Let’s first give a proof under a stronger assumption: each
\(f_n'(x)\) is continuous on
\([a, b]\text{,}\) and let’s denote
\(\lim_{n\to\infty} f_n'(x)\) by
\(g(x)\text{.}\) Then
\(g(x)\) is continuous on
\([a, b]\text{,}\) \(f_n(x)=f_n(x_0)+\int_{x_0}^x f_n'(t)\, dt\text{,}\) and
Theorem 2.2.7 implies that
\(\int_{x_0}^x f_n'(t)\, dt\to \int_{x_0}^x g(t)\, dt\text{,}\) so
\(\lim_{n\to\infty} f_n(x)\) exists---denote it as
\(f(x)\) and
\begin{equation*}
f(x)=\lim_{n\to\infty} f_n(x_0)+ \int_{x_0}^x g(t)\, dt\text{.}
\end{equation*}
It follows that
\begin{equation*}
f'(x)= g(x)=\lim_{n\to \infty} f_n'(x).
\end{equation*}
Furthermore, the convergence above is uniform over \(x\in [a, b]\text{,}\) as
\begin{equation*}
\sup_{x\in [a, b]} \left| \int_{x_0}^x f_n'(t)\, dt- \int_{x_0}^x g(t)\, dt \right|
\le \int_a^b \left| f_n'(t)-g(t)\right|\, dt \to 0
\end{equation*}
as \(n\to \infty\text{.}\)
We now do a proof in the general case. First, we prove that \(\{f_n(x)\}\) converges uniformly over \([a, b]\text{.}\) Under the assumption of uniform convergence of \(\{f_n'(x)\}\) over \([a, b]\text{,}\) for any \(\epsilon >0\text{,}\) there exists \(N\) such that for \(n, m\ge N, x\in [a, b]\text{,}\) we have
\begin{equation}
|f_n'(x)-f_m'(x)| \lt \epsilon.\tag{2.4.1}
\end{equation}
Applying the theorem of the mean to \(f_n(x)-f_m(x)-\left[ f_n(x_0)-f_m(x_0)\right]\text{,}\) we get
\begin{equation*}
f_n(x)-f_m(x)-\left[ f_n(x_0)-f_m(x_0)\right]=\left(f_n'(x^*)-f_m'(x^*)\right)(x-x_0)
\end{equation*}
for some \(x^*\) depending on \(n, m, x\text{.}\) This leads to
\begin{equation*}
\left| f_n(x)-f_m(x)\right| \le \left| f_n(x_0)-f_m(x_0)\right| + \epsilon (b-a).
\end{equation*}
Since \(f_n(x_0)-f_m(x_0) \to 0\) as \(n, m \to \infty\text{,}\) this shows that \(\{f_n(x)\}\) satisfies the Cauchy Criterion for Uniform Convergence on \([a, b]\text{,}\) therefore it converges uniformly to some \(f(x)\) on \([a, b]\text{.}\)
Next for any
\(x\in (a, b)\) we define
\(g_n(h)=\left[f_n(x+h)-f_n(x)\right]/h\) for
\(0 \lt |h| \lt \delta\) for some
\(\delta >0\) (when
\(x=a\) or
\(b\text{,}\) we restrict
\(h\) to have appropriate sign). Note that
\(g_n(h)\to f_n'(x)\) as
\(h\to 0\) and that
\(g_n(h)\to \left[f(x+h)-f(x)\right]/h\) as
\(n\to \infty\text{.}\) We next show that this convergence is uniform for
\(0 \lt |h| \lt \delta\text{.}\)
We apply the theorem of the mean to \(g_n(h)-g_m(h)\) to get
\begin{equation*}
g_n(h)-g_m(h)=f_n'(x^*)-f_m'(x^*)
\end{equation*}
for some
\(x^*\) between
\(x\) and
\(x+h\) depending on
\(n, m, x, h\text{.}\) But
(2.4.1) holds for
\(n, m\ge N, x\in [a, b]\text{,}\) so we get
\(| g_n(h)-g_m(h) | \lt \epsilon\) for
\(n, m\ge N\) uniformly in
\(0 \lt |h| \lt \delta\text{.}\) This shows that
\(g_n(h)\) converges to
\(\left[f(x+h)-f(x)\right]/h\) uniformly in
\(0 \lt |h| \lt \delta\) as
\(n\to \infty\text{.}\) Now
Theorem 2.2.1 applies to
\(g_n(h)\) to conclude that
\begin{equation*}
\lim_{h\to 0} \left[f(x+h)-f(x)\right]/h = \lim_{n\to \infty} f_n'(x).
\end{equation*}
Namely, \(f'(x)\) exists and equals \(\lim_{n\to \infty} f_n'(x)\text{.}\)
