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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?

Thanks

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2 Answers 2

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.

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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 $k=(k_1,...,k_n)\in \mathbb Z^n$.

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