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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.

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Thanks

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    $a = 0$, $b=1$?2012-06-01

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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.)

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    Thank you very much sir. I was confused with how to show $x(a) = x(b)$?2012-06-01
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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.
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    @JonasMeyer BTW $Y$ *is* a subspace.2012-06-01