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 ?



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


