If $\int_0^1 f(y)\sin(xy) dy = 0$ for every $x$, then $f = 0$ almost everywhere.

Question

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 $B = \{ y : f(y) > 0 \}$.

Answer 1

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).