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Let $f_n$ be a sequence of differentiable functions on $[0, 1]$ with

1. $f_n \to f$ uniformly for some (necessarily) continuous $f$.
2. $f'_n - g \to 0$ in $L^{\infty}$ for some measurable $g$.

Is it true that $f$ is differentiable everywhere, with $f' = g$ almost everywhere?

Some comments: An almost everywhere version of the same question is answered negatively here. I expected to have an easy affirmative answer if $f$ are assumed instead everywhere differentiable, but to my surprise this seems to be much more subtle. Note that we do not assume that $f’_n$ are in $L^1$, nor that $f_n$ are absolutely continuous, so that the fundamental theorem of calculus does not apply.

Update:

The answer by CityHunter below shows that we have that $f$ is differentiable a.e. with $f’ = g$ a.e., under the weaker condition that $f’_n - g \to 0$ in $L^1$. Thus the remaining question, and the one I expect to be hardest is the question of whether $f$ is differentiable everywhere. To isolate the essential difficulty of the problem of everywhere differentiability, we could maybe consider first the following version with stronger convergence assumptions.

Assume $f, f_n \in W^{1, \infty}$ with $f_n$ everywhere differentiable, and $f_n \to f$ in $W^{1, \infty}$. Then is $f$ everywhere differentiable?

Even this weaker version seems to be nontrivial.

Update 2: It seems the weaker version is true, so it is left to either remove the regularity requirements as in the original post, or to try to derive a multidimensional version of the latter claim. I have added an outline of a proof strategy for the multidimensional case in the answers below.

Let $f_n$ be a sequence of differentiable functions on $[0, 1]$ with

1. $f_n \to f$ uniformly for some (necessarily) continuous $f$.
2. $f'_n - g \to 0$ in $L^{\infty}$ for some measurable $g$.

Is it true that $f$ is differentiable everywhere, with $f' = g$ almost everywhere?

Some comments: An almost everywhere version of the same question is answered negatively here. I expected to have an easy affirmative answer if $f$ are assumed instead everywhere differentiable, but to my surprise this seems to be much more subtle. Note that we do not assume that $f’_n$ are in $L^1$, nor that $f_n$ are absolutely continuous, so that the fundamental theorem of calculus does not apply.

Update:

The answer by CityHunter below shows that we have that $f$ is differentiable a.e. with $f’ = g$ a.e., under the weaker condition that $f’_n - g \to 0$ in $L^1$. Thus the remaining question, and the one I expect to be hardest is the question of whether $f$ is differentiable everywhere. To isolate the essential difficulty of the problem of everywhere differentiability, we could maybe consider first the following version with stronger convergence assumptions.

Assume $f_n \to f$ in $W^{1, \infty}$ with $f_n$ everywhere differentiable. Then is $f$ everywhere differentiable?

Even this weaker version seems to be nontrivial.

Update 2: It seems the weaker version is true, so it is left to either remove the regularity requirements as in the original post, or to try to derive a multidimensional version of the latter claim. I have added an outline of a proof strategy for the multidimensional case in the answers below.

Let $f_n$ be a sequence of differentiable functions on $[0, 1]$ with

1. $f_n \to f$ uniformly for some (necessarily) continuous $f$.
2. $f'_n - g \to 0$ in $L^{\infty}$ for some measurable $g$.

Is it true that $f$ is differentiable everywhere, with $f' = g$ almost everywhere?

Some comments: An almost everywhere version of the same question is answered negatively here. I expected to have an easy affirmative answer if $f$ are assumed instead everywhere differentiable, but to my surprise this seems to be much more subtle. Note that we do not assume that $f’_n$ are in $L^1$, nor that $f_n$ are absolutely continuous, so that the fundamental theorem of calculus does not apply.

Update:

The answer by CityHunter below shows that we have that $f$ is differentiable a.e. with $f’ = g$ a.e., under the weaker condition that $f’_n - g \to 0$ in $L^1$. Thus the remaining question, and the one I expect to be hardest is the question of whether $f$ is differentiable everywhere. To isolate the essential difficulty of the problem of everywhere differentiability, we could maybe consider first the following version with stronger convergence assumptions.

Assume $f, f_n \in W^{1, \infty}$ with $f_n$ everywhere differentiable, and $f_n \to f$ in $W^{1, \infty}$. Then is $f$ everywhere differentiable?

Even this weaker version seems to be nontrivial.

