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Let $\mu$ be a measure on Borel-sigma algebra on $[0,1]$ which has the property that for every $f$ continuously differentiable, we have

$$ \bigg| \int f' \ d\mu \bigg| \leq \left (\int_{0}^{1} f(x)^2 \ dx \right )^{1/2} $$

Then prove that :

(i) $\mu$ is absolutely continuous w.r.t Lebesgue measure on $[0,1]$.

(ii) If $g$ is the Radon-Nikodym derivative of $\mu$ w.r.t Lebesgue measure, then there exists a constant $c$ such that $$ |g(x)-g(y)| \leq c |x-y|^{1/2} $$

Can anyone help me with this ? I am unable to proceed in any fruitful direction.

If I have $m(A)=0$, who do I use the above condition to show $\mu(A)=0$ ? All I have is an integral over $[0,1]$.

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marked as duplicate by PhoemueX, saz, Ali Caglayan, tired, Rory Daulton Nov 7 '15 at 17:33

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