Suppose $\{f_j\}$ converges to $f_0$ in $L^2$ sense. Show that the functions $F_j (x) = \int_{0}^x f_j(t) dt$ converges uniformly to $\int_{0}^x f_0(t) dt$ for all $x \in [-\pi,\pi]$.
My proof: Let $\epsilon > 0$. We want to find $N$ such that whenever $j>N$, \begin{equation} \bigl| \int_{0}^x f_j (t) dt - \int_{0}^x f_0 (t) dt \bigr| < \epsilon \text{ for all } x \in [-\pi,\pi] \end{equation} Apply Cauchy Schwarz,
$$\bigl| \int_{0}^x f_j (t) dt - \int_{0}^x f_0 (t) dt \bigr| \le \int_{0}^x |f_j (t) - f_0 (t) | dt\\ \le \left(\int_{0}^x (f_j(t) - f_0(t))^2 dt\right)^\frac{1}{2} \left( \int_{0}^x 1 dt\right)^\frac{1}{2}\\ \le \left(\int_{-\pi}^\pi (f_j(t) - f_0(t))^2 dt\right)^\frac{1}{2} \left(\int_{-\pi}^\pi 1 df\right)^\frac{1}{2}\\ = \|f_j - f_0\|_{L^2} \sqrt{2\pi}$$ Since $f_j$ converges to $f_0$ in $L^2$ sense, the result follows.
I want to make sure I did this proof correctly. Thanks.