Convergence of a sequence of functions

[radou]

Homework Statement

Let the sequence of functions fn : [0, 1] --> R be defined with fn(x) = x^n. Show that the sequence (fn(x)) converges for each x from [0, 1], but that it doesn't converge uniformly.

The Attempt at a Solution

Now, let x be from [0, 1>. Indeed, fn converges f(x) = 0, since for every ε > 0 one can find a positive integer N such that |0 - x^n| = |x^n|< ε. If x equals 1, the sequence converges to f(x) = 1 trivially.

Now, from the conclusion above and the definition of uniform convergence, it follows that the sequence cannot converge uniformly, since it converges to one limit for x in [0, 1>, and to another one for x = 1.

I hope I did this right, thanks in advance for any help.

 

[ystael]

You have the idea that leads to a correct argument, but not all of a correct argument. In particular, your second paragraph is missing the crucial point (you merely say "from ... the definition of uniform convergence").

It is true, as you say, that the pointwise limit $$f(x) = \lim_{n\to\infty} f_n(x)$$ is the function $$f(x) = 0$$ for $$x\in[0,1)$$, $$f(1) = 1$$. However, exactly how does this show that $$(f_n)$$ does not converge uniformly to $$f$$? If you say "by the definition of uniform convergence", explain how the definition applies in this instance.

 

[radou]

ystael, thanks for the reply.

Indeed, my argument was a bit fuzzy.

One says that a sequence of functions fn converges uniformly to the function f if, given any ε > 0, there exists a positive integer N such that d(fn(x) - f(x)) < ε for all n >= N and all x in X, where X is the domain of both fn and f, and d is the metric on Y.

What's confusing to me is that I have to "assume" the limit function. So, for f(x) = 0, and for x = 1, we can't find any integer such that |1^n| < ε, for any ε > 0. So, fn doesn't converge uniformly to f(x) = 0.

Here's another point - all the functions fn are continuous. Assume they converge to some function f. Then f must be continuous. But there doesn't seem to be any such continuous function (it always "breaks down" at x = 1), hence fn doesn't converge uniformly.

 

[ystael]

That last paragraph is a correct argument by itself: the uniform limit of continuous functions is continuous; the pointwise limit $$f$$ of $$(f_n)$$ is not a continuous function; therefore the convergence is not uniform.

(You need the observation that uniform convergence is strictly stronger than pointwise convergence, that is, if a sequence of functions converges uniformly then its uniform limit is its pointwise limit. Your second to last paragraph betrays some confusion on this point: $$(f_n)$$ cannot possibly converge uniformly to $$g(x) = 0$$, because it does not even converge pointwise to this function.)

You can also produce a correct argument by writing out the epsilonics in detail, a bit more carefully than you do in your second to last paragraph. That would go something like this: I claim the convergence of $$(f_n)$$ to its pointwise limit $$f(x) = 0$$ for $$x \in [0,1)$$, $$f(1) = 1$$, is not uniform. Let $$\epsilon = \textstyle\frac12$$. Then for any fixed natural number $$N$$, there is $$n > N$$ and $$x_0 \in [0,1)$$ so that $$f_n(x_0) > \textstyle\frac12$$ (on a homework set I was grading, I would expect a student who took this tack to compute explicitly $$n$$ and $$x_0$$ given $$N$$). Therefore there is no $$N$$ so that $$|f_n(x) - f(x)| < \textstyle\frac12$$ for all $$x\in[0,1], n > N$$, and $$(f_n)$$ does not converge uniformly to $$f$$.

Obviously the argument which uses the concept of continuity is easier! However, the explicit epsilonic argument proves the slightly stronger statement that the restrictions of $$f_n$$ to $$[0,1)$$ do not converge uniformly to zero on $$[0,1)$$, which you can't prove by complaining that the limit isn't continuous -- it is.

 

[radou]

Thanks a lot for your help, it's much more clear now.