If we have a compactly supported, integrable sequence of functions $g_i$ over $\mathbb{R}^n$ with their integrals $\int_{\mathbb{R}^n}g_i=1\ \forall i
$. Then, if $f$ be a uniform continuous function, can we say that $g_i*f\to f$ uniformly over $\mathbb{R}^n$? When can it be possible if it is not possible with the given conditions?
I have a theorem that says if $g_i$ satisfy the conditions as above and $f$ be bounded and integrable over $\mathbb{R}^n$, then $g_i*f\to f$ over $\mathbb{R}^n$ with the convergence being uniform over all compact subsets of $\mathbb{R}^n$. Now, if we make $f$ uniform continuous, how can we ensure uniform convergence throughout $\mathbb{R}^n$? Any hints? Thanks beforehand.
Best Answer
What is $g_i$ ? Did you mean $g_i(x)=i^n g_1(i x)$ ? Otherwise you'll need that the supports of the $g_i$ "tend" to $\{0\}$ and $A=\sup_i \|g_i\|_{L^1}<\infty$.
The proof is that we have a sequence $c_i\to 0$ such that the support of $g_i$ is in $|x|<c_i$ so that $$g_i\ast f(y)-f(y)=\int_{\Bbb{R}^n} g_i(x ) (f(y-x )-f(y))dx\le A \|f(y-x )-f(y)\|_{L^\infty(|x|<c_i)}$$