# Common Problems while showing Uniform Convergence of functions

All, I am having some trouble in checking whether a sequence of functions converges pointwise or uniformly. The problem is, sometimes, my intuition is right and sometimes its wrong. Finding the limit function is the most difficult task I would say.

I am able to understand some simple examples, such as a pointwise convergence of $f_{n}(x)= \frac{x}{n}$ over $\mathbb{R}$. But when given complex problems, I am having trouble in guessing the limit function. For example, I had trouble in solving this.

• Test the convergence of $f_{n}(x) = n \Bigl[ (x)^{1/n} - 1 \Bigr]$ over $X=[1,t]$ for $t>1$.

Given such functions, how can I improve my ability of guessing the limit function, and then proceeding to check whether the convergence is uniform or not. One method which I found extremely useful is the following theorem:

• $f_{n} \to f$ uniformly on $X$ iff $y_{n} =\sup \{|f_{n}(x)-f(x)| : x \in X \} \to 0$. But as I see with some examples, this result is not always successful.
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Is that a theorem? That is almost the definition. If I'm not sure, I try to draw some graphs for a few $n$ and then an Ansatz. – Jonas Teuwen Nov 18 '10 at 20:38
@Arturo:Whats the problem with my TeXing. You seem to have edited something. – anonymous Nov 18 '10 at 20:40
@Chandru1: Not your TeXing, your grammar. The first person singular pronoun is I, not i. – Arturo Magidin Nov 18 '10 at 20:47
@Arturo: You are so strict :x) – anonymous Nov 18 '10 at 20:48
@Chandru1: It's called a en.wikipedia.org/wiki/Pet_peeve – Arturo Magidin Nov 18 '10 at 20:49

Intuition is developed by doing examples and becoming familiar with these things, so in that respect you are heading in the right direction.

As far as "guessing the limit function", that's not what we want to do! We don't want to guess, we want to figure it out.

Certainly if there is going to be uniform convergence, then there has to be pointwise convergence. So you can begin by considering the limit of $f_n(x)$ for fixed $x$; if it does not exist, you're done: no convergence, pointwise or uniform. If it always exists, then the limit function must be the function $$f(x_0) = \lim_{n\to\infty}f_n(x_0)$$ for each $x_0$. I hope that by "guessing the function" you actually mean "recognize it as a function I may already know". This is not always possible in any case.

For the functions $f_n(x)=n\left((x)^{1/n}-1\right)$, you will first want to consider the convergence at a given point $a$ that lies in $[1,t]$. Does the limit exist for a fixed $a$? That is, does $$\lim_{n\to\infty} n\left(a^{1/n}-1\right)$$ exist? If so, you want to try to express it as a function of $a$. This is a problem in sequences (or limits). This is an $\infty\times 0$ indeterminate, so you can try using L'Hopital's Rule to find it: \begin{align*} \lim_{n\to\infty}n\left(a^{1/n}-1\right) &= \lim_{x\to\infty}\frac{a^{1/x}-1}{\frac{1}{x}} = \lim_{x\to\infty}\frac{e^{\ln(a)/x}-1}{\frac{1}{x}}\\ &= \lim_{x\to\infty}\frac{-\frac{1}{x^2}\ln(a)e^{\ln(a)/x}}{-\frac{1}{x^2}}\\ &= \lim_{x\to\infty}\ln(a)e^{\ln(a)/x} = \ln(a). \end{align*} So there is a pointwise limit, and $\lim\limits_{n\to\infty}f_n(a) = \ln(a)$.

This tells you that the function you are converging pointwise to is $f(x)=\ln(x)$. Now that you have a target function, you can check to see if the convergence is uniform in any of the many ways we have of checking of the convergence is uniform.

Added: Mind you, just like we have ways of determining that a sequence converges without actually determining what it converges to (for example, showing that the sequence is Cauchy), we also have ways of determining if a sequence of functions converges pointwise without actually determining to what. And we have ways of determining if a sequence of functions converges uniformly to something without having to first determine if it converges pointwise, or what it converges to. You could show that the sequence is "uniformly Cauchy" or "Cauchy in the sup norm": you define $$||f_n-f_m||_{\infty} = \sup\{ |f_n(x)-f_m(x)|\}.$$ We say the sequence $\{f_n\}$ is Cauchy in the sup norm if and only if for every $\epsilon\gt 0$ there exists $N\gt 0$ such that if $n,m\geq N$, then $||f_n-f_m||_{\infty}\lt \epsilon$. It is straightforward to show that if $\{f_n\}$ is Cauchy in this sense, then for each $a$ you have $\{f_n(a)\}$ is Cauchy (in the usual sense), so there must be pointwise convergence (to something); and you can also prove that if the sequence $\{f_n\}$ is Cauchy in the sup norm, then the convergence will be uniform (to whatever it is that it converges). This is to figuring it if $\{f_n\}$ converges uniformly to a given $f$ like showing a sequence $\{a_n\}$ is Cauchy is to figuring out if it converges to a given $L$.

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Very nice explanation. Exactly the one i was looking for. – anonymous Nov 18 '10 at 20:49
I think i have done that result. chandrumath.wordpress.com/2010/11/05/… – anonymous Nov 18 '10 at 21:02
Anyway, once again thanks for taking time to explain. – anonymous Nov 18 '10 at 21:03
@Chandru1: The points is: one can show the sequence is uniformly convergent without first determining what it coverges to (without looking for a "target candidate"). But if you want to recognize the limit as some known function, that's going to depend very much on your sequence of functions, and it will not always be possible; there are vastly more functions than there are "known" functions, so your odds are pretty slim for "random" sequences. – Arturo Magidin Nov 18 '10 at 21:05
Ok, let me go right now, and work some more exercises on uniform convergence and see whether i have improved or not. – anonymous Nov 18 '10 at 21:07