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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?

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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.

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