I'm trying to figure out this excercise about convergence and could really use some help. I have partial answers that I'd like for you to check and would really appreciate hints for the others.
Let $f_k\colon [a,b] \to \mathbb{R}$ a sequence of continuously differentiable functions on $[a,b]$, and $f_k \to f $ on $[a,b]$. Which of the following statements are true?
(a) $f$ is continuous on $[a,b]$.
(b) $(f_k)$ converges uniformly to $f$ on $[a,b]$.
(c) The sequence $(f_k')$ is point-wise convergent on $[a.b]$.
(d) If $f$ is continuously differentiable, then $(f_k')$ is point-wise convergent on $[a,b]$.
(e) If $f$ is continuously differentiable and if $(f_k'(x))$ converges for each $x$ in $[a,b]$, then $(f_k')$ is point-wise convergent to $f'$ on $[a,b]$.
Answers:
(a) Here I state that this is false since the sequence $f_k = x^k$ on $[0,1]$ with $k \in \mathbb{N}$ converges to $f(x) = \begin{cases} 0, & x<1 \\ 1,& x=1\end{cases}$
which of course is discontinuous.
(b) Using the same example as for (a) and choosing $\epsilon = \frac{1}{2}$, this is false.
(c) Again by the same example, for $f_k'$ to converge point-wisely on $[a,b]$ it should converge to $f'$, but since $f$ is not necessarily continuous, $f'$ doesn't necessarily exist, and thus the statement must be false.
(d) Here is when I ran into some trouble. I have a feeling this one is true but haven't been able to prove it. Any hint here would be helpfull.
(e) Ditto.
Thanks in advance.
For part (e) consider the following sequence on $\mathbb{R}$ or any interval containing $0$,
$$f_n(x) = \frac{x}{1+nx^2}.$$
Then
$$f'_n(x) = \frac{1-nx^2}{(1+nx^2)^2}$$
Note that $f_n(x) \rightarrow f(x) \equiv 0$ and $f'(0) = 0$.
Also, $(f'_n(x))$ converges pointwise, but $f'_n(0) \rightarrow 1 \neq f'(0).$
Hence, (e) is false.