Implication of Convolution with Gaussian Function Vanishing on Interval

Question

Suppose $g$ is the density of $N(0,1)$ and $f:\mathbb{R}\to\mathbb{R}$ is arbitrary. Suppose that the convolution $[f\ast g](x)=\int_{-\infty}^{\infty} f(x-t)g(t)dt=0$ for all $x$ on a closed interval $I$.

If $I=\mathbb{R}$, then we can apply the Convolution Theorem $\mathcal{F}(f\ast g)=\mathcal{F}(f)\cdot \mathcal{F}(g)$ and use fact that $\mathcal{F}(g)>0$ to deduce that $\mathcal{F}(f)=0$, and thus $f=0$ almost everywhere. But what if $I=[a,b]$ where $-\infty<a<b<+\infty$? Is it still true that $f=0$ almost everywhere?

Answer 1

Let $h=g*f$ and assume that $f$ is integrable.

Here are some facts:

1. Convolution preserves analyticity. Therefore, $h$ is an analytic function on $\mathbb{R}$ since the Gaussian function is analytic.
2. Using the identity theorem for analytic functions we have that $h(x)=0,x \in [a,b] \rightarrow h(x)=0, x\in \mathbb{R}$.

Now you can use the Fourerits method to show that $f=0$ everwhere.