On this answer, the function $f_n(x)=x^n$ in the interval $[0,1]$ is given as a pathologic example with pointwise convergence.
Can I say that this Cauchy sequence does not (pointwise) converge because the limit of the sequence is a function like this (not continuous):
without specifying any particular norm? I read that pointwise convergence, doesn't imply $d_\infty$ (uniform) convergence, and that uniform convergence implies pointwise convergence. But does lack of pointwise convergence negate uniform convergence?
Does this contradict in any way (or under certain norms) the fact that $C[a,b]$ with respect to $\Vert f \Vert_{\infty}$ is a Banach space? In other words, why is not an example of a Cauchy sequence that does not converge to some $f\in C[0,1]$?
Best Answer
The notion of a sequence being "Cauchy" is inherently a metric concept. When you talk about a Cauchy sequence, you need to be talking about convergence in a metric. Pointwise convergence does not correspond to any metric (this can be proven). If you use the $\|f\|_{\infty}$ metric, the sequence isn't Cauchy. For any finite $m$, there is some point $x_m$ sufficiently close to $1$ for which $x_{m}^{m} \approx 1$. However, if you take $n$ sufficiently large, $x_{m}^{n} \approx 0$, so that $\|x^{n} - x^{m}\|_{\infty} \approx 1$. Since the sequence isn't Cauchy, it won't converge in our metric.
As for the question: if the pointwise limit is not continuous, can the sequence converge in the $\|f\|_{\infty}$ metric? The answer here is no: if it did converge in $\|f\|_{\infty}$, the limit would be continuous (because our space is Banach, so complete) , and since the pointwise limit has to be the same as the uniform limit, we would arrive at a contradiction.