$C[a,b]$ is not Banach under Lp norm 
$\|f\|_p=\biggr(\int_{a}^b|f|^pdx\biggr)^{1/p}$

-Generally we choose $\mathbb R$ as field-
I am trying to show $C[a,b]$ is not Banach under $\|.\|_p$ norm. 1) First I need to show 
$$ f_n(x) =
\begin{cases}
0,  & \text{if $a\le x\le c-(1/n)$} \\
nx-nc+1, & \text{if $c-(1/n)\lt x \le c$}\\
1,  & \text{if $c\lt x \le b$}
\end{cases}$$
is a Cauchy Sequence i.e. $$\biggr(\int_{a}^b|f_n-f_m|^pdx\biggr)^{1/p} \lt \varepsilon$$  for $\forall \varepsilon \lt 0$
Let $m \gt n\Rightarrow 1/n \gt 1/m \Rightarrow c-(1/m) \gt c-(1/n) $ 
$$\|f_n-f_m\|_p=\biggr(\int_{a}^b|f_n(x)-f_m(x)|^pdx\biggr)^{1/p}=$$
$$\biggl(\int_{c-(1/n)}^{c-(1/m)}|nx-nc+1|^pdx+\int_{c-(1/m)}^{c}|(m-n)c+(n-m)x|^pdx\biggl)^{1/p}$$
I have calculated
$$\int_{c-(1/n)}^{c-(1/m)}|nx-nc+1|^pdx =\frac{(-1-(n/m)^{p+1}}{n(p+1)}$$
Since $|nx-nc+1|= nc-nx+1$  $(x\le c)$ but I am not sure about other absolute value. $\biggr(|(m-n)c+(n-m)x|=?\biggr)$
After calculating how can I show 
$$\biggl(\int_{c-(1/n)}^{c-(1/m)}|nx-nc+1|^pdx+\int_{c-(1/m)}^{c}|(m-n)c+(n-m)x|^pdx\biggl)^{1/p} \lt \varepsilon$$  ??
2) My other question about that:
we know that $f_n(x)\rightarrow g(x)$ where 
  $$ g(x) =
\begin{cases}
0,  & \text{if $a\le x\le c$} \\
1,  & \text{if $c\lt x \le b$}
\end{cases}$$
and $g(x) \notin C[a,b]$
I think, I should show $\biggr(\int_{a}^b|f_n(x)-g(x)|^pdx\biggr)^{1/p} \to 0$ 
There is a note under question "use Minkowski inequality". I guess I should use this inequality for showing $\biggr(\int_{a}^b|f_n(x)-g(x)|^pdx\biggr)^{1/p} \to 0$.
Should I write $\biggr(\int_{a}^b|f_n(x)-g(x)|^pdx\biggr)^{1/p} \le\biggr(\int_{a}^b|f_n(x)|^pdx\biggr)^{1/p} + \biggr(\int_{a}^b|-g(x)|^pdx\biggr)^{1/p}$ or should I write integrals for $[a,c-(1/n)],[c-(1/n),c-(1/m)]$ etc. ?
Could you fix my mistakes and warn me about unnecessary things?I have functional analysis midterm tomorrow.Your helps are very important for me
Thanks
 A: Before I discuss your proof in more detail, let me give some general advice. Start by drawing a picture of your function $f_n$ and remember that, as is often the case in analysis, working with loose upper bounds can often give you the estimates you want far more easily than exact calculations will! Now onto your proof...
You've begun by checking that $f_n$ is a Cauchy sequence in $L^p$. I would argue that this step isn't necessary for a couple of reasons.
Firstly, as long as you are allowed to assume the standard result that $L^p[a,b]$ is a Banach space for the usual norm, it is enough to check that $C[a,b]$ isn't closed in $L^p[a,b]$. Indeed, a subspace of a Banach space is complete for the induced norm if and only if it is closed. As a result, we could just check that $f_n$ is a sequence in $C[a,b]$ converging in $L^p$ to $g \not \in C[a,b]$ and hence $C[a,b]$ is not closed in $L^p[a,b]$.
Additionally, the fact that $f_n$ is a Cauchy sequence for the $L^p$-norm will follow from the fact that it converges in $L^p[a,b]$ to $g$ since convergent sequences are always Cauchy.
However, I will also show that $f_n$ is Cauchy in $L^p$ directly since it illustrates some ideas well. First notice that for every $n,m$ and $x$, $|f_n(x) - f_m(x)| \leq 1$. Now, without loss of generality, suppose that $n \geq m$. Then $f_n(x) = f_m(x)$ whenever $x \not \in [c - \frac{1}{m}, c]$. As a result, we can bound
\begin{align*}
\|f_n - f_m \|_p^p = \int_a^b |f_n(x) - f_m(x)|^p dx = \int_{c - \frac{1}{m}}^c |f_n(x) - f_m(x)|^p dx \leq \int_{c - \frac{1}{m}}^c 1 dx = \frac{1}{m}.
\end{align*}
Since this goes to $0$ as $n,m \to \infty$, we have that $f_n$ is Cauchy in $L^p[a,b]$.
We can show that $f_n \to g$ in $L^p$ using a similar trick. We have that
\begin{align*}
\|f_n - g\|_p^p = \int_a^b |f_n(x) - g(x)|^p dx = \int_{c-\frac{1}{n}}^c |f_n(x) - g(x)| dx \leq \int_{c-\frac{1}{n}}^c 1 dx = \frac{1}{n}
\end{align*}
and hence $f_n \to g$ in $L^p$-norm.
I'll finish by noting that I think the hint to use the Minkowski inequality is potentially misleading as it really isn't necessary here. However, a good indication that you shouldn't write 
\begin{align*} \|f_n - g \|_p = \biggr(\int_{a}^b|f_n(x)-g(x)|^pdx\biggr)^{1/p} &\le\biggr(\int_{a}^b|f_n(x)|^pdx\biggr)^{1/p} + \biggr(\int_{a}^b|-g(x)|^pdx\biggr)^{1/p} \\& = \|f_n\|_p + \|g\|_p \end{align*} is that the right hand side is bounded below by $\|g\|_p > 0$ and hence you definitely can't show $\|f_n - g \|_p \to 0$ in this way.
Good luck for the midterm!
A: 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 $x\in[a,b]\setminus(c-\frac{h}{n},c)$.
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!
