# State on classical space

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• Jun 22nd 2011, 10:58 PM
Mauritzvdworm
State on classical space
I am looking for an example of a state (positive linear functional of norm 1) on the space $\displaystyle C(\mathbb{T}^2)$ of continuous functions on the torus $\displaystyle \mathbb{T}^2$.

My first guess would be something like

$\displaystyle \omega(f)=\int^{2\pi}_{0}\int^{2\pi}_{0}f(x,y)dxdy$

with norm

$\displaystyle \|\omega\|=\sup_{f\in C(\mathbb{T}^2)}\left|\int^{2\pi}_{0}\int^{2\pi}_{ 0}f(x,y)dxdy\right|$

with $\displaystyle f\in C(\mathbb{T}^2)$. However, if we take the function $\displaystyle h$ to map any element of the torus identically to 1 the above definition would imply that

$\displaystyle \omega(h)=\int^{2\pi}_{0}\int^{2\pi}_{0}h(x,y)dxdy$

$\displaystyle =\int^{2\pi}_{0}\int^{2\pi}_{0}dxdy$

$\displaystyle =4\pi^2$

so this idea does not work. Any ideas?
• Jun 23rd 2011, 07:01 PM
ojones
Re: State on classical space
What text are you using?
• Jun 23rd 2011, 10:33 PM
Jose27
Re: State on classical space
Following your ideas, |w(f)| < 4(pi)^2 ||f|| where the last norm is the one in your space so what about v(f)=w(f)/4(pi)^2?
• Jun 23rd 2011, 11:36 PM
Mauritzvdworm
Re: State on classical space
I don't think that will work. It will be fine for that particular function, but we can just as well consider a function that maps any element of the torus to say 2, then the integral will have twice that value. I think I have an idea how to construct such a state, will post it if it works.
• Jun 24th 2011, 10:22 AM
Jose27
Re: State on classical space
I was assuming there was a typo in your definition of the norm of the functional since by linearity that quantity will never be bounded. So I assumed you meant $\displaystyle \| \omega \| = \sup_{\| f \| =1} |\omega (f)|$ and for this definition my example does work.