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Assume $f$ is a twice differentiable function on $\Bbb R $ and $ f'' $ is continuous. Assume further that $ f(-1) = f(1) = 0$ then prove that $$\int\limits_{-1}^1 f' ^ 2\leq \frac{1}{2} \left( \int\limits_{-1} ^ 1 f^2 + \int _{-1}^ 1 (f'')^2 \right) $$

I thought of applying Rolle's theorem on $f$ but I'm not sure how to use it to prove this inequality. I also thought of using A.M. - G.M. inequality but it does not seem fruitful. I am stuck and unable to start this problem. Thank you.

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  • $\begingroup$ Are you familiar with Sobolev inequalities? $\endgroup$ Commented Jun 8, 2020 at 3:12
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    $\begingroup$ No but give me a link of any material related to it which is easy to read by beginner. . I will go through it $\endgroup$ Commented Jun 8, 2020 at 3:17

1 Answer 1

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We use integration by parts.

\begin{align*} \int_{-1}^1 (f'(x))^2 \, dx &= \int_{-1}^1 f'(x) \frac{df(x)}{dx} \, dx\\ &= \left[f'(x)f(x)\right]_{-1}^1 - \int_{-1}^1 f(x) f''(x) \,dx\\ &= -\int_{-1}^1 f(x)f''(x)\,dx\\ &\leq \int_{-1}^1 |f(x)||f''(x)|\,dx\\ &\leq \frac 12 \int_{-1}^1 f(x)^2 + f''(x)^2 \, dx \text{ (by AM-GM)}\\ &= \frac 12 \left(\int_{-1}^1 f(x)^2 \, dx + \int_{-1}^1 f''(x)^2 \, dx\right), \end{align*} as required.

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