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