I have some questions about the Danny Pak-Keung Chan 's answer given in this post $C[a,b]$ is not Banach under Lp norm to prove $C[a,b]$ is not complete with the p-norm. I am copying the answer and inserting my questions in the middle:
Let $p\in[1,\infty)$. We go to show that $C([a,b])$ is not complete with respect to the $||\cdot||_{p}$-norm, where $a,b\in\mathbb{R}$ with $a<b$. Prove by contradiction. Suppose the contrary that $C([a,b])$ is complete. Let $h=\frac{(b-a)}{10}$ and $c=(a+b)/2$. For each $n\in\mathbb{N}$, define a piecewise linear function $f_{n}\in C([a,b])$ by $$ f_{n}(x)=\begin{cases} 0, & \mbox{ if }x\in[a,c-\frac{h}{n}]\\ 1+\frac{n}{h}(x-c), & \mbox{ if }x\in(c-\frac{h}{n},c)\\ 1, & \mbox{ if }x\in[c,b] \end{cases}. $$ Note that for any $m,n\in\mathbb{N}$ with $m\leq n$ and $x\in[a,b]$, $|f_{n}(x)-f_{m}(x)|\leq2$. Moreover $|f_{n}(x)-f_{m}(x)|=0$ if $x\in[a,b]\setminus(c-\frac{h}{m},c)$. Therefore \begin{eqnarray*} & & ||f_{n}-f_{m}||_{p}^{p}\\ & = & \int_{a}^{b}|f_{n}(x)-f_{m}(x)|^{p}dx\\ & \leq & 2^{p}\cdot\frac{h}{m}. \end{eqnarray*} It follows that $(f_{n})$ is a Cauchy sequence with respect to $||\cdot||_{p}$-norm. By assumption, there exists $f\in C([a,b])$ such that $f_{n}\rightarrow f$ in $||\cdot||_{p}$. That is \begin{eqnarray*} & & \int_{a}^{b}|f_{n}(x)-f(x)|^{p}dx\\ & = & ||f_{n}-f||_{p}^{p}\\ & \rightarrow & 0, \end{eqnarray*} as $n\rightarrow\infty$. On the other hand, define $g:[a,b]\rightarrow\mathbb{R}$ by $g(x)=\begin{cases} 0, & \mbox{ if }x\in[a,c)\\ 1, & \mbox{ if }x\in[c,b] \end{cases}.$ By direct calculation, we also have \begin{eqnarray*} & & \int_{a}^{b}|f_{n}(x)-g(x)|^{p}dx\\ & \leq & \frac{h}{n}\\ & \rightarrow & 0, \end{eqnarray*} as $n\rightarrow0$ because $|f_{n}(x)-g(x)|^{p}\leq1$ and $|f_{n}(x)-g(x)|^{p}=0$ for all $x\in[a,b]\setminus(c-\frac{h}{n},c)$.
My question is why can't I end the proof here?, we have found a cauchy $f_n $sequence, we guessed g is the limit in $\|\cdot\|_p$ norm is $g$(graphically by drawing the $f_n$ functions) and proved the guess by calculating $||f_n -g||_p\to 0$, that is the limit is something not continuous.
I would also avoid assuming as done above that there exists $f\in C([a,b])$ such that $f_{n}\rightarrow f$ in $||\cdot||_{p}$, since it is later proved explicitly that the limit is the non continuous g, so what is the point of doing all the stuff below ?. In fact the other answer, Rhys Steele's, which is the accepted one, skips that part
Note that \begin{eqnarray*} & & \sqrt[p]{\int_{a}^{b}|f(x)-g(x)|^{p}dx}\\ & = & \sqrt[p]{\int_{a}^{b}|[f(x)-f_{n}(x)]+[f_{n}(x)-g(x)]|^{p}dx}\\ & \leq & \sqrt[p]{\int_{a}^{b}|f(x)-f_{n}(x)|^{p}dx}+\sqrt[p]{\int_{a}^{b}|f_{n}(x)-g(x)|^{p}dx}\\ & \rightarrow & 0, \end{eqnarray*} by Minkowski inequality. Therefore $\int_{a}^{b}|f(x)-g(x)|^{p}dx=0$ and hence $f(x)=g(x)$ for $x$-a.e. (We assume the result $\int|\phi(x)|dx=0\Rightarrow\phi(x)=0$ for $x$-a.e. without proof). Let $A=\{x\in[a,b]\mid f(x)\neq g(x)\}$. Since $f=g$ a.e., we have $m(A)=0$, where $m$ is the usual Lebesgue measure. We assert that $f(x)=0$ for all $x\in[a,c)$. Prove by contradiction. Suppose that there exists $x_{0}\in[a,c)$ such that $f(x_{0})\neq0$. By continuity of $f$ at $x_{0}$, there exists $\delta>0$ such that $f(x)\neq0$ for all $x\in[a,c)\cap(x_{0}-\delta,x_{0}+\delta)$. It follows that $[a,c)\cap(x_{0}-\delta,x_{0}+\delta)\subseteq A$. However, $m([a,c)\cap(x_{0}-\delta,x_{0}+\delta))>0$, which is a contradiction. Similarly, we can prove that $f(x)=1$ for all $x\in(c,b]$. Now $\lim_{x\rightarrow c-}f(x)=0\neq1=\lim_{x\rightarrow c+}f(x)$, so $f$ is discontinuous at $c$. This is a contradiction!