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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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    @Theorem: is it useful here? I'll check it.2012-06-14

2 Answers 2

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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.2012-06-14
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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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    @ 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 : $\|f\|^2 \geq \|p_F(f) \|^2 =4$.2012-06-14