# [SOLVED] Cauchy-Schwartz and norms

• Nov 29th 2009, 08:40 AM
nmatthies1
[SOLVED] Cauchy-Schwartz and norms
Let $\displaystyle ||-||_{1}$ and $\displaystyle ||-||_{2}$ be the $\displaystyle L^1$-norm and the $\displaystyle L^2$-norm on $\displaystyle C[a,b]$ the space of continuous real valued functions on the closed interval $\displaystyle [a,b]$: explicitly
$\displaystyle ||f||_{1}=\int_{a}^{b} |f|$, $\displaystyle ||f||_{2}=\sqrt{\int_{a}^{b} |f|^2}$.

Prove that $\displaystyle ||f||_{1} \leq ||f||_{2}$ for all $\displaystyle f \in C[a,b]$ (1). Hint: Remember the Cauchy-Schwartz inequality for integrals.

Cauchy-Schwartz:

$\displaystyle \int_{a}^{b} f(x)^{2} dx \int_{a}^{b} g(x)^{2} dx \geq (\int_{a}^{b}f(x) g(x) dx)^{2}$ for two real integrable functions in an interval [a,b].

Applying this to $\displaystyle \int_{a}^{b} |f| = \int_{a}^{b} |f| \cdot 1$ I get $\displaystyle \int_{a}^{b} |f| dx \leq \sqrt{b-a} \sqrt{\int_{a}^{b} |f|^2 dx}$ but this is not what I want to show. Am I missing something here?
• Nov 29th 2009, 09:15 AM
gibsion
Yes you are missing something. In lectures John Jones said that there was an error in the question and that you should replace [a,b] with [0,1].
• Nov 29th 2009, 09:23 AM
nmatthies1
Oh ok, thanks!! I've been ill for 2 weeks so I've probably missed that...thanks!
• Nov 29th 2009, 09:42 AM
gibsion
you're welcome.