Metric space of continuous functions
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.
Reza Habibi
Updated on August 01, 2022Comments
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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?
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Reza Habibi about 7 yearsfor the direct proof, i think the proof is like bounded function with metric d?
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6005 about 7 years@RezaHabibi Yes -- it's going to be the exact same proof :)
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Ѕааԁ almost 6 yearsPlease use MathJax to format.