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Let be $f:[0,1] \longrightarrow R $, $f$ is an integrable function such that:

$$\int_{0}^{1} f(x) \space dx = \int_{0}^{1} xf(x) \space dx=1$$

I need to prove that:

$$\int_{0}^{1} f^2(x) \space dx\geq4$$

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is it possible to apply cauchy's inequality ? – Theorem Jun 14 '12 at 13:06
@Theorem: is it useful here? I'll check it. – user 1618033 Jun 14 '12 at 13:07
up vote 9 down vote accepted

Note that if $h(x)=-2+6x$ then $$\int_0^1 h(x)\, dx = \int_0^1 xh(x)\, dx =1.$$ Moreover $$ \int_0^1 (h-f)^2 dx\ge 0 $$
The rest is simple. Also Cauchy--Schwarz works with a slightly modified proof.

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thanks. That's an interesting approach, but at the same time very simple. – user 1618033 Jun 14 '12 at 13:38

A geometric reading of this question is to consider the space $E=\mathcal C([0,1],\mathbb R)$ with inner product : $$\langle f,g \rangle = \int_0^1 f(t) g(t) dt$$ and say : Show that forall $f \in F^{\bot}$ , we have : $\| f \| \geq 2 $ where $F= \text{Span} \{x \mapsto 1, x \mapsto x \} $. By Gram-Schmidt orthogonalization procedure, we obtain $(x \mapsto 1, x \mapsto \sqrt 3(2x-1))$ as orthonormal basis of $F$ who gives one expression of $p_F(f)$ projection of $f$. By Pythagore theorem we have : $\|p_F(f)\| \leq \|f\|$ and since the basis is orthonormal wa have : $$\|P_F(f)\|^2=(\langle f,1 \rangle )^2 + (\langle f,\sqrt 3(2x-1) \rangle )^2 = 1^2 + \sqrt 3^2 = 4$$

All this can explain the provenace of the function $x \mapsto 6x-2 = 1 + 3(2x-1) $ used by Unoqualunque.

Edit : $x \mapsto 6x-2 = 1 + 3(2x-1) $ instead of $x \mapsto 6x-2 = 3(2x-1) $

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i think he mainly looked after a polynomial that satisfies the above relation and then he used it as an auxiliary function such that he may obtain the above inequality. (i think - - - i try to see things in an simpler manner) – user 1618033 Jun 14 '12 at 14:11
@Mohamed: <<....function $x↦6x−2=3(2x−1)$ used...>>?? – Unoqualunque Jun 14 '12 at 14:26
@ Unoqualunque Sorry I wont say : $ 6x -2 = 1+ 3(2x-1) $ who is in $\text{Span}\{ \sqrt{3} (2x-1),1\}=F^{\bot} $ . Precesely: since $(1,\sqrt 3(2x-1))$ is an orthonormal basis of $F^{bot}$, we have coordinates of $p_F(f)$ expessions : $\langle f,1 \rangle = 1$ and $\langle f, \sqrt 3(2x-1) \rangle = \sqrt 3$, so : $ \|p-F(f) \|^2 = 4$ , Pythagore gives : $\|f\|^2 \geq \|p_F(f) \|^2 =4$. – Mohamed Jun 14 '12 at 16:05

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