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$$ \left( {\int\limits_0^1 {f^2(x)\ \text{d}x} }\right)^{\frac{1} {2}} \ \geqslant \quad \int\limits_0^1 {\left| {f(x)} \right|\ \text{d}x} $$

I can't prove it )=

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Do you know Cauchy-Schwarz inequality? – Fabian Jul 2 '11 at 21:17
Do you know Jensen's inequality? – GEdgar Jul 2 '11 at 22:04
up vote 10 down vote accepted

$$\int_0^1 |f(x)| \, dx = \int_0^1 |1||f(x)| \, dx \leq \sqrt{\int_0^1 1 \, dx} \sqrt{\int_0^1 |f(x)|^2 \, dx} = \sqrt{\int_0^1 |f(x)|^2 \, dx}$$

By Cauchy-Schwarz.

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Jonas' post generalizes to the result that if p<q and there are no sets of infinite measure in S, then $L^q<L^p$, for functions in a total ambient set ( in the measure triple) S. – gary Jul 2 '11 at 21:52

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.

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@gary: "Jonas' post generalizes to the result ...", but Shai Covo's answer is more fundamental. – puresky Jul 3 '11 at 6:22
@puresky: Can you explain why it is more fundamental? I surely like it, but for me this is more a probabilistic argument and mine more analytic. – Jonas Teuwen Aug 23 '11 at 22:33
@Jonas Teuwen:your and Shai Covo's proofs can be induced frome finite situations, i.e. $\sum=>\int$, where Cauchy-Schwarz inequality $(\sum {a_i b_i})^2\leqslant (\sum {a_i^2}) (\sum {b_i^2})$ is more difficult to prove than the inequality $((\sum {a_i})/n)^2\leqslant (\sum {a_i^2})/n\Leftrightarrow ((a+b)/2)^2\leqslant (a^2+b^2)/2$. While my former comment is merely to understand the proof:) – puresky Aug 25 '11 at 8:41

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