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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.

1 Answers 1

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

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    @Na$t$eEldredge. Sure, that would also do it.2012-04-23