I have a question about Theorem 7.17 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $\left\{ f_n \right\}$ is a sequence of functions, differentiable on $[a, b]$ and such that $\left\{ f_n \left( x_0 \right) \right\}$ converges for some point $x_0$ on $[a, b]$. If $\left\{ f_n^\prime \right\}$ converges uniformly on $[a, b]$, then $\left\{ f_n \right\}$ converges uniformly on $[a, b]$, to a function $f$, and
$$\tag{27} f^\prime(x) = \lim_{n \to \infty } f_n^\prime(x) \qquad \qquad (a \leq x \leq b). $$
Rudin's proof is as follows:
Let $\varepsilon > 0$ be given. Choose $N$ such that $n \geq N$, $m \geq N$, implies
$$ \tag{28} \left\lvert f_n \left( x_0 \right) – f_m \left( x_0 \right) \right\rvert < \frac{\varepsilon}{2} $$
and
$$ \tag{29} \left\lvert f_n^\prime(t) – f_m^\prime(t) \right\rvert < \frac{\varepsilon}{2(b-a)} \qquad \qquad (a \leq t \leq b). $$
If we apply the mean value theorem 5.19 to the function $f_n – f_m$, (29) shows that
$$ \tag{30} \left\lvert f_n (x) – f_m (x) – f_n (t) + f_m (t) \right\rvert \leq \frac{ \lvert x-t \rvert \varepsilon }{ 2(b-a) } \leq \frac{ \varepsilon }{ 2 } $$
for any $x$ and $t$ on $[a, b]$, if $n \geq N$, $m \geq N$. The inequality
$$ \left\lvert f_n (x) – f_m (x) \right\rvert \leq \left\lvert f_n (x) – f_m (x) – f_n \left( x_0 \right) + f_m \left( x_0 \right) \right\rvert + \left\lvert f_n \left( x_0 \right) – f_m \left( x_0 \right) \right\rvert $$
implies, by (28) and (30), that
$$ \left\lvert f_n(x) – f_m(x) \right\rvert < \varepsilon \qquad \qquad (a \leq x \leq b, n \geq N, m \geq N), $$
so that $\left\{ f_n \right\}$ converges uniformly on $[a, b]$. Let
$$ f(x) = \lim_{n \to \infty} f_n (x) \qquad (a \leq x \leq b). $$
Let us now fix a point $x$ on $[a, b]$ and define
$$ \tag{31} \phi_n(t) = \frac{ f_n(t) – f_n(x) }{ t-x}, \qquad \phi(t) = \frac{ f(t) – f(x) }{ t-x} $$
for $a \leq t \leq b$, $t \neq x$. Then
$$ \tag{32} \lim_{ t \to x } \phi_n (t) = f_n^\prime(x) \qquad (n = 1, 2, 3, \ldots). $$
The first inequality in (30) shows that
$$ \left\lvert \phi_n(t) – \phi_m(t) \right\rvert \leq \frac{ \varepsilon}{2(b-a) } \qquad ( n \geq N, m \geq N), $$
so that $\left\{ \phi_n \right\}$ converges uniformly, for $t \neq x$. Since $\left\{ f_n \right\}$ converges to $f$, we conclude from (31) that
$$\tag{33} \lim_{n \to \infty } \phi_n (t) = \phi(t) $$
uniformly for $a \leq t \leq b$, $t \neq x$.
If we now apply Theorem 7.11 to $\left\{ \phi_n \right\}$, (32) and (33) show that
$$ \lim_{t \to x} \phi(t) = \lim_{n \to \infty} f_n^\prime(x); $$
and this is (27), by the definition of $\phi(t)$.
How is the claim on line 33 proven? From my understanding, Rudin has shown only that ${\phi_n(t)}$ converges uniformly by the Cauchy criterion for uniform convergence, but it is unclear to me as to why it converges in particular to $\phi(t)$.
Best Answer
(31) shows that $\phi_n(t)\to \phi (t)$ for each $t\neq x$. Now let $m \to \infty$ in the inequality $$|\phi_n(t)-\phi_m (t)| \leq \frac {\epsilon} {b-a}$$ to see that $$|\phi_n(t)-\phi (t)| \leq \frac {\epsilon} {b-a}$$ for $n \geq N$ for $t \neq x$.