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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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    Maybe one comment worth mentioning is that you can't use the usual $L^{2}(\mathbb{R})$ Cauchy-Schwarz inequality, since the constant function $1$ is not in $L^{2}(\mathbb{R})$. But instead, for each fixed $x,y\in \mathbb{R}$ you may consider $f$ and the constant function $1$ restricted to $[x,y]$ and the Cauchy-Schwarz related to $L^{2}([x,y])$. But anyways, it doesn't really change anything, just a technical detail.2012-04-22
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    I agree, I should precise I work for fixed $x$ and $y$, then show that the constant $C$ doesn't depend on them.2012-04-22
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    Should I check that $F(x)-F(y)$ is well defined?2012-04-22
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    Probably yes, but in fact just check that $F(x)$ is well-defined for each $x$.2012-04-22
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    @ThomasE.: Alternatively, write the integral as $\int_{\mathbb{R}} f(t) 1_{[x,y]}(t)\,dt$ and use Cauchy-Schwarz in $L^2(\mathbb{R})$.2012-04-23
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    @NateEldredge. Sure, that would also do it.2012-04-23