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$X=(C[0,1],\rho_\infty)$ where $\rho_\infty$ is the uniform norm. $M\in(0,\infty)$, define $A=\{f\in X:f(0)=0, f\;\mathrm{differentiable\;on}\;(0,1)\;,|f^\prime(x)|\leq M\;\;\forall x\in(0,1)\}$. I was trying to prove that $A$ is compact, I wanted to use Ascoli-Arzelà, so I proved that $A$ is equicontinuous and bounded, but I don't know how to prove that it's closed, what is giving me problems is that if $(f_n)\subset A$ and $f_n\rightarrow f$ then $f$ is differentiable, I'm not even sure that it's true. I mean, in general it's not true, but I don't know if the other conditions of $A$ force this to be true, could you help me?

EDIT: following the hint of yoyo I think that we should take the sequence $f_n(x)=1/2-|x-1/2|$ if $|x|\geq 1/2+1/n$, and $f_n(x)=1/2-n/2(x-1/2)^2-1/(2n)$ if $|x|\leq 1/2+1/n$, but I'm having problems to prove that this sequence converges uniformly to $1/2-|x-1/2|$, any hints?

EDIT EDIT: following the hint of Jonas I think I solve the problem: modify the function as Jonas said, then $f(x)-f_n(x)=-|x-1/2|+n/2(x-1/2)^2+1/(2n)$ if $|x-1/2|\leq 1/n$ and zero elsewhere. But $-1/(2n)=-1/n+0+1/(2n)\leq-|x-1/2|+n/2(x-1/2)^2+1/(2n)\leq0+n/2\cdot1/n^2+1/(2n)=1/n$ and so $|f(x)-f_n(x)|\leq 1/(2n)$ and the convergence is uniform. Am I right?

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drawing something like absolute value and differentiable approximations with derivatives between $\pm1$ makes me think it's not closed –  yoyo Feb 2 '12 at 3:28
    
Where you wrote $|x|\geq (\leq) 1/2+1/n$, I think you mean $|x-1/2|\geq (\leq) 1/n$. –  Jonas Meyer Feb 2 '12 at 4:28
    
It seems you actually showed $|f(x)-f_n(x)|\leq \frac{1}{n}$, but otherwise it looks good. Another way to think of it: for $x$ in $[1/2-1/n,1/2+1/n]$, you have $1/2-1/n \leq f_n(x)\leq 1/2$ and $1/2-1/n\leq f(x)\leq 1/2$, and since $f_n(x)$ and $f(x)$ both lie in the interval $[1/2-1/n,1/2]$, their distance is at most $1/n$. (And outside $[1/2-1/n,1/2+1/n]$ you have $f_n=f$.) –  Jonas Meyer Feb 2 '12 at 5:41
    
you're right $|f(x)-f_n(x)|\leq 1/n$ –  Alex M Feb 2 '12 at 6:04

1 Answer 1

No, $A$ is not closed. For example, if $f_n(x)=1-\left|2\left(x-\frac{1}{2}\right)\right|^{1+1/n}$, then $f_n$ converges uniformly to $f(x)=1-2\left|x-\frac{1}{2}\right|$.

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why $f'_n$ exist in $(0,1)$? –  leo Feb 2 '12 at 4:12
    
$f_n'(1/2)=0$ for all $n$, $f_n'(x)=-2(1+1/n)(2(x-1/2))^{1/n}$ when $x>1/2$, $f_n'(x)=2(1+1/n)(2(1/2-x))^{1/n}$ when $x<1/2$. –  Jonas Meyer Feb 2 '12 at 4:22

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