Prove subspace in $L^1(\mathbb{R})$ equal to $L^1(\mathbb{R})$

Question

Let $f(x)=\frac{1}{1+x^2}$ . Then linear closed subspace $V \subset L^1(\mathbb{R})$ including $\{f\}$ is $L^1(\mathbb{R})$.

I want to prove $C_0(\mathbb{R})\subset V$ because $C_0(\mathbb{R})$ is dense in $L^1(\mathbb{R})$.But I have no idea.

Answer 1

It suffices to prove that if $g$ is a square integrable function such that for all $t\in\mathbb R$, $\langle f_t,g\rangle=0$, then $g\equiv 0 $. Indeed, in this case, for all $f\in V $, $\langle f,g\rangle=0$ hence the orthogonal of $V$ would be reduced to the null function.

So let $g$ be square integrable function such that for all $t\in\mathbb R$, $\langle f_t,g\rangle=0$. Rewriting this in terms of convolution product, we have that $h_0\star g=0$, where $h_0\colon x\mapsto 1/(1+x^2)$. Taking the Fourier transform, we get that the Fourier transform of $g$ vanishes identically, hence so does $g$.