Theorem 7.17 in Baby Rudin: How is a shorter proof possible if the continuity of the $f_n^\prime$ is assumed in addition?

Question

Here is 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 $$\tag{27} f^\prime(x) = \lim_{n \to \infty } f_n^\prime(x) \qquad \qquad (a \leq x \leq b). $$

And, here is Rudin's proof:

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 $\phi(t)$.

Here is Theorem 5.19 in Baby Rudin:

Suppose $\mathbf{f}$ is a continuous mapping of $[a, b]$ into $\mathbb{R}^k$ and $\mathbf{f}$ is differentiable in $(a, b)$. Then there exists $x \in (a, b)$ such that $$ \lvert \mathbf{f} (b) - \mathbf{f} (a) \rvert \leq (b-a) \left\lvert \mathbf{f}^\prime (x) \right\rvert. $$

And, here is Theorem 7.11:

Suppose $f_n \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that $$ \tag{15} \lim_{ t \to x } f_n (t) = A_n \qquad (n = 1, 2, 3, \ldots). $$ Then $\left\{ A_n \right\}$ converges, and $$ \tag{16} \lim_{t \to x} f(t) = \lim_{n \to \infty} A_n. $$ In other words, the conclusion is that $$ \tag{17} \lim_{t \to x} \lim_{ n \to \infty} f_n(t) = \lim_{ n \to \infty} \lim_{t \to x} f_n(t). $$

Now here is the Remark following the proof of Theorem 7.17 in Baby Rudin:

If the continuity of the functions $f_n^\prime$ is assumed in addition to the above hypotheses, then a much shorter proof of (27) can be based on Theorem 7.16 and the fundamental theorem of calculus.

Here is Theorem 7.16 in Baby Rudin, 3rd edition:

Let $\alpha$ be monotonically increasing on $[a, b]$. Suppose $f_n \in \mathscr{R}(\alpha)$ on $[a, b]$, for $n = 1, 2, 3, \ldots$, and suppose $f_n \to f$ uniformly on $[a, b]$. Then $f \in \mathscr{R}$ on $[a, b]$, and $$ \tag{23} \int_a^b f \ \mathrm{d} \alpha = \lim_{n \to \infty} \int_a^b f_n \ \mathrm{d} \alpha. $$ (The existence of the limit is part of the conclusion.)

And, here is Theorem 6.21 in Baby Rudin (i.e. the fundamental theorem of calculus):

If $f \in \mathscr{R}$ on $[a, b]$ and if there is a differentiable function $F$ on $[a, b]$ such that $F^\prime = f$, then $$ \int_a^b f(x) \ \mathrm{d} x = F(b) - F(a). $$

Although I've understood the proof of Theorem 7.17, I do not know how to give the much shorter proof of (27) that Rudin has asserted in the Remark above?

Answer 1

Let each $f_n^\prime$ be continuous and let $\left(f_n^\prime \right)_n$ converge uniformly on $[a,b]$ to $g$. And let $\left( f_n \left(x_0 \right) \right)_n$ converge with $x_0 \in [a , b]$. Then $g$ is continuous, and $$ \begin{align} & \ \ \ \lim_{ n \to \infty } \sup_{x \in [a,b] } \left\lvert \int_{x_0}^x g(t) \ \mathrm{d} t - \left[ f_n(x) - f_n \left( x_0 \right) \right] \right\rvert \\ &= \lim_{ n \to \infty } \sup_{x \in [a , b ] } \left\lvert \int_{x_0}^x g(t) \ \mathrm{d} t - \int_{x_0}^x f_n^\prime (t) \ \mathrm{d} t \right\rvert \\ &= \lim_{ n \to \infty } \sup_{x \in [a , b ] } \left\lvert \int_{x_0}^x \left( g(t) - f_n^\prime (t) \right) \ \mathrm{d} t \right\rvert \\ &\leq \lim_{ n \to \infty } \sup_{ x \in [a , b ] } \int_{x_0}^x \left\lvert g(t) - f_n^\prime (t) \right\rvert \ \mathrm{d} t \\ &\leq \lim_{ n \to \infty } \int_a^b \left\lvert g(t) - f_n^\prime (t) \right\rvert \ \mathrm{d} t \\ &=0. \end{align} $$

I will leave the rest to you. We can use the integrals because $g$, and each $f_n$, is continuous.

Answer 2

If the continuity of the functions $f_n^{\prime}$ is added to the hypothesis of Theorem 7.17, we can prove it based on Theorem 7.16 and fundamental theorem of calculus as follows.

Since the functions $f_n^{\prime}$ are continuous, we get $f_n^{\prime}∈\mathcal{R}$. Then suppose $f_n^{\prime}→F^{\prime}$ uniformly. It follows, by Theorem 7.12, that $f^{\prime}∈\mathcal{R}$. By Theorem 7.16 and we get $$∫_x^{x+h}\;F^{\prime} (t)dt=\lim_{n→∞}⁡∫_x^{x+h}\;f_n^{\prime} (t)dt\ \ ∀x∈[a,b].$$ By the fundamental theorem of calculus, we have $$F(x+h)-F(x)=\lim_{n→∞} [f_n (x+h)-f_n (x)] ∀x∈[a,b].$$ It follows that ${f_n}$ converges. Letting $f_n→f$, we get $$F(x+h)-F(x)=f(x+h)-f(x)\; \; ∀x∈[a,b].$$ Thus, $$f^{\prime} (x)=F^{\prime} (x)=\lim_{n→∞}f_n^{\prime} (x).$$