If $\int_0^1 f(y)\sin(xy) dy = 0$ for every $x$, then $f = 0$ almost everywhere. Can someone please give me a hint on this question, I have no idea where to start.  Let $f \in L^p$ for some $1 \leq  \infty$.  Assume for all $x \in [0,1]$ that $$\int_0^1 f(y)\sin(xy) dy = 0$$ Show that $f = 0$ almost everywhere.  All I got is that I should probably define the sets $A = \{ y : f(y) < 0\}$ and $B = \{ y : f(y) > 0 \}$.
 A: Recall the power series representation
$$
\sin\left(xy\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}\cdot\left(xy\right)^{2n+1}
$$
and note that for each fixed $x\in\mathbb{R}$, we have
\begin{eqnarray*}
 &  & \int_{0}^{1}\sum_{n=0}^{\infty}\left|f\left(y\right)\right|\left|\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}\cdot\left(xy\right)^{2n+1}\right|\,{\rm d}y\\
 & \leq & \sum_{n=0}^{\infty}\int_{0}^{1}\left|f\left(y\right)\right|\frac{\left|x\right|^{2n+1}}{\left(2n+1\right)!}\,{\rm d}y\\
 & \leq & \sum_{m=0}^{\infty}\frac{\left|x\right|^{m}}{m!}\cdot\int_{0}^{1}\left|f\left(y\right)\right|\,{\rm d}y<\infty.
\end{eqnarray*}
Hence, we can interchange summation and integration even without the
absolute value (dominated convergence) and we get
\begin{eqnarray*}
0 & = & \int_{0}^{1}f\left(y\right)\cdot\sin\left(xy\right)\,{\rm d}y\\
 & = & \sum_{n=0}^{\infty}\underbrace{\frac{\left(-1\right)^{n}\cdot\int_{0}^{1}f\left(y\right)\cdot y^{2n+1}{\rm d}y}{\left(2n+1\right)!}}_{=:a_{n}}\cdot x^{2n+1}
\end{eqnarray*}
for all $x\in\mathbb{R}$, where the series is a convergent power
series.
By uniqueness of power series, we conclude $a_{n}=0$ and hence
$$
0=\int_{0}^{1}f\left(y\right)\cdot y^{2n+1}\,{\rm d}y\qquad\left(\dagger\right)
$$
for all $n\in\mathbb{N}_{0}$.
Now let $g\in C_{c}\left(\left(0,1\right)\right)$ be arbitrary. Define
$$
h\left(y\right):=\frac{g\left(\sqrt{y}\right)}{y}\qquad\text{ for }y\in\left[0,1\right]
$$
and note that this defines a continuous function, because $g$ is
non-negativ and has compact support in $\left(0,1\right)$ (i.e. it
vanishes identically in a neighborhood of $0$, where the denominator
vanishes).
By the (Stone-)Weierstraß theorem, there is a sequence of polynomials
$\left(p_{n}\right)_{n\in\mathbb{N}}$ with $p_{n}\to h$ uniformly
on $\left[0,1\right]$. Let $p_{n}\left(y\right)=\sum_{m=0}^{M_{n}}a_{m}^{\left(n\right)}y^{m}$.
Then
$$
y\cdot p_{n}\left(y^{2}\right)=y\cdot\sum_{m=0}^{M_{n}}a_{m}^{\left(n\right)}y^{2m}=\sum_{m=0}^{M_{n}}a_{m}^{\left(n\right)}y^{2m+1},
$$
which implies $\int_{0}^{1}f\left(y\right)\cdot\left(y\cdot p_{n}\left(y^{2}\right)\right)\,{\rm d}y=0$
by linearity and $\left(\dagger\right)$. But $y\cdot p_{n}\left(y^{2}\right)\to y\cdot h\left(y^{2}\right)=g\left(y\right)$
uniformly, which implies $\int_{0}^{1}f\left(y\right)\cdot g\left(y\right)\,{\rm d}y=0$.
As $g\in C_{c}\left(\left(0,1\right)\right)$ was arbitrary, we get
$f\equiv0$ almost everywhere by standard results (often known as
the fundamental lemma of the calculus of variations, see http://planetmath.org/fundamentallemmaofcalculusofvariations).
