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Questions on uniform convergence

  1. Nov 22, 2007 #1
    I have several questions

    1. Find an example of two sequences of functions, f_n and g_n such that they both converge uniformly to f, g on some set E but such that f_n * g_n does not converge uniformly on E.

    Let f_n = x^2 for all n. and g_n=sinx/(xn). Since sinx/x is bounded by 1 it is easy to show that g_n converges uniformly to the zero function. however, the product is (xsinx)/n. This new function converges pointwise to the zero function. But, for any fixed n we can make (xsinx)/n as large as we please. So f_n * g_n is not uniformly convergent to the zero function on the real line. But f_n and g_n DO converge uniformly on the real line.

    Is this correct?

    2. consider f(x)=infinite sum from n=1 to infinity of {1/(1+(n^2)x)}

    By the comparison test it is easy to show that the series converges absolutely for any positive nonzero x. For x=-1/(n^2) for any n, it's clear that f(x) doesn't converge. Otherwise the comparison test can be used to show that f(x) converges absolutely.

    On what interval(s) does the series converge to f uniformly?

    I put E= [c,infinity) with c>0. Then for all x in E we see that (n^2)x+1>(n^2)c+1>(n^2)c so that 1/((n^2)c)>1/((n^2)c+1)>1/((n^2)x+1). By convergence of p-series (p=2) and the wieirstrass m-test we see that the series converges to f uniformly on E. I take it that although c was arbitrarily chosen it does NOT follow that the series converges uniformly on (0,infinity)? This because f(x) is unbounded on (0,infinity) (b/c it blows up as it approaches 0) and the partial sums of the series are bounded on the nonzero positive reals. Since uniform convergence on the nonzero positive reals of a sequence (of partial sums of functions) of bounded functions implies a bounded limit function on the nonzero positive reals which is not the case we have here. So the convergence cannot be uniform on all of the positive numbers.

    is this correct?

    Please DO NOT tell me any answers, i ONLY want to know if the arguments I have given are correct. Thanks for your time.
  2. jcsd
  3. Nov 23, 2007 #2
    continuing on the question: which intervals does the series converge to f uniformly?

    fix c<-1 and then the partial sums are defined for all n on E=(-infinity, c]. A similar argument as above (showing the same thing for [c,infinity) except with c>0) shows that the series converges uniformly to f on E. This time however we use 1/(|c|n^2)>|1/(1+xn^2| for all x in E when we apply the wieirstrass M-test.

    We now consider any interval with -1/(p^2) for some natural number p, as an endpoint. Let s_n (x) denote the nth partial sum of functions, we see that for all n>p the s_n (x) grows without bound near the endpoint -1/(p^2). Hmm, it seems that the convergence is still uniform on such an interval as long as f(x) is defined on the interval (ie no point -1/(n^2) for some positive integer n is CONTAINED in the interval).

    So basically s_n converges to f uniformly on E if E is a interval that doesn't have 0 as an endpoint nor does it contain any any point x=-1/(n^2) for some natural number n.

    from this we gather that f is continuous wherever the series converges. also that f cannot be bounded.

  4. Nov 23, 2007 #3


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    I think '1' is correct, although it wouldn't hurt to state formally why f*g isn't u.c.

    I haven't looked at '2' yet.
  5. Nov 23, 2007 #4
    1 is wrong. You claimed that g_n = sin(x)/(nx) is U.C. on the real line. But g_n cannot even be defined continuously at x = 0, and it's not bounded there, so it's easy to see g_n does not converge uniformly on (0,e] or [-e,0). But #2 in Chapter 7 only asks "on some set E", so you can modify your argument correctly by taking E = [1,infty).

    2 is correct. f is not defined as is for x = -1/k^2, for any positive integer k, and the sum of f_n(0) diverges. |f_n(x)| <= |f_n(c)| on [c,infinity) so f is U.C. on [c,infty) for any c > 0. It's not U.C. on (0,e]. Nonetheless, f is continuous on [c,infty) for c > 0, thus f is continuous on (0,infty). But clearly not bounded on (0,c], because by choosing x small enough you can make the first N denominators (1+n^2x) <= 1 + e, so you'll add 1/(1+e) N times, for arbitrary e > 0, where the only dependence is x <= e/N^2.
  6. Nov 23, 2007 #5


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    Thank you for correcting my oversight.
  7. Nov 23, 2007 #6
    i thought the limit of sinx/x equals 1? times by 1/n for any fixed n and we see that g_n is never greater than 1.

    am i missing something?
  8. Nov 23, 2007 #7
    Crap, my mistake. I had that sin(1/x)/x example in my head. Yes, sin(x)/x -> 1. So correction, your problem #1 is right!
  9. Nov 27, 2007 #8


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    Thank you for the recorrection.
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