Let X be the set of functions $f\in C(\mathbb{R})$ such that $\exists\, r> 0$ such that $f(x) = 0$ when $|x|\geq r$.
With $n\in \mathbb{N}$ let $\displaystyle\Lambda(f)(x) = \frac{n}{\sqrt{\pi}} \int_{\mathbb{R}} f(x-y)e^{-(ny)^2}\,dy.$
How do I prove that $\Lambda_n(f)$ is uniformly continuous on $\mathbb{R}$ $\forall n?$
First, a bit of jargon: the elements of $X$ are continuous functions with compact support.
Now, two hints.
Is a well-known fact that a continuous function with compact domain is uniformly continuous, so $f\vert_{[-r,r]}$ is uniformly continuous. Now, you can prove easily that this plus $f(x) = 0$ for $|x|\ge r\implies$ $f$ is uniformly continuous in $\Bbb R$.
Finally, using the uniform continuity of $f$, you can bound $$ \Lambda_n(f)(x_1) - \Lambda_n(f)(x_2) = \frac{n}{\sqrt{\pi}} \int_{\mathbb{R}}(f(x_1 - y) - f(x_2 - y))e^{-(ny)^2}\,dy. $$ Btw, the constant $\frac{n}{\sqrt{\pi}}$ is irrelevant.