sequence of functions and continuity

Question

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?)

Answer 1

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.

Answer 2

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.

Q2: let

$$f_n(x)=\begin{cases}x\in\mathbb Q\to\frac1n,\\x\notin\mathbb Q\to0.\end{cases}$$ The $f_n$ are nowhere continuous yet the sequence converges to a differentiable function.

Q3: your question is circular. If the sequence of functions is convergent everywhere, then the sequence is pointwise convergent.