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Statement of problem: For f $\in B_1$ prove that there is a sequence $(g_n)$ in $C[0,1]$ s.t. $(g_n)$ converges to f pointwise on [0,1] and for each n we have $||g_n|| \le ||f||$. $B_1$ is the set of all bounded real valued functions on [0,1] which are a pointwise limit of continuous functions. Also ||f|| defined as $sup_{t\in [0,1]} |f(t)| $ i.e. the sup norm.

Context: Real Analysis by Carothers and Principles of Analysis by Rudin This was given as extra credit on a previous homework (already past due) and I would like to see how to tackle this problem to get ready for finals.

Work so far: We know that if f $\in B_1$ that it is continuous at on a dense set of points (Baire-Osgood Theorem). Also by definition we know there is a sequence of functions in $(f_n) \in C[0,1]$ converging to f $\in B_1$. My method of attack would be to prove by construction by taking this sequence and taking a subsequence for which the inequality applies but I am having a hard time justifying the viability of such a construction.

I'd appreciate any insight or help. Thank you.

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The Baire-Osgood theorem is somewhat more generous than this: It says that $f$ is continuous on a co-meagre set of points. This is clearly much stronger than merely being dense which is satisfied by for example the rationals in the unit interval. – kahen Mar 25 '15 at 23:22
up vote 1 down vote accepted

Let $\{f_n\}$ be a sequence in $C[0,1]$ pointwise convergent to $f$. We may assume that $\|f\|>0$. Then $g_n$ can be obtained by truncating $f_n$ as follows: if $|f_n(x)|< \|f\|$, let $g_n(x)=f_n(x)$; if $|f_n(x)|\ge \|f\|$, let $g_n(x)=\|f\|\frac{f_n(x)}{|f_n(x)|}$.

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Thanks for your reply. I see how it satisfies the inequality but how do you justify that $g_n$ converges pointwise? – jimmywho Nov 27 '12 at 5:12
@jimmywho: If $|f(x)|<\|f\|$, then when $n$ is large, $g_n(x)=f_n(x)$; if $|f(x)|=\|f\|$, then $\lim_{n\to\infty}|f_n(x)|=\|f\|$. Can you see the pointwise convergence now? – 23rd Nov 27 '12 at 5:31
Yes I see now, thank you. – jimmywho Nov 27 '12 at 5:35
@jimmywho: You are welcome! – 23rd Nov 27 '12 at 5:57

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