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Borel lemma states that for $x_0 \in \mathbf{R}$ and for a real sequence $(a_n)_{n \in \mathbf{N_0}}$ there exists a smooth function $f: \mathbf{R} \rightarrow \mathbf{R}$ such that $f^{(n)}(x_0)=a_n$ for $n \in \mathbf{N_0}$.

However is it true for $x_1, x_2 \in \mathbf{R}, x_1 \neq x_2$, and for real sequences $(a_n)_{n \in \mathbf{N_0}}, (b_n)_{n \in \mathbf{N_0}}$ there exists a smooth function $f: \mathbf{R} \rightarrow \mathbf{R}$ such that $f^{(n)}(x_1)=a_n$, $f^{(n)}(x_2)=a_n$ for $n \in \mathbf{N_0}$ ?


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By using bump functions and Borel's lemma, we find a function $f_1$ whose derivatives at $x_1$ match $a_n$, and which is identically zero outside of a small neighorhood (small enough that it does NOT contain $x_2$), and similarly a $f_2$ that does what we want at $x_2$ and is $0$ outside of a small neighborhood. Then $f_1+f_2$ does what we want. In general, bump functions and partitions of unity allow you to make smooth functions that have specified local behavior. –  Aaron Sep 21 '11 at 20:13
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2 Answers 2

up vote 4 down vote accepted

Yes. Take a bump function $\zeta_1$ centered at $ x_1$ and a bump function $\zeta_2$ centered at $x_2$ such that $\mathrm{supp}(\zeta_1) \cap \mathrm{supp}(\zeta_2)=\varnothing$. Use Borel's lemma to find $f_1, f_2 \in C^\infty(\mathbb{R})$ s.t. $f_1^{(n)}(x_1)=a_n, f_2^{(n)}(x_2)=b_n$. Then $f=\zeta_1f_1+\zeta_2f_2$ has the desired property.

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Since Giuseppe has already given an answer, I will give the details of the extension.

Let $A\subset \mathbb R$ a set which consist of isolated points and for each $a\in A$, $\{x(a)_n\}$ a sequence of real numbers. We can find a smooth function $f$ such that for all $a\in A$ and $n\in\mathbb N$, we have $f^{(n)}(a)=x(a)_n$.

Since $\mathbb R$ is separable, this set has to be countable, hence we can write $A=\{a_n,n\in\mathbb N\}$. We can write $b_{n,k}=x(a_n)_k$ for each $n$ and $k$. For each $n$, we consider $r_n $such that $\left]a_n-3r_n,a_n+3r_n\right[\cap A=\{r_n\}$. Let $g_n$ be a bump function, $g_n=1$ on $\left]-r_n,r_n\right[$ and $\mathrm{supp}g_n\subset \left[-2r_n,2r_n\right]$. Put $$f_{n,j}(x)= \frac{b_{n,j}}{j!}(x-a_n)^jg_n\left(\frac{x-a_n}{\varepsilon_{n,k}}\right)$$ where $\varepsilon_{n,k}=\frac 1{1+|b_{n,k}|}\frac 1{4^kM_{n,k}}$, with $\displaystyle M_{n,k}:=\max_{0\leq j\leq k}\sup_{x\in\mathbb R}|g_n^{(j)}(x)|$. Note that $f_{n,k}$ has a compact support, and putting for all $n$: $\displaystyle h_n(x):=\sum_{k=0}^{+\infty}f_{n,k}(x)$, $h_n$ is a smooth function. Indeed, for $n$ fixed and $k\geq d+1$ we have $\sup_{x\in\mathbb R}|f_{n,k}^{(d)}|\leq \frac 1{2^k}$, hence the normal convergence of the series on the whole real line for all the derivatives. A (boring but not hard) computation which uses Leibniz rule gives that we have $h_n^{(k)}(a_n)=b_{n,k}$.

Now, put $$f(x):=\sum_{n=0}^{+\infty}\sum_{j=0}^{+\infty}f_{n,j}(x).$$ Since the supports of $h_n$ are pairwise disjoint, the series in $n$ is convergent to a smooth function.

We will note that we cannot extend the result to sets $S$ which contain a point which is not isolated, because the continuity of the derivative gives a restriction on the sequence $\{x(s)_n\}$ for the $s$ in a neighborhood of the non-isolated point. Namely, if $s_0$ is a non-isolated point and $\{s_n\}\subset S$ a sequence which converges to $s_0$ then the sequence $\{x(s_n)_0\}$ has to converge to $x(s_0)_0$ (hence we can't get all the sequences).

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Maybe I misunderstood something. I have one doubt. Let's assume that $A$ has in $\mathbb R$ an accumulation point, say $x_0$ ($x_0$ is not by assumption in $A$). The sum on RHS of the formula $f(x_0)=\sum_{n=0}^\infty h_n(x_0)$ is finite but in arbitrary neighbourhood $U$ of $x_0$ there is infinite many nonzero terms. For differentiability of $f$ in $x_0$ we need uniformly convergence on $U$ of all series $\sum_{n=0}^\infty f_n^(k)(x)$ ($k=0,1,2,...$). –  Alex Oct 28 '13 at 17:01
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