Is it necessarily true for $(f_n)$ to converge uniformly on $[a,b]$

Question

Let $(f_n)$ be sequence of real valued functions on $[a,b]$. If $f_n'(c)$ exists for all $c\in(a,b)$ and $n\in N$, and $(f_n(c))$ converges for all $c \in[a,b]$ and $(f_n')$ converges uniformly on $[a,b]$, then is it necessarily true for $(f_n)$ to converge uniformly on $[a,b]$?

I know that, it is given that $f_n$ to some $f$ pointiwse on $[a,b]$

and $f_n' \to $ some $g $ uniformly on $[a,b]$

and $f_n$ are differentiable,

then $f$ must be differentiable and $f'=g$

But how do I prove/disprove $f_n$ converges uniformly on $[a,b]$

Answer 1

In fact, it's enough to assume that $(f_n(x_0))_n$ converges for some $x_0 \in (a,b)$ instead of the whole $[a,b]$:

Let's define a new function $$F_{nm}(x)=f_n(x)-f_m(x)$$ Let's consider the interval $[x_0,x]$ (the argument is the same for $[x,x_0]$). $F_{nm}$ is differentiable in it's interior, so we can apply the MVT: $\exists c \in (x_0,x)$ so that $$F_{nm}(x)-F_{nm}(x_0)=F'_{nm}(c)(x-x_n)$$ And we can rewrite it as $$f_n(x)-f_m(x)=f_n(x_0)-f_m(x_0)+[f'_n(c)-f'_m(x)](x-x_0)$$ Now you just need to get an uniform bound. Can you continue?

Answer 2

$f_n(x)=f_n(c)+\int_c^{x}f_n'(t)dt$. If $f_n' \to g$ uniformly then this gives $f_n(x) \to l+\int_c^{x}g(t)dt$ uniformly where $l =\lim f_n(c)$.

You can take any particular $c$ (in particular, $c=a$).

EDIT: if you are not assuming that $f_n$ are continuously differentiable then you apply a similar argument to show that $f_n-f_m \to 0$ uniformly as $n,m \to \infty$. This implies that $(f_n)$ converges uniformly.