Use of MVT in this proof

Question

I cannot seem to understand the use of mean value theorem in the proof of this theorem:

$\mathbf{Theorem}$. Let$(a,b)$ be a bounded interval and suppose that $f_n$ is a sequence of functions which converges at some $x_0 \in (a,b)$. If each $f_n$ is differentiable on $(a,b)$, and $f'_n$ converges uniformly on $(a,b)$ as $n \to \infty$, then $f_n$ converges uniformly on $(a,b)$ and $$\lim_{n\to\infty} f'_n(x)=(\lim_{n\to\infty}f_n(x))'$$ for each $x\in(a,b)$.

Here is how they started the proof:

Proof: Fix $c\in (a,b)$ and define $$g_n(x)= \left\{ \begin{array}& \frac{f_n(x)-f_n(c)}{x-c} & x\neq c \\ f'_n(c) & x=c\\ \end{array} \right. $$for $n\in\mathbb{N}$. We claim that for any $c\in(a,b)$, the sequence $g_n$ converges uniformly on $(a,b)$. Let $\epsilon>0,n,m\in\mathbf{N}$, and $x\in(a,b)$ with $x\neq c.$ By the Mean Value Theorem, there is a $\lambda $ between $x$ and $c$ such that $$g_n(x)-g_m(x)=\frac{f_n(x)-f_m(x)-(f_n(c)-f_m(c))}{x-c}=f' _n(\lambda)-f' _m(\lambda)$$ Now, why is this the case? I thought MVT is used for functions evaluated at different points. In this case, they are using different functions evaluated at the same point. I fail to draw its connection with the Mean Value Theorem. Can someone explain how MVT is used in this case? What function(s) are they even trying to use MVT on?

Answer 1

They apply Cauchy's mean value theorem to the functions $\;f(x)=f_n(x)-f_m(x)$ and $g(x)=x$.

This version asserts this:

If two functions $f$ and $g$ are continuous on the closed interval $[a,b]$ and differentiable in the open interval $(a,b)$, there exists $c\in(a,b)$ such that $$\bigl(f(b)-f(a) \bigr)g'(c)=\bigl(g(b)-g(a) \bigr)f'(c). $$

If $g(a)\ne g(b)$ and $g'(c)\ne 0$, this implies $$\frac{f(b)-f(a)}{g(b)-g(a)} =\frac{f'(c)}{g'(c)}.$$

Answer 2

I get it now. $$g_n(x)-g_m(x)=\frac{f_n(x)-f_m(x)-(f_n(c)-f_m(c))}{x-c}=f' _n(\lambda)-f' _m(\lambda)$$ The first equality is trivial because it follows by definition. It has nothing to do with the MVT.

The second equality uses MVT on the function $f_n(x)-f_m(x)$ at points $x=x,x=c$. We don't need Cauchy's MVT either.