$ \left( {\int\limits_0^1 {f^2(x)\ \text{d}x} }\right)^{\frac{1} {2}} \ \geqslant \quad \int\limits_0^1 {\left| {f(x)} \right|\ \text{d}x} $
I can't prove it )=
$ \left( {\int\limits_0^1 {f^2(x)\ \text{d}x} }\right)^{\frac{1} {2}} \ \geqslant \quad \int\limits_0^1 {\left| {f(x)} \right|\ \text{d}x} $
I can't prove it )=
$\int_0^1 |f(x)| \, dx = \int_0^1 |1||f(x)| \, dx \leq \sqrt{\int_0^1 1 \, dx} \sqrt{\int_0^1 |f(x)|^2 \, dx} = \sqrt{\int_0^1 |f(x)|^2 \, dx}$
By Cauchy-Schwarz.
L^q– 2011-07-02, for functions in a total ambient set ( in the measure triple) S.
Define $ \mu = \int_0^1 {|f(x)|\,dx} $ and $ \sigma^2 = \int_0^1 {(|f(x)| - \mu )^2 \,dx} . $ Then $ \sigma^2 = \int_0^1 {f^2 (x)\,dx} - 2\mu \int_0^1 {|f(x)|\,dx} + \mu ^2 = \int_0^1 {f^2 (x)\,dx} - \mu ^2 . $ Since $\sigma^2 \geq 0$, $ \int_0^1 {f^2 (x)\,dx} \geq \mu ^2. $ Taking square roots of both sides yields the desired result: $ \bigg(\int_0^1 {f^2 (x)\,dx} \bigg)^{1/2} \ge \int_0^1 {|f(x)|\,dx}. $
EDIT: The idea used here is that for a random variable $Y$ with finite second moment, $ {\rm Var}(Y) := {\rm E}[Y - {\rm E}(Y)]^2 \geq 0. $ So, $ {\rm Var}(Y) = {\rm E}(Y^2) - 2{\rm E}(Y){\rm E}(Y)+ {\rm E}^2 (Y) = {\rm E}(Y^2) - {\rm E}^2 (Y), $ and hence $ {\rm E}(Y^2) \geq {\rm E}^2 (Y). $ To relate this to the question at hand, let $X$ be a uniform$[0,1]$ random variable, and let $Y=|f(X)|$ (for $f$ a square-integrable function on $[0,1]$). Then $ {\rm E}(Y^2) = {\rm E}[f^2 (X)] = \int_0^1 {f^2 (x)\,dx} $ and $ {\rm E}^2 (Y) = {\rm E}^2 (|f(X)|) = \bigg(\int_0^1 {|f(x)|\,dx} \bigg)^2 . $ Hence $ \int_0^1 {f^2 (x)\,dx} \geq \bigg(\int_0^1 {|f(x)|\,dx} \bigg)^2 , $ which gives the desired result after taking square roots.