Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Omega \subseteq \mathbb{R}^n$ be a domain.

How can I prove that $C_c^\infty(\Omega)$ (with the usual topology) is not sequentially complete?

I don't think I have ever seen a proof of this claim. I guess the idea is to construct a sequence of functions with increasing supports and small differences, but I don't know how to do that.

share|improve this question
It may be easier to draw a sequence of pictures to convince yourself. For example, here is a sequence of smooth functions on $[0,1]$ converging to a discontinuous function on $[0,1]$: mathworks.com/matlabcentral/fx_files/23972/18/content/chebfun/… –  Neal Mar 15 '13 at 3:59
@Neal: This sequence will not be a Cauchy sequence in $C_{c}^\infty((0,1))$ since the derivative and all the higher order derivatives explode near the origin. –  Martin Mar 15 '13 at 7:53

4 Answers 4

Let us do it for $C^\infty_c(\mathbb{R})$ first. With the help of $e^{-\frac{1}{x^2}}$, we can construct, for every $n$, a $C^\infty_c$ function $f_n$ such that $f_n(x)=0$ for $|x|\geq n+1$, and $f_n(x)=1$ for $|x|\leq n$.

Now for every compact $K$ in $\mathbb{R}$, $K$ is bounded so there exists $N$ such that $K\subseteq [-N,N]$. So for all $n,m\geq N$ and all $x\in K$, we have $f_n(x)=f_m(x)=1$. Hence $\sup_K|f_n^{(j)}(x)-f_m^{(j)}(x)|=0$ for all $j$ and all $n,m\geq N$.

So this a Cauchy sequence for the topology of compact convergence. But the pointwise limit is the constant function equal to $1$, which is not compactly supported. So $C^\infty_c(\mathbb{R})$ is not sequentially complete.

Working with $e^{-\frac{1}{\|x\|^2}}$ in $\mathbb{R}^n$ and with balls instead of intervals yields a very similat proof for $C^\infty_c(\mathbb{R}^n)$.

To do the case $C^\infty_c(\Omega)$, we need a compact exhaustion of $\Omega=\bigcup_{l\geq 1}K_l$ such that $K_l$ is contained in the interior of $K_{l+1}$ for every $l$. Then we construct for each $l$ a $C^\infty_c$ function $f_l$ such that $f_l=1$ on $K_l$ and $0$ on $K_{l+1}^c$. This is more difficult. It can be done by convolution. Still with the help of the exponential function. This sequence is Cauchy again, but converges pointwise to the constant function equal to $1$.

share|improve this answer
Thanks. I have one more question: is $\{f_n\}$ Cauchy with respect to the topology induced by the seminorms $||\cdot||_N$ (where $||f||_N$ is the sum of supremums of derivatives of order $\leq N$)? –  comp Mar 16 '13 at 18:01
@comp Take $\Omega=\mathbb{R}$ and $f_n$ as above. Then if you are talking of the sup over all $\mathbb{R}$, then the sequence is clearly no Cauchy as $\|f_{n+2}-f_n\|_\infty =1$ for all $n$. You need to take these seminorms over compact subsets of $\Omega$. You can restrict to countably many of them by exhaustion. Then yes, as I said above, the sequence is Cauchy with respect to this topology. –  1015 Mar 16 '13 at 18:08
and how can I construct a non-converging Cauchy sequence with respect to the topology induced by the seminorms $||\cdot||_N$? (adding it to your above answer would be highly appreciated) –  comp Mar 16 '13 at 18:15
@comp What is $\|\cdot\|_N$? Sum of sups over $\Omega$ of all $\|D^\alpha f \|$ for $|\alpha|\leq N$? –  1015 Mar 16 '13 at 18:20
Yes, this is what I mean. –  comp Mar 16 '13 at 18:21

The answer depends on what you mean by ''the usual topology''. Dieudonne and Schwartz invented LF-spaces to have a complete locally convex toplogy on $\mathscr D(\Omega)=C^\infty_c(\Omega)$.

Anyway, $C^\infty_c(\Omega)$ endowed with the topology of uniform convergence on compact sets of all derivatives is not complete because it is dense in $C^\infty(\Omega)$ (with the same topology) but clearly different.

share|improve this answer

Convolve a continuous non-differentiable function with a $C^{\infty}$ family of approximations to the identity. Each convolution will be $C^{\infty}$ but the limit won't be.

share|improve this answer
Why will this sequence of convolutions form a Cauchy sequence in $C^\infty$? –  Martin Mar 15 '13 at 7:57

We illustrate it in the case $\Omega=\mathbb{R}$. Take a nonzero function $\varphi\in C^\infty_c(\mathbb{R})$ whose support is small and concentrated near $0$, and consider the sequence \begin{equation} \varphi_k(x) = \varphi(x) + 2^{-1}\varphi(x-1) + \ldots + 2^{-k}\varphi(x-k), \qquad k=1,2,\ldots. \end{equation} Obviously, this sequence is Cauchy with respect to the seminorms $\|\cdot\|_{C^m(\mathbb{R})}$, but the support of the limit function is not compact.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.