Prove that $\Lambda (f)$ converges uniformly on $\mathbb{R}$. Let $X=C(\mathbb{R})$ the set of all continuous functions with compact support such that there exist $r>0$ such that $f(x)=0$ when $|x|\geq r$.
For any $f\in X$ and $n\in\mathbb{N}$ define $\Lambda_n(f)(x)=\frac{n}{\sqrt\pi}\int_\mathbb{R}f(x-y)e^{-(ny)^2}dy$.
I've been trying to show that $\lim_{n\rightarrow\infty}\Lambda_n(f)(x)=f(x)$.
I've tried to use the product rule for integration, but no luck there. Tried to use the definition of uniform convergence, but nothing there as well. Also tried to use the $\sup$ method.
I would really appreciate clues!
BTW: I may use that $\frac{n}{\sqrt\pi}\int_{-\infty}^{\infty}e^{-(ny)^2}dy=1$ and $\lim_{n\rightarrow\infty}\frac{n}{\sqrt\pi}\int_{-R}^{R}e^{-(ny)^2}dy=1$.
 A: Hint: 
\begin{align*} 
|\Lambda_n(f)(x)-f(x)| &= \Bigg|\int_{\mathbb{R}}{(f(x-y)-f(x))\frac{ne^{-n^2y^2}}{\pi^{1/2}}\,dy\,}\Bigg| \\ 
&\leq \int_{\mathbb{R}}{\left(\sup_{|r-x| \leq n^{-1/2}}\,|f(r)-f(x)|\right)\frac{ne^{-n^2y^2}}{\pi^{1/2}}\,dy} + \int_{|y| \geq n^{-1/2}}{2\|f\|_{\infty}\frac{ne^{-n^2y^2}}{\pi^{1/2}}\,dy} \\
&\leq \left(\sup_{|r-x| \leq n^{-1/2}}\,|f(r)-f(x)|\right) + 2\|f\|_{\infty}\int_{|y| \geq n^{1/2}}{e^{-y^2}\,dy}.
\end{align*}
A: The other answer provides the solution, but I will add another, in hopes of clearing up 
the questions posed by the OP in the comments. 
$1).\ $ Uniform convergence/convergence in sup norm: a sequence $(f_n)$ converges to $f$ uniformly on a set $S\subseteq \mathbb R$ if and only if to each $\epsilon>0$, there is an integer $N$ such that $n>N\Rightarrow |f_n(x)-f(x)|<\epsilon$ $\textit{for all}\ x\in S.$ This means that if $n>N,$ we must have $\sup_{x\in S}|f_n(x)-f(x)|\le \epsilon.$ Now, by definition, the sup norm on $S$ is $\|f\|_{\infty}=\sup_{x\in S}|f_n(x)-f(x)|$. The argument is reversible: if we start with the definition of the sup norm, we arrive back at the definition of uniform convergence of $(f_n).$  So the two notions of convergence are the same. 
$2).$ The sequence functions $(K_n(x))=\left(\frac{ne^{-n^2x^2}}{\pi^{1/2}}\right )$ is an $\textit{approximate identity};\ $ i.e., it satisfies the following conditions:
$i).\ \int K_n(x) dx = 1$ for every $n$
$ii).\  $ there exists a constant $C ≥ 0$ such that $\int |K_n(x)| dx ≤ C$ for every $n$;
$iii).\ \lim_{n\to \infty} \int_{|x|≥\delta}|K_n(x)| dx = 0$ for every $\delta >0.$
Now, if we use these observations about our sequence, it is easy to prove the claim. Let $\epsilon>0$ and suppose $f$ is supported on the compact set $K$. Since $f$ is continuous on $K$ and $K$ is compact, $f$ is also uniformly continuous there. This means that there exists a number $\delta>0$ such that $|f(x) − f(x − y)| < \epsilon$ for every $x \in K$ and every $y\in \mathbb R$
that satisfies $|y| < \delta$. For this $\delta$, now choose $N$ so large that $\int_{|y|≥\delta}|Kn(y)|dy < \epsilon$ for $n>N.$ Then, if $x\in K$, and $n>N,$  we have 
$\Lambda_n(f)(x)-f(x)\le$
$ \int_{|y|<\delta} |f(x) − f(x − y)||K_n(y)| dy
+ \int_{|y|≥\delta}|f(x) − f(x − y)||K_n(y)| dy \le$
$C\epsilon + 2\epsilon \|f\|_{\infty}.$
