Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Borel theorem says that for arbitrary sequence $(f_n)_{n=0}^\infty$ of smooth functions $f_n : \mathbb R\rightarrow \mathbb R$ with compact supports there exists a smooth function $F: \mathbb R^2 \rightarrow \mathbb R$ such that $\frac{\partial^n F}{\partial x^n}(0,y)=f_n(y)$ for $n=0,1,2,\ldots$, $y \in \mathbb R$.

In Wikipedia (see here) is stated that the assumption that all $f_n$ have compact support is not important and it could be obtained from the case when all $f_n$ have compact support by using partition of unity.

I have a problem with understanding it. How to do it?


share|cite|improve this question
up vote 2 down vote accepted

Let $(f_n)_{n=0}^\infty$ be an arbitrary sequence of smooth functions $f_n:\mathbb R\to\mathbb R$ (not necessarily with compact support.) Let $\mathcal U$ be an open cover for $\mathbb R$, defined by $$\mathcal U=\{(k-1,k+1)|\; k\in\mathbb Z\}$$ to keep things nice and let $$\{\phi_U:\mathbb R\to[0,1]|\;U\in\mathcal U\}$$ be a partition of unity subordinate to $\mathcal U$. Each of these functions $\phi_U$ has compact support in $U$ and $$\sum_{U\in\mathcal U}\phi_U(x) = 1$$ for each $x\in\mathbb R$. Then Borel's theorem holds for each sequence of functions $(f_n \phi_U)_{n=0}^\infty$, since these have compact support. So, for each $U\in\mathcal U$, there is a function $F_U:\mathbb R^2\to\mathbb R$ such that $$\frac{\partial^n F_U}{\partial x^n}(0,y)=f_n(y)\phi_U(y)$$ Now we want to sum these functions somehow. But first we have to ensure the sum is well defined. To do this, we modify the functions $F_U$ a bit. We define $$G_U(x,y)=F_U(x,y)\psi_U(y)$$ where $\psi_U:\mathbb R\to \mathbb R$ is an arbitrary smooth function such that $\psi_U(x)=1$ for $x\in U$ and $\psi_U$ has support in $\tilde U$, where for $U=(k-1,k+1)$, we write $\tilde U=(k-2,k+2)$. Now, we may define $$F(x,y):=\sum_{U\in\mathcal U}G_U(x,y)$$ This sum is locally finite because of our construction of the functions $G_U$: each of these is supported in $\mathbb R\times\tilde U$. Furthermore, because $G_U=F_U$ on $\mathbb R\times U$, we have: $$\frac{\partial^n G_U}{\partial x^n}(0,y)=\frac{\partial^n F_U}{\partial x^n}(0,y)$$ Because of the linearity of partial derivatives and because all of the sums are locally finite, the following calculation now makes sense: $$\frac{\partial^n F}{\partial x^n}(0,y)=\sum_{U\in\mathcal U}\frac{\partial^n G_U}{\partial x^n}(0,y)=\sum_{U\in\mathcal U}f_n(y)\phi_U(y) = f_n(y)$$ This completes the proof.

share|cite|improve this answer

I think that in the case $f_n: V \rightarrow \mathbb R$, for $n=0,1....$, proof goes in the same way as in the answer above. Instead $ \{(k-1,k+1)\}_{k\in \mathbb Z}$ you could take the family of all intersections $V\cap Q_k$ , where $Q_k$ denotes the cube, with sides parallel do coordinates axes and of length 2, with center in $k=(k_1,...,k_n)\in \mathbb Z^n$. Instead of $\{ (k-2,k+2)\}_{k \in \mathbb Z}$ take
$V\cap \tilde{Q_k}$ , where $\tilde{Q_k}$ denotes the cube, with sides parallel do coordinates axes and of length 4, with center in $k=(k_1,...,k_n)\in \mathbb Z^n$.

share|cite|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.