Does the Sequence Converge Uniformly Across Different Intervals?

[twoflower]

Hi,

next one I've little problems with:

$$f_{n} = \frac{2nx}{1+n^{2}x^{2}}$$

$$\mbox{a) } x \in [0, 1]$$

$$\mbox{b) } x \in (1, \infty)$$

First the pointwise convergence:

$$\lim_{n \rightarrow \infty} \frac{2nx}{1+n^{2}x^{2}} = 0$$

I computed the derivative of my function to know where to get supremum and I got that maximum is in $x=\frac{1}{n}$.

I used it to get supremum on the first interval and I got

$$\sup \left\{\left|\frac{2nx}{1+n^{2}x^{2}}\right|, x \in [0,1]\right\} = 1 \Rightarrow f_{n} \mbox{ doesn't converge uniformly on } [0,1]$$

On the second interval $(1, \infty)$ I cannot use the point $x = \frac{1}{n}$ and since the function is decreasing on this interval I will take the leftmost point to compute the supremum. It will be

$$\sup \left\{\left|\frac{2nx}{1+n^{2}x^{2}}\right|, x \in (1,\infty)\right\} \leq \left|\frac{2n}{1+n^2}\right| \longrightarrow 0 \mbox{ as n goes to \infty}$$

Thus

$$f_{n} \rightrightarrows 0 \mbox{ on (1, \infty)}$$

According to "official" results this is not complete result. Would you help me please to finish it?

Thank you.

 

[Galileo]

The only error I see is that the supremum on [0,1] is not 2, but 1.
All else looks right to me. The function converges pointwise to the zero function and. It doesn't converge uniformly on [0,1], but it does on the interval $[\delta,\infty)$ for any $\delta>0$.

 

[twoflower]

The only error I see is that the supremum on [0,1] is not 2, but 1.

You're right, my mistake.

...but it does on the interval $[\delta,\infty)$ for any $\delta>0$.

That's what I was asking for - how can you see that? I can't see it so far - I proved it doesn't converge uniformly on some interval ($[0,1]$) but I can't see the possibility it could still converge uniformly on its part.

Btw, would $(0, \infty)$ be the same as you wrote? (I mean just not using delta here).

 

[quasar987]

I didn't say anything.

 

[Galileo]

You're right, my mistake.


That's what I was asking for - how can you see that? I can't see it so far - I proved it doesn't converge uniformly on some interval ($[0,1]$) but I can't see the possibility it could still converge uniformly on its part.

Btw, would $(0, \infty)$ be the same as you wrote? (I mean just not using delta here).

You already showed that the function $f_n$ is monotonically decreasing after the point 1/n, so this maximum shifts to the left (if you visualize the graph) as n increases. That means if you have the interval $[\delta, \infty)$ the point at which $f_n$ has a maximum will eventually leave this interval (provided $\delta>0$). So the function will be monotonically decreasing in this interval, so you can take the supremum to be the left endpoint and show that it goes to zero, just as before.

It may seem surprising at first, but the although the sequence converges uniformly on $[\delta, \infty)$ it does NOT converge uniformly on $(0,\infty)$. That is because the spike at 1/n will always lie in the interval (0,->) for any n.

 

[twoflower]

Thank you very much Galileo, it helped me a lot.

Then, would this be correct formal finish of the problem?

$$\delta > 0,\mbox{ then \\} \sup \left\{\left|\frac{2nx}{1+n^2x^2}\right|, x \in [\delta, \infty)\right\} = \frac{2\delta n}{1+\delta^2n^2} \longrightarrow 0 \mbox{ as n goes to \infty}$$

I already understood this particular problem, but I'm afraid it won't always be that obvious so that I can guess on which interval it could still converge uniformly. Is there any common approach how to find interval, on which it could still converge uniformly when I prove it doesn't on some others?

 

[Galileo]

Almost right. Ofcourse I wasn't totally precise on this point either, but the supremum of $f_n$ on the interval $[\delta,\infty)$ is not always the left endpoint. What matters is that from a given moment on the spike will leave the interval. I.e. there exists an $N>0$ such that the supremum is the value of the function at the left endpoint (as in your expression) for all n>N (in this case $n>1/\delta$. So it is still true that
$$\lim_{n\to \infty} \sup_{x\in D}|f_n(x)-f(x)|=0$$.

Where in this case ofcourse $D=[\delta,\infty)$, $f(x)=0$ (the limit function).

Usually the reason why a sequence of functions doesn't converge uniformly is because of the limiting behaviour at a specific point (in this case at x=0).
For a sequence of $\arctan(nx)$ functions, the problem also lies at x=0 and this also converges uniformly on $[\delta, \infty)$ for any $\delta>0$. I think practice is the best way to get a feel for these things. I personally always visualize the functions and see how the graphs change as n increases, because I have a very visual way of thinking.
Also see the analogy between (continuity - uniform continuity) and (convergence - uniforn convergence).
they are expressed in epsilons and delta's (or more commonly N's for the latter), but with continuity, you first specify a point x in the domain of f, then you any $\epsilon>0$ and then find your $\delta>0$ to match, so your $\delta$ may depend on the point and the chosen epsilon $\delta=\delta(x,\epsilon)$, while for uniform continuity you can find a $\delta$ which works (uniformly) for all points x in the region of interest and depends only on $\epsilon$.
Likewise, for pointwise convergence your take the pointwise limit ($f_n(x)\to f(x)$ for each point $x$), so the $N$ in your limit is chosen after specifying the point and the $\epsilon$, while for uniform convergence you can find a $N(\epsilon)$ which works (uniformly) or ALL points in the region of interest.

Anyway, when you've found the supremum of |fn(x)-f(x)| and you find that it depends on n, you should check whether this point will remain in the region of interest.