Question 9.T.5: Suppose (fn) is a differentiable sequence of functions on [a...

Let ε > 0. From the convergence of (f_{n}(x_{0})) and the uniform convergence of (f^{\prime}_{n}), we deduce that there is an N ∈ \mathbb{N} such that

m, n ≥ N ⇒ \left|f^{\prime}_{n}(x) − f^{\prime}_{m} (x)\right| < ε  for all x ∈ [a, b],

and

m, n ≥ N ⇒ |f_{n}(x_{0}) − f_{m}(x_{0})| < ε.

Given any two points x, t ∈ [a, b], we know from the mean value theorem, applied to f_{n} − f_{m}, that we can find a point c between x and t such that

f_{n}(x) − f_{m}(x) − [f_{n}(t) − f_{m}(t)] = (x − t)[f^{\prime}_{n}(c) − f^{\prime}_{m}(c)].

If m, n ≥ N then

|f_{n}(x) − f_{m}(x) − [f_{n}(t) − f_{m}(t)]| < ε |x − t| ,        (9.14)

hence

|f_{n}(x) − f_{m}(x)| ≤ |f_{n}(x) − f_{m}(x) − [f_{n}(x_{0}) − f_{m}(x_{0})]|

+ |f_{n}(x_{0}) − f_{m}(x_{0})|

< ε |x − x_{0}| + ε

≤ ε(b − a + 1) = Cε,  x ∈ [a, b].

This proves that (f_{n}) satisfies Cauchy’s criterion for uniform convergence, and must therefore converge uniformly to some limit f.

For any fixed point x ∈ [a, b], define

g_{n}(t) = \frac{f_{n}(t) − f_{n}(x)}{t − x},  t ∈ [a, b]\setminus \left\{x\right\},

g(t) = \frac{f (t) − f (x)}{t − x},  t ∈ [a, b]\setminus \left\{x\right\}.

Clearly g_{n} → g as n → ∞, and we shall now show that this convergence is uniform. If m, n ≥ N then, using (9.14),

|g_{n}(t) − g_{m}(t)| < ε  for all t ∈ [a, b]\{x},

hence, by Cauchy’s criterion, g_{n} \overset{u}{\rightarrow} g on [a, b]\{x}. Since

\underset{t→x}{\lim}  g_{n}(t) = f^{\prime} _{n} (x)  for all n ∈ \mathbb{N},

we can use Theorem 9.3 to conclude that

\underset{t→x}{\lim}  g(t) = \underset{n→∞}{\lim} f^{\prime}_{n} (x).

But since \lim_{t→x} g(t) = f^{\prime}(x), this proves that f is differentiable at x, and that

f^{\prime}(x) = \underset{n→∞}{\lim} f^{\prime}_{n} (x).