Solving equations of the form $y(x) f(x) =0$ When speaking with my advisor recently, we were led in the course of a physics problem to an equation of the form $$y(x) \ f(x) = 0$$ with $f(x)$ known and $y(x)$ unknown. My immediate instinct was to conclude that $y(x) = 0$ for all $x$. Clearly, this must be the right conclusion if there is no $x$ such that $f(x)=0$.
However, my advisor claims that a more general solution for any $f(x)$ is $$y(x) = \sum_{n=1}^m c_n \delta(x-x_n) $$ where the $c_n$ are arbitrary, $\delta()$ is the dirac distribution, and the $x_n$ satisfy $f(x_n) = 0$. 
Intuitively, I understand why this is reasonable, but can it be justified rigorously? Is there any reason to prefer this solution over my instinctual one? The case of $f(x) =0$ seems particularly troublesome for this idea.
Follow-up: Thanks to Martin for providing a counter-example for arbitrary $f$. Does the claim hold if we require that $f$ is continuous and vanishes at a finite number of points?
 A: It completely depends on what $f$ looks like. Your advisor's suggestion seems to imply that the thinks that $f=0$ on a finite number of points $x_1,\ldots,x_m$. But unless you specify this as a condition, it is not necessarily the case: consider for instance
$$
f(x)=\begin{cases}x^2,&\mbox{ if }x\geq0 \\ 0,&\mbox{ if } x<0  \end{cases}
$$
and $y(x)=f(-x)$. Then both $f$ and $y$ are differentiable, $y(x)f(x)=0$ for all $x$, and $y$ is not of the form you mention. 
A: Check this example: $\displaystyle{{\rm y}'\left(x\right) = 0}$ with
$\displaystyle{{\rm y}\left(0\right) = y_{0}}$. It has the obvious solution
$\displaystyle{{\rm y}\left(x\right) = y_{0},\ \forall\ x}$. Let's assume we insist ($~\small\it\mbox{it's like killing an cockroach with a gun machine}~$) to use a Fourier transform to solve this equation:
\begin{equation}
{\rm y}\left(x\right)
=
\int_{-\infty}^{\infty}{{\rm d}k \over 2\pi}\,\tilde{\rm y}\left(k\right)
{\rm e}^{{\rm i}kx}.
\quad
\mbox{We get}\phantom{A}
{\rm i}k\tilde{\rm y}\left(k\right)
=
0
\quad\Longrightarrow\quad
\tilde{\rm y}\left(k\right) = A\,\delta\left(k\right)
\qquad\qquad\left(\Large\tt\mbox{I}\right)
\end{equation}
where $A$ is a constant. Then
$$
{\rm y}\left(x\right)
=
\int_{-\infty}^{\infty}{{\rm d}k \over 2\pi}\,
A\,\delta\left(k\right){\rm e}^{{\rm i}kx}
=
{A \over 2\pi}
\quad\Longrightarrow\quad
{\rm y}\left(0\right) = y_{0} = {A \over 2\pi}
\quad\Longrightarrow\quad
{\rm y}\left(x\right) = y_{0}\,,\ \forall\ x
$$
If you set $\tilde{\rm y}\left(k\right) = 0$ in step
$\left(\Large\tt\mbox{I}\right)$, you get the wrong result ${\rm y}\left(x\right) = 0$.
