Norms

**nmatthies1** · November 28th 2009, 04:13 PM

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

$||f||_{1}=\int_{a}^{b} |f|$ , $||f||_{2}=\sqrt{\int_{a}^{b} |f|^2}$ .

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

(b) For $n \geq 1$ , define $f_{n}:[0,1] \to \mathbb{R}$ by

$f_{n}(x)= \begin{matrix} n & if 0 \leq x \leq \frac{1}{n} \\ & \\ 2n-n^{2}x & if \frac{1}{n} \leq x \leq \frac{2}{n} \\ & \\ 0 & if \frac{2}{n} \leq x \leq 1\end{matrix}$

Caculate $||f_{n}||_{1}$ and $||f_{n}||_{2}$ .

So the thing is that I've been ill for over a week now and couldn't go to lectures resulting in me not having lecture notes and no notion of norms.

I have found hat Cauchy-Schwartz states that $\int_{a}^{b} f(x)^{2} dx \int_{a}^{b} g(x)^{2} dx \geq (\int_{a}^{b}f(x) g(x) dx)^{2}$ for two real integrable functions in an interval [a,b]. I don't really know how to use this to prove (1).

As for (b) I am confused by the sequence of functions which I don't really know how to interpret (what is it geometrically?) or plug into $||f||_{1}=\int_{a}^{b} |f|$ , $||f||_{2}=\sqrt{\int_{a}^{b} |f|^2}$ .

Now, if someone could just start me off that would be great!

**Krizalid** · November 28th 2009, 04:39 PM

for the first one.

if $f\equiv0,$ it's a trivial case, so suppose $b>a,$ then $\int_{a}^{b}{\left| f \right|}=\int_{a}^{b}{\left| f \right|\cdot 1}\le \sqrt{\left( \int_{a}^{b}{\left| f \right|^{2}} \right)\left( \int_{a}^{b}{1^2} \right)},$ you can finish that now.

**nmatthies1** · November 28th 2009, 05:00 PM

When I do that I end up with $\int_{a}^{b} |f| dx \leq \sqrt{b-a} \sqrt{\int_{a}^{b} |f|^2 dx}$ but this does not necessarily imply $\int_{a}^{b} |f| dx \leq \sqrt{\int_{a}^{b} |f|^2 dx}$ , does it?
I'm really tired so I might be missing something really obvious.

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