Problem 1

Exercise 5.1 (A vanishing argument for odd functions)   Let $f : [-1,1] \to\mathbb{R}$ be a continuous odd function and suppose

$$\int_{-1}^1 f(x)x^{2k-1}dx=0$$ (5.1)

for all $k>0$. Then $f(x)\equiv 0$.

Proof. Include $f\in L^2$ by noticing that $f$ being uniformly continuous implies $f^2$ is uniformly continuous. Then consider the subspace of odd functions in $L^2$ which also contains $f$:

$$F = \{g \in L^2 \;\vert\; g(-x)=-g(x) \quad \forall x\in[-1,1]\}$$ (5.2)

which has a countable dense subset, namely, $\{x, x^3, x^5, \dots\}$ by a similar argument to the Weierstrass approximation theorem. The inner product in this space naturally arises as

$$\langle f, g\rangle = \int_{-1}^1 f(x)g(x)dx$$ (5.3)

and we know $\langle f, x^m\rangle=0$ for any basic element $x^m$. Therefore, $f\equiv 0$. $\qedsymbol$