If a function integrates to zero against every even function, then it is odd I'm in the process of figuring this one out as well:
Let $f : [−1, 1] \to\mathbb R$ be a smooth function. Show that if
$$ \int_{0}^{1}f(x)v(x)dx = 0 $$
for every smooth function $v : [0, 1] \to \mathbb{R}$ for which $v(x) = v(−x)$, then $f(x) = −f(−x)$ for all $x \in [−1, 1]$.
Not sure where to begin. Where can I go with $ \int_{0}^{1}f(x)v(x)dx = 0 $ and $v(x) = v(−x)$? Reminds me of even and odd.
 A: By hypothesis
$$
\begin{align}
0
&=\int_{-1}^1f(x)\overbrace{\left(\frac{f(x)+f(-x)}2\right)}^{\text{even function}}\,\mathrm{d}x\\
&=\int_{-1}^1\left(\frac{f(x)+f(-x)}2+\frac{f(x)-f(-x)}2\right)\left(\frac{f(x)+f(-x)}2\right)\,\mathrm{d}x\\
&=\int_{-1}^1\left(\frac{f(x)+f(-x)}2\right)^2\,\mathrm{d}x+\int_{-1}^1\left(\frac{f(x)-f(-x)}2\right)\left(\frac{f(x)+f(-x)}2\right)\,\mathrm{d}x\\
&=\int_{-1}^1\left(\frac{f(x)+f(-x)}2\right)^2\,\mathrm{d}x+\frac14\int_{-1}^1\left(f(x)^2-f(-x)^2\right)\,\mathrm{d}x\\
&=\int_{-1}^1\left(\frac{f(x)+f(-x)}2\right)^2\,\mathrm{d}x
\end{align}
$$
Therefore,
$$
\frac{f(x)+f(-x)}2=0
$$
and thus
$$
\begin{align}
f(x)
&=\frac{f(x)+f(-x)}2+\frac{f(x)-f(-x)}2\\
&=\frac{f(x)-f(-x)}2
\end{align}
$$
which is an odd function.
A: (Assuming user164587's comment is correct:)
Let $g$ be an arbitrary smooth function, and consider $v(x)=g(x)+g(-x)$. We have
\begin{eqnarray}
0&=&\int_{-1}^1 f(x)\left(g(x)+g(-x)\right)dx\\
&=& \int_{-1}^1 f(x)g(x) \, dx + \int_{-1}^1 f(x)g(-x)\, dx \\
&=& \int_{-1}^1 f(x)g(x) \, dx + \int_{-1}^1 f(-x)g(x)\, dx\\
&=& \int_{-1}^1 g(x) \left(f(x)+f(-x)\right)dx \, .
\end{eqnarray}
Since this holds for any smooth $g$,  $f(x)+f(-x)$ is identically zero and so $f$ is odd.
A: There's probably a slicker way to do this, but here's the brute-force method. Suppose there is some $x > 0$ for which $f(x) \neq -f(-x)$. Without loss of generality, say $f(x) > -f(-x)$. Then this is still true on some small interval $I$ around $x$, by continuity. Construct $g$ to be $1$ on $I$ and $-I$, and zero elsewhere; this is clearly even. Then
$$ \int_{-1}^{1} f(x)g(x) = 0$$
which means that the integral of $f$ over $I$, plus the integral of $f(-x)$ over $-I$, is equal to zero. But this is impossible, because $f(x) > -f(-x)$ on $I$. 
