TOPOLOGIES OF TEST/DISTRIBUTION SPACES
Mark asked me for some references on the topologies of these spaces, I am still
thinking about a good one but in the meantime here are some notes.
Let me add a little more background. The conventional wisdom is that you
do not really need to know about these topologies, that it is enough to know the
estimates – derived below – for continuity of linear functionals (i.e. the definition
of distributions) and continuity of linear maps between the various spaces. On
the other hand I realize that enquiring minds want to know and understand these
things, so here goes.
First, we started with the space S(Rn ) which has a countable collection of norms
that fix the topology – not surprisingly these are called countably-normed spaces.
Having got this far we know actually that we can take Hilbert norms instead of the
weighted supremum type we started with since for u ∈ S(Rn ),
(1)
sup |Dα [(1 + |x|2 )k/2 u]| ≤ Ck k(1 + |ξ|2 )p (1 +\
|x|2 )p/2 ukL2
|α|≤k
≤ Cp sup |(1 + |x|2 )n/2 Dα [(1 + |x|2 )p/2 u]|, p > k + n/2.
|α|≤p
Here I have used the Sobolev embedding theorem to bound the C k norm in terms of
a Sobolev norm and then Cauchy-Schwartz to bound that in terms of a supremum
norm again. Increasing the power from k to p inside the integral uses boundedness
properties we know as does the fact that the last norm is bounded again by a higher
version of the first norm.
So, either way S(Rn ) can be written as an intersection of Hilbert or Banach
spaces. The completeness is not really important – you can replace the spaces by
Hilber/Banach space by S(Rn ) again just keeping the norms. Still, it means that
S(Rn ) can be written as a ‘directed limit’ – a projective limit – of Hilbert spaces
H (k) = {u ∈ L2 (Rn ); (1 + |ξ|)k/2 (1 +\
|x|2 )k/2 u ∈ L2 },
(2)
. . . −→ H (N ) −→ H (N −1) −→ .
All the maps here are injections and S(Rn ) is the intersection of the spaces but that
means an element of S(Rn ) can be identified with a sequence (u0 , u1 , . . . ) such that
the image of uN is always uN −1 under these injections. The topology on S(Rn )
is a metric topology where a set O ⊂ S(Rn ) is open if for each point (sequence)
in O there is an N and a ball around the corresponding uN contains all sequences
(points in S(Rn ) with u0N in that ball are in O. Transferring the norms to S(Rn )
as is sensible this is what we know to be the topology from the metric
X
kukk
(3)
d(u, v) =
2−k
1 + kukk
k
n
and S(R ) is complete with respect to this metric.
Now, why have I made this more complicated? To try to make things more
systematic as regards the other spaces. If this is not helping, maybe stop reading!
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TOPOLOGIES OF TEST/DISTRIBUTION SPACES
For C ∞ (Ω) we can write things in a similar way, starting with an exhaustion by
compact subsets Kj b Ω which are increasing, Kj+1 ⊃ Kj and such that for any
K b Ω, there is a j such that K ⊂ Kj . In fact it is better to assume that Kj is
the closure of its interior. You can arrange this by replacing one exhaustion by
the closures of their interiors and you will get another. In fact it is even better to
arrange that the Kj have smooth boundaries, that can be done but let’s not bother.
So we can define C ∞ (Kj ) as the space of smooth functions on the interior of Kj
all the derivatives of which are bounded (this is a nicer space if Kj has smooth
boundary, like a closed ball). These are complete countably normed spaces rather
like S(Rn ), with C k norms and so on. In fact we can replace C ∞ (Kj ) by C j (Kj )
with the fairly obvious definition. Again C ∞ (Ω) is a directed limit since we have
restriction maps
. . . −→ C ∞ (KN ) −→ C ∞ (KN −1 ) −→ or
(4)
. . . −→ C N (KN ) −→ C N −1 (KN −1 ) −→ .
Again an element of C ∞ (Ω) can be identified with a sequence which is consistent
under these restriction maps. The major difference is that the maps are not injective
(in general, you could insist that each Kj is contained in the interior of Kj+1 and
then it is always true). This means that the norms on C N (KN ) correspond to
seminorms on C ∞ (Ω) but apart from that things are pretty much the same. The
topology has the same definition and is a complete metric topology.
These are Fréchet spaces – complete metric spaces where the metric comes from
a countable number of seminorms with no common null space. You can always
assume the seminorms grow with j by replacing they jth by the sum of the first j.
