If the convolution of two functions $f\star g$ is equal to $g$, $f$ is even with compact support and $g$ is bounded, implies that $g$ is constant?

Question

Let $f$ be an even continuous function with compact support such that $$ \int f(t)\,\mathrm{d}t=1, $$ and let $g$ be a bounded continuous function such that the convolution $f\star g$ satisfies the following equality $$ (f\star g)(x)=g(x). $$ How to prove that if $g$ has global minimum then $g$ is constant?
I thought about using the Convolution Theorem, but it seems it doesn't work in this case. Maybe, without the requirement that $g$ has global minimum, there is a way to prove that $g$ is linear?

Answer 1

The sought-after statement is wrong: $g$ can be non-constant. Fourier transforming your conditions we see that the Fourier transform $\hat{g}$ is supported at points where $\hat{f}$ is equal to $1$. We also see $\hat{f}(0)=1$ and $\hat{f}(-\xi)=\hat{f}(\xi)$. This motivates the following condition.

It should be trivial to construct an even function $f$ with compact support and such that $\int f(x)dx=1$ and $\int f(x)\cos(x)\,dx=1$. Then take $\hat{g}$ to be supported at $0$ and $\pm 2\pi$, for instance $g(x)=\cos(x)$. Then $$ (f\star g)(y) = \int f(x) g(y-x) dx = \cos(y) \int f(x) \cos(x)\,dx + \sin(y) \int f(x)\sin(x)\,dx = \cos(y) = g(y) $$ where I used $\int f(x)\cos(x)\,dx = 1$ by construction of $f$, and $\int f(x)\sin(x)\,dx = 0$ because $f$ is even.