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.


1 Answer 1


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$.

  • $\begingroup$ 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? $\endgroup$
    – Vicky
    Commented Nov 19, 2020 at 8:41
  • $\begingroup$ 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 $\lambda >\lambda_0$. $\endgroup$ Commented Nov 19, 2020 at 8:50

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