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Let $u=\int e^{\dot{\imath}K(x,y)} f(y) dy$. My advisor told me that we can disprove an integrability estimate $$\|u\|_{L^p}\lesssim \|f\|_{L^{1}}\label{1}\tag{1}$$ by disproving it when $f=\delta$, the Dirac Delta distribution.

When I asked him for the reasoning for this, he told me $\delta$ is the limit of a sequence of $L^{1}$ functions with norm 1, in the sense of distributions. Indeed, if $f\in L^{1}(\mathbb{R}^{d})$ with $\|f\|_{L^{1}(\mathbb{R}^{d})}=1$, we can define $f_{n}(x)=n^d\,f(nx)$, and then we can change variables and apply the dominated convergence theorem to show that $$\int_{\mathbb{R}^{d}}f_{n}(x)\phi(x)dx=\int_{\mathbb{R}^{d}}f(x)\phi\left(\frac{x}{n}\right)dx\longrightarrow \phi(0)\int_{\mathbb{R}^{d}}f(x)dx=\delta(\phi)$$ for every test function $\phi$.

My question is: how this convergence in the sense of distributions justify/implies that if the estimate \eqref{1} is false when $f=\delta$ then \eqref{1} is also false for a general $f\in L^1$ ?

I mean if the convergence were in $L^{1}$ norm, then the claim is obvious. So I guess my question is, does there exist a sequence $f_n$ of (normalized) $L^{1}$ functions such that $$\int_{\mathbb{R}^{d}}|f_{n}-\delta|\rightarrow 0\qquad\qquad ?$$

Obviously, by the argument above, we have $$\int_{\mathbb{R}^{d}}f_{n}(x)\phi(x)dx\rightarrow \int_{\mathbb{R}^{d}}\delta(x)\phi(x)dx\Longrightarrow \int_{\mathbb{R}^{d}}[f_{n}(x)-\delta(x)]\phi(x)dx\rightarrow0$$ for every test function $\phi$. Where to go from here ?

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  • $\begingroup$ Not a justification for the $\operatorname L^1$ versus distributional convergence, but a point of language: What I would understand from your advisor's claim is that an estimate that fails for the $\delta$ function must fail for some integrable $f$, not for a generic integrable $f$ (in whatever sense that's meant). $\endgroup$
    – LSpice
    Jul 19, 2020 at 8:17

1 Answer 1

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First of all, observe that your "subquestion" is ill-posed because $\int_{\mathbb{R}^{d}}|f_{n}-\delta|$ does not make any sense: the Dirac delta distributions is of course not an $L^1$ function, it is merely a distribution $\mathcal D'$.

The right answer goes as follows (well, more or less, you should actually give more information about the kernel $K$, but let me sketch out the idea nonetheless). Just as you correctly pointed out that $\int f_n \phi\to <\delta,\phi>=\phi(0)$, it is easy to see that $u_n(x)=\int e^{i K(x,y)}f_n(y)\,dy$ converges pointwise a.e. to $u(x)=e^{iK(x,0)}=<e^{iK(x,\cdot),\delta}>$. Regardless of the kernel $K$, this function of $x$ has modulus $\left|e^{iK(x,0)}\right|\equiv 1$ and is therefore not in $L^p$, so clearly this tends to violate $\|u\|_p\lesssim \|f\|_1$.

In order to conclude completely rigorously, one can argue as follows:

  1. I first claim that pointwise a.e convergence can be improved to distributional convergence. Indeed, from the pointwise bound $|u_n(x)|=\left|\int e^{-iK(x,y)}f_n(y)\,dy\right|\leq \int\left| e^{-iK(x,y)}f_n(y)\,dy\right|=\int |f_n(y)|\,dy=1$ we see that $\|u_n\|_\infty\leq 1$. Hence for any test function $\phi\in C_c$, and from the previous ponitwise a.e. convergence $u_n(x)\to u(x)$, an easy application of Lebesgue's dominated convergence theorem yields $$ <u_n,\phi>=\int u_n(x)\phi(x)\to \int u(x)\phi(x)\,dx=<u,\phi>. $$
  2. assume now by contradiction that your estimate $\|u\|_p\lesssim\|f\|_1$ holds. Then taking $f=f_n$ with $\|f_n\|_1=1$ you see that $\|u_n\|_p\lesssim 1$ would be bounded. By the Banach-Alaoglu-Bourbaki theorem you conclude that, up to a subsequence, there would exist some $v\in L^p$ such that $u_n\rightharpoonup v$ weakly in $L^p$. But since weak $L^p$ convergence is stronger than distributional convergence, and by uniqueness of the limit in the sense of distributions, step 1 implies $u=v$. This is impossible since $u$ is not $L^p$ but $v$ would be (as a weak $L^p$ limit.)

Note: for $p=+\infty$ the Banach-Alaoglu-Bourbaki still applies, and $u_n\overset{*}{\rightharpoonup} v$ weakly star. The weak-* convergence is still better that distributional convergence, so we're good. The case $p=1$ might be more tricky, I've never thought of that (I'm not from harmonic analysis but I'm sure this must be a classical issue)

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  • $\begingroup$ Thanks a lot. I appreciate the time and effort you put into making the answer so clear. $\endgroup$
    – Medo
    Jul 19, 2020 at 15:48

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