Convolution of a probability distribution with another function is a quadratic polynomial

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Convolution of a probability distribution function(which is never zero) with an analytic function, f(x), is a quadratic polynomial. can we say that f(x) must be polynomial too? If not can you come up with a counterexample? I have to mention that we do not know if f(x) has Fourier transform or not.

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This is an answer to the original version of the question, regarding the convolution of a probability distribution with an analytic function.

Say $X$ is a random variable with $P(X=0)=P(X=\pi)=0$. So the distribution of $X$ is $\mu=(\delta_0+\delta_\pi)/2$, where $\delta_t$ denotes a point mass at $t$. Let $f(t)=\sin(t)$. Then $\mu*f=0$.

In detail: $$\mu*f(x)=\int f(x-t)\,d\mu(t)=(f(x)+f(x-\pi))/2=0.$$