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Product of Uniformly Continuous Functions

  1. Oct 30, 2005 #1
    "Let [tex]h[/tex] and [tex]g[/tex] be uniformly continuous on [tex]I\subset\mathbb{R}[/tex] and are both bounded. Show that [tex]hg[/tex] is uniformly continuous."

    h and g are uniformly continuous and bounded on I implies that h and g are continuous and bounded on I. This implies that hg (h times g) is continuous and bounded on I. Let f(x) = h(x)g(x).

    Assume that f is continuous on I but not uniformly, then there exists e>0 such that for all d>0, there exists x and y in I such that |x-y|<d and |f(y)-f(x)|>e. In particular, for every n>0, there exists x_n and y_n in I such that |y_n - x_n|<1/n and |f(y_n)-f(x_n)|>e. Since {x_n} is a sequence in I, by the Bolzano-Weierstrass theorem, there exists a subsequence x_n_p converging to z in I. As y_n_p = x_n_p + (y_n_p - x_n_p) for every n, and {y_n_p - x_n_p} converges to 0, {y_n_p} also converges to z. Since f is continuous at z, {f(y_n_p)-f(x_n_p)} converges to f(z)-f(z) = 0. This contradicts the assumption that |f(y_n_p)-f(x_n_p)|>e for all n. Hence, f must be uniformly continuous on I.

    Does anyone see any flaws in that argument? I did it without referring to the fact that h and g are both bounded.
  2. jcsd
  3. Oct 30, 2005 #2
    You can prove this "directly" from the definition. This will be a useful identity:

    |f(x)g(x) - f(y)g(y)| = |f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)|.

    The problem with your proof is this:

    You assume that I is bounded (and later you assume that it's closed too). Which you really can't.
    Last edited: Oct 30, 2005
  4. Oct 30, 2005 #3
    Darn it. I thought something was amiss. Thanks.
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