# Convergence of convolution under $\| \cdot \|_\infty$ with a weaker kernel

Let $$(K_\delta)_{\delta > 0}$$ be a family of integrable functions so that there is a constant $$C \in (0 ,\infty)$$ such that $$\int K_\delta = 1$$, $$\int | K_\delta | \leq C$$ for every $$\delta > 0$$ and for every $$\varepsilon > 0$$ we have $$\begin{equation*} (*) \quad \underset{\delta \to 0+}{\lim} \int_{| x | \geq \varepsilon} | K_\delta(x) | dx = 0 \end{equation*}$$ Let $$p \in [1, \infty]$$. Prove that for every $$f \in L^p(\mathbb{R}^d)$$ we have $$f * K_\delta \to f$$ in $$L^p(\mathbb{R}^d)$$ as $$\delta \to 0+$$.

My thoughts:

I solved it for $$p\in [1, \infty)$$, but I believe there is a problem with $$p = \infty.$$ So here is my attempt at a counterexample:

let $$f(x) = \begin{cases} -1 & \text{ if } x\in (-\infty, 0] \\ 1 & \text{ if } x\in (0, \infty) \\ \end{cases}$$ and $$K_\delta$$ a non-negative smooth even function. Then $$\begin{equation*} (*) \quad f * K_\delta(0) = \int_{\mathbb{R}} f(-t)K_\delta(t) dt = \int_{\mathbb{R}_{\leq 0}} K_\delta dt - \int_{\mathbb{R}_{> 0}} K_\delta dt = 0 \end{equation*}$$ Using $$(*),$$ we have \begin{align*} f*K_\delta(0) - f(0) &= \int_\mathbb{R} (f(-t) + 1)K_\delta(t) \\ &= \int_\mathbb{R} f(-t)K_\delta(t) dt + \int_\mathbb{R} K_\delta(t) dt \\ &= 0 + 1 = 1 \quad \text{ by (*).} \end{align*} Thus, $$\begin{equation*} \| f*K_\delta - f \|_\infty \geq 1 \end{equation*}$$ For an example, consider $$\begin{equation*} K_\delta(x) = \frac{1}{\pi} \frac{\delta}{x^2 + \delta^2} \quad \text{ for }x\in \mathbb{R} \end{equation*}$$

Any feed back is much appreciated.

I'm not actually sure if the example I give at the end fits. It was an example given in Stein's Real Analysis of a kernel.

• $|h(0)| \geq 1$ does not imply that $\|h\|_{\infty} \geq 1$. (If $h$ is continuous then this implications is true). But I do suspect that the result is not true for $p=\infty$. – Kavi Rama Murthy Apr 8 '19 at 7:22

The result is false for $$p=\infty$$. To see this, assume that $$K$$ is smooth and compactly supported, and let $$(K_\delta)_{\delta>0}$$ be the approximate identity generated by $$K$$. Then for any $$f\in L^\infty(\mathbb{R})$$, $$K_\delta*f$$ is a continuous function (in fact uniformly continuous). But let $$E\subset[0,1]$$ be a fat Cantor set (i.e. a Cantor set of positive measure), and let $$f$$ be the indicator function of $$E$$. Then $$K_\delta*f$$ cannot converge in $$L^\infty$$ to $$f$$. For each $$K_\delta*f$$ is continuous, and for continuous functions convergence in $$L^\infty$$ is equivalent to uniform convergence. But then $$K_\delta*f$$ must be a sequence of continuous functions converging uniformly, and hence the limit must be continuous. Call this limit $$f_\infty$$; then $$\|f_\infty - f\|_\infty > 0$$. For $$f$$ cannot be equal a.e. to a continuous function: such a continuous function would be $$0$$ on a dense subset of $$[0,1]$$, and therefore $$0$$ everywhere, but also $$1$$ on a positive measure subset of $$[0,1]$$, which is absurd.
What is true is that if $$f\in L^\infty$$, then $$K_\delta*f$$ will converge to $$f$$ on the set where $$f$$ is continuous. The convergence will be uniform on the set where $$f$$ is uniformly continuous.