Space $Y\subset C[0,1]$ consisting of all $x \in C[a,b]$ such that $x(a) = x(b)$ is complete.

Question

This is problem from kreyszig functional analysis: I have to show that space $Y\subset C[a,b]$ consisting of all $x \in C[a,b]$ such that $x(a) = x(b)$ is complete.

I am struggling with this problem for over an hour. Even I am not able to figure out such space.I need help to solve this problem.

Edited:

Thanks

score 5 · Accepted Answer · answered Jun 01 '12 at 05:30

I hope it's ok to post a more detailed version, maybe OP (or other users) will benefit from it.

$\newcommand{\norm}[1]{\lVert{#1}\rVert}\newcommand{\abs}[1]{|{#1}|}$I suppose you work with sup norm $\norm{x}=\sup_{t\in[a,b]}\abs{x(t)}$, since this is the usual meaning of $C(X)$.

We know that $C[a,b]$ is complete (it is a Banach space). Recall that a subspace of a complete metric space is complete if and only if it is closed. (A proof of this result can be found at proofwiki.) So it suffices to show that $Y$ is closed in $C[0,1]$.

Now I think that it is quite clear that if a sequence of functions $(x_n)$ converges to $x$ uniformly and $x_n(a)=x_n(b)$, then $x(a)=x(b)$. (Even pointwise convergence would be sufficient for this last argument.)

Thank you very much sir. I was confused with how to show $x(a) = x(b)$? — Srijan, Jun 01 '12 at 05:37

score 3 · Answer 2 · answered Jun 01 '12 at 05:16

3

What you probably already know: $C[0,1]$ is complete.

What you can show if you haven't yet:

$Y$ is a closed subspace of $C[0,1]$.
A closed subspace of a complete space is complete.

answered Jun 01 '12 at 05:16

Jonas Meyer

53,602

How to show $Y$ is closed in $C[0,1]$? I am sorry,I have just started reading metric space so my question may sound elementary. – Srijan Jun 01 '12 at 05:26
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@srijan Whenever $x_n\in Y$ converges to $x\in C[0,1]$, you must prove $x\in Y$. – AD - Stop Putin - Jun 01 '12 at 05:29
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@srijan: Are you familiar with the fact that if $A$ is a subset of a metric space $X$, then $A$ is closed if and only if for each sequence $(a_n)$ in $A$ that converges to some $x\in X$, $x$ is also in $A$? – Jonas Meyer Jun 01 '12 at 05:29
@JonasMeyer yes I am familiar with this fact. – Srijan Jun 01 '12 at 05:30
@srijan: Good; I recommend proving that $Y$ is closed using this sequence method, as mentioned in AD.'s first comment and elaborated by Martin's answer. – Jonas Meyer Jun 01 '12 at 05:34
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@JonasMeyer Thank you very much sir. I got valuable information form here. – Srijan Jun 01 '12 at 05:36
@JonasMeyer BTW $Y$ is a subspace. – AD - Stop Putin - Jun 01 '12 at 05:36

Space $Y\subset C[0,1]$ consisting of all $x \in C[a,b]$ such that $x(a) = x(b)$ is complete.

2 Answers2