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Question: Prove that $(B, \|-\|_{\infty})$ is complete. B the set of bounded real valued functions on [0,1] which are pointwise limit of continuous functions on [0,1].

Context: Old exam problem I'm using to study. Real Analysis by Carothers.

I've attempted to avoid a direct proof from the definition using Cauchy sequences by appealing to the fact that ($B_{\infty}, \|-\|_{\infty}$) is a normed space where $B_{\infty}$ is the set of bounded real valued functions on $\mathbb{R}$ that are pointwise limits of continuous functions. So $B \in B_{\infty}$ and by a Theorem a normed spacve is complete if and only if every absolutely summable series in $B$ is summable.

I'm having trouble making a solid case for proving the condition in the last part and also how to say that $B$ is indeed a normed space.

Thank you in advance.

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    $\begingroup$ If I were you, I'd go with the definition of Cauchy sequence. Maybe you can use the fact that $\mathbb{R}$ is complete. $\endgroup$ Nov 27, 2012 at 1:34
  • $\begingroup$ I understand the argument that you are suggesting. I was thinking that this method would be easier since if $\sum ||f_n||_{\infty} < \infty$ and since $||f_n|| \le M_n$ for $f_n \in B$ this implies $\sum M_n$ converges but this is an upper bound for $\sum f_n(t)$ for t $\in [0,1]$. $\endgroup$
    – aawaldrop
    Nov 27, 2012 at 1:59

2 Answers 2

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(I will use $\mathcal{B}$ to denote the vector space of bounded functions that are pointwise limits of continuous functions.) Suppose $B_n$ is a Cauchy sequence in $\mathcal{B}$. Passing to a subsequence, we may assume that $\|B_{n+1} - B_n\| < \frac{1}{2^{n+1}}$. It follows that if we set $C_n = B_{n+1} - B_n$, then $\|C_n\| < \frac{1}{2^{n+1}}$ and that $C_n$ is the pointwise limit of continuous functions.

Now, since we have $|C_n(x)| < \frac{1}{2^{n+1}}$ holds for each $x$, the series $\sum_{i=1}^\infty C_n(x)$ converges, say to the value $C(x)$. This defines a function $C$, which is bounded. The tricky part is to show that there is a sequence of continuous functions converging to $C$. For a proof of this, see lemma 2 of : http://www.whitman.edu/mathematics/SeniorProjectArchive/2007/huh.pdf

(The $C_n$ are Baire one functions, bounded uniformly by $M_n = \frac{1}{2^{n+1}}$. )

To finish, simply note that $B_1(x) + \sum_{i=1}^n C_i(x) = B_{n+1}(x)$, and so $B_n$ converges in the norm of $\mathcal{B}$ to $B_1 + C$, and this is a Baire one function.

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This might be better as a comment than as an answer, but Rudin did prove this in the end of Chapter 3 in his book Real and Complex Analysis.

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