# Complete metric space

• May 3rd 2010, 01:21 PM
javeiro
Complete metric space
Let X be a open subset on Rn.

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

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 PM
Drexel28
Quote:

Originally Posted by javeiro
Let X be a open subset on Rn.

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

Is (S,rho) a complete metric space?

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

Thanks,
JP

What do you think? I assume you know that $\displaystyle \mathcal{C}^2[X,\mathbb{R}]$ is a Banach space with $\displaystyle \|f\|_2=\sup_{x\in X}|f(x)|+\sup_{x\in X}|f'(x)|$. What do you think the deletion of the second term does?
• May 3rd 2010, 03:20 PM
javeiro
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 PM
Drexel28
Quote:

Originally Posted by javeiro
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...

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.