Deje $(f_n)_{n=0}^\infty$ ser una secuencia arbitraria de las funciones lisas $f_n:\mathbb R\to\mathbb R$ (no necesariamente con soporte compacto.) Deje $\mathcal U$ ser una cubierta abierta de a $\mathbb R$, definido por $$\mathcal U=\{(k-1,k+1)|\; k\in\mathbb Z\}$$
para mantener las cosas agradables y dejar $$\{\phi_U:\mathbb R\to[0,1]|\;U\in\mathcal U\}$$ 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 $x\U$ 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: $$\frac{\partial^n F}{\partial x^n}(0,y)=\sum_{U\in\mathcal U}\frac{\partial^n G_U}{\partial x^n}(0,y)=\sum_{U\in\mathcal U}f_n(y)\phi_U(y) = f_n(y)$$
Esto completa la prueba.
