$$ \left( {\int\limits_0^1 {f^2(x)\ \text{d}x} }\right)^{\frac{1} {2}} \ \geqslant \quad \int\limits_0^1 {\left| {f(x)} \right|\ \text{d}x} $$
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.
– 2011-07-02
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.