Let $\phi _n : \mathbb{R} \to \mathbb{R}$ be a sequence of functions with the following properties:

1) For all $n \in \mathbb{N}$ we have $\phi _n (x) \geq 0$ for all $x \in \mathbb{R}$ and $\phi_n (x) = 0$ for all $|x| \geq 1$.

2) $\int_{- \infty}^{\infty} \phi_n (x) \text{ } dx = 1$ for all $n \in \mathbb{N}$.

3) For every $\delta >0$ we have $\lim_{n \to \infty} \int_{|x| > \delta} \phi _n (x) \text{ } dx = 0.$

Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function so that $f(x) = 0$ for all $|x| \geq 1$. Prove that $\{f * \phi _n\}$ uniformly converges to $f$.

$(f*p)(x) = \int_{-\infty}^\infty f(y) p(x-y) dy$ is convolution.

First note that $f$ uniformly continuous on $\mathbb{R}$. Given $\epsilon >0$ there exists $\delta >0$ so that $|t-s| \leq \delta$ implies $|f(t)-f(s)| < \epsilon$. Apply (3) with this delta and choose $N$ so that for all $n \geq N$ we have $$\left| \int_{|x| > \delta} \phi _n (x) \text{ } dx \right| < \epsilon.$$ Let $n \geq N$.

I think we need to compute $|f * \phi _n (x) - f(x)|$ for arbitrary $x$. How can we finish this proof?


Continuing with the $\delta$ and $\varepsilon$ you defined :

$f$ is bounded, let say by $M$ (from the hypothesis : as a continuous function, $f$ is bounded over the compact set $\{u, \ |u|\leq 1\}$, and null outside this set); since $\int \varphi_n =1$, \begin{align*} |f * \phi _n (x) - f(x)|&= \bigg|\int_{-\infty}^\infty f(x-y) \phi_n(y) dy- f(x)\underbrace{\int_{-\infty}^\infty \phi_n(y) dy}_{=1}\bigg| \\ & =\bigg|\int_{-\infty}^\infty f(x-y)-f(x) \phi_n(y) dy\bigg| \\ & \leq \int_{-\infty}^\infty |f(x-y)-f(x)| \phi_n(y) dy \\ &= \int_{|y|>\delta} \underbrace{|f(x-y)-f(x)|}_{\leq 2M} \phi_n(y)dy +\int_{|y|\leq\delta} |f(x-y)-f(x)| \phi_n(y)dy \\ &\leq 2M \varepsilon +\int_{|y|\leq\delta} \underbrace{|f(x-y)-f(x)|}_{\leq \varepsilon} \phi_n(y) \\ & \leq 2M \varepsilon + \varepsilon\underbrace{\int_{-\infty}^\infty \phi_n(y) dy}_{=1} \\ &\leq (2M+1)\varepsilon, \qquad \forall x\in \mathbb{R} \end{align*} so the convergence is uniform.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.