Prove that a sequence of functions converges pointwise and uniformly.

[mahler1]

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.

 

[Dick]

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.

 

[dirk_mec1]

Note that:

\frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2} \leq \frac{2}{N}

Now choose

N= \frac{2}{ \epsilon}

 

[mahler1]

Note that:

\frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2} \leq \frac{2}{N}

Now choose

N= \frac{2}{ \epsilon}

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 $\frac{ 2}{1+(n+1)^2}$.

 

[jbunniii]

Note that:

\frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2}

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.

 

[dirk_mec1]

Yes, you're right then I choose the easy way:

The entire fraction is smaller than 2:\frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2 x^2} \leq 2n \leq 2N

Choose N = \frac{\epsilon}{2}

 

[jbunniii]

Yes, you're right then I choose the easy way:

The entire fraction is smaller than 2:


\frac{(1+x^2)x}{1+(n+1)^2x^2} \leq \frac{ 2}{1+(n+1)^2 x^2} \leq 2n

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.

 

[mahler1]

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.

 

[jbunniii]

Looks good!

 

[mahler1]

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 $0$.

 

[Dick]

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.

 

[jbunniii]

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.

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