# Thread: [SOLVED] Cauchy-Schwartz and norms

1. ## [SOLVED] Cauchy-Schwartz and norms

Let $||-||_{1}$ and $||-||_{2}$ be the $L^1$-norm and the $L^2$-norm on $C[a,b]$ the space of continuous real valued functions on the closed interval $[a,b]$: explicitly
$||f||_{1}=\int_{a}^{b} |f|$, $||f||_{2}=\sqrt{\int_{a}^{b} |f|^2}$.

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

Cauchy-Schwartz:

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

Applying this to $\int_{a}^{b} |f| = \int_{a}^{b} |f| \cdot 1$ I get $\int_{a}^{b} |f| dx \leq \sqrt{b-a} \sqrt{\int_{a}^{b} |f|^2 dx}$ but this is not what I want to show. Am I missing something here?

2. 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].

3. Oh ok, thanks!! I've been ill for 2 weeks so I've probably missed that...thanks!

4. you're welcome.