Uniform convergence of $f'$ on an interval implies locally uniform convergence of $f$

Question

Let $(f_{n})_{n}$ be a sequence of differentiable functions defined on an interval $I \subseteq \mathbb{R}$ with values in $\mathbb{R}$. Suppose that $(f'_{n})_{n}$ converges uniformly on $I$ and that there exists an $x_{0} \in I$ for which $(f_{n}(x_{0}))_{n}$ converges. Show that $(f_{n}(x))_{n}$ converges for every $x \in I$.

This question is already asked before, but without an answer, only a hint.(Uniform convergence of derivative implies pointwise convergence of sequence of functions). I just realy don't see how to solve it.

An additional question is if $I$ is bounded, show that $(f'_{n})_{n}$ converges uniformly.

Thanks in advance

Answer 1

Pick an arbritrary $x\in I$ and $\varepsilon>0$. Then, there exists an $n_0\in\mathbb N$, such that for all $x\in I$, $$ m,n\ge n_0\quad\Longrightarrow\quad |\,f_m'(x)-f_n'(x)|<\frac{\varepsilon}{2|x-x_0|}, $$ and $$ m,n\ge n_0\quad\Longrightarrow\quad |\,f_m(x_0)-f_n(x_0)|<\frac{\varepsilon}{2}. $$ Hence $$ \big|\,\big(\,f_m(x)-f_n(x)\big)-\big(f_m(x_0)-f_n(x_0)\big)\big|=|x-x_0||\,f'_m(\xi)-f_n'(\xi)|<\frac{\varepsilon}{2|x-x_0|} |x-x_0|=\frac{\varepsilon}{2}, $$ for some $\xi\in(x_0,x)$, due to MVT. Thus $$ \big|\,\big(\,f_m(x)-f_n(x)\big)\big|<\frac{\varepsilon}{2} +\big|\big(\,f_m(x_0)-f_n(x_0)\big)\big|<\varepsilon. $$ Hence the sequence $\{f_n(x)\}$ is Cauchy and thus converges. This is true for every $x\in I$.

Note. The derivatives of the $f_n$'s DO NOT have to be integrable! Also, the requirement that the sequence $\{\,f_n'\}$ converges uniformly in $I$ can be relaxed to locally uniformly in $I$.

Answer 2

Hint: Let $x\in I.$ The idea is to show $f_n(x)$ is Cauchy. Note

$$f_n(x)- f_m(x) = [f_n(x)- f_m(x)- (f_n(x_0)- f_m(x_0))] + (f_n(x_0)- f_m(x_0)).$$

Apply the MVT to the expression in brackets on the right and use the hypothesis on $f_n'$ to see this is small for large $n,m.$. The other term on the right is Cauchy by hypothesis.