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
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 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.