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As the title suggests. I want the function to be from $[a, b]$ to $\mathbb R$. By unbounded derivative, I mean if $f_n$ is a sequence of functions and $M_n$ is the sup norm of $f_n'$ on $[a, b]$, then $\{M_n\}$ is an unbounded sequence of reals. I really have no idea how to do these kind of questions (the ones where you are asked to give an example). Can anyone help me out?

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Is this really what you wanted to ask? If so, consider $f_n=0$. –  1015 Feb 21 '13 at 23:48
    
oh oops sorry. i mean the derivatives are unbounded. –  Aden Dong Feb 21 '13 at 23:49
    
What about $f_n(x)=\dfrac{e^{nx}}{n}$? –  MyUserIsThis Feb 21 '13 at 23:53
    
What is your definition of smooth on $[a,b]$? –  1015 Feb 21 '13 at 23:54
1  
@HalilDuru Yes, of course, I understood the question like that at first since it is not properly stated. But apparently the OP means: $\sup |f'_n|\longrightarrow +\infty$ as $n\rightarrow +\infty$. –  1015 Feb 22 '13 at 0:46

1 Answer 1

up vote 1 down vote accepted

Consider the function $f$ defined on $[0,1]$ by $$ f(x)=x\sin\left(\frac{1}{x}\right) $$ for $x\neq 0$ and $$ f(0)=0. $$

Then for each $n$, take a smooth function $\phi_n$ such that $$ \phi_n(x)=0\quad\forall x\in[0,\frac{1}{ n}]\quad \phi_n(x)=1\quad\forall x\in[\frac{2}{n},1] $$ and $$ 0\leq \phi_n(x)\leq 1\quad \forall x\in [\frac{1}{n},\frac{2}{n}]. $$ This can be achieved with appropriate modifications of $e^{-1/x^2}$.

Now consider the sequence $$ f_n=f\phi_n. $$

I believe this works.

1) Equicontinuity:

First fix $x$ in $(0,1]$ and a small $r>0$. For $n$ large enough, namely $2/n<x-r$, we have $f_n(y)=f(y)$ for all $y\in(x-r,x+r)$.

Take $\epsilon>0$. By continuity of $f$ at $x$ there is $0<\delta<r$ such that $$ |f_n(y)-f_n(x)|=|f(y)-f(x)|<\epsilon \qquad \forall n\geq N=\lfloor\frac{2}{x-r}\rfloor+1. $$ Now $f_1,f_2,\ldots,f_{N-1}$ are all continuous at $x$ and in finite number, so there exist a common $\delta'>0$ such that $|f_n(y)-f_n(x)|\epsilon$ for these $n$ and all $y\in(x-\delta',x+\delta')$.

It only remains to take the minimum of $\delta$ and $\delta'$ to get what you want and equicontinuity at $x$ follows.

Now take $x=0$. We have $|f_n(y)|\leq y\sin (1/y)\leq y$ for all $y\in[0,1]$ and all $n$. So equicontinuity at $0$ follows.

By compactness of $[0,1]$, we even have uniform equicontinuity.

2) Uniform unboundedness of the derivatives:

Compute $f'(x)$ and check it is unbounded as $x$ approaches $0$. The claim follows easily.

3) Smoothness is obvious.

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Can you elaborate a little bit why $f_n$ is equicontinuous at $x \in (0, 1]$? Sorry if this is a stupid question, but it doesn't seem immediately obvious to me. –  Aden Dong Feb 22 '13 at 16:46
    
Also, I'm having trouble understanding why f_n is smooth because I believe $f'$ has no limit as $x \to 0$. –  Aden Dong Feb 22 '13 at 16:47
    
@AdenDong Ok, forget the previous comment, I was in a rush and I don't know what I was thinking. I constructed this example precisely to have $\sup|f'_n$ finite for each $n$, but tending to $+\infty$ as $n\rightarrow+\infty$. On $[0,1/n]$, $f_n=0$ so it is smooth as you mean it, with limits of the derivatives at $0$ equal to $0$. On $[1/n,1]$ it is equalt to the smooth $\phi_nf$ which is smooth with bounded derivatives and limits of the derivatives at $1$ exist also, even if they are nonzero. So it does the job. Regarding equicontinuity, I will add a few words. Let me know if it's better. –  1015 Feb 22 '13 at 21:05
    
thank you for the thorough explanation! –  Aden Dong Feb 23 '13 at 5:33
    
@AdenDong You're welcome. Glad it helped. –  1015 Feb 23 '13 at 10:46

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