May 2000#5

Problem Statement: For every n\in\mathbb{N} and x\in(0,1) define f_n(x)=\dfrac{1}{nx+1}. Find the function f(x), f:(0,1)\rightarrow\mathbb{R} that \{f_n(x)\} converges to pointwise. Is the converges uniform?

Proof: Fix x\in(0,1). Then \lim_{n\rightarrow\infty}\dfrac{1}{nx+1}=0 and so \{f_n(x)\} converges to 0 pointwise. 

Claim: This convergence is not uniform. 

If the convergence is uniform then it follows that for every \varepsilon>0 we may find an N\in\mathbb{N} such that |f_n(x)|<\varepsilon for every x\in (0,1) and for every n>N. Let \varepsilon=\dfrac{1}{3} and n>N. Since x=\dfrac{1}{n}\in (0,1) this implies that |f(n)(\dfrac{1}{n})|=|\dfrac{1}{n\dfrac{1}{n}+1}|=|\dfrac{1}{1+1}|=\dfrac{1}{2}<\dfrac{1}{3}=\varepsilon. This is clearly a contradiction and so we have found a counterexample such that the sequence breaks uniform continuity. Thus, \{f_n(x)\} converges to 0 pointwise but not uniformly.


Reflection: This one tripped me up a bit. I was able to get the piecewise convergence but I could not decide whether it converged uniformly or not. My intuition told me that it didn’t, but I couldn’t think of a counter example like above where it broke uniform convergence. While trying to find a counter example I saw that as x got closer to 0 the N\in\mathbb{N} needed to make f_n(x) converge got larger and larger. This made me think that we couldn’t find a universal N that would work for all x\in(0,1). After seeing the above counter example it makes a lot more sense. I think what I need to keep in mind when trying to find a counter example for these kind of things is that I don’t necessarily want to find a specific x value where it breaks, but that I should tailor my s value to what I know, i.e. the N.

This entry was posted in Analysis, Math, Sequence of Functions, Uniform Convergence. Bookmark the permalink.

One Response to May 2000#5

  1. Allan Boone says:

    The way that I identified immediately, that the convergence would not be uniform, is that I thought about extending the function to [0,1].

    But when x=0, fn(x) = 1. And so you would have a sequence of continuous functions converging to something discontinuous on a compact space! So the convergence couldn’t have been uniform.

    Also, as you said, you should always look for a breaking point that depends on n. No specific point could work due to pointwise convergence.

    Alternatively, using a continuity argument, you could say that fn(0)=1 and fn(1) is less than or equal to 1/2 for any n. So by the intermediate value theorem, there exists x in (0,1) s.t. fn(x)=.75. And since such an x exists for any n, convergence is not uniform. (The advantage of this is that you don’t even need to find any values of x)

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