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In Principles of Mathematical Analysis ( I have the third version, the theorem is at 152 p.)

7.17 Theorem: Suppose {$f_n$} is a sequence of functions, differentiable on $[a,b]$ and such that {$f_n(x_0)$} converges for some point $x_0$ on $[a,b]$. If {$f_{n}^{'}$} converges uniformly on $[a,b]$, then {$f_n$} converges uniformly on $[a,b]$, to a function $f$, and $$f'(x)= \lim f_{n}^{'}$$ $$a \leq x \leq b $$

The proof goes as follows:

Let $\epsilon >0$ be given. Choose $N$ such that $n \geq N$, $m \geq N$, implies $$|f_n(x_0)-f_m(x_0)| \leq \frac{\epsilon}{2}$$ and $$|f_n^{'}(t)-f_m^{'}(t)| \leq \frac{ \epsilon }{2(b-a)}$$ $$a\leq t \leq b $$ If we apply the mean value theorem to the function $f_n -f_m$, we have $$|f_n(x)-f_m(x)-f_n(t)+f_m(t)| \leq \frac{|x-t| \epsilon}{2(b-a)} \leq \frac{\epsilon}{2} $$

Could someone explain how exactly is applied the Mean Value theorem? I am unable to comprehend how we get the right-hand result.

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Applying the Mean Value Theorem on $h:=f_n -f_m\colon [x,t] \to \mathbf R$ gives that there exists a $\xi \in (x,t)$ with $$h(x)-h(t)= h'(\xi)(x-t).$$ Now take the absolute value of this equality and use the inequality for $|h'|$. If $a\leq x,t \leq b$ we have $|x-t|\leq b-a$ what shows the last inequality.

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