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.
<em>Let us prove that any norm defines a metric.</em>
Assume that is a norm in a linear space . Define
and let us prove that is a metric.
The first to axioms of metric are trivial:
- for all . Moreover, and if and only if .
- .
As usual, the triangular inequality is less simple, but not so hard in this case:
and this holds for every . Recall that from the definition of norm we already have a triangular inequality: .
Now, we are in conditions to prove that the supremum norm, is a norm.
<em>The supremum norm is a norm on </em>.
The supremum norm is defined as , where is a continuous function over the set . Next, we are going to prove the three axioms.
(N1): .
(N2):
(N3):
from here we get