my method is to assume such a function exists and show that it must be zero. I'm trying to follows the solution for this similar problem, but I can't seem to get anywhere.

What I've been doing is that by Stone-Weierstrass there exists $p_k(x)$, a sequence of polynomials that converges to uniformly to $f$, then I can write $$ \int_{-2}^3 f(x)f(x)\,dx = \int_{-2}^3 f(x)\lim_{k\rightarrow \infty} p_k(x) \, dx = \lim_{k\rightarrow \infty}\int_{-2}^3 f(x) p_k(x) \, dx $$ But of course I can't conclude this integral is $0$ from this. So I'm not sure where to go from here

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    The subalgebra $\mathfrak{A}$ generated by $(x^{5n})_{n\geq 0}$ (i.e., polynomial functions of the form $\sum_k a^k x^{5k}$) contains a non-zero constant function and separates points, and so, is dense in $C([-2,3],\mathbb{R})$ by the Stone-Weierstrass Theorem. So you can pick $p_k\in\mathfrak{A}$ that converges uniformly to $f$ on $[-3,2]$ and the rest follows from your argument. – Sangchul Lee Jul 29 '20 at 22:33
  • Without further clarification this has an obvious counterexample: any continuous function that is $0$ on $[-2,3]$ – Ninad Munshi Jul 29 '20 at 23:39

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Consider $p_k$ a sequence of polynomials that converges uniformly to $f\left(\sqrt[5]x\right)$ on $\left[-2^5,3^5\right]$ instead, hence $$\int_{-2}^3(f(x))^2\,dx=\int_{-2}^3 f(x)\lim_{k\to\infty} p_k(x^5)\,dx=\lim_{k\to\infty}\int_{-2}^3 f(x)p_k(x^5)\,dx=0$$