Dear Colleagues,

Could you please help me in solving the following problem:

Show that the subspace $\displaystyle Y\subseteq[a,b]$ consisting of all $\displaystyle x\in C[a,b]$ such that $\displaystyle x(a)=x(b)$ is complete.

Best Regards.

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- Apr 7th 2011, 04:28 AMraedA complete Metric Space
Dear Colleagues,

Could you please help me in solving the following problem:

Show that the subspace $\displaystyle Y\subseteq[a,b]$ consisting of all $\displaystyle x\in C[a,b]$ such that $\displaystyle x(a)=x(b)$ is complete.

Best Regards. - Apr 7th 2011, 05:29 AMOpalg
- Apr 7th 2011, 06:03 AMraed
Okay I understand that but how can I prove that x(a)=x(b) ?

- Apr 7th 2011, 06:29 AMOpalg
- Apr 7th 2011, 12:43 PMDrexel28
Perhaps simpler than pure definition (though not much) is to notice that the linear functional $\displaystyle \varphi:C[a,b]\to F:f\mapsto f(a)-f(b)$ is continuous and thus $\displaystyle \ker\varphi$ (our set) is a closed subspace of the complete space $\displaystyle C[a,b]$ and thus complete (this is easier assuming that you know (which I've stated to you before and which is a very simple fact) that the evaluation functionals $\displaystyle \varphi_z:C[a,b]\to F:f\mapsto f(z)$ is continuous and $\displaystyle \varphi=\varphi_a-\varphi_b$ thus continuous).