Let $\phi:\mathbb R\mapsto\mathbb R$ is defined by $\phi(x)=\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}$ for all $x\in\mathbb R$ and $$ {\cal L}_0^2(\mathbb R)=\left\{f:\mathbb R\mapsto\mathbb R\ |\ \int_{\mathbb R}f(x)\phi(x)dx=0,\textrm{ and }\int_{\mathbb R}[f(x)]^2\phi(x)dx<\infty\right\}. $$ Note that ${\cal L}_0^2(\mathbb R)$ is a Banach space if it is completed with norm $\|.\|$ defined by $\|f\|=\int_{\mathbb R}[f(x)]^2\phi(x)dx.$
Now, suppose that we have a sequence of functions $f_n\in {\cal L}_0^2(\mathbb R)$ converging to $f\in {\cal L}_0^2(\mathbb R)$ in the sense $\|f_n-f\|\to 0.$
I wish to have $\{F_n\}$ converges uniformly to $F$ where $F_n(x)=\int_{-\infty}^xf_n(t)\phi(t)dt$ and $F(x)=\int_{-\infty}^xf(t)\phi(t)dt$.
What I have tried so far is:
Since $\{f_n\}$ and $f$ are bounded, then according the Dominated Convergence Theorem, we have $\int_{-\infty}^xf_n(t)\phi(t)dt\to \int_{-\infty}^xf(t)\phi(t)dt$ that means $\{F_n(x)\}\to F(x)$ for every $x\in\mathbb R$. However, it only proved $\{F_n\}$ converges to $F$ pointwise, didn't it? My question is: how will I improve it to uniformly convergence of $\{F_n\}?$ Any suggestion? Any help would be appreciated.
You have
$$\lvert F_n(x) - F(x)\rvert \leqslant \int_{-\infty}^x \lvert f_n(t) - f(t)\rvert\phi(t)\,dt \leqslant \int_{-\infty}^\infty \lvert f_n(t)-f(t)\rvert\phi(t)\,dt.$$
Use the Cauchy-Schwarz inequality and $\lVert f_n-f\rVert\to 0$.