# Convergence in $C([0, 1])$

Let's consider the space of continuous functions on the interval $$[0, 1]$$. We'll denote it by $$C([0, 1])$$.
Now we can define a sequence: $$f_n(x) = x(1-x^n).$$ It's easy to find it's pointwise limit: $$\lim_{n \to \infty} f_n(x) = g(x) = \begin{cases} x,\text{if } x\in[0, 1)\\0, \text{if } x = 1\end{cases}$$ I would like to show that the sequence does not converge in norm. $$\left\lVert f_n- g \right\rVert = \max_{x \in [0,1]} |f_n(x)-g(x)|$$ I don't know where to go from here. I would appreciate any tips or hints.

• The sequence doesn't converge since $\|f_n - g\| = 1 \not\to 0$. Commented Oct 23, 2018 at 21:00

Notice that for $$x=1,$$ $$|f_n(x) - g(x)| = 0$$. Hence $$\sup_{x\in[0,1]}|f_n(x)-g(x) | = \sup_{x\in[0,1)}|f_n(x)-g(x)| = \sup_{x\in[0,1)} |x(1-x^n)-x| = \sup_{x\in[0,1)} |x^n| = 1,$$ for all $$n$$, so $$f_n \not\to g$$ uniformly.
• It's quite strange for me that $\sup_{x \in [0, 1)} |x^n| = 1$. In Weierstrass theorem we need to have a closed subset $[a, b]$. Commented Oct 23, 2018 at 21:08
• @Hendrra Notice that the limit function $g$ is not continuous, so that you can no longer guarantee that it attains its extrema (indeed, it does not, which is why I wrote $\sup$ instead of $\max$).
• Thanks! One more question - $\sup_{x \in [0, 1)} |x^n| = 1$. That means that for a tiny $\epsilon$ we would have $(1 - \epsilon)^n = 1$. That's very strange for me. What is the reason why $\sup_{x \in [0, 1)} |x^n| = 1$? Commented Oct 23, 2018 at 21:16
• @Hendrra This means that $1$ is the least upper bound of $|x^n|$ on $[0,1)$. In other words, you can get arbitrarily close to $1$ from below, i.e. for any $\epsilon > 0$, there is an $x \in [0,1)$ such that $1 - \epsilon < |x^n| \leq 1$.
• Thank you! I think I was confused because I forgot that $\forall_{n \in \mathbb{N}} x^n$ is continuous, only $g$ the limit function is not continuos. Am I right? Commented Oct 23, 2018 at 21:37