Metric space of continuous functions

2,709

It sounds like you already know that the set of bounded functions, $B[a,b]$, is a metric space under the metric $d$. In that case, you made a very nice observation that since continuous functions are bounded, $C[a,b]$ is contained in $B[a,b]$. And any subset of a metric space is a metric space! Well done.

You can also show it directly -- nonnegativity and symmetry are almost immediate, and then you only have to show triangle inequality.

Share:
2,709
Reza Habibi
Author by

Reza Habibi

Updated on August 01, 2022

Comments

  • Reza Habibi
    Reza Habibi over 1 year

    Let $C[a,b]$ denote all continuous functions $:[a,b] \rightarrow \mathbb{R}$, with the metric

    $$d(f,g)=\sup |f(x) - g(x)| \text{ for } a\leq x \leq b.$$

    Show that $d(f,g)$ is a metric space.

    I have started with this: Since continuos function is bounded, so i use the same proof from it: " $B[a,b]$ denote for all bounded functions $:[a,b] \rightarrow \mathbb{R}$, with

    $d(f,g)=\sup |f(x) - g(x)|$ for $a\leq x \leq b$ "

    to prove those (continuos function is metric space). Is it right? Or there is another way?

  • Reza Habibi
    Reza Habibi about 7 years
    for the direct proof, i think the proof is like bounded function with metric d?
  • 6005
    6005 about 7 years
    @RezaHabibi Yes -- it's going to be the exact same proof :)
  • Ѕааԁ
    Ѕааԁ almost 6 years
    Please use MathJax to format.