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This question already has an answer here:

If

$u(x) \in C([a, b]), u(a) = 0,\; u(x) = \int_{a}^{x}u^{'}(t)dt$

then

$\int_{a}^{b} |u|^{2} dx \le \frac{1}{2}(b - a)^{2}\int_{a}^{b}|u^{'}(t)|^{2}dt$

The book said it can be proved using cauchy-schwarz-inequality, but I cannot make it.

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marked as duplicate by Martin R, mrtaurho, idm, Macavity inequality Dec 22 '18 at 18:45

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One long line proves it: $$\int_a^b |u(t)|^2 dt=\int_a^b\left|\int_a^t 1\cdot u'(t')dt'\right|^2 dt\le\int_a^b\left[\int_a^t 1^2 dt'\cdot\int_a^t |u'(t')|^2dt'\right] dt\\\le\int_a^b\left[(t-a)\cdot\int_a^b |u'(t')|^2dt'\right] dt=\frac{(b-a)^2}{2}\int_a^b |u'(t')|^2dt'.$$The first $\le$ uses Cauchy-Schwarz; the second replaces an $\int_a^t dt'$ with $\int_a^b dt'$.

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