Let $f \in L^{1}(\mathbb R)$ such that the Fourier transform of $f$ equals to $0$ for a $\xi_{0} \in \mathbb R$: $\mathcal Ff(\xi_{0}) = 0$.
Let $\tau_{s}f(x) = f(x-s)$, $s \in \mathbb R$. I want to show that the vector space made of all the translations $\tau_{s}f$ of $f$ isn't dense in $L^1(\mathbb R)$.
That is to show that $ \exists \epsilon > 0:$ $|| \tau_{s}f - f||_{1} > \epsilon$, when $s \to 0$.
To do this, I use the continuous linear form $\phi : L^1(\mathbb R) \to \mathbb R$, where $\phi(f) = \mathcal Ff(\xi_{0})$.
Therefore, I have to show that $| \phi(\tau_{s}f) - \phi(f)| > \epsilon$ when $s \to 0$.
I end up having: $|\phi(\tau_{s}f) - \phi(f) | = |\int_{\mathbb R} f(x)dx |$, once I made $s \to 0$.
Here, I don't know how to conclude. Obviously, I want to have $|\int_{\mathbb R} f(x)dx | >0$, and $|\int_{\mathbb R} f(x)dx | = 0$ if $f \equiv 0$, but I think there is an argument that excludes this case thanks to the Fourier transform, and I can't recall what it is.
Any help please ?
The Fourier transforms of $\tau_sf$ also vanish at $\xi_0$. If $(g_n)$ is a sequence of $L^1$ functions with $\mathcal{F}g_n(\xi_0)=0$ converging in $L^1$ norm to $G$, then $$|\mathcal{F}g_n(\xi_0) -\mathcal{F}G(\xi_0)|\le\|f_n-G\|_1$$ and so $\mathcal{F} G(\xi_0)=0$. But there are $L^1$ functions whose Fourier transforms vanish nowhere...