Suppose that $f_n : (a,b)\to \mathbb R $ is a sequence of differentiable functions such that there is $c \in (a, b)$ satisfying $\lim_{n \to \infty} f_n(c) = L \in \mathbb R$. Suppose further that given $[\alpha, \beta] \subset (a, b)$, the sequence $(f'_n)$ converges uniformly to the null function in $[\alpha, \beta]$ . Show that in each $[\alpha, \beta] \subset (a, b)$ the function $f_n$ converges uniformly to the constant function equal to $L$.
What I've managed to do so far is this
Fix $\epsilon \gt 0$,and choose $N$ such that $m, n \ge N$ implies $|f_n(x_0) − f_m(x_0)| \lt \frac{\epsilon}{2}$ and $|f'_n(t) − f'_m(t)| \lt \frac{\epsilon}{2(b-a)}$ for all $t \in [a, b]$. So lets say $g = f_n −f_m$ is small on the entire interval $[a, b]$, now applying the Mean Value Theorem to $g$, we get $$|g(x) − g(t)| = |x − t||f'(c)| \le |x − t| \frac{\epsilon}{2(b − a)} \le \frac{\epsilon}{2}$$ valid for all $x$, $t \in[a, b]$ and all $m$, $n \ge N$. Now for all $x \in [a, b]$ $$|g(x)| \le |g(x) − g(x_0)| + |g(x_0)| \lt \epsilon$$ This shows that the sequence $f_n$ is uniformly Cauchy, hence uniformly converges to some function, $f$ on $[a, b]$
I'm in the right way?
Thanks in advance for any help.
I think you are going in the right direction, but the approach can be simplified. Let $[\alpha,\beta]$ be given. Then for $x\in [\alpha,\beta],$ the MVT leads to
$$|f_n(x)-L|\le |f_n(x)-f_n(c)| +|f_n(c)-L| =|f_n'(c_n)(x-c)| +|f_n(c)-L|.$$
In absolute value the right hand side is no more than $$(\sup_{[\alpha,\beta]}|f_n'|)|b-a| + |f_n(c)-L|.$$ Notice the last expression $\to 0$ as $n\to \infty,$ and is independent of $x\in [\alpha,\beta].$ This implies the desired uniform convergence of $f_n$ to $L$ on $[\alpha,\beta].$