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let $a,b \in \mathbb{R}$.

$C(a,b)$ is the space of continuous functions defined on $(a,b)$, and it can be equipped with $\|\cdot\|_{\infty}$.

So does it mean that $C(a,b)=C[a,b]$? Since $\|\cdot\|_{\infty}$ eliminates the functions blowing up at boundary $a,b$?

It should be an elementary question but I've overlooked until recently studying embedding theorems.

Any suggestions?

---------------------------UPDATE---------------------------

The question is occured from the formulation of Hölder space $C^\gamma(\bar\Omega)$. The norm on this space can be equipped with $$\|x\|_\gamma:=\|x\|_\infty+[x]_\gamma$$ where $[x]_\gamma :=sup_{a,b\in\Omega, a\neq b}\frac{|x(a)-x(b)|}{|a-b|^\gamma}$.

Since $\|\cdot\|_\infty$ exists for $x\in C(\Omega)$, so in this context, $C(\Omega)$ actually is $C_b(\Omega)$, i.e. space of bounded continuous functions?

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6  
Let $a=0$ and have a look at $\sin(1/x)$. –  user20266 May 20 '12 at 13:03
    
@Thomas Yes, it has oscillation near 0. So it is no longer continuous at 0. –  newbie May 20 '12 at 13:10
1  
The tile doesn't give a good impression, "simple question"! –  Gigili May 20 '12 at 13:34

2 Answers 2

up vote 0 down vote accepted

To answer the updated question, I am used to the following notation $$C_b(\Omega) := \{f\in C(\Omega) \, | \,\, ||f||_\infty < \infty\}$$ By just writing $f\in C(\Omega)$ you do not imply $f$ is bounded.

(And neither of both $ C_b(\Omega) $ or $C(\Omega)$ equals $C( \overline{\Omega}),$ when $\Omega$ is open, which is what I'm assuming here).

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Note $[a,b]$ is compact so $C[a,b]$ is a set of all bounded continuous functions whereas $C(a,b)$ is a set of all continuous functions on $(a,b)$.

Example: Take a=o, b=1 then $\frac {1}{x}\in C(0,1)$ but $\notin C[0,1]$ so easily you can conclude $C[a,b]$ can be equipped with sup norm but $C(a,b)$ can't be. (Usual study is first study the set then consider whether there is a distance between any two element of that set or not)

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