# Integral inequality

• Apr 14th 2010, 07:17 AM
ejgmath
Integral inequality
Hi, I need to show that $\|f\|_1=\int_{0}^{1}|f(x)|dx\leq(\int_{0}^{1}|f(x) |^2dx)^\frac{1}{2}=\|f\|_2$

I think I need to use the Cauchy Schwarz inequality:

$||\leq\|x\|_2\|y\|_2$

Any help would be great. Thanks
• Apr 14th 2010, 07:22 AM
simplependulum
$\left(\int_0^1 |f(x)g(x)|~dx \right )^2 \leq \left(\int_0^1 [f(x)^2] ~dx \right )\left(\int_0^1 [g(x)^2] ~dx\right )$

Let $g(x) = 1$ so we have
$

\left(\int_0^1 |f(x))|~dx\right )^2 \leq \left(\int_0^1 [f(x)^2] ~dx\right )\left(\int_0^1 dx\right ) = \int_0^1 [f(x)^2] ~dx
$