# Norm piecewise continuous function

Let be $f:\mathbb{R} \rightarrow \mathbb{C}$. Consider the space of piecewise linear curves, with support in the interval [-1,1], sucht that $f(x)= A-|x|$ if $|x|\leq 1$; $f(x)= 0$ otherwise. For this space show that expression $p(f)=\int_{-\infty}^{\infty}|f(x)|^2dx$ is a norm.

I am show that $p(af)=|a|p(f)$, $\forall a \in \mathbb{C}$. How I will be able to demonstrate the other two properties?

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Think, which integrals you have to sompute. They are not that hard. – Hagen von Eitzen Oct 17 '12 at 16:53
How did you prove $p(af) = |a|p(f)$? $$p(af) = \int_{-\infty}^{\infty}|af(x)|^2dx = |a^2| \int_{-\infty}^{\infty}|f(x)|^2dx = |a^2| p(f)$$ – Rudy the Reindeer Oct 17 '12 at 18:21
What is $A$? ${}$ – Rudy the Reindeer Oct 17 '12 at 18:22
$A>0 \in \mathbb{R}$ – juaninf Oct 17 '12 at 18:23
@Juan I suggest you do the following: you have already shown that what your teacher wrote is not a norm since $p(af) = a^2 p(f)$. Write this on your homework sheet. Then write: but if we define $p$ with the square root then it is indeed a norm. And then you write the three parts of the proof that it is indeed a norm, the first one is $p(af) = |a|p(f)$. The second will be to verify that it's $\geq$ zero and zero if and only if $f=0$ and the third one is the triangle inequality. – Rudy the Reindeer Oct 19 '12 at 6:14

Let $L$ denote the space of piecewise linear functions $f: \mathbb R \to \mathbb C$ that is, $f(x) = \begin{cases} A - |x| & x \in [-1,1] \\ 0 & \text{ otherwise} \end{cases}$ where $A \in \mathbb R_{>0}$.

Define $p(f) := \sqrt{\int_{-\infty}^\infty |f(x)|^2 dx}$.

Claim: $p: L \to [0,\infty)$ defines a norm on $L$.

Proof:

(i) $p \geq 0$ is clear. If $f=0$ then clearly $p(f) = 0$. Now assume $p(f)=0$ and show that $f=0$. By contradiction, if $f$ was non-zero at some point then since it is continuous it would have to be non-zero on an open interval and then the integral would also be non-zero. Contradiction. So $f$ is the zero function.

(ii) $p(af) = \sqrt{\int_{-\infty}^\infty |af(x)|^2 dx} = |a|^2 \sqrt{\int_{-\infty}^\infty |f(x)|^2 dx}= |a|^2 p(f)$

(iii) Here you want to show $p(f+g) \leq p(f) + p(g)$.

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Hope this helps. – Rudy the Reindeer Oct 19 '12 at 6:25
thanks @Matt N. – juaninf Oct 19 '12 at 16:15
@Juan Any time! : ) – Rudy the Reindeer Oct 19 '12 at 17:59