Limit of inner product for mollifiers

Question

If $K(x)$ is a bounded, continuous, and $\phi_n\mathop\rightarrow\limits^{n\rightarrow \infty} \delta$ (weakly in space of tempered distributions) where $\phi_n$ is Schwartz function. Is it true that:

$ \int_{R} \phi_n(x) (K(x)-K(0))dx \mathop\rightarrow\limits^{n\rightarrow \infty} 0 $ ?

If $K(x)$ is Schwartz function, then it is obvious. But is it true if $K$ is only bounded and continuous?

Answer 1

EDIT. I thought it was almost a triviality but the comments of reuns clearly show that this is not the case.

ANSWER. It is not true. Here's the counterexample of reuns: suppose that $K$ is continuous and bounded, and there is a sequence $\lambda_n\to \infty$ such that $$ \lambda_n^hK(\lambda_n)\text{ does not tend to }0, $$ for some $h\in\mathbb N$. Letting $$ \phi_n(x):=n\eta(nx)+(\lambda_n)^{h+1}\eta(\lambda_n(x-\lambda_n)),$$ where $\eta$ is a standard mollifier, we define a sequence such that $\phi_n\to \delta$ in the sense of tempered distributions, but $$ \int_{\mathbb R} \phi_n(x)K(x)\, dx = K(0)+ \lambda_n^h K(\lambda_n)+o(1), $$ which does not tend to $K(0)$.

---

To make it true, we need two more additional assumptions. The first is a condition on $K$; $$\tag{1}\lim_{x\to \pm \infty} x^h K(x)=0, \qquad \forall h\in\mathbb N.$$ We also need the following uniform $L^1$ boundedness of $\phi_n$; $$ \tag{2} \lVert \phi_n\rVert_{L^1}\le C <\infty.$$

Indeed, from (1) we infer that there is a Schwartz function $K_\epsilon$ such that $$ \lVert K-K_\epsilon\rVert_\infty \le \epsilon;$$ we can take $K_\epsilon:=K\ast \eta_\epsilon$, where $\eta_\epsilon$ is a usual mollifier. We can now conclude by observing that (assuming $K(0)=K_\epsilon(0)=0$ to ease typing) $$\begin{split} \left\lvert \int_{\mathbb R} \phi_n(x)K(x)\, dx\right\rvert &\le \left\lvert \int_{\mathbb R} \phi_n(x)(K-K_\epsilon)(x)\, dx \right\rvert + o(1; n\to\infty),\\ &\le C\epsilon+o(1; n\to \infty),\end{split} $$ where we used (2). Thus, $$ \operatorname*{limsup}_{n\to \infty} \left\lvert \int_{\mathbb R} \phi_n(x)K(x)\, dx\right\rvert \le C\epsilon, $$ from which the claim follows.

---

The previous version of this answer contained a flaw. I "proved" that (2) is a consequence of the convergence of $\phi_n$ in the sense of tempered distributions. This is not true; the counterexamples of reuns converge in tempered distributional sense, but $\lVert\phi_n\rVert_{L^1}\to \infty$. The flawed arguement is the following.

Now, by assumption, $\phi_n\in L^1(\mathbb R)$ for all $n$, and it converges in $\mathcal{S}'$. We claim that this implies the uniform bound $$\tag{*}\lVert \phi_n\rVert_{L^1}\le C<\infty.$$ Indeed, there must be constants $C_{h,k}>0$ such that $$\lvert\langle \phi_n, \psi\rangle \rvert \le C_{h,k} \lVert x^h \tfrac{d^k}{dx^k}\psi\rVert_\infty , \qquad \forall \psi\in \mathcal S;$$ see Wikipedia (this is not stated explicitly, but it is a consequence of the seminorm characterization of the convergence in $\mathcal{S}'$). Letting $h=k=0$ we have a uniform bound in $(L^\infty)'$, which proves the bound (*) by duality.

Where's the mistake? The problem is that I implicitly used that $\mathcal S\subset L^\infty$ is dense. This is false.