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I have this sequence of function $f_n(x)=\frac{1}{1+x^n}\quad x\in(0,\infty)$ study convergence in $L^\infty$

My professor said in this way: if $f\in C^0(U)$ the space of continuous function on open set we have that infinity norm is supremum norm
(and so far ok) but now the limit function is not continuous so there is not convergence in $L^\infty $ and here I have problems : I think that she uses the property of completeness fof the space of continuous function to say so buy in my opinion the space where I am looking for convergence is always $L^\infty $ not $C^0(0,\infty)$

and second point :I know $C^0[a,b]$ is complete and I don't think C^0(0,\infty) is complete under sup norm (I think I can find counterexamples with sequences going to 0 ...)

Dsrksidemath
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    $C^0\big((0, \infty)\big)$ is complete if you impose the additional condition that they are bounded (which your $f_n$ fulfil). See here. Also, for continuous functions, the $\sup$ equals the $\mathrm{esssup}$. There are a lot of references on this site as well for this. So your recognition that the pointwise limit is discontinuous, tells you everything you need. – Hyperbolic PDE friend Sep 20 '24 at 18:38
  • @HyperbolicPDEfriend thanks for your answer my doubt is : now I know that $f_n$ does not converge in $C^0(0,\infty)$ but why we don't talk anymore of $L^\infty (0,\infty)$? In theory I am still looking my sequence in $L^\infty$ not in $C^0$ this is my doubt are L^\infty with sup norm equivalent in C^0 sup norm or something ? – Dsrksidemath Sep 20 '24 at 18:49
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    The norms are not only equivalent, but they are equal, because the $\mathrm{esssup}$ is the $\sup$ for continuous functions. – Hyperbolic PDE friend Sep 20 '24 at 18:52
  • @HyperbolicPDEfriend ok but sorry for my stupid question . the space in which I am studying convergence is $L^\infty$ and not $C^0$ I have a sequence of function that does not converge on $C^0$ and that is ok . But why also in $L^\infty$ ? the space does not matter? – Dsrksidemath Sep 20 '24 at 18:56
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    The norms (and hence the definitions of convergence) are the exact same. – Hyperbolic PDE friend Sep 20 '24 at 19:03
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    @HyperbolicPDEfriend ok really thanks – Dsrksidemath Sep 20 '24 at 19:10

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