X is the set of functions $f\in C(\mathbb{R})$ such that $\exists \hspace{2pt} r>0$ such that $f(x) = 0$ when $|x|\geq r$. Hints: $\forall n\in \mathbb{N}$:
$\frac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-(ny)^2)}dy =1$ & $\forall R>0$: $\underset{n\to \infty}{lim}\frac{n}{\sqrt{\pi}}\int_{-R}^{R} e^{-(ny)^2)}dy =1$
I'm supposed to show that any $f\in X$ is uniformly continuous.
I don't see how I'm supposed to start solving this.
I don't understand why you've written down all these integrals. What you've written is just a general fact about functions with the property you wrote down. You're dealing with functions which are $0$ off a compact set. But a continuous function on a compact set is uniformly continuous. Since $f$ is uniformly continuous on $(-r,r)$ and trivially uniformly continuous on the complement, you can take the minimum of the two $\delta$'s of uniform continuity to see that if $x,y$ are both in $(-r,r)$ or both outside of it, uniform continuity holds. You only need to treat the case where one is inside and one is outside then.
This question has probably been asked before on this site, so this is a hint - keep searching if you can't work it out further.
Another approach to this problem is to use continuity, and the fact that the function is eventually $0$ to reduce to the following also easy-to-prove fact: A continuous function with limit at infinity is uniformly continuous. This question is asked for rays here, and you can easily adapt to your case.