Update 2: It seems the weaker version is true, so it is left to either remove the regularity requirements as in the original post, or to try to derive a multidimensional version of the latter claim.

Let $f_n$ be a sequence of differentiable functions on $[0, 1]$ with

1. $f_n \to f$ uniformly for some (necessarily) continuous $f$.
2. $f'_n - g \to 0$ in $L^{\infty}$ for some measurable $g$.

Is it true that $f$ is differentiable everywhere, with $f' = g$ almost everywhere?

Some comments: An almost everywhere version of the same question is answered negatively here. I expected to have an easy affirmative answer if $f$ are assumed instead everywhere differentiable, but to my surprise this seems to be much more subtle. Note that we do not assume that $f’_n$ are in $L^1$, nor that $f_n$ are absolutely continuous, so that the fundamental theorem of calculus does not apply.

Update:

The answer by CityHunter below shows that we have that $f$ is differentiable a.e. with $f’ = g$ a.e., under the weaker condition that $f’_n - g \to 0$ in $L^1$. Thus the remaining question, and the one I expect to be hardest is the question of whether $f$ is differentiable everywhere. To isolate the essential difficulty of the problem of everywhere differentiability, we could maybe consider first the following version with stronger convergence assumptions.

Assume $f, f_n \in W^{1, \infty}$ with $f_n$ everywhere differentiable, and $f_n \to f$ in $W^{1, \infty}$. Then is $f$ everywhere differentiable?

Even this weaker version seems to be nontrivial.

Update 2: It seems the weaker version is true, so it is left to either remove the regularity requirements as in the original post, or to try to derive a multidimensional version of the latter claim. I have added an outline of a proof strategy for the multidimensional case in the answers below.

Let $f_n$ be a sequence of differentiable functions on $[0, 1]$ with

1. $f_n \to f$ uniformly for some (necessarily) continuous $f$.
2. $f'_n - g \to 0$ in $L^{\infty}$ for some measurable $g$.

Is it true that $f$ is differentiable everywhere, with $f' = g$ almost everywhere?

Some comments: An almost everywhere version of the same question is answered negatively here. I expected to have an easy affirmative answer if $f$ are assumed instead everywhere differentiable, but to my surprise this seems to be much more subtle. Note that we do not assume that $f’_n$ are in $L^1$, nor that $f_n$ are absolutely continuous, so that the fundamental theorem of calculus does not apply.

Update:

The answer by CityHunter below shows that we have that $f$ is differentiable a.e. with $f’ = g$ a.e., under the weaker condition that $f’_n - g \to 0$ in $L^1$. Thus the remaining question, and the one I expect to be hardest is the question of whether $f$ is differentiable everywhere. To isolate the essential difficulty of the problem of everywhere differentiability, we could maybe consider first the following version with stronger convergence assumptions.

Assume $f, f_n \in W^{1, \infty}$ with $f_n$ everywhere differentiable, and $f_n \to f$ in $W^{1, \infty}$. Then is $f$ everywhere differentiable?

Even this weaker version seems to be nontrivial.

Let $f_n$ be a sequence of differentiable functions on $[0, 1]$ with

1. $f_n \to f$ uniformly for some (necessarily) continuous $f$.
2. $f'_n - g \to 0$ in $L^{\infty}$ for some measurable $g$.

Is it true that $f$ is differentiable everywhere, with $f' = g$ almost everywhere?

Some comments: An almost everywhere version of the same question is answered negatively here. I expected to have an easy affirmative answer if $f$ are assumed instead everywhere differentiable, but to my surprise this seems to be much more subtle. Note that we do not assume that $f’_n$ are in $L^1$, nor that $f_n$ are absolutely continuous, so that the fundamental theorem of calculus does not apply.

Update:

The answer by CityHunter below shows that we have that $f$ is differentiable a.e. with $f’ = g$ a.e., under the weaker condition that $f’_n - g \to 0$ in $L^1$. Thus the remaining question, and the one I expect to be hardest is the question of whether $f$ is differentiable everywhere. To isolate the essential difficulty of the problem of everywhere differentiability, we could maybe consider first the following version with stronger convergence assumptions.

Assume $f, f_n \in W^{1, \infty}$ with $f_n$ everywhere differentiable, and $f_n \to f$ in $W^{1, \infty}$. Then is $f$ everywhere differentiable?

Even this weaker version seems to be nontrivial.

Update 2: It seems the weaker version is true, so it is left to either remove the regularity requirements as in the original post, or to try to derive a multidimensional version of the latter claim.