# Metric space of continuous functions

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?

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.

• for the direct proof, i think the proof is like bounded function with metric d? – Reza Habibi Sep 13 '16 at 9:55
• @RezaHabibi Yes -- it's going to be the exact same proof :) – 6005 Sep 13 '16 at 9:56

if C(X,Y) denote all continuous functions :X→Y, Y complet metric space, X compact

, with the metric

d(f,g)=sup|f(x)−g(x)| for x in X. so C(X,Y) with uniform norm d, complet metric space or Banach space? and if Y only metric space, do you know show c(X,Y) Banach space?

• Please use MathJax to format. – Saad Jan 30 '18 at 11:43