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 [tex]\|\cdot\|[/tex] is a norm in a linear space [tex]E[/tex]. Define
[tex]d(x,y)=\|x-y\|[/tex]
and let us prove that [tex]d[/tex] is a metric.
The first to axioms of metric are trivial:
As usual, the triangular inequality is less simple, but not so hard in this case:
[tex]d(x,y) = \|x-y\| = \|x-z+z-y\| \leq \|x-z\| + \|z-y\| = d(x,z)+d(z,y)[/tex]
and this holds for every [tex]x,y,z\in E[/tex]. Recall that from the definition of norm we already have a triangular inequality: [tex]\| x+y\| \leq \|x\| + \|y\|[/tex].
Now, we are in conditions to prove that the supremum norm, is a norm.
The supremum norm is a norm on [tex]\mathcal{C}(X)[/tex].
The supremum norm is defined as [tex]\|f\|_\infty=\sup_{x\in[a,b]}|f(x)|[/tex], where [tex]f[/tex] is a continuous function over the set [tex] X[/tex]. Next, we are going to prove the three axioms.
(N1): [tex]\displaystyle f\equiv 0\Rightarrow\sup_{x\in[a,b]}|f(x)|=0[/tex].
[tex]\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[/tex]
(N2): [tex]\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[/tex]
(N3): [tex]\displaystyle \|f+g\|_\infty=\sup_{x\in X}|f(x)+g(x)|[/tex]
[tex]\forall x\in X\;\;|f(x)+g(x)|\leq|f(x)|+|g(x)|[/tex] from here we get
[tex]\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[/tex]
