Question

2. Let SCR be a measurable regular region, and let L(S) be the vector space of bounded continuous functions f : S-> R satisfying el Prove that |is a nor on L(R)

Answer 1

In order to prove that $||\cdot||_1$ is a norm on $L^1_{bc}(\mathbb R)$ , we need to prove that

a) $||f||_1\geq 0,~\mbox{and}~||f||_1=0~\Rightarrow~f=0~~\mbox{for all}~f\in L^1_{bc}(\mathbb R)$

b) $||\lambda\cdot f||_1=|\lambda|\cdot||f||_1~~\mbox{for all}~f\in L^1_{bc}(\mathbb R),~\lambda\in\mathbb R$

c) $||f+g||_1\leq ||f||_1+||g||_1~~\mbox{for all}~f,g\in L^1_{bc}(\mathbb R)$

Proof of a)

For $f\in L^1_{bc}(\mathbb R)$ , since $|f(x)|\geq 0$ for all $x\in\mathbb R$ , we conclude that

$||f||_1=\int_{\mathbb R}|f(x)|~dx\geq \int_{\mathbb R}0~dx=0$

Suppose that $f\in L^1_{bc}(\mathbb R)$ is such that $|f(x_0)|=2\epsilon>0$ for some $x_0\in\mathbb R$ . Then, by continuity, there is $\delta>0$ such that $|x-x_0|<\delta~\Rightarrow~|f(x)|>\epsilon$ . Hence,

$||f||_1=\int_{\mathbb R}|f(x)|~dx\geq \int_{x\in\mathbb R,~|x-x_0|<\delta}\epsilon dx=\epsilon\delta>0$

Thus, if $f\in L^1_{bc}(\mathbb R)$ is nonzero, then $||f||_1>0$ . This proves a):

$||f||_1\geq 0,~\mbox{and}~||f||_1=0~\Rightarrow~f=0~~\mbox{for all}~f\in L^1_{bc}(\mathbb R)$

b) For all $f\in L^1_{bc}(\mathbb R), ~\lambda\in\mathbb R$ , we have

$||\lambda\cdot f||_1=\int_\mathbb R|\lambda f(x)|~dx=|\lambda|\int_\mathbb R|f(x)|~dx=|\lambda|\cdot||f||_1$

This proves b):

$||\lambda\cdot f||_1=|\lambda|\cdot||f||_1~~\mbox{for all}~f\in L^1_{bc}(\mathbb R),~\lambda\in\mathbb R$

c) Let $f,g\in L^1_{bc}(\mathbb R)$ ; then, by usual triangle inequality of real numbers, we get

$|f(x)+g(x)|\leq |f(x)|+|g(x)|$

for all $x\in\mathbb R$ . Therefore, using property of integration we get

$||f+g||_1=\int_\mathbb R|f(x)+g(x)|~dx\leq \int_\mathbb R|f(x)|~dx+\int_\mathbb R|g(x)|~dx=||f||_1+||g||_1$

This proves c):

$||f+g||_1\leq ||f||_1+||g||_1~~\mbox{for all}~f,g\in L^1_{bc}(\mathbb R)$

Therefore, $||\cdot||_1$ is a norm on $L^1_{bc}(\mathbb R)$ .