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> **Proposition:** Suppose $f$ is continuous and $\int\_a^bf(x)x^ndx = 0$ for all $n$. Then $f$ is zero on $[a,b]$.
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This can be proven by uniformly approximating $f$ with polynomials via the Weierstrass theorem. Before I heard this answer, I was trying to prove it by constructing a sort of polynomial bump function that would peak where $f(x)$ was nonzero, and be small outside that region. 

Suppose $f \neq 0$ on $[a,b]$ and suppose WLOG that $f(y)>0$. Choose $(c,d)$ such that $|f(x)|>|f(y)|/2$ on $(c,d)$ and choose $(\gamma,\delta)\supseteq (c,d)$ such that $f(x)>0$ on $(\gamma, \delta)$.

I want a polynomial that is $\geq 1$ on $(c,d)$ and uniformly small on $[a,b]\setminus (\gamma, \delta)$. Is such possible? How could I construct it?