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Uniform Convergence question

  1. Jan 17, 2015 #1
    So I'm reading "An Introduction to Wavelet Analysis" by David F. Walnut and it's saying that the following sequence

    "[itex] (x^n)_{n\in \mathbb{N}} [/itex] converges uniformly to zero on [itex] [-\alpha, \alpha] [/itex] for all [itex]0 < \alpha < 1[/itex] but does not converge uniformly to zero on [itex](-1, 1) [/itex]"

    My problem is that isn't this interval [itex] [-\alpha, \alpha] [/itex] for all [itex]0 < \alpha < 1[/itex] and [itex] (-1,1) [/itex] the same thing? Am I missing some key analysis fact?

    I keep thinking of the example where the sequence functions given by [itex] f_n(x) = x + \dfrac{1}{n} [/itex] converges uniformly to [itex] f(x) = x [/itex] for all [itex]x \in \mathbb{R} [/itex] and [itex]\mathbb{R}[/itex] is clopen. Do intervals have to be closed and bounded for uniform convergence to work?
  2. jcsd
  3. Jan 18, 2015 #2


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    The answer to the second question has the trivial counter of a constant sequence or a linear
    map y=ax+b. For the first part, we need to see if

    there is an integer N so that, for any fixed ## \epsilon>0 ## , we have ## \alpha^N < \epsilon ## for all ##n>N## .

    ## \alpha^n < \epsilon \implies \alpha < \epsilon^{1/n}## But you run into trouble towards the endpoints, when ## \alpha## becomes indefinitely small.
  4. Jan 19, 2015 #3


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    [itex] [-\alpha, \alpha] [/itex] for all [itex]0 < \alpha < 1[/itex] is not one interval, but infinitely many, one for each ##\alpha##. If we choose one arbitrary such ##\alpha##, the convergence is uniform on ##[-\alpha,\alpha]##, but the convergence is not uniform on the union of all these intervals (which is ##(-1,1)##)
  5. Jan 19, 2015 #4
    That makes so much more sense Erland. I didn't look at it that way.
  6. Jan 20, 2015 #5


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    Sorry, did not make my point very clearly. As you go to the endpoints, say you have an n that works for , say, 0.999 , so that

    ##(0.999)^n < \epsilon ##. Then you can go further out into, say, ## 0.9999999 ## Then you will need a larger n. This is not

    quite a full proof, but notice that EDIT for fixed a , ## lim_n \rightarrow {\infty} (1- \frac{a}{n})^n =e^{-a} \rightarrow 1 \neq 0## , i.e., the closer your number 1-x (set x =a/n) gets to 1 , the harder it is to make ##(1-x)^n \rightarrow 0 ##. This is an obstacle to uniform convergence.

    Re your first question,
    It seems you were trying to argue that ## \cup_{0 < \alpha <1 } [-\alpha, \alpha] =(-1,1) ##
    Is that what you were stating?
    Last edited: Jan 20, 2015
  7. Jan 20, 2015 #6
    Yes, but I think Erland hit that point, which makes more sense. I totally feel you on your "not quite proof". Do you have a physical meaning for uniform convergence vs. pointwise convergence by any chance?
  8. Jan 20, 2015 #7


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    Well, uniforme convergence means that all functions in the sequence converge at the same rate, i.e., there is a single n so that all functions f_n in the family are within the same epsilon for this same n. Let me think if I can come up with something more physical.
  9. Jan 20, 2015 #8
    The picture is that there's a little tube around the limiting function of width 2 epsilon, such that all the functions stay within the tube for large n. Convergence can be non-uniform because you might have to go arbitrarily far into the sequence before you can get the functions all into the tube, even though they do get there eventually.
  10. Jan 20, 2015 #9
    I really like this example, but I was thinking in the sense of a more physics like interpretation if that makes any sense.
  11. Jan 20, 2015 #10


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    I am throwing this out there without knowing enough physics to make a good judgment of how good the analogy is:

