# A complete Metric Space

• Apr 7th 2011, 04:28 AM
raed
A complete Metric Space
Dear Colleagues,

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 AM
Opalg
Quote:

Originally Posted by raed
Dear Colleagues,

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.

Where are you having difficulties with this? You need to take a sequence $\displaystyle (x_n)$ in $\displaystyle Y$ which converges in $\displaystyle X$. If $\displaystyle x_n\to x$ then you have to show that $\displaystyle x\in Y$.
• Apr 7th 2011, 06:03 AM
raed
Okay I understand that but how can I prove that x(a)=x(b) ?
• Apr 7th 2011, 06:29 AM
Opalg
Quote:

Originally Posted by raed
Okay I understand that but how can I prove that x(a)=x(b) ?

Well you know that $\displaystyle x_n\in Y$, and so $\displaystyle x_n(a) = x_n(b)$, for each n. Doesn't that help?
• Apr 7th 2011, 12:43 PM
Drexel28
Quote:

Originally Posted by raed
Dear Colleagues,

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.
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).