Question

does
\[d(f,g)=max_{0\leq t\leq 1}|f(t)-g(t)|\] define a metric on
\[C[0,1]\] , and can continuous functions on [0,1] be replaced by bounded functions on [0,1]?

Answer 1

yes, it does define a metric and quite a good one. It's not hard to show that
- d(f,f) = 0
- d(f,g) = d(g,f)
- d(f,g) \leq d(f,h) + d(h,g)

Answer 2

I dont think i'm as far as you in math but im in College Calculus 1. can you help?

Answer 3

does "bounded" make any difference?

Answer 4

It should also to be a metric on the bounded functions

Answer 5

ty

Answer 6

Provided it's truly bounded, and not almost everywhere bounded, yes.

Answer 7

I'm in serious need of help with logarithmic differentiation

Answer 8

@twig: post your question on the left and Im sure there's a few people who can chime in.

Answer 9

\[d(f,g)\leq d(f,h)+d(h,g)\] ?

Answer 10

it is and has been .... no hits yet

Answer 11

The triangle inequality, one of the four key axioms for a metric.

Answer 12

yes thank you again

Answer 13

I've listed three, the other one is just that
d(f,g) \geq 0 for all f, g

Answer 14

...which is trivially obvious for this metric.

Answer 15

if i prove for bounded, that will take care of continuous as well correct?

Answer 16

Yes. Note by the way that as you're dealing with functions on a closed interval, their image will also be bounded, a theorem you may or may not have proven yet, but critical here so that d(f,g) is defined.

In other words, if you were dealing with C(0,1), this d is not a metric because d(f,g) can be arbitrarily large.

Answer 17

actually, it's not just that d(f,g) can be arbitrarily large; it's that the max may not exist.

Answer 18

yes i see. thank you once again.

Answer 19

E.g., the function f(x) = 1/x.

Answer 20

sure. these sorts of questions are the fun ones for me.

Answer 21

i promise more.