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Another nontrivial (trick) question in Chapter 7

  1. Nov 26, 2007 #1
    Problem 13 in Chapter 7 of baby Rudin states:

    Assume [tex]\{f_n\}[/tex] is a sequence of monotonically increasing functions on R with [tex]0 \leq f_n(x) \leq 1[/tex] for all x and all n.
    (a) Prove that there is a function f and a sequence [tex]\{n_k\}[/tex] such that [tex]f(x) = \lim_{k \rightarrow \infty}f_{n_k}(x)[/tex] for every x in R.
    (b) If, moreover, f is continuous, prove that [tex]f_{n_k} \rightarrow f[/tex] uniformly on R.

    Anyone else notice a problem with (b)?
  2. jcsd
  3. Nov 26, 2007 #2
    I have boldy decided to post what I think is a misleading error in baby Rudin. And it's slightly less trivial than the error in Definition 1.5 (ii). I urge anyone to correct me if I'm wrong.

    Nonetheless, I can even strengthen the hypothesis with [tex]f_1 \leq f_2 \leq ...[/tex] and get a counterexample to (b). Set [tex]f_n(x) = 0, x \leq -n, f_n(x) = 1, x > -n[/tex]. I.e. [tex]f_n(x) = I(x+n)[/tex] is a sequence of step functions. In particular, step functions are monotonic increasing, so they satisfy the hypothesis. This is a pointwise convergent sequence, every subsequence converges to the same thing, [tex]f(x) = 1[/tex]. f is continuous, but the convergence is not uniform.

    However, it converges uniformly on compact subsets. In summary, I think that's a typo. (b) should read [tex]f_{n_k} \rightarrow f[/tex] uniformly on compact subsets of R.

    P.S. You can also construct a sequence of strictly increasing functions that will yield the same counterexample. I think you can modify [tex]f_n(x) = tan^{-1}(x+n)[/tex] appropriately.
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