# Showing $||k_{\lambda} * f - f||_{\infty} \to 0$ as $\lambda \to \infty$

Let $$f$$ be bounded and uniformly continuous on $$\mathbb{R}$$ and let $$(k_{\lambda})_{\lambda \in \mathbb{N}}$$ be a Dirac family on $$\mathbb{R}$$. I would like to show that $$||k_{\lambda} * f - f||_{\infty} \to 0$$ as $$\lambda \to \infty$$. Here, $$*$$ is a convolution operator and $$||g||_{\infty} = \operatorname{ess sup}_{x \in \mathbb{R}}|g(x)|$$.

My attempt:

We have, by triangle inequality for integral, $$|k_{\lambda} * f - f| \leq \int_{\mathbb{R}}|f(x - t)- f(x)||k_{\lambda}(t)|\ dt.$$ By uniform continuity of $$f$$, we know that $$\forall \varepsilon > 0, \exists \delta > 0$$, such that $$|f(x-t) - f(x)| < \varepsilon$$. Also, by boundedness of $$f$$, we can find $$M > 0$$ such that $$|f(x)| \leq M$$ for any $$x \in \mathbb{R}$$. So, \begin{align*} |k_{\lambda} * f - f| &\leq \int_{|t|< \delta}|f(x - t)- f(x)||k_{\lambda}(t)|\ dt + \int_{|t| \geq \delta}|f(x - t)- f(x)||k_{\lambda}(t)|\ dt\\ &< \varepsilon\int_{|t|< \delta}|k_{\lambda}(t)|\ dt + 2M\int_{|t| \geq \delta}|k_{\lambda}(t)|\ dt. \end{align*} Since $$(k_{\lambda})_{\lambda \in \mathbb{N}}$$ is a Dirac family, we know that for any $$\delta > 0$$, $$\lim_{\lambda \to \infty}\int_{|x| > \delta}|k_{\lambda}(x)|dx = 0$$. So $$2M\int_{|t| \geq \delta}|k_{\lambda}(t)|\ dt \to 0$$ as $$\lambda \to \infty$$. What is left is $$\varepsilon\int_{|t|< \delta}|k_{\lambda}(t)|\ dt$$. I am not sure what I can do with it. I know that, a Dirac family on $$\mathbb{R}$$ is a sequence $$(k_{\lambda})_{\lambda \in \mathbb{N}}$$ of continuous functions in $$L^1(\mathbb{R})$$, which satisfies

1. $$\forall \lambda \in \mathbb{N} : \int_{\mathbb{R}}k_{\lambda}(x)\ dx = 1,$$
2. $$\limsup_{\lambda \to \infty}||k_{\lambda}|||_1 < \infty,$$
3. $$\forall \delta > 0: \int_{|t| \geq \delta}|k_{\lambda}(t)|\ dt \to 0$$ as $$\lambda \to \infty$$.

I have used property 3. I don't know how to use the other properties, so I can show that $$||k_{\lambda} * f - f||_{\infty} \to 0$$ as $$\lambda \to \infty$$. Any idea would be appreciated.

You are almost done. By 2) $$\lim \sup_{\lambda \to \infty} \int_{|t|<\delta} |k_{\lambda} (t)| dt \leq \lim \sup_{\lambda \to \infty}\|k_{\lambda}\|_1 <\infty$$.
• Right, I was thinking of the second property. So we have $$|k_{\lambda} * f - f| < \varepsilon\limsup_{\lambda}||k_{\lambda}||_1$$ for any $\varepsilon > 0$. But, why can I conclude that $||k_{\lambda}* f - f||_{\infty} \to 0$ from that result? Commented Nov 19, 2020 at 8:41
• Well, given $\eta >0$ choose $\epsilon$ such that $\epsilon \lim \sup \|k_{\lambda}\|_1 <\eta /2$. There exists $\lambda_0$ such that the second term in your inequality is less than $\eta /2$ and $\epsilon \|k_{\lambda}\|_1 <\eta /2$ for $\lambda >\lambda_0$. We then have $\|k_{\lambda} *f-f\| <\eta$ for $\lambda >\lambda_0$. Commented Nov 19, 2020 at 8:50