Let $\{f_n\}$ be a sequence of differentiable functions on $\mathbb R$ such that $f_n'$ is bounded for each $n$ ; if $\{f_n'\}$ converges uniformly to $f$ on $\mathbb R$ then is it true that $f$ is bounded and the derivative of some function ?
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Using $f$ as the limit of $f'_n$ is a little confusing, I will use $g$ as the limit function.
Let $h_n(x) = \int_0^x f'_n(t) dt$. Then $h_n(0) = 0$, and $h'_n = f'_n$.
Pick some interval $[a,b]$ that contains $0$. Using the result here Uniform convergence of derivatives, Tao 14.2.7. we see that $h_n$ converge uniformly (on $[a,b]$) to some limit $h$ such that $h' = g$.
Since the interval was arbitrary, we see that $h'=g$ everywhere.
Boundedness follows from uniform convergence to $g$ and the fact that each $f'_n$ is bounded.
copper.hat
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I think the interval $[a,b]$ should contain the point $x=0$ for which the sequence $h_n(0) \to 0=:L$. Otherwise, I like your solution. – Svetoslav Mar 14 '16 at 16:43
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@Svetoslav: That was my intent, thanks for catching that. The solution works anyway, but requires extra steps. – copper.hat Mar 14 '16 at 17:18
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We don't know $f_n'$ is Riemann integrable. Fix: Use $h_n= f_n-f_n(0).$ – zhw. Mar 14 '16 at 21:30
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@zhw.: We know that $f'_n$ is measurable and since it is bounded it is Lebesgue integrable. Also, the definition above gives the same value as in your comment. – copper.hat Mar 14 '16 at 23:26
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Yes that's true, but why go through all that when you could just let $h_n = f_n-f_n(0)?$ – zhw. Mar 14 '16 at 23:55
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@zhw.: True. ${}{}{}$ – copper.hat Mar 14 '16 at 23:59
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$f$ is indeed bounded. By uniform convergence, there is some $N$ such that $\forall x\in \mathbb R, |f(x)-f'_N(x)|\leq 1$.
Then $\forall x\in \mathbb R, |f(x)|\leq 1+|f'_N(x)|\leq 1+M_N$
Gabriel Romon
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