I am confused, the text am reading says that in function space $L$ with its respective norm (say for example the Euclidien $L^2$ norm), a Cauchy sequence $\textbf{x}_n \in l$ is s.t. $\|\textbf{x}_n-\textbf{x}_m\| < \epsilon$, for all $n,m > N$, for any $\epsilon > 0$.
But what if we have the sequence (similar to typewriter sequence).
$$
x_1 = I_{(0,1/2]} \quad x_2=I_{(1/2,1]} \quad x_3 = I_{(0,1/4]} \quad x_4=I_{(1/4,1/2]} \quad x_5 = I_{(1/2,3/4]}…
$$
In this case, given the norm, we have that the sequence above converges in the norm and should meet the definition of Cauchy (as per my text), but it does not converge pointwise… this is a common example in probability for a sequence that converges in the norm but does not converge almost everywhere (or pointwise)
Does this mean that there can be Cauchy sequences in a complete normed sequence space that do not converge pointwise?
Best Answer
The point here is that you are confusing what it means to converge in an abstract normed vector space. While it's true that in this case the vectors are themselves functions, it is not necessarily true that convergence in this space is equivalent to pointwise convergence. You can have a convergent sequence of functions which do NOT converge pointwise, depending on which norm you select.