Let $\{f_n\} \subset C([0,1])$ be a sequence of functions with $f_n(x) = x^n$ for $n = 1, 2, \ldots$
-
Prove that $\{f_n\}$ does not have any uniformly convergent subsequence.
-
By (1) to prove that the closed unit ball in $C([0,1])$ is not compact.
By Arzela-Ascoli Theorem, I think $\{f_n\}$ seems to contain a uniformly convergent subsequence since $\{f_n\}$ is equicontinuous and pointwise bounded. Is it right?
Best Answer
Note that any subsequence of $(f_n)_{n\in\mathbb N}$ converges pointwise the function $$f(x)\equiv\begin{cases}0&\text{if $x\in[0,1)$,}\\1&\text{if $x=1$.}\end{cases}$$ Hence, if uniform convergence occurred, then the uniform limit would have to agree with the pointwise limit. However, the pointwise limit $f$ is not continuous. Since the uniform limit of continuous functions is continuous, it follows that no subsequence can converge uniformly.
Now, $\|f_n\|=\sup_{x\in[0,1]}|x^n|=1$ for each $n\in\mathbb N$, so that the sequence $(f_n)_{n\in\mathbb N}$ is contained in the unit ball of $C([0,1])$ (when this space is endowed with the uniform norm). If the unit ball were compact, there would exist a uniformly convergent subsequence, which, as one has seen, is not possible.
The Arzelà–Ascoli theorem doesn't work here because the sequence is not equicontinuous at $x=1$. To see this, pick any $\varepsilon\in(0,1)$. If the sequence were equicontinuous, there would exist some $\delta>0$ such that if $y\in(\max\{1-\delta,0\},1]$, then $$|f_n(1)-f_n(y)|<\varepsilon\quad\forall n\in\mathbb N.$$ But this is impossible, as $$|f_n(1)-f_n(y)|=|1-y^n|\to 1\quad\text{as $n\to\infty$}$$ whenever $y\in(0,1)$, so $|f_n(1)-f_n(y)|$ eventually exceeds $\varepsilon<1$.
It may be worth keeping in mind that, in general, if $(X,\|\cdot\|)$ is any infinite-dimensional normed vector space, then its closed unit ball is never compact with respect to the norm topology.