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Prove that a sequence of functions converges pointwise and uniformly.

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Homework Statement .
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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  • #2
Dick
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Homework Statement .
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.
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.
 
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  • #3
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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]
 
  • #4
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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]
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}##.
 
  • #5
jbunniii
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Note that:

[tex] \frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2}[/tex]
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.
 
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  • #6
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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]
 
  • #7
jbunniii
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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[/tex]
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.
 
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  • #8
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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.
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.
 
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Looks good!
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##.
 
  • #11
Dick
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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##.
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.
 
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  • #12
jbunniii
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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.
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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