We consider the sequence $(f_n)_{n \in \mathbb{N}}$ defined by $f_n : [0, 2] \rightarrow \mathbb{R}$ as follows:
$f_n(x) =\begin{cases} x^n, & \text{if } 0 \leq x \leq 1 \\ 1, & \text{if } 1 < x \leq 2 \end{cases}$ for $n \in \mathbb{N}$.
Show that the sequence $(f_n)_{n \in \mathbb{N}} \in (C^0([0,2]),||\cdot||_1)$, is a Cauchy-sequence, but does not converge.
Note: $||f_n(x)||_1 = \int_0^2 |f_n(x)|dx$.
(a) Show that $(f_n)_{n \in \mathbb{N}}$ is a Cauchy-Sequence:
We have to show $\forall \epsilon >0 \exists N \in \mathbb{N} : \forall m,n > N: ||f_n(x)-f_m(x)||_1 < \epsilon $. We assume w.l.o.g. $n<m$. Choose $\epsilon >0$ arbitrary, then by choosing a corresponding $N(\epsilon) = \frac{1}{\epsilon}$ we can show that: $$||f_n(x)-f_m(x)||_1 = \int_0^2 |f_n(x)-f_m(x)|dx$$ $$\int_0^1 |f_n(x)-f_m(x)|dx + \int_1^2|f_n(x)-f_m(x)|dx$$ Since $|f_n(x)-f_m(x)|$ for all $x \in (1,2]$ is zero we can drop that part. Also, we can neglect the absolute value, due to $n<m$. $$ ´\int_0^1 f_n(x)-f_m(x)dx = \int_0^1 f_n(x)dx - \int_0^1 f_m(x)dx = \frac{x^{n+1}}{n+1} \bigg|_0^1 - \frac{x^{m+1}}{m+1} \bigg|_0^1 $$ $$\frac{1}{n+1} - \frac{1}{m+1} < \frac{1}{n+1} < \frac{1}{N} = \epsilon \quad \square $$
(b) Show the sequence $(f_n)_{n \in \mathbb{N}}$ does not converge:
I'm currently working on an exercise with two parts. In part (a), I need to demonstrate that the sequence forms a Cauchy sequence. I would greatly appreciate it if someone could review my work to ensure its accuracy. For part (b), I've hit a roadblock. I'm familiar with two different types of convergence: pointwise and uniform. Unfortunately, the problem doesn't specify which one to use. My intuition suggests that the sequence converges uniformly, because I can make the difference $||f_n(x)-f(x)||$ as small as I like. Could someone tell me why I am wrong? If so, I'd appreciate a hint on how to proceed with proving it. Thank you very much for your assistance.