Is the integral function of a $L^2[0,1]$ function continuous? [duplicate]

My problem is:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function (w.r.t. the Lebesgue measure) that is in $L^2$. Show that the function

$$F(x)=\int_0^x f(t)\,dt$$

satisfies $|F(x)-F(y)| \leq C|x-y|^{\frac{1}{2}}$

I don't really know where to start. I feel like it's kind of because $f(t)$ isn't allowed to grow more quickly than $(x-t)^{-\frac{1}{2}}$ near $x$, but I don'tknow how to make anything out of this.

It's a consequence of Cauchy-Schwarz inequality: for fixed $x,y\in\mathbb R$ we have $$|F(x)-F(y)|=\left|\int_{[x,y]}f(t)\cdot 1 dt\right|\leq \sqrt{\int_{[x,y]}f(t)^2dt}\sqrt{\int_{[x,y]}dt}\leq \sqrt{|x-y|}\lVert f\rVert_{L^2}.$$