Let $C[0,1]$ have the following norm for finite $p$:
$$||f||_p=\left(\int_0^1|f(x)|^pdx\right)^{1/p}$$
I want to show that $C[0,1]$ is not complete for this norm.
How do I do this?
What I know:
So I want to find a Cauchy sequence that does not converge in $C[0,1]$.
My idea was
$$f_n(x) := \begin{cases}
0 & \text{if $0 \le x < \frac1{2}-\frac{1}{n}$}\\
\frac{n}{2}(x-\frac{1}{2}+\frac{1}{n}) & \text{if $\frac {1}{2}-\frac{1}{n} \le x \le \frac {1}{2} + \frac {1}{n}$}\\
1 & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $}
\end{cases}$$
It is clear that this converges to
$$f(x) := \begin{cases}
0 & \text{if $0 \le x < \frac1{2}$}\\
1 & \text{if $ \frac {1}{2} <x \leq 1 $}
\end{cases}$$
which is not continuous. The point is now to prove that $f_n$ is Cauchy, but I have no idea how to prove this. Can you help?
Edit:
According to me, $||f_n-f_m||_p=\left(\int_{\frac1{2}-\frac{1}{m}}^{\frac1{2}+\frac{1}{m}}|f_n(x)-f_m(x)|^pdx\right)^{1/p}$. BUt how do I continue from here?
Best Answer
Let $n,m\in\mathbb{N}$ and $n\geq m$.
Then $|f_n(x)-f_m(x)|\le 1$ on $[\frac{1}{2}-\frac{1}{m},\frac{1}{2}+\frac{1}{m}]$, since $0\le |f_i(x)|\le 1$ for all $i$.
This means that $$||f_n-f_m||_p=\left(\int_{\frac1{2}-\frac{1}{m}}^{\frac1{2}+\frac{1}{m}}|f_n(x)-f_m(x)|^pdx\right)^{1/p}\le \left(\int_{\frac1{2}-\frac{1}{m}}^{\frac1{2}+\frac{1}{m}}1dx\right)^{1/p}=\left(\frac{2}{m}\right)^{1/p}.$$ For $n,m\rightarrow\infty$, we see that $||f_n-f_m||_p\rightarrow 0$. Thus $(f_n)$ is Cauchy.
Suppose $f_n$ has a limit $f$ in $C[0,1]$. Then $\int_{0}^{1}|f(x)-f_n(x)|^pdx\leq||f-f_n||_p\rightarrow 0$ as $n\rightarrow\infty$. This gives us that $$f(x) = \begin{cases} 0 & \text{if $0 \le x < \frac1{2}$}\\ 1 & \text{if $ \frac {1}{2} \le x \leq 1 $} \end{cases}$$ which is not continuous.