Differentiation and Uniform Convergence

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I've just read the following theorem in some lecture notes:

Let $f_n : [a,b] \to\mathbb{R}$ be differentiable with $f_n'$ continuous for all $n$. Suppose $f_n \to f$ pointwise and $f_n'$ converges uniformly on $[a,b]$. Then $f:[a,b] \to \mathbb{R}$ is differentiable and $f'(x) = \lim_{n \to \infty} f_n'(x)$ for all $x \in [a,b]$.

I'm struggling to understand what $f'(x) = \lim_{n \to \infty} f_n'(x)$ for all $x \in [a,b]$ means in this context.

Does it mean

a) $f_n'(x)$ tends to $f'(x)$ just as a sequence of real numbers for each $x \in [a,b]$

b) $f_n'(x)$ converges pointwise to $f'(x)$

c) $f_n'(x)$ converges uniformly (and so also pointwise) to $f'(x)$

I would be very grateful for any help,

Jack

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When the author says, they mean to say (a): $f'(x) = \lim_{n\to\infty} f_n'(x)$. The "for all $x$ is implied". (a) and (b) are definitionally the same, so (b) is also true.

But wait! $f'_n$ converges uniformly to something by hypothesis. Suppose $f_n'$ converges uniformly to $g$. Then $f_n'$ converges pointwise to $g$ and to $f'$ so $g = f'$ by uniqueness of pointwise limits. In other words, $f_n'$ uniformly converges to $f'$.

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The sequence of derivatives converge uniformly (and hence to their pointwise limit). So the answer is (c) and hence in particular (b) and (a) follow.