Smoothness of limit of sequence of functions when sequence of derivatives converges to a continuous function This is a follow-up to my previous question on MSE. As in the previous set-up, let $\{ f_n \}$ be a sequence of differentiable real functions over the reals which converges pointwise to some function $f$. We no longer assume that $f$ is smooth. Now let $D: C^\infty (\mathbb{R}) \to C^\infty (\mathbb{R})$ be a smooth differential operator. We had taken $D = \frac{d}{dx}$ earlier.
Assume that $\{ Df_n \}$ converges pointwise to a continuous function $g$, as opposed to being differentiable.

My new question now is does this imply that $Df$ exists and is continuous?

I am interested in the case of a general $D$ but would be satisfied with $\frac{d}{dx}$, and I would then work to generalize to $D$.
I tried looking up regularity properties of limits of sequences of smooth functions, but nothing came up that seemed relevant. I know that if $Df$ exists and is continuous, then we get that $Df = g$, this is due essentially to the same argument as in the previous question but appropriately generalized. I'm not sure how this would help in solving the problem, though. I've also been thinking about a weak-derivative approach as used in functional analysis but I'm not very knowledgeable in that field.
Any help would be appreciated!
 A: I've found a counterexample to my question. Let $$f_n(x) = 
     \begin{cases}
       n^2 \frac{x^2}{2} &\quad\text{if } x \in [0, \frac{1}{2n}] \\
       nx-n^2\frac{x^2}{2} -\frac{1}{4} &\quad\text{if } x \in (\frac{1}{2n}, \frac{1}{n}] \\
       \frac{1}{4} &\quad\text{if } x \in (\frac{1}{n},1] 
     \end{cases},
$$
which is in $C^1 (\mathbb{R})$. (I believe this can be bumped to a $C^\infty$ sequence with the use of mollifiers.) We then get
$$f'_n(x) = 
     \begin{cases}
       n^2 x &\quad\text{if } x \in [0, \frac{1}{2n}] \\
       n-n^2x &\quad\text{if } x \in (\frac{1}{2n}, \frac{1}{n}] \\
       0 &\quad\text{if } x \in (\frac{1}{n},1] 
     \end{cases},
$$
which is continuous.
Now $f_n (x)$ converges pointwise to $\frac{1}{4}H(x)$, where $H$ is the Heavyside function, which in particular is NOT continuous (but continuous almost everywhere!), while $f'_n(x)$ converges pointwise to $0$, obviously continuous.
We can't take the derivative of $H(x)$ but we know its weak derivative is $0$ (no it's not, see comments). So while this is a counterexample to my question, it (no longer) supports as a valid example of a weaker version of the question:

Under the original conditions of the question, does $f' = g$ as a weak derivative? And then generalized to $Df = g$ weakly?

