# How to prove that $C^k(\Omega)$ is not complete

Let $\Omega \subset\mathbb{R}^n$ be some bounded domain. And Consider the set of all k-times differentiable functions $C^k(\Omega)$. I want to prove that this set is not complete with the inner product $\langle f,g\rangle=\int\limits_{\Omega}f\cdot g\text{ } dx$.

First of all, is my hypothesis right? I'm not sure about this. Furthermore i need help in finding a nice Cauchy-Sequence, which has no limit.

Maybe it is easy to consider the special case $\Omega =(0,1)$ and $k=1$. But couldn't find a good C-S. yet. I hope you can help me.

Regards

• It is not complete even for the case $k=0$. Here $\Omega=(0,1)$. Consider functions$$f_n(t)=\begin{cases}0\qquad\qquad\qquad\qquad t\in\left(0,\frac{1}{2} -\frac{1}{2n}\right)\\\frac{1}{2}+n\left(t-\frac{1}{2}\right)\qquad t\in\left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right]\\1\qquad\qquad\qquad\qquad t\in\left(\frac{1}{2}+\frac{1}{2n},1\right)\end{cases}$$ – Norbert Feb 27 '12 at 18:12
• Classical example. Norbert you should answer your comment. – Patrick Da Silva Feb 27 '12 at 19:09

Ok, I will try to write a complete explanation of classical example given in the comment above. Consider functions $$f_n(t)=\begin{cases}0\qquad\qquad\qquad\qquad t\in\left(0,\frac{1}{2} -\frac{1}{2n}\right)\\\frac{1}{2}+n\left(t-\frac{1}{2}\right)\qquad t\in\left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right]\\1\qquad\qquad\qquad\qquad t\in\left(\frac{1}{2}+\frac{1}{2n},1\right)\end{cases}$$ It is easy to check that $\{f_n\}_{n=1}^\infty\subset C^0((0,1))$. Moreover this is a Cauchy sequence. Indeed $$\lim\limits_{N\to\infty}\sup\limits_{m,n>N}\Vert f_n-f_m\Vert_2 =\lim\limits_{N\to\infty}\sup\limits_{m,n>N}\frac{|m-n|}{(24mn\max(m,n))^{1/2}} \leq\lim\limits_{N\to\infty}\frac{1}{(24N)^{1/2}}=0$$ Assume that there exist $f\in C^0((0,1))$ such that $\lim\limits_{n\to\infty}\Vert f_n-f\Vert_2=0$.
Assume that there exist $t_0\in\left(\frac{1}{2},1\right)$ such that $f(t_0)\neq 1$. Since $f$ is continuous then there exist neighborhood $U_\delta(t_0)$ such that for all $t\in U_\delta(t_0)$ we have $|1-f(t)|\geq\frac{|1-f(t_0)|}{2}$. Now take $\delta'=\min\left(\delta,t_0-\frac{1}{2}\right)$, so we can assume $U_{\delta'}(t_0)\subset \left(\frac{1}{2},1\right)$. Define $N=\frac{4(t_0-\delta')-1}{2}$, then for all $n>N$ we have $U_{\delta'}(t_0)\subset \left(\frac{1}{2}+\frac{1}{2n},1\right)$. Note that for all $n>N$ and $t\in U_{\delta'}(t)$ we have $f_n(t)=1$, so $$\Vert f_n-f\Vert_2^2= \int\limits_{(0,1)}|f_n(t)-f(t)|^2dt\geq \int\limits_{(t_0-\delta',t_0+\delta')}|f_n(t)-f(t)|^2dt=$$ $$\int\limits_{(t_0-\delta',t_0+\delta')}|1-f(t)|^2dt\geq \int\limits_{(t_0-\delta',t_0+\delta')}\left(\frac{1-f(t_0)}{2}\right)^2dt\geq \frac{\delta'(1-f(t_0))^2}{2}>0$$ and as the consequence $$0=\lim\limits_{n\to\infty}\Vert f_n-f\Vert_2^2\geq\frac{\delta'(1-f(t_0))^2}{2}>0$$ Contradiction, therefore $f(t_0)=1$ for all $t_0\in\left(\frac{1}{2},1\right)$. Similarly one can show that for all $t_0\in\left(0,\frac{1}{2}\right)$ we also have $f(t_0)=1$. Now note that $$\lim\limits_{t\to 1/2-0}f(t)=0\qquad\qquad\lim\limits_{t\to 1/2+0}f(t)=1$$ so $f$ is not continuous! Contradiction.
The smooth functions of compact support ${\mathcal C}^\infty_c(\Omega)$ are dense in ${\mathcal L}^2(\Omega)$. Hence any subspace of continuous or continuously differentiable bounded functions is also dense in ${\mathcal L}^2(\Omega)$. A subset of a complete metric space is itself complete if and only if it is a closed subset.
Since there are plenty of discontinuous functions in ${\mathcal L}^2(\Omega)$, we are done.