I have to demonstrate that the space of continuous real functions on a compact set, with respect to the norm $\|\cdot\|_{1}$ ($\mathcal{C}^{(0)}([a,b],\|\cdot\|_1)$), is not a complete space.
I used this argument: let $\{f_n\}_{n\in\mathbb N}$ be a sequence of real functions $$ f_n:[a,b]\rightarrow\mathbb{R}; \qquad f_n = \frac{x-a}{x-a+1/n} $$ $f_n$ converges to $f$, where $f(x)=\frac{x-a}{x-a}=1\;\forall x\in(a,b]$. In fact $$ \|f_n-f\|_1 = \int_{a}^{b}|1-\frac{x-a}{x-a+1/n}|dx = \frac{ln((b-a)n+1)}{n}\xrightarrow{n\rightarrow+\infty}0 $$ $f$ is not defined in $a$ (not sure about this), so $f_n$ is a Cauchy sequence that doesn't converge in $\mathcal{C}^{(0)}([a,b],\|\cdot\|_1)$ $\Rightarrow$ the space is not complete.
Is this demonstration correct? Thanks in advance!
No, that is not correct. The sequence $(f_n)_{n\in\Bbb N}$ converges in $\mathcal{C}^{(0)}([a,b])$ to the constant function $1$.
You can take$$f_n(x)=\begin{cases}0&\text{ if }x<\frac{a+b}2-\frac1{2n}\\n\left(x-\frac{a+b}2\right)+\frac12&\text{ if }\frac{a+b}2-\frac1{2n}\leqslant x\leqslant\frac{a+b}2+\frac1{2n}\\1&\text{ if }x>\frac{a+b}2+\frac1{2n}.\end{cases}$$It is a Cauchy sequence in $\mathcal{C}^{(0)}([a,b])$, but it doesn't converge.