Question

Prove c(x) with the supremum norm is a metric space showing the supremum nom is a metric. 1.

Answer 1

Answer:

Step-by-step explanation:

The best way to solve this problem is to prove that any norm defines a metric, and then proving that the supremum norm is, in fact, a norm.

Let us prove that any norm defines a metric.

Assume that $\|\cdot\|$ is a norm in a linear space $E$. Define

$d(x,y)=\|x-y\|$

and let us prove that $d$ is a metric.

The first to axioms of metric are trivial:

• $\|x-y\|\geq 0$ for all $x,y \in E$. Moreover, $0=d(x,y)=\|x-y\|$ and $\|x-y\|=0$ if and only if $x-y=0$.

• $d(x,y)=\|x-y\| = \|y-x\| =d(y,x)$.

As usual, the triangular inequality is less simple, but not so hard in this case:

$d(x,y) = \|x-y\| = \|x-z+z-y\| \leq \|x-z\| + \|z-y\| = d(x,z)+d(z,y)$

and this holds for every $x,y,z\in E$. Recall that from the definition of norm we already have a triangular inequality: $\| x+y\| \leq \|x\| + \|y\|$.

Now, we are in conditions to prove that the supremum norm, is a norm.

The supremum norm is a norm on $\mathcal{C}(X)$.

The supremum norm is defined as $\|f\|_\infty=\sup_{x\in[a,b]}|f(x)|$, where $f$ is a continuous function over the set $X$. Next, we are going to prove the three axioms.

(N1): $\displaystyle f\equiv 0\Rightarrow\sup_{x\in[a,b]}|f(x)|=0$.

                   $\displaystyle\sup_{x\in X} |f(x)|=0\Rightarrow\forall x\in E\;\;|f(x)|\leq 0 \Rightarrow f(x)=0\Rightarrow f\equiv 0$

(N2): $\displaystyle\|\lambda f\|_\infty=\sup_{x\in X}|\lambda f(x)|=\sup_{x\in E}|\lambda| |f(x)|=|\lambda|\sup_{x\in X}|f(x)|=|\lambda|\|f(x)\|_\infty$

(N3): $\displaystyle \|f+g\|_\infty=\sup_{x\in X}|f(x)+g(x)|$

                   $\forall x\in X\;\;|f(x)+g(x)|\leq|f(x)|+|g(x)|$ from here we get

               $\displaystyle\sup_{x\in X}|f(x)+g(x)|\leq\sup_{x\in X}(|f(x)|+|g(x)|) \leq \sup_{x\in X}|f(x)|+\sup_{x\in X}|g(x)|=\|f\|_\infty+\|g\|_\infty$