Why doesn't the lack of pointwise convergence of $f_n(x)=x^n$ contradict the fact that $C[0,1]$ is Banach?

Question

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 $f\in C[0,1]$?

Answer 1

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.

Answer 2

After some research on the topic, I think that the answer is no (as canonically answered by Elchanan), and the critical bit is his sentence, "Pointwise convergence does not correspond to any metric (this can be proven)."

But there remains the meta-questions of what is happening with that "pathologic" function, and what words would give the uninitiated some sense of direction.

The problem with the sequence $\{x^n\}_{n=1}^\infty$ in the interval $[0,1]$ is that it is pointwise convergent, but not uniformly convergent.

The sequence $f_n \to f$ pointwise on $[0,1]$ if $f_n(x) \to f(x)$ as $n \to \infty$ for every $x \in [0,1].$ There is no $\varepsilon!$

The limit of a pointwise convergent sequence of continuous functions does not have to be continuous, and does not generally preserve boundedness. In this case:

$$\lim_{n\to\infty} f_n(x) = f(x) = \begin{cases} 0 & (0\leq x\lt 1) \\ 1 & (x=1) \end{cases}$$

In uniform convergence, the uniform norm is introduced. Uniform convergence implies pointwise convergence, but not the other way around.

A sequence of functions converges uniformly if for every $\varepsilon>0$ there is an $N\in \mathbb N$ such that for all $n≥N$ and all $x \in [0,1],$ $d(f_n(x),f(x))<\varepsilon,$ where $d(f, g) = \sup_{x\in [0,1]} |f_n(x) - f(x)|.$

The space $C[0,1]$ with the infinity metric is complete (Banach), but the sequence $x^n$ is not Cauchy.

The link to Cauchy sequences is that $$\small \text{uniform convergence}\iff \text{Cauchy under the }\Vert \cdot \Vert_\infty \text{ uniform or infinity norm}. $$

The lack of uniform convergence is explained in Elchanan's answer, as well as here by contradiction:

If $f_n(x)$ converges uniformly, then the limit function must be $f(x)=0$ for $x∈[0,1)$ and $f(1)=1.$ Uniform convergence implies that for any $ϵ>0$ there is an $Nϵ \mathbb N$ such that $|x^n−f(x)|<ϵ$ for all $n≥ \mathbb N$ and all $x∈[0,1].$ Assuming this is indeed true we may choose $ϵ,$ in particular, we can choose $ϵ=1/2.$ Then there is an $N∈ \mathbb N$ such that for all $n≥N$ we have $|x^n−f(x)|<1/2.$ We may choose $n$ and $x.$ Let us choose $n=N,$ and $x=(\frac 3 4)^{1/N}.$ Then we have $f(x)=0$ and thus

$$|f_N(x)−f(x)|=x^N−0=\frac3 4>\frac 1 2, $$

a contradiction.