Norm of convolution operator

Question

By Young's inequality for any $f\in L^p(\mathbf{R})$ the map $T_f:g\mapsto f\star g$ is a continuous operator from $L^q(\mathbf{R})$ to $L^r(\mathbf{R})$ where $1\leq p,q,r\leq \infty$ satisfy $1+\frac1r=\frac1p+\frac1q$ and we even have

\begin{align*} \|T_f\|_{p\rightarrow r} \leq \|f\|_q. \end{align*} If I am not mistaking in general $\|T_f\|_{p\rightarrow r}$ and $\|f\|_{q}$ are not equivalent :

• When $r=q=2$ and $p=1$ we have by Plancherel's formula (for a correctly normalized Fourier transform) $\| T_f(g)\|_2 =\| \hat{f}\hat{g}\|_2$ from which we get $\|T_f\|_{2\rightarrow 2}=\|\hat{f}\|_\infty$, and $\|f\|_1\lesssim\|\hat{f}\|_\infty$ is just not reasonable.
• On a more sophisticated level, on the torus I know that the partial Fourier Series $S_N(f)$ corresponding to the Dirichlet kernel $D_N$ converge in $L^p(\mathbf{T})$ for non extremal values of $p$. The Dirichlet kernel being unbounded in $L^1(\mathbf{T})$, $\|D_N\|_1\lesssim\|T_{D_N}\|_{p\rightarrow p}$ is not possible because of the Banach-Steinhauss Theorem.

On the other hand, one can prove that $\|T_f\|_{1\rightarrow 1}$ and $\|T_f\|_{\infty\rightarrow\infty}$ are both equivalent to $\|f\|_1$.

My questions :

1. Are they any other cases of exponents for which this equivalence holds ?
2. When the equivalence does not hold, is there any description of $\|T_f\|_{p\rightarrow r}$ (with emphasis on the case $p=r$) ?
3. Is there any elementary (= not as above) proof that $\|T_f\|_{p\rightarrow p}$ is not equivalent to $\|f\|_1$ when $p\notin\{1,2,\infty\}$ ?

I found several results in the literature linked to this question but they either treat the optimality of the Young inequality as the continuity of the operator $(f,g)\mapsto f\star g$ (not interested) or precisely state the equivalence if $f$ is non negative.

Answer 1

You can take $f(x) = e^{- \alpha |x|^{2}}$ (Gaussians) to be a "test function" in order to prove that the one of the equivalences is not true.