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Let $f : [0, 1] \to \mathbb{R}$ a differentiable function with continuous derivative and $\int_0^1 f(x) dx=\int_0^1 x f(x) dx=1$. Prove that $\int_0^1 |f'(x)|^3 dx \ge \left(\frac{128}{3\pi}\right)^2$

My try is to write $f$ in terms of the shifted Legendre polynomials $\tilde{L}_n(x)=L_n(2x-1)$ as $$ f(x) = \sum_{n=0}^{+\infty} a_n\,\tilde{L}_n(x), $$ The two conditions on $f$ gives $a_0=1$ and $a_1=3$. But I didn't find a way to get the given inequality.

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    Can you provide some more context for where this problem comes from? – 2017-02-22
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    (+1) Nice question. The exponent $3$ in $\left|f'(x)\right|^3$ makes a direct application of Parseval's identity almost impossible. What about combining Holder's inequality with Parseval's identity? – 2017-02-22
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    Holder's inequality is enough. See below. – 2017-05-05

1 Answers 1

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Note that $$\int_0^1 x(1-x) f'(x) dx=\left[x(1-x) f(x)\right]_0^1-\int_0^1 (x-x^2)' f(x) dx\\=0-\int_0^1 f(x) dx+2\int_0^1 xf(x) dx=-1+2=1.$$ Hence, by Holder's inequality, $$1=\int_0^1 x(1-x) f'(x) dx\leq \left(\int_0^1 (x(1-x))^{3/2}dx\right)^{2/3} \left(\int_0^1 |f'(x)|^3 dx\right)^{1/3}.$$ Finally $$\int_0^1 |f'(x)|^3 dx\geq \left(\int_0^1 (x(1-x))^{3/2}dx\right)^{-2}= B(5/2,5/2)^{-2}=\left(\frac{\Gamma(5)}{\Gamma(5/2)^2}\right)^{2} =\left(\frac{128}{3\pi}\right)^{2}$$ where $B(x,y)$ is the Beta function.