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.


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.

  • $\begingroup$ 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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