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Thread: Sequence of differentiable functions, non-differentiable limit

  1. #1
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    Sequence of differentiable functions, non-differentiable limit

    I am trying to find a sequence of differentiable functions which converge uniformly on [-1,1] but such that the uniform limit is NOT differentiable on (-1,1).

    I figure I'd like to make something converge to f(x)=|x| (which is not differentiable at 0, so not differentiable on (-1,1) ). I have an idea, but haven't been able to hack through the details. The idea is to define each member of the sequence such that outside of [-1,1] each function is identically |x|, but on (-1,1) I want to mutate x^2 somehow so that the sequence converges uniformly to |x|...and them's the details I haven't hacked through.
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  2. #2
    Senior Member JaneBennet's Avatar
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    Try $\displaystyle f_n(x)=|x|^{\frac{n+1}n}.$
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  3. #3
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    Thank you very much! That is so much nicer than my attempt. I'm going to apply that sequence for sure, but still I am curious to know if my attempt could be made to work.

    Hopefully I'll have time to give it some mind and if I do I'll put up my efforts (but I'm a few weeks behind in coursework so for now I'll just move on).
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  4. #4
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    Here is another possiblity. For all n, $\displaystyle f_n(x)= 0$ for x< 0, $\displaystyle f_n(x)= nx$ if $\displaystyle 0\le x\le 1/n$, f(x)= 1 for x> 1/n, is continuous for all x (but NOT differentiable at x= 1/n). The sequence converges to f(x)= 0 for x< 0, 1 for $\displaystyle x\ge 0$ which is not continuous at x= 0.

    So integrate that: $\displaystyle F_n= 0$ for x< 0, $\displaystyle F_n(x)= \frac{n}{2}x^2$ for $\displaystyle 0\le x\le 1/n$, $\displaystyle F_n(x)= x$ for x> 1/n is differentiable for all n (in particular, its derivative at 0 is 0, and at 1/n is 1) but its limit, F(x)= 0 for x< 0, x for $\displaystyle x\ge 0$ which is not differentiable at x= 0.
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