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Prove that a sequence of functions converges pointwise and uniformly.

  1. Nov 11, 2013 #1
    The problem statement, all variables and given/known data.
    Given ##f_n=\frac {x} {1+x^2}-\frac {(1+x^2)x} {1+(n+1)^2x^2}## , prove that ##\{f_n\}_{n \in \mathbb N}## converges pointwise and uniformly to a continuous function on the interval ##[0,1]##

    The attempt at a solution.

    It's easy to prove that this sequence tends to the function ##f(x)=\frac {x} {1+x^2}## pointwise, I just calculated the limit for a fixed ##x## in the interval. Now, I am trying to prove by hand (without using theorems of sequences of functions defined on compact sets or anything of the sort) that the sequence tends to ##f## uniformly. So, to prove this I have to prove that for a given ##ε>0##, there is some ##N \in \mathbb N## such that for all ##n≥N## ##|f-f_n|<ε##.##|f-f_n|=|\frac {(1+x^2)x} {1+(n+1)^2x^2}|=\frac {(1+x^2)x} {1+(n+1)^2x^2}≤\frac {(1+x^2)x} {(n+1)^2x^2}=\frac {(1+x^2)} {(n+1)^2x}##. So, if I could choose ##N## such that for every ##x \in [0,1]##, ##\frac {(1+x^2)} {x}≤ε(n+1)^2##, I would be done, the problem is I don't know which ##N## would work.
     
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  3. Nov 11, 2013 #2

    Dick

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    No value of N will work. ##\frac {(1+x^2)} {x}## goes to infinity as x goes to zero. You need to get a better estimate. It should depend only on n not on x.
     
    Last edited: Nov 12, 2013
  4. Nov 12, 2013 #3
    Note that:

    [tex] \frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2} \leq \frac{2}{N} [/tex]

    Now choose

    [tex]N= \frac{2}{ \epsilon} [/tex]
     
  5. Nov 12, 2013 #4
    Yes, I've already try that, but are you sure that estimate is correct? I mean, in the denominator, when ##x \to 0##, the denominator tends to ##1##, so the whole expression is bigger than ##\frac{ 2}{1+(n+1)^2}##.
     
  6. Nov 12, 2013 #5

    jbunniii

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    This inequality is incorrect, as you can see by plugging some example values for ##x## and ##n##.

    To get a bound that works, try the following trick. The denominator is not quite a perfect square, but if we insert a term we can make it so:
    $$1 - 2(n+1)x + (n+1)^2 x^2 = (1 - (n+1)x)^2$$
    And of course since this expression is a square, it must be nonnegative:
    $$1 - 2(n+1)x + (n+1)^2 x^2 \geq 0$$
    Try using this to get a bound for your expression.
     
  7. Nov 12, 2013 #6
    Yes, you're right then I choose the easy way:

    The entire fraction is smaller than 2:


    [tex]\frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2 x^2} \leq 2n \leq 2N [/tex]

    Choose [tex] N = \frac{\epsilon}{2} [/tex]
     
  8. Nov 12, 2013 #7

    jbunniii

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    That's not what I get. Indeed, an upper bound of ##2n## wouldn't help at all because that bound grows as ##n## increases. You need a bound that converges to zero as ##n\rightarrow \infty##.

    Take another look at the inequality I suggested earlier:
    $$1 - 2(n+1)x + (n+1)^2 x^2 \geq 0$$
    This is equivalent to
    $$1 + (n+1)^2 x^2 \geq 2(n+1)x$$
    Since both sides are positive when ##x > 0##, we may rearrange this to get
    $$\frac{1}{1 + (n+1)^2 x^2} \leq \frac{1}{2(n+1)x}$$
    So that gives you an inequality involving the denominator. How about the numerator? Also note that you'll have to handle the ##x=0## case separately, but that should be easy enough.
     
  9. Nov 12, 2013 #8
    Thanks very much, that inequality works well: ##x \in [0,1] \implies x≤1##, so the numerator ##(x^2+1)x≤(1^2+1)x=2x##. Putting all those inequalities together, we get ##\frac {(x^2+1)x} {1+(n+1)^2x^2}≤\frac {2x} {2(n+1)x}=\frac {1} {n+1}##. Then, given ##ε>0##, if ##\frac {1} {n+1}<ε##, that means, if we choose ##n## such that ##n>\frac {1} {ε}-1##, we have that ##|f_n-f|<ε##. It follows that ##f_n \to f## uniformly.
     
  10. Nov 12, 2013 #9

    jbunniii

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    Looks good!
     
  11. Nov 12, 2013 #10
    I forgot to separate the case ##x=0##, but that one is pretty easy since ##f_n=0##, every natural number will work for ##0##.
     
  12. Nov 12, 2013 #11

    Dick

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    If you want a little less clever solution than jbunniii's but maybe more general, you can try to just use calculus to find the max of ##\frac{(1+x^2)x}{1+(n+1)^2x^2}##. Differentiate with respect to x and set that equal to zero. In this case, that turns out to give you something too hard to solve. So try and simplify it a little by using (1+x^2)<=2. So try ##\frac{2x}{1+(n+1)^2x^2}##. That one you can just mechanically solve. Turns out to give you the same upper bound of 1/(n+1). That's the way I would have tackled it. I'm not too good at clever tricks.
     
  13. Nov 12, 2013 #12

    jbunniii

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    I was going to suggest the same thing, but I got a nasty expression when I tried it because your trick didn't occur to me. Fortunately, spending a few extra minutes to look for a different trick paid off in this case. Sometimes laziness is a virtue. :tongue:

    Also, a trick that you can apply more than once is no longer a trick, it's a technique:

    https://www.physicsforums.com/showthread.php?t=720807
     
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