Hi, I need to show that $\displaystyle \|f\|_1=\int_{0}^{1}|f(x)|dx\leq(\int_{0}^{1}|f(x) |^2dx)^\frac{1}{2}=\|f\|_2$
I think I need to use the Cauchy Schwarz inequality:
$\displaystyle |<x,y>|\leq\|x\|_2\|y\|_2$
Any help would be great. Thanks
Hi, I need to show that $\displaystyle \|f\|_1=\int_{0}^{1}|f(x)|dx\leq(\int_{0}^{1}|f(x) |^2dx)^\frac{1}{2}=\|f\|_2$
I think I need to use the Cauchy Schwarz inequality:
$\displaystyle |<x,y>|\leq\|x\|_2\|y\|_2$
Any help would be great. Thanks
$\displaystyle \left(\int_0^1 |f(x)g(x)|~dx \right )^2 \leq \left(\int_0^1 [f(x)^2] ~dx \right )\left(\int_0^1 [g(x)^2] ~dx\right )$
Let $\displaystyle g(x) = 1 $ so we have
$\displaystyle
\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
$