Prob. 8, Chap. 5, in Rudin's PMS, 3rd ed: If $f^\prime$ is continuous on $[a, b]$, then $f$ is uniformly differentiable on $[a,b]$ Here is Prob. 8, Chap. 5, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f^\prime$ is continuous on $[a, b]$ and $\varepsilon > 0$. Prove that there exists $\delta > 0$ such that
$$ \left\lvert \frac{f(t)-f(x)}{t-x} - f^\prime(x) \right\rvert < \varepsilon 
$$
whenever $0 < |t-x| < \delta$, $a \leq x \leq b$, $a \leq t \leq b$. (This could be expressed by saying that $f$ is uniformly differentiable on $[a, b]$ if $f^\prime$ is continuous on $[a, b]$.)


Does this hold for vector-valued functions too?

My Attempt:

As $f^\prime$ is continuous on $[a, b]$ and as $[a, b]$ is compact, so $f^\prime$ is uniformly continuous on $[a, b]$. So, for any real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that
$$ 
\left\lvert f^\prime(x) - f^\prime(y) \right\rvert < \varepsilon \tag{1} $$
for all $x, y \in [a, b]$ for which $\lvert x-y \rvert < \delta$.


Now suppose $a \leq t \leq b$, $a \leq x \leq b$, and $0 < \lvert t-x \rvert < \delta$. Then by the Mean Value Theorem there is some point $y$ between $t$ and $x$ such that
$$ f(t) - f(x) = (t-x) f^\prime(y), \tag{2} $$
and also $$ \lvert y-x \rvert < \lvert t-x \rvert  < \delta; \tag{3} $$
moreover, as $\lvert t-x\rvert > 0$, so $t \neq x$, and from (2) we can write
$$
\frac{ f(t) - f(x)}{t-x} = f^\prime(y),$$
which together with (3) and (1) yields
\begin{align}  
\left\lvert \frac{f(t)-f(x)}{t-x} - f^\prime(x) \right\rvert &= \left\lvert f^\prime(y) - f^\prime(x) \right\rvert \\ 
&< \varepsilon.
\end{align}

Am I right?
Now for vector-valued functions.

Suppose $$\mathbf{f} = \left( f_1, \ldots, f_k \right) $$ be a mapping of $[a, b]$ into some $\mathbb{R}^k$ and suppose that
$$ 
\mathbf{f}^\prime = \left( f_1^\prime, \ldots, f_k^\prime \right) 
$$
is continuous on $[a, b]$. Then each of the component functions $f_1^\prime, \ldots, f_k^\prime$ is also continuous on $[a, b]$. So, given any real number $\varepsilon > 0$, we can find real numbers $\delta_i$, for $i = 1, \ldots, k$, such that
$$
\left\lvert  \frac{ f_i(t) - f_i(x) }{ t-x } - f_i^\prime(x) \right\rvert  < \frac{ \varepsilon }{\sqrt{k}} $$ whenever $0 < \lvert t-x \rvert < \delta_i$, $a \leq t \leq b$, and $a \leq x \leq b$.


Now let
$$\delta := \min \left\{ \delta_1, \ldots, \delta_k \right\}.$$ Therefore, If $a \leq t \leq b$, $a \leq x \leq b$, and $0 < \lvert t-x \rvert  < \delta$, then for each $i = 1, \ldots, k$, we obtain $0 < \lvert t-x \rvert < \delta_i$ and so $$
\left\lvert  \frac{ f_i(t) - f_i(x) }{ t-x } - f_i^\prime(x) \right\rvert < \frac{ \varepsilon }{\sqrt{k}}, $$
which then implies that
\begin{align}
& \ \ \ \left\lvert  \frac{ \mathbf{f}(t) - \mathbf{f}(x) }{ t-x } - \mathbf{f}^\prime(x) \right\rvert \\ 
&= \left\lvert   \left( \frac{f_1(t) - f_1(x)}{t-x}, \ldots, \frac{f_k(t) - f_k(x) }{t-x} \right) - \left( f_1^\prime(x), \ldots, f_k^\prime(x) \right) \right\rvert  \\ 
&= \left\lvert \left( \frac{ f_1(t) - f_1(x)}{t-x} - f_1^\prime(x), \ldots, \frac{ f_k(t) - f_k(x)}{t-x} - f_k^\prime(x) \right) \right\rvert  \\
&= \sqrt{ \sum_{i=1}^k \left\lvert \frac{ f_i(t) - f_i(x)}{t-x} - f_i^\prime(x) \right\rvert^2 } \\
&< \sqrt{ \sum_{i=1}^k \frac{\varepsilon^2}{k} } \\
&= \varepsilon.
\end{align}
Thus the above result holds for vector-valued functions as well.

Am I right?
Is my reasoning correct in each of the above two cases? If not, then where have I erred?

Yes, I verify that your attempt is correct.
To summarize, the key ideas for proving the scalar case are indeed uniform continuity and the mean value theorem:

*

*the mean value theorem allows one to replace the Newton quotient in the conclusion with a derivative;

*the uniform continuity of $f'$ gives a desired $\delta$.

For the vector-valued case, the key idea is that in finite-dimensional normed vector space, one can "control" the whole vector by "controlling" each component. Adding up finitely many $O(\varepsilon)$ quantities, one still has a $O(\varepsilon)$ quantity.

The proof of the scalar case could be written in a concise way as follows, omitting less important formal details.
Proof. Since $f'$ is continuous on a compact interval, it is uniformly continuous. Let $\delta$ be such that $\left|f^{\prime}(x)-f^{\prime}(u)\right|<\varepsilon$ for all $x, u \in[a, b]$ with $|x-u|<\delta$. $\dagger$
By the mean value theorem, for $t,x\in[a,b]$ with $0<|t-x|<\delta$, there exists $u$ between $t$ and $x$ such that
$$
\frac{f(t)-f(x)}{t-x}=f^{\prime}(u)\;.
$$
Hence, since $|u-x|<\delta$,
$$
\left|\frac{f(t)-f(x)}{t-x}-f^{\prime}(x)\right|=\left|f^{\prime}(u)-f^{\prime}(x)\right|<\varepsilon .
$$
(Since this result holds for each component of a vector-valued function $f(x)$, it must hold also for $\mathbf{f}$.)

$\dagger$ Notes. The  $\varepsilon$ is already given in the statement of the exercise.
