We have $ 1 \leq p \leq \infty$ and $f \in L^p(\mathbb{R})$. We note $T_f(h)=f*h$ for $h \in L^1(\mathbb{R})$.
The linear map $T_f : L^1(\mathbb{R}) \rightarrow L^p(\mathbb{R})$ is continuous and $||T_f|| \leq ||f||_p$.
How to prove that $||T_f|| = ||f||_p$ by testing $T_f$ on a mollifier (approximations to the identity) $(a_n)_{n \in \mathbb{N}}$.
Could someone help me ?
You need to use the fact that $f\ast\varphi_{\epsilon}\rightarrow f$ in $L^{p}$ for a family of approximate identity $\{\varphi_{\epsilon}\}_{\epsilon>0}$. Usually, we take a smooth and compactly supported function $\varphi\geq 0$, and make $\varphi_{\epsilon}=\epsilon^{-n}\varphi(\cdot/\epsilon)$ also with that $\displaystyle\int\varphi(x)dx=1$.
With such, $\|\varphi_{\epsilon}\|_{L^{1}}=1$ and $\|f\ast\varphi_{\epsilon}\|_{L^{p}}\rightarrow\|f\|_{L^{p}}$, so $\|T_{f}\|=\sup_{\|g\|_{L^{1}}=1}\|T_{f}(g)\|_{L^{p}}\geq\|T_{f}(\varphi_{\epsilon})\|_{L^{p}}\rightarrow\|f\|_{L^{p}}$.