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Let $f(x)=e^{-x^2/2}$ and $g(t)$ be some symmetric, positive and bounded function and let the convolution of $f(x)$ and $g(x)$ be defined as follows: \begin{align} h(x)= \int f(x-t)g(t)\, dt. \end{align}

My questions is: Suppose we choose some value of $y=c$. How many times does the function $h(x)$ equals $c$ on some interval $[-a,a]$?

In other words, can we say something about the number of zeros of the function on $[-a,a]$ \begin{align} F(x)= h(x)-c. \end{align} Let $S_a(F)$ denote the set of zeros of $F$ on $[-a,a]$.

Partial Answer: Because convolution "increases" analyticity we have that $h(x)$ and $F(x)$ are analytic. Therefore, by the standard identity theorem argument, we have that $F(x)$ can have finitely many zeros on any interval $[-a,a]$. So, $S_a(F)$ is a finite set.

My question: Can we say more about the cardinality of the set of zeros $S_a(F)$? In particular, can we give a bound on it? Most likely a uniform bound is impossible. However, I think we can give a bound that depends on $h(x)$.

Comment I would also appreciate a reference if this question been addressed before. Also, if you can think of some keywords that would be great too.

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The Fourier transform of the convolution is the product of the Fourier transforms; since the Fourier transform $e^{-x^2/2}$ is again a Gaussian, the convolution of $g$ with $f$ still contains all the same frequencies as $g$, just at smaller amplitudes for larger frequencies. So, for instance, if you take $g(x)=A+B\cos(kx)$, then $h(x)=A'+B'_k\cos(kx)$, which will cross $A'$ a number of times proportional to $k$ on an interval. Since $k$ can be arbitrarily large, so can the number of crossings.

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  • $\begingroup$ Would things change if we impose that $h(x)$ is zero outside of an interval $[-c,c]$ and $\int_{-c}^c h(x) dx<\infty$? $\endgroup$ – Lisa Jun 1 '18 at 14:39

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