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Let $f_n \in C[a; b]$, where $\{f_n\}$ converges uniformly to $f$.

Is it true that $f \in C[a; b]$ too ? How do I prove or disprove it?

Pedro
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RiaD
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1 Answers1

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For $\varepsilon>0$, there is $n$ such that $|f(z)-f_n(z)|<\dfrac{\varepsilon}{3}$ for all $z$, take one such $n$, and as $f_n$ are continuous at $a$, for the same $\varepsilon$ there is $\delta>0$ s.t. $|x-a|<\delta\,\Rightarrow |f_n(x)-f_n(a)|<\dfrac{\varepsilon}{3}$. So, we have

$|f(x)-f(a)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|<\varepsilon$.

Ofni Chadesch
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Berci
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  • Why not $\varepsilon/3$ instead? Ends up looking nicer =) – Pedro Mar 28 '13 at 22:01
  • Hi@PeterTamaroff my question was that if $(f_k)$ (sequence of continuous functions) converges---->f in $C(\Bbb R)$. Can we say $(f_k)$ will converge to f in $\mathcal D'(\Bbb R)$? – Sara Tancredi May 20 '13 at 20:38