Limit of Sequence of Differentiable Functions and Sequence of Derivatives

Question

I have a two Cauchy sequences: $$ \{u_{n} \} \ \text{and} \ \{u'_{n}\} $$ Where $u_{n}$ is a continuously differentiable and bounded function on $\mathbb{R}$, and the sequences are Cauchy with respect to the norm:

$$ ||u|| = ^{\text{sup}}_{x \in \mathbb{R}} |u| $$

So far, I have shown that both of these sequences converge uniformly to functions which are bounded and continuous, however, I have yet to show that the limit of $u_{n}$ with respect to $n$ is differentiable, and then that the derivative is equal to the limit of $u'_{n}$.

I have seen it written that the fact that $u'_{n}$ converges uniformly is sufficient to show this, but I would like to show why that is so. So far, I have struggled with the limit definition of derivatives and attempted to swap the $^{\text{lim}}_{h \rightarrow 0}$ and the $^{\text{lim}}_{n \rightarrow \infty}$ but I have not made any progress.

Could someone please let me know if there is a better approach. Just a hint as to where to start should be enough. Thank you very much!

Answer 1

Let $u=\lim u_n$ and $v=\lim u_n'$. The easiest way to see $u'=v$ is to use the fundamental theorem of calculus: $$ u_n(x)-u_n(0)=\int_0^x u_n'(t)\,dt. $$ The left side converges to $u(x)-u(0)$, while uniform convergence of $(u_n')$ to $v$ implies that the right side converges to $\int_0^x v(t)\,dt$. Thus $u$ is continuously differentiable and $u'=v$.

In fact, this proof shows that it suffices to assume that $(u_n)$ converges pointwise and $(u_n')$ converges locally uniformly (uniformly on bounded intervals).

Answer 2

Suppose that $u_n\to u$ uniformly and $u_n'\to v$ uniformly (on $\Bbb R$), then we you need to show is that $u'(x)=v(x)$ for all $x\in \Bbb R$. It is enough to prove this for all $x\in[-M,M]$ where $M>0$ is arbitrary.

Hint 1: Write $u_n(x) = u_n(0) + \int_0^x u'_n(t) dt$.

Hint 2: If $f_n:[a,b]\to\Bbb R$ is a sequence of continuous function converging uniformly to $f$, then we have $$ \int_a^b f(t)dt = \lim_{n\to\infty} \int_a^b f_n(t)dt. $$