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Uniform convergence in [0,1]

  1. Jan 8, 2009 #1
    Let fn(x)=nx^2/1+nx ; x lies in [0,1]

    Is the convergence uniform?

    Since lim as n-->infinity of fn is x, I can see that fn(x) converges pointwise to f(x)= x
    But I get stuck when I try to show the convergence is uniform or not.
     
  2. jcsd
  3. Jan 8, 2009 #2

    Dick

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    Since you know the limit is f(x)=x, look at |fn(x)-f(x)|. What's the maximum of that function on the interval [0,1]?
     
  4. Jan 8, 2009 #3
    I think the maximum of |fn(x)-f(x)| on [0,1] is 1/1+n. So can I conclude that the convergence is uniform because there is an "N" that doesn't depend on x such that for all n> or eq. to N, 1/1+n gets arbitrarily small?
     
  5. Jan 8, 2009 #4

    Dick

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    Can you make that conclusion? If I give you an e>0 can you find an N such that 1/(1+n)<e for all n>N? Sure you can. If you're not sure you'd better figure out how pick a corresponding N.
     
  6. Jan 8, 2009 #5
    that N should be > than 1-e/e right?
     
  7. Jan 8, 2009 #6

    Dick

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    Sure. Use parentheses when you write something like (1-e)/e, ok? 1-e/e at first glance looks like 1-(e/e), which looks like 0.
     
  8. Jan 8, 2009 #7
    Right, thanks.
     
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