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Say we have that $\mu$ is a measure on the Borel $\sigma$-algebra on $[0, 1]$ and for every $f$ that is real-valued and continuously differentiable we have$$\left| \int_0^1 f'(x)\,d\mu(x)\right| \le \sqrt{\int_0^1 f(x)^2\,dx}.$$Is $\mu$ absolutely continuous with respect to Lebesgue measure on $[0, 1]$?

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  • $\begingroup$ Is $\mu$ a positive measure? $\endgroup$ – zhw. Oct 14 '16 at 16:27
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Hint/idea: Let $0\le a < b\le 1.$ Define $f$ to be the piecewise linear function whose graph connects the points $(0,0), (a,0), (b,1),(1,1).$ This $f$ is not $C^1$ but I think it shows the way. We can think of $f'$ as equal to $0$ on $[0,a),$ equal to $1/(b-a)$ on $[a,b],$ and equal to $0$ on $(b,1].$ Then

$$|\int_0^1 f'\, d\mu| = |\int_{[a,b]} f'\, d\mu| = |\frac{\mu([a,b])}{b-a}| \le (\int_0^1 f^2 )^{1/2} \le 1.$$

Thus $|\mu([a,b])| \le b-a.$ This is true for any $a,b$ and thus shows $\mu$ is AC with respect to Lebesgue measure.

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