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The classical Borel Lemma states that for an arbitrary sequence $(v_n)_{n \in \mathbb{N}_0}$ of complex numbers there is a smooth function $f\colon \mathbb{R} \longrightarrow \mathbb{C}$ with Taylor coefficients at $0$ given by the $v_n$. Some generalizations work for functions of $d$ variables and also for values taken in an arbitrary Fréchet space $V$ instead of the complex numbers.

The proofs I know use essentially the fact that $V$ has a countable set of seminorms defining the topology. On the other hand, taking the space of compactly supported smooth functions with its usual LF topology as $V$ and $v_n$ with increasing support gives easily a counter-example that for this sequence we can not have a smooth $f\colon \mathbb{R} \longrightarrow C^\infty_0(\mathbb{R})$ with $v_n$ being the Taylor coefficients, unless the $v_n$ are all in same $C^\infty_0(K)$ for a fixed compact subset $K$.

So my question is to which lcs one actually can extend the Borel lemma? Are Fréchet spaces the end of the story?

EDIT: There is of course a stupid way to extend it beyond Fréchet: whenever you have a coarser lc topology on $V$ then every smooth function with respect to the orignal one is also smooth with respect to the coarser one. So if $V$ is Fréchet then every coarser topology on $V$ will also have a valid Borel Lemma. Examples are e.g. the operator topologies on the bounded operators on a Hilbert space (soooorry for overlooking this in the first try).

So the refined question is: are there other lcs with topologies not dominated by a Fréchet one for which the Borel Lemma holds?

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up vote 2 down vote accepted

Yes, there are other spaces in which Borel's Theorem holds (property (BT)).

Example 1. Note that every cartesian product of locally convex spaces with (BT) has (BT). In particular, any uncountable power ${\mathbb R}^I$ has (BT). Yet, ${\mathbb R}^I$ cannot be given a topology ${\mathcal O}$ which makes it a Fréchet space and is finer than the product topology ${\mathcal T}$.

In fact, for any such topology ${\mathcal O}$, we could choose a decreasing sequence of absolutely convex $0$-neighbourhoods $U_n$ in $({\mathbb R}^I,{\mathcal O})$ which form a basis of $0$-neighbourhoods. Let $C_n$ be the closure of $U_n$ in $({\mathbb R}^I,{\mathcal T})$.

Now $({\mathbb R}^I,{\mathcal T})$ is a Baire space (see Oxtoby, J.C., Cartesian products of Baire spaces, Fundam. Math. 49 (1961), 157-166).

As in the usual proof of the open mapping theorem, we see that each $C_n$ is a $0$-neighbourhood in $({\mathbb R}^I,{\mathcal T})$. Since each $0$-neighbourhood in $({\mathbb R}^I,{\mathcal T})$ contains some $C_n$, we see that $({\mathbb R}^I,{\mathcal T})$ would be first countable, which is absurd.

This contradiction shows that ${\mathcal O}$ cannot exist.

Example 2. Recall that the usual proof of Borel's Theorem (for $V$ a Fréchet space) furnishes a smooth function $f\colon {\mathbb R}^d\to V$ of the form $f(x)=\sum_{\alpha \in{\mathbb N}_0^d} \;v_\alpha h(m_\alpha x)x^\alpha$, where $h\colon {\mathbb R}^d\to[0,1]$ is a cut-off function around $0$ and the $m_\alpha$ are positive constants. It can always be achieved that $m_\alpha\to\infty$ as $|\alpha|\to\infty$. As a consequence, the above sum is a finite sum (almost all summands vanish) for each fixed $x\in{\mathbb R}^d$, and likewise for all partial derivatives of the summands. It is clear from this observation that the completeness of $V$ does not play a role: The construction works just as well if $V$ is any (not necessarily complete) metrizable locally convex space.

Let $V$ be a metrizable locally convex space whose dimension as an abstract vector space is infinite and countable. By the preceeding, $V$ has (BT). Yet, there is no finer vector topology on $V$ making it a Fréchet space.

Let me mention that the product topology on ${\mathbb R}^I$ from Example 1 cannot be refined to a non-complete metrizable locally convex vector topology ${\mathcal O}$ either (by the same proof as above).

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Thanks a lot for these (class of) examples. – Stefan Waldmann Apr 21 at 5:59

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