Let X be the set of continuous functions on $\mathbb{R}$ with compact support, ie, the set of functions $f\in C(\mathbb{R})$ such that $\exists$ $r>0$ such that $f(x) = 0$ when $|x|\geq r$. And any $f\in X$ is uniformly continuous.
Hint: $\frac{n}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} e^{-(ny)^2}dy = 1$ (*) for any $n\in \mathbb{N}$.
Let $\Lambda_n(f)(x) = \frac{n}{\sqrt(\pi)}\int_{\mathbb{R}} f(x-y)e^{-(ny)^2}dy$
And I'm to prove that $\Lambda_n(f)$ is uniformly continuous on $\mathbb{R}$ for all n.
So I'm thinking let $x,x_2 \in \mathbb{R}$, then
$|\Lambda(f)(x)-\Lambda(f)(x_2)| = |\frac{n}{\sqrt{\pi}} \int_{\mathbb{R}} [f(x-y)-f(x_2-y)]e^{-(ny)^2}dy|$
$=\epsilon|\frac{n}{\sqrt{\pi}} \int_{\mathbb{R}} e^{-(ny)^2}dy|$, (for any $\epsilon>0$, bc f is uni cont, I'm almost certainly wrong about this step)
$= 1\cdot\epsilon$, by (*)
I'm very much lost and any help would be appreciated.
Fix $\epsilon> 0$. Let $x, x_2\in \mathbb{R}$, then \begin{align} |\Lambda_nf(x) - \Lambda_nf(x_2)| &= \left|\frac{n}{\sqrt{\pi}} \int_{\mathbb{R}} [f(x-y)-f(x_2-y)]e^{-(ny)^2}dy\right|\\ &\leq \frac{n}{\sqrt{\pi}} \int_{\mathbb{R}} |f(x-y)-f(x_2-y)|e^{-(ny)^2}dy. \end{align} Since $f$ is uniformly continuous, there exists $\delta > 0$ such that $|x - x_2| < \delta$ implies $|f(x - y) - f(x_2 - y)| < \epsilon$. Hence, $$|\Lambda_nf(x) - \Lambda_nf(x_2)| < \epsilon \frac{n}{\sqrt{\pi}} \int_{\mathbb{R}}e^{-(ny)^2}dy = \epsilon$$ as long as $|x - x_2| < \delta$. This completes the proof. I'm not sure what your first paragraph about $X$ is for though.