# A problem regarding a sequence in $C[0;1]$

Given normed space $C[0;1]$ (the space of continous functions defined on $[0;1]$) with the norm $||.||$ defined as following

$$||x|| = \max\limits_{x \in [0;1]} |f(x)|$$ Prove that the sequence $\{x_n\}_n$ doesn't converge in $(C[0;1],||.||)$ with $$x_n(t) = \left\{ \begin{array}{ll} nt,& 0 \le t \le \dfrac{1}{n} \\ 1,& \dfrac{1}{n} < t \le 1 \end{array} \right.$$

In the attempt of solving this problem, I found out that the sequence converges by individual points to the function $x$ (means for every $t \in [0;1]$, $\lim\limits_{n \to \infty} x_n(t) = x(t)$) with $$x(t) = \left\{ \begin{array}{ll} 0,& t = 0 \\ 1,& 0 < t \le 1 \end{array} \right.$$ which isn't an element of $C[0;1]$.

This might be a point in the solution. However, I don't know how to rigorously proceed with the correct definition of the coverage in a normed space (as $x_n - x$ isn't an element of the normed space). Please help me. Thank you for reading.

## 3 Answers

You were nearly there. Suppose that $(x_n)_{n\in\mathbb N}$ converges in $C\bigl([0,1]\bigr)$ to some function $x$. Then, for each $t\in[0,1]$,$$x(t)=\lim_{n\to\infty}x_n(t)=\begin{cases}1&\text{ if }x\in(0,1]\\0&\text{ otherwise.}\end{cases}$$But there is no such function in $C\bigl([0,1]\bigr)$.

Note that the equality $x(t)=\lim_{n\to\infty}x_n(t)$ follows from the fact that you're using the norm that you are using. If, say, $\|x\|=\int_0^1\bigl|x(t)\bigr|\,\mathrm dt$, this would not work.

• In that alternate (integration) norm, the sequence does actually converge to the function $x(t)=1$. – Arthur May 4 '18 at 15:37
• @Arthur Nice remark. – José Carlos Santos May 4 '18 at 15:38
• Yes, sir. That is the second problem regarding this sequence. I have solved it successfully just as you said. – ElementX May 4 '18 at 15:44
• Sir @JoséCarlosSantos, is that equality because $g(u) = \max(u)$ a continous function? Am I understading it correctly? – ElementX May 4 '18 at 15:50

Here is an alternative formal approach. Take any function $f\in C[0,1]$, and prove that $$\lim_{n\to\infty}\|f-x_n\|\geq \frac12$$ by dividing into two cases: either $f(0)<\frac12$, or $f(0)\geq\frac12$. Thus $x_n$ cannot converge to $f$.

You can prove it by contradiction:

If there exist $\xi$ such that $x_n \to \xi$ then:

• As uniform convergence implies point-wise convergence for all $t$, $\xi(t)=x(t)$, and $\xi=x$ is not continuous.

• As a uniform limit of continuous function is continuous, $\xi$ is continuous.

So there is a contradiction.