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Can a series converge neither pointwise nor uniformly, or are they the only two 'options' for convergence? Clearly uniformly $\implies$ pointwise, but can a series be neither?

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    $\begingroup$ A series that is not pointwise convergent is simply divergent. I assume that you are talking about series of functions; in that case, let $f_n(x) = x/n$. Then $\sum_{n = 1}^\infty f_n$ is obviously neither uniformly nor pointwise convergent. $\endgroup$ Apr 5, 2012 at 10:53
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    $\begingroup$ There is also a concept of convergence in the mean (integral of the square of the difference) - Apostol's Mathematical Analysis sect 9.13 discusses this and gives an example which converges in the mean to $f(0)=0$ on [0,1], but does not converge pointwise anywhere. Essentially the value is 1 on an interval of length $\frac 1{2^k}$ and zero everywhere else. The functions are created by marching the interval along from 0 to 1, and then going back to zero and halving the length. $\endgroup$ Apr 5, 2012 at 11:42

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The question seems to be answered by Johannes Kloos' comment. Since we don't like leaving questions unanswered, I'm reposting his comment as a CW-answer.

A series that is not pointwise convergent is simply divergent. I assume that you are talking about series of functions; in that case, let $f_n(x) = x/n$. Then $\sum_{n = 1}^\infty f_n(x)$ is obviously neither uniformly nor pointwise convergent.

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Note that pointwise as well as uniform convergence relate to an explicitly given or tacitly understood set $A$ where this convergence should occur.

A "formal" function series $\sum_{k=1}^\infty f_k(x)$ is pointwise convergent on a set $A$ if for each point $p\in A$ the constant series $\sum_{k=1}^\infty f_k(p)$ converges.

Therefore such a series is not pointwise convergent on $A$ if one can find at least one point $p\in A$ where the constant series $\sum_{k=1}^\infty f_k(p)$ diverges.

An example: The series $$\sum_{k=1}^\infty {e^{-k x^2}\over k}$$ is not pointwise convergent on ${\mathbb R}$, but is pointwise convergent on $\dot{\mathbb R}$.

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