Let $f: (a,b) \rightarrow \Bbb R$ be a differentiable function in $(a,b)$. Calculate the pointwise limit of:

$$f_n(x)=n(f(x+1/n)-f(x)), x\in(a, b-1/n). $$

Let $E_n$ be a countable family of sets:

$$E_n = \{x \in (a,b) \mid |f_i(x)-f_j(x)| \le 1 \forall i,j\ge n \};$$

What is $\cup_n E_n$?

Infer that there exists $(c,d)\subset (a,b)$ in which $f'(x)$ is bounded for every $x\in(c,d)$.


$$f_n(x)\rightarrow f'(x) \forall x \in (a,b).$$

$\{f_n(x)\}$ is a Cauchy sequence in $\Bbb R $ so $\cup_n E_n=(a,b)$.

Then I took $[\alpha, \beta] \subset (a,b)$. It is closed, so it is a complete metric space and I can use Baire's Theorem.

$\cup_n A_n=[\alpha, \beta]$ like before.

There is a non-nowhere dense $A_n$ and I can pick an inteval like requested.


To the point that you say "so $\bigcup_n E_n=(a,b)$", your arguments seem to be correct, but you didn't specify what $A_n$ is. I would do as follows:

Take $[\alpha,\beta]\subset(a,b)$, $\alpha<\beta$. If $1/N<b-\beta$, then $f_n$ is defined in $[\alpha,\beta]$ for $n\geq N$, and, since each $f_n$ is continuous (in its domain), then $E_n\cap [\alpha,\beta]$ is closed for $n\geq N$. Also, since $E_n\subseteq E_{n+1}$, we have $[\alpha,\beta]=\bigcup_{n=N}^\infty E_n\cap[\alpha,\beta]$. By Baire's theorem, some $E_n\cap[\alpha,\beta]$ has nonempty interior (in $[\alpha,\beta]$), so this $E_n\cap[\alpha,\beta]$ contains some interval $(c,d)$. Finally, $f'$ is bounded in $E_n\cap[\alpha,\beta]$: We have $|f_i(x)|\leq|f_n(x)|+1$ for $i\geq n$ and $x\in E_n$. Taking $i\rightarrow\infty$, we have $|f'(x)|\leq|f_n(x)|+1$, and $\sup_{x\in E_n\cap[\alpha,\beta]}|f_n(x)|<\infty$, since $E_n\cap[\alpha,\beta]$ is compact and $f_n$ is continuous. This solves the problem.

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  • $\begingroup$ This is what I thought, sorry for not having specified what $A_n$ were to me, but you got the point. Thank you so much $\endgroup$ – Nicola Ferro Feb 6 '14 at 19:14

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