# If an estimate is false on $L^{1}$, then it is false for the $\delta$ distribution?

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 ?

• 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). Jul 19, 2020 at 8:17

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 $$u(x)=e^{iK(x,0)}=$$. 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 $$=\int u_n(x)\phi(x)\to \int u(x)\phi(x)\,dx=.$$
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)

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