    Say you have a countable collection of water molecules in a glass of water; the temperature of each molecule is a function. You put this water glass in a refrigerator, so that the temperature of each molecule is eventually zero. Given a fixed temperature t>0 , is it true, for any t, that there is a point in time after which all molecules are at a temperature t or lower? If yes, then you may say the temperature converges uniformly to 0 , and that it does not otherwise. I don't know if this is the type of example you are looking for and I don't know if it is realistic.
  12. Jan 20, 2015 #11
    The problem with a physical interpretation is that analysis uses the assumption of continuity. In physics (here I mean nature and not the models of it that we use), there's not necessarily any such thing as an arbitrarily small epsilon or even an infinite sequence of functions. This may have something to do with why a few mathematicians reject the idea of an infinite set altogether. Physically, it's not clear that any such thing exists. Mathematically, there's nothing stopping us from thinking that way, and it can be useful in modeling physics because things do get very small. So, the experimental analogues of these things would involve throwing away the tail ends of the sequences arbitrarily far down the sequence and rounding any epsilon small enough to not make any measurable difference down to 0. But if you do that, uniform convergence doesn't really have a meaning any more because it's just a question of whether all the functions end up in a tube close enough to the function not to make any measurable difference. What might matter to you is how fast the functions end up in the tube, so that you don't have to wait forever for the sequence to end up there. Speed of convergence. So, these are some of the issues that you are confronted with if you are doing numerical analysis, rather than analysis proper. If your sequence doesn't converge fast enough to compute the answer, physically speaking, it's not going to help you. You can use it to reason theoretically, but you can't use it to compute the answers and results.

    There are cases in which uniform convergence is relevant to physical problems, especially if you want to do it rigorously, but physically, it's hard to tell what's going on there physically. The issue is that continuous mathematics deals with whether things give you the right answer when extrapolated to infinitely fine distances and so on. That can be quite a different thing from sequences that are produced using some physical process. Practically speaking, you care about how fast things converge to the answer, rather than whether it's theoretically going to the answer.

    If you ignore this kind of thing, basically, all you have to do is give you functions a physical interpretation, like temperature. You can't look at individual molecules because they will have different speeds. Temperature is more a measure of total kinetic energy, so all the energies are getting lumped together. So, you'd just have to say the temperature at each point is less than epsilon from the limiting temperature distribution. This makes sense in the theory of heat transfer, according to the mathematical model, but physically, it's going to break down if you zoom in enough because temperature is really a macroscopic concept that doesn't make sense microscopically for individual molecules.
  13. Jan 20, 2015 #12
    I see, very cool, I mean I always felt the "speed of convergence thing" but that does not really stick to my head. It definitely will now. I definitely like the numerical analysis as opposed to analysis proper mention. That definitely cleared the air a bit. I don't know if I felt the water example WWGD because I couldn't see the idea of going out far enough into the sequence that we can throw it away, in that example. But thanks for the help everyone.
  14. Jan 20, 2015 #13
    There's one more point that I can make. While there's not much physical meaning, so far as I can see for a sequence of uniformly convergent functions, the idea of failure to converge uniformly makes a little more sense. The negation of uniform convergence is that there is some epsilon for which the functions will never all be within the tube, no matter how big n is. And that epsilon might be big enough to make a physical difference. So, no matter how far in the sequence you went, you'd always have more functions that will get out of the tube, if you go further down in the sequence. You run into issues of infinity here, with the sequence being infinite, but it is possible to have an infinite sequence of things in finite time, in theory. For example, the first function could appear at t = 1/2, then next at t = 3/4, the next at t = 7/8, etc. No matter how close to t = 1 you get, you'd still be out of the epsilon-tube eventually, so it would always be possible to measure a temperature that's not close enough to the limit by if you just wait a bit longer. Of course, it's not clear if 1/2^googloplex seconds has any physical meaning, so the tail end of the sequence still doesn't make complete physical sense if you get past the smallest measurable time interval.

    You'd end up getting sequences that are uniformly convergent for all practical purposes, since the sequence will always get within any measurably big tube, but they might not converge uniformly in theory, since they might fail to get uniformly inside an epsilon-tube that is too small to be measurable.
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