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.

  • $\begingroup$ "Multiplier" might be a good keyword to search for. See for example here: en.wikipedia.org/wiki/Multiplier_(Fourier_analysis) $\endgroup$ Aug 15 '19 at 23:09
  • $\begingroup$ 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. $\endgroup$ Sep 6 '19 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.

  • 2
    $\begingroup$ Could you expand a bit ? I am surprised the counter-example comes from a Gaussian as I found some results stating the equivalence when the function is assumed non-negative. $\endgroup$ Sep 6 '19 at 18:40
  • $\begingroup$ In some cases, in Harmonic analysis, and in PDE, when we are working whit validity of inequalities we can to construct counter-examples come from "scaling arguments" . In some cases, we use the Schwartz functions because of the nice properties under convolution and Fourier transforms. The Gaussian is used sometimes because we have invariance in relation to Fourier transform and it is a radial function (which is good in Spherical Coordinates). You can found examples in some books as Elias M. Stein or Loukas Grafakos. $\endgroup$
    – Marcelo Ng
    Sep 7 '19 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.