# sequence of functions and continuity [closed]

I have a couple of questions on sequences of functions to help me understand the concept better.

If you have a function f, what must be required of the function to be able to build a sequence of convergent functions. Is it just that the sequence has to converge to f?

If you have a differentiable function f, then will the sequence of functions that converges to f all be continuous (and differentiable)? And if they are all continuous then will the sequence be pointwise convergent?

(I have seen examples of sequences of functions that were all continuous but the sequence wasn't uniformly continuous so I know continuity does not imply uniform continuity, but what about pointwise?)

## closed as unclear what you're asking by Namaste, José Carlos Santos, Did, Mohammad Riazi-Kermani, user284331Apr 3 '18 at 0:13

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• Are you asking this: If $f_n \to f$ uniformly, and $f$ is differentiable, must each $f_n$ be continuous? Not at all. You can infer information about the limit from information about the terms in a sequence, but not vice versa. – Matthew Leingang Apr 2 '18 at 18:54
• No what I was asking was: if f is differentiable, does that tell us anything about the sequence fn that converges to it. I was thinking it would mean that the functions in fn are continuous. – gunnersFc Apr 2 '18 at 19:07
• Hm, that sounds like exactly the question as I rephrased it. In any case, I am pretty sure the answer is no. If I have a convergent sequence of continuous functions, I can alter each of them at a point (a different one in each case), rendering them discontinuous, but having the same limit. – Matthew Leingang Apr 2 '18 at 19:18
• But if you know that the limit of the sequence of functions is continuous wouldn't the functions in the sequence also be continuous? You are going from the sequence to the limit, but I want to know about going from the limit to the sequence. – gunnersFc Apr 2 '18 at 19:33
• I will add an answer to show why I don't think this is true. – Matthew Leingang Apr 2 '18 at 19:48

For each natural number $n$, define $f_n\colon[0,1] \to \mathbb{R}$ by $$f_n(x) = \begin{cases} \frac{1}{n} & x \neq \frac{1}{n} \\ \frac{2}{n} & x = \frac{1}{n} \end{cases}$$ Then none of the functions $f_n$ are continuous. However, the sequence $(f_n)$ converges uniformly to the zero function, and that function is continuous, differentiable, etc.
Q1: there are no requirements for $f$. For any given $f(x)$, you can always build a sequence of functions $f_n(x)$ that converges to it. There is no relation between the values of the functions (neither $f$ nor $f_n$) at neighboring $x$s.
$$f_n(x)=\begin{cases}x\in\mathbb Q\to\frac1n,\\x\notin\mathbb Q\to0.\end{cases}$$ The $f_n$ are nowhere continuous yet the sequence converges to a differentiable function.