Let X be a open subset on Rn.

Let S be the set of all real, bounded, andfunctions on X, and rho is the sup norm.continuously diferentiable

Is (S,rho) a complete metric space?

Could you give me a counter-example if it is not?

Thanks,

JP

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- May 3rd 2010, 01:21 PMjaveiroComplete metric space
Let X be a open subset on Rn.

Let S be the set of all real, bounded, andfunctions on X, and rho is the sup norm.__continuously diferentiable__

Is (S,rho) a complete metric space?

Could you give me a counter-example if it is not?

Thanks,

JP - May 3rd 2010, 02:04 PMDrexel28
- May 3rd 2010, 03:20 PMjaveiro
Hum...

I guess it would be incomplete.

I was wondering if (abs(x))^(1+1/n) would be a counter example.

It converges to abs(x), but i'm failing to show that it is Cauchy... - May 3rd 2010, 04:23 PMDrexel28
What you said makes no sense. If you believe that this space is incomplete then you would want to find a Cauchy sequence that doesn't converge. Also, every Cauchy sequence is convergent.

But, let's assume you meant to take the open subset of $\displaystyle \mathbb{R}$ to be $\displaystyle (-1,1)=U$ and the function $\displaystyle |x|^{1+\frac{1}{n}}\in\mathcal{C}^1\left[U,\mathbb{R}\right]$. Is that even differentiable at $\displaystyle 0$?

I haven't checked it out, but what about $\displaystyle f_n(x)=\begin{cases}x^{1+\frac{1}{n}}\sin(x) & \mbox{if}\quad x\ne 0 \\ 0 & \mbox{if}\quad x=0\end{cases}$ with $\displaystyle X=(-1,1)$. This is differentiable and continuous, and I'm pretty sure Cauchy but it "wants" to converge to $\displaystyle x\sin(x)$ which is only differentiable at $\displaystyle 0$, not continuously differentiable.