I'm struggling trying to find a real function sequence $\{f_n\}_n$ such that
- $\forall n: f_n$ is defined on an open and limited interval $(a,b)$;
- $\forall n: f_n$ is everywhere differentiable (wrt $x$) on $(a,b)$;
- $\exists x_0\in (a,b)$ such that $\{f_n(x_0)\}_n$ converges;
- $\{f'_n\}_n$ is uniformly convergent on $(a,b)$;
- $\{f_n\}_n$ does not converges uniformly on $(a,b)$.
Now, conditions 2), 3), 4) are sufficient to guarantee pointwise convergence of $\{f_n\}_n$ on $(a,b)$, let's say to a function $f:(a,b)\to\mathbb{R}$, and the chance to interchange limit with differentiation in the sense that $$\forall x\in (a,b): f'(x)=\lim_n f'_n(x).$$ Moreover, under those same supposititions, we can have uniform convergence on every compact subinterval of $(a,b)$. My goal is to find a function sequence that satisfies together all those five conditions: I suppose that the problem must come from openess of the functions' domain that messes up with the uniform convergence of the $\{f_n\}_n$ : unfortunately, all my attempts failed so I'm here to ask you some help to find a sequence like that. Any ideas?
Best Answer
As long as the interval is bounded, uniform convergence of $(f_n)$ must hold.
By the mean value theorem for all $x \in (a,b)$ there exists $\xi_x$ between $x$ and $x_0$ such that
$$f_m(x) - f_n(x) = f_m(x_0) - f_n(x_0) + [f'_m(\xi_x) - f'_n(\xi_x)](x- x_0),$$
and
$$\sup_{x \in (a,b)}|f_m(x) - f_n(x) | \leqslant |f_m(x_0) - f_n(x_0)| + \sup_{x \in (a,b)}|f'_m(\xi_x) - f'_n(\xi_x)|\sup_{x \in (a,b)}|x - x_0| \\ \leqslant|f_m(x_0) - f_n(x_0)|+ (b-a)\sup_{x \in (a,b)}|f'_m(x) - f'_n(x)| $$
By pointwise convergence of $(f_n(x_0))$ and uniform convergence of $(f'_n)$ on $(a,b)$ we can find $N(\epsilon)$ such that the RHS is less than $\epsilon$ for all $m > n > N(\epsilon)$. Therefore, $f_n$ must converge uniformly on $(a,b)$ (even though the interval is open).