# Norm of convolution operator

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.

• "Multiplier" might be a good keyword to search for. See for example here: en.wikipedia.org/wiki/Multiplier_(Fourier_analysis) – Christian Remling Aug 15 at 23:09
• Does this mean that there is no general answer ? I searched in this direction Especially for the point 3. above, I am surprised not to find a direct proof. Thanks anyway. – Ayman Moussa Sep 6 at 18:43

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.