I need to prove that the linear operator $$B:L^p(\mathbb{R})\rightarrow L^p(\mathbb{R})$$ defined as follows for all $f \in L^p$ $B(f)=1_{[-L,L]}f*h$ such $h\in L^1$ and $L>0$.
$B $ is clearly linear and bounded (convolution property: $||f*h||_p\leq ||h||_1||f||_p)$.
So I need to prove that $B$ is compact (proves that $B(B(0,1)$ is relavely compact in $L^P$, with $B(0,1)$ is the unit ball in the space $L^p$) so I have tried to use Fréchet-Kolmogorov theorem so in order to prove that $B(B(0,1))$ is relatively compact I have to prove that:
$1-$ $B(B(0,1))$ is bounded.
$2-$ $\lim_{r\rightarrow +\infty}\int_{|x|>r}|g|^pdx=0$ for all $g \in B(B(0,1))$ uniformly
$3-$ $\lim_{a\rightarrow 0}||\tau_ag-g||_p=0$ for all $g \in B(B(0,1))$ uniformly
I proved that $ B(B(0,1))$ is bounded, due to the inequality mentionned before and we take $f$ such $||f||_p \leq 1$.
and the second point, is verifed because it's the rest of a convergent integral.
But I could not prove the third point.