Let $f\in C(\mathbb{R})$.
Let $\Lambda_n(f)(x) = \frac{n}{\sqrt{\pi}}\int_{\mathbb{R}}f(x-y)e^{-(ny)^2}dy.$ Show that $\Lambda_n(f)$ is uniformly continuous on $\mathbb{R}$ for all n, given that : $$ \frac{n}{\sqrt{\pi}}\int_{\mathbb{R}}e^{-(ny)^2}dy =1$$ for all $n \in \mathbb{N}$
I am not understand i fully understand the notation. Can one assume $f(x-y)$ can be evaluated as some function $f(y) + K, K\in \mathbb{R}$, so that one can look at the integral: $$ \frac{n}{\sqrt{\pi}}\int_{\mathbb{R}}(f(y) +K)e^{-(ny)^2}dy $$ After looking at this for quite some time, it does not seem like simply looking at $\epsilon - \delta$ on $\Lambda(f)(x)$ gets me very far without some usful simplification. If anybody has a suggestion to a usful theorem or result that can get me started, I would be very greatful.