Show that
\(f_n: [a, b]\to \RR\) are continuously differentiable with derivatives \(f_n'\)
The sequence of derivatives \(f_n'\) converges uniformly to some function \(g\)
There exists at least one point \(x_0\) such that \(\lim_n f_n(x_0)\) exists,
Then \(f_n \to f\) uniformly to some differentiable \(f\), and \(f' = g\).
Solution
1. \(g\) is continuous on \([a,b]\).
\(g\) is the uniform limit of the continuous functions \(f_n'\).
2. For every \(x \in [a,b]\), \(f_n(x) = f_n(x_0) + \int_{x_0}^x f_n'(t) \, dt\).
This is the fundamental theorem of calculus, since \(f_n\) is continuously differentiable.
3. Let \(c = \lim_n f_n(x_0)\) and \(f(x) \coloneqq c + \int_{x_0}^x g(t)\,dt\). Then \(f_n \to f\) uniformly on \([a,b]\).
1.By step 2, for every \(x \in [a,b]\),\[ \abs{f_n(x) - f(x)} \leq \abs{f_n(x_0) - c} + \abs{\int_{x_0}^x (f_n'(t) - g(t))\,dt} \leq \abs{f_n(x_0) - c} + (b-a)\norm{f_n' - g}_\infty, \]and the right-hand side does not depend on \(x\) and tends to \(0\).
4. \(f\) is differentiable and \(f' = g\).
Since \(g\) is continuous by step 1, the fundamental theorem of calculus applied to \(f(x) = c + \int_{x_0}^x g(t)\,dt\) gives \(f'(x) = g(x)\) for every \(x \in [a,b]\). In particular \(f \in C^1[a,b]\).
5. Q.E.D.
Step 3 gives \(f_n \to f\) uniformly, and step 4 gives that \(f\) is differentiable with \(f' = g\).