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Let $(f_{n})_{n}$ be a sequence of functions $f_{n}: \mathbb{R} \to \mathbb{R} $. They converge pointwise to a function $f: \mathbb{R} \to \mathbb{R} $. We know that the convergence is uniform on every interval of the form $[-M,M]$ with $M \in \mathbb{R}^+$

a) Let $f_{n}$ be continuos in a real number $a \in \mathbb{R}$ for every n. Can we assume that $f$ is continuous in $a$?

b) Let $f_{n}$ be uniformly continuous on $\mathbb{R}$ for every n. Can we assume that $f$ is uniformly continuous on $\mathbb{R}$?

I have thought about this question for quite some time but I can't figure out the answer. Is there someone who can help me?

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  • $\begingroup$ What is the definition of "row of functions" ? If this means that $f_n$ is a sequence of functions, it's wrong that $f_n$ converges (and also wrong that $f_n$ converges uniformly on every compact interval). Could you clarify it ? $\endgroup$ – user171326 Jun 14 '17 at 14:26
  • $\begingroup$ I mean sequence, I was confused for a moment because in my language, we use the word 'row' for a sequence, sorry $\endgroup$ – Mee98 Jun 14 '17 at 14:28
  • $\begingroup$ Ok thanks ! ${{}}$ $\endgroup$ – user171326 Jun 14 '17 at 14:33
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$a)$ This is true. Given $\varepsilon>0$ there is $n$ such that $|f_n(x)-f(x)|<\varepsilon/3$ for all $x\in[-a-1,a+1]$, and there is some $\delta>0$ such that $|x-a|<\delta$ implies $|f_n(a)-f_n(x)|<\varepsilon/3$. Then for $x\in[-a-1,a+1]$ such that $|x-a|<\delta$ we have \begin{align}|f(x)-f(a)|&\leq|f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|\\&<\varepsilon/3+\varepsilon/3+\varepsilon/3\\&=\varepsilon. \end{align}

$b)$ This is not true. Consider the functions $f_n:\mathbb{R}\to\mathbb R$ defined by $$f_n(x)=\left\{ \begin{array}{lcc} x^2&:&|x|\leq n \\ n^2&:& |x|>n. \end{array} \right. $$ Then each $f_n$ is uniformly continuous, $f_n$ converges pointwise (and uniformly on compact sets) to $f:\mathbb R\to\mathbb R$ defined by $f(x)=x^2$, which is not uniformly continuous on $\mathbb R$.

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