Now, for the third space Cc∞ (Ω) ⊂ Cc∞ (Rn ) consisting of the smooth functions on
n
R which vanish outside a compact subset of Ω. This is a union over any compact
exhaustion
[
(5)
Cc∞ (Ω) =
Cc∞ (Kj ), Cc∞ (K) = {u ∈ Cc∞ (Rn ); supp(u) ⊂ K}.
j
Cc∞ (Kj )
are closed subsets of S(Rn ) so they are already countably normed
Here the
Fréchet spaces. Again there are maps but this time the ‘go the other way’
(6)
Cc∞ (K0 ) −→ Cc∞ (K1 ) −→ . . . −→ Cc∞ (Kj ) −→ . . .
and are again continuous injections (since they are bigger closed susbsets of the
same space). Now we are looking at a inductive limit since the spaces are getting
bigger. An element of C ∞ (Ω) can be thought of as a consistent sequence going to
the right and just starting somewhere, not at the beginning. This is just the settheoretic union in this case. A set in I(Ω) is open if it meets each of the Cc∞ (Kj )
in an open set. The result is not a metric space. Sometimes these are call ‘LF’ or
‘ILF’ spaces as ‘inductive limits of Fréchet spaces’.
These definitions unravel when you start to look at the dual spaces – the spaces
of continuous linear maps U : F −→ C. Of course this just means that the inverse
image of the unit ball is open, but since the map is linear it just means (using the
linearity again to translate around) that 0 ∈ F is an interior point of U −1 (|z| < 1).
For a Fréchet space this means that there is some seminorm and constant such that
(7)
kU (f )kk ≤ Ckf kk
TOPOLOGIES OF TEST/DISTRIBUTION SPACES
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which we already knew. It is perfectly reasonable just to take this as the definition
of a tempered distribution (for F = S(Rn )) or a distribution with compact support
in Ω (for F = C ∞ (Ω)) and move on.
For the LF space such as Cc∞ (Ω) continuity reduces, from the definition of an
open set, to continuity of U : Cc∞ (Ω) −→ C reduces to continuity of every restriction
U : Cc∞ (Kj ) −→ C and this means that for each K (it is enough to take Kj so there
are really only countably many conditions) there is a k and C such that
(8)
|U (f )| ≤ Ckf kk ∀ f with supp(f ) ⊂ K.
So in some sense we always get the same condition but you have to watch the
quantifiers carefully!
Now, we need to know about continuity of linear maps between these spaces
L : F1 −→ F2 . The result is pretty much the same but I better write it out unless
someone would like to do it for me – this is a matter of examining open sets and
what happens to them and again it is only an issue of deciding when 0 is an interior
point of the inverse image but now of a ‘basis’ of open sets containing 0 in the
image space, so that you get every open set containing 0 in the end.
What about topologies on the dual spaces, what are they like? We already talked
about S 0 (Rn ). From the discussion above each U ∈ S 0 (Rn ) is continous with respect
to some (admissible, i.e. continuous) norm on S(R). So we can let B−k be those
which satisfy
(9)
|U (f )| ≤ CU kf kk , kU k−k =
sup
|U (f )| < ∞
kf kk ≤1;f ∈S(Rn )
defines a norm on B−k with respect to which it is a Banach space. In our case if
we take the weighted Sobolev norms above, these are Hilbert spaces and you can
see that they are (negatively weighted/order) Sobolev spaces. Then
[
(10)
S 0 (Rn ) =
B−k
k
is written as an inductive limit of Banach (in this case Hilbert) spaces. So it is a
simpler case than the LF spaces above – in fact not significantly simpler. So this
gives S 0 (Rn ) a topology with respect to which it is complete.
There is however more to the story than that, even for S 0 (Rn ), since there are
other topologies. For instance there is the weak topology. Each of the elements
f ∈ S(Rn ) fixes a seminorm on S 0 (Rn ) by pairing
(11)
S 0 (Rn ) 3 U 7−→ |U (f )|.
So, now we have a space with uncountably many seminorms, and none is really
stronger than the others so there is no hope of dropping down to a countable
subcollection. What to do? Well, they define a topology. Declare the sets formed
by a finite intersection of semi-norm balls with centres at Uj to be open
(12)
{V ∈ S 0 (Rn ); |(V − Uj )(fj )| < j , f1 , . . . , fN ∈ S(Rn ), j > 0}
to be open. Any finite intersection of these is then also open since it is of the same
form. It follows that arbitrary unions of these sets form a topology – since the finite
intersection of such an arbitray union is a union of finite intersections (check basis
of topology ...).
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TOPOLOGIES OF TEST/DISTRIBUTION SPACES
Exercise 1. Show that the distributional pairing
(13)
S(Rn ) × S 0 (Rn ) 3 (f, U ) −→ U (f ) ∈ C
is continuous with respect to either the product of the Fréchet and weak topologies
or the products of the Fréchet and inductive limit topologies and that it in both
cases it is a ‘perfect pairing’ in that the restricted to either factor it gives all
continous linear maps on the other.
Meaning of course that the continous linear functionals on S 0 (Rn ) with respect
to the weak topology or the inductive limit topology are precisely the elements of
S(Rn ).