I want to prove this theorem:

Let $(f_n)$ be a sequence of differentiable functions defined on the closed interval $[a, b]$, and assume $(f_n')$ converges uniformly on[a, b]. If there exists a point $x_0 ∈[a, b]$ where $f_n(x_0)$ is convergent, then $(f_n)$ converges uniformly on $[a, b]$.

The text suggests to use the MVT:

Let $x \in [a, b]$ and assume, without loss of generality, that $x > x_0$ . Applying the Mean Value Theorem to the function $f_n -f_m$ on the interval $[x_0 , x]$, we get that there exists a point $\alpha$ such that $(f_n(x) - f_m(x)) - (f_n (x_0) - f_m(x_0))$ = $(f'_n(\alpha)- f'_m(\alpha))(b - a) \space \space \space \space \space \space \space \space \space \space \space \space(1)$

My question is that, since we are using the MVT over an arbitrarily constructed interval $[x_0,x]$, does this not suggest that $(1)$ should be multiplied by $(x-x_0)$ as opposed to $(b-a)$?

And also, just wanted to confirm (whilst seemingly obvious, however not stated in the text) if it is true that $\alpha \in [x_0,x]$

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  • $\begingroup$ There's a problem with the statement. In the first sentence, you say "assume $\{f_n\}$ converges uniformly on $[a,b]$". Then given a point $x_0$ for which $\{f(x_0)\}$ you want to conclude that $\{f_n\}$ converge uniformly. Looks like you are trying to prove a statement that you are already assuming. $\endgroup$ – Tim Raczkowski Mar 5 '15 at 3:47
  • $\begingroup$ Well picked up on my friend, it was a typo which I have since rectified. The assumption is that $(f'_n)$ converges. Apologies for the mix up. $\endgroup$ – elbarto Mar 5 '15 at 4:04
  • $\begingroup$ Yes, I believe (1) should be multiplied by $x-x_0$. Also, looks like you should say "there exists a point $\alpha$", and yes, $\alpha$ will be, in fact, in $(x_0,x)$. $\endgroup$ – Tim Raczkowski Mar 5 '15 at 4:15
  • $\begingroup$ Thanks Tim! Missed that one too $\endgroup$ – elbarto Mar 5 '15 at 4:32
  • $\begingroup$ Lol. I've sometimes edited my own posts 4 or 5 times. $\endgroup$ – Tim Raczkowski Mar 5 '15 at 4:34

You are right in both of your suspicions (though precisely speaking, $\alpha\in]x,x_0[$. But that's not too big of a deal, since you want to bound $f_n(x)-f_m(x)$:

$$\begin{align*} |f_n(x)-f_m(x)|&=|(f_n(x_0)-f_m(x_0))+(f_n'(\alpha)-f_m'(\alpha))(x-x_0)| \\&\leq |f_n(x_0)-f_m(x_0)|+|f_n'(\alpha)-f_m'(\alpha)||x-x_0| \\&\leq |f_n(x_0)-f_m(x_0)|+|f_n'(\alpha)-f_m'(\alpha)|M \end{align*}$$

where $M:=|b-a|.$

The first summand in the last expression can be made arbitrarily small since $\{f_n(x_0)\}_n\subseteq\mathbb{R}$ is convergent. The second summand too can be made arbitrarily small by the uniform convergence of the derivative. Hence we are done (I'll leave it to you to determine which epsilon(s) should be chosen for a clean exposition).

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  • $\begingroup$ Uniform convergence is needed to make the second estimate independent of the actual location of $α$. $\endgroup$ – Lutz Lehmann Mar 6 '15 at 2:10

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