Partial Differential Equations/Test function spaces
Before we dive deeply into the chapter, let's first motivate the notion of a test function. Let's consider two functions which are piecewise constant on the intervals and zero elsewhere; like, for example, these two:
Let's call the left function , and the right function .
Of course we can easily see that the two functions are different; they differ on the interval ; however, let's pretend that we are blind and our only way of finding out something about either function is evaluating the integrals
for functions in a given set of functions .
We proceed with choosing sufficiently clever such that five evaluations of both integrals suffice to show that . To do so, we first introduce the characteristic function. Let be any set. The characteristic function of is defined as
With this definition, we choose the set of functions as
It is easy to see (see exercise 1), that for , the expression
equals the value of on the interval , and the same is true for . But as both functions are uniquely determined by their values on the intervals (since they are zero everywhere else), we can implement the following equality test:
This obviously needs five evaluations of each integral, as .
Since we used the functions in to test and , we call them test functions. What we ask ourselves now is if this notion generalises from functions like and , which are piecewise constant on certain intervals and zero everywhere else, to continuous functions. The following chapter shows that this is true.
In order to write down the definition of a bump function more shortly, we need the following two definitions:
Let , and let . We say that is smooth iff all the partial derivatives
exist in all points of and are continuous. We write .
Let . We define the support of , , as follows:
Now we are ready to define a bump function:
is called a bump function iff and is compact.
These two properties make the function really look like a bump, as the following example shows:
Example 3.4: The standard mollifier , given by
, where , is a bump function.
As for the bump functions, in order to write down the definition of Schwartz functions shortly, we first need two helpful definitions.
Let be an arbitrary set, and let be a function. Then we define the supremum norm of as follows:
For a vector and a -dimensional multiindex we define , to the power of , as follows:
Now we are ready to define a Schwartz function.
We call a Schwartz function iff the following two conditions are satisfied:
By we mean the function .
Example 3.8: The function
is a Schwartz function.
Every bump function is also a Schwartz function.
This means that for example, the standard mollifier is also a Schwartz function.
Let be a bump function. Then, by definition of a bump function, . By the definition of bump functions, we choose such that
, as in , a set is compact iff it is closed & bounded. Further, for arbitrary,
Topological vector spaces
Topological vector spaces are sets which are vector spaces and at the same time equipped with a topology (we will also require a certain compatibility between the two). Before we define topological vector spaces in a precise way, let's shortly repeat the notions of topologies and vector spaces, the topological definition of continuity and topological fields.
Let be a completely arbitrary set. Let , where is the power set of . We say that is a topology if and only if it satisfies the following three properties:
- The whole set and the empty set are in ()
- Arbitrary unions of sets in are again in (for a family of sets such that holds that )
- Finite intersections of sets in are again in ()
The elements of are called open sets. A set is called closed set iff it's complement is an open set, i. e. .
Let be a topology, and let equipped with be an arbitrary topological space. That then arbitrary intersections of closed sets are closed again.
where is an arbitrary set and are arbitrary closed sets. We have
Since is open as the union of open sets, is closed.
Let be a field. A set is called a vector space over iff it is endowed with a vector addition
and a scalar-vector multiplication
such that the following conditions are fulfilled:
- (we call the zero element)
- (we call ‘minus ’)
- If is the multiplicative identity in , then
Let , be two sets with topology, and let be a function. We say that is continuous iff for all open sets , is open in .
Let , be two sets with topology, and let be a function. is continuous iff for all closed sets , is closed in
Let be any set. Then by definition of closed sets, is closed iff is open. Therefore, is continuous iff for all closed sets , we have that is open. But
, and therefore is continuous iff for all closed sets , we have that is closed.
Let , be sets with a topology. We consider the cartesian product . The topology on with the fewest open sets such that both projections
are continuous is called the product topology on .
We can also describe the product topology with respect to the closed sets of the topological spaces and :
Let , be sets with a topology. We consider the cartesian product . The topology on which induces the fewest closed sets such that both projections
are continuous is the product topology on .
This follows directly from the fact that for every open set, there is a closed set and vice versa. Therefore, in order for a topology to have fewer open sets than another topology, the first topology must have fewer closed sets.
A field is called a topological field iff it is endowed with a topology in which addition , multiplication and inversion are continuous, where we endow with the product topology.
By inversion we mean mapping a nonzero number to it's multiplicative inverse.
Now, having defined five different things, we are finally ready for the definition for topological vector spaces.
A vector space over a topological field is called a topological vector space iff it is endowed with a topology in which vector addition and scalar-vector multiplication are continuous with respect to the product topology of or respectively.
Can you identify each of the definitions 3.10, 3.11, 3.12, 3.14, 3.15 in the above definition?
Example 3.19: (see exercises 2 and 3) Let's equipp with the usual topology, i. e. a set is open iff
Then is a topological field. Further, if we equipp , which is a vector space over , with the usual topology, which means that a set is open iff
, then is a topological vector space over , given that is equipped with this usual topology.
In this section, we develop a framework to construct topological vector spaces out of the set of bump functions and the set of Schwartz functions, respectively. First, we give the definition of a notion of convergence.
Let be any set. We say that has a notion of convergence iff for every sequence with elements in and every , we have either or .
If we have two sets and which both have a notion of convergence, we may define sequential continuity for functions .
Let , be two sets with a notion of convergence and let be a function. is defined to be sequentially continuous iff
In order to show that a vector space space with a topology is a topological vector space, we have to show the continuity of vector-vector addition and scalar-vector multiplication. Sequential continuity would be easy to verify, but would continuity then follow from sequential continuity? The answer is yes in the cases we will consider, and this follows from the spaces we will consider being sequential spaces.
Let be a topological space which has a notion of convergence. We call a sequential space iff for all sets
Now let's show that for functions from a sequential space into a sequential space, sequential continuity implies continuity.
Let be sequential spaces, and let be a function. Then we have
Let be sequentially continuous. Let be an arbitrary closed set. We show that is closed, thereby showing continuity (due to Lemma 3.13, which showed that in the definition of continuity, we may replace open with closed sets). Let be a sequence such that and . We have
, because else would by definition of the preimage not be in . Since is sequentially continuous, we have
Due to the closedness of and due to the fact that is a sequential space it further follows that
and thus, again by definition of the preimage,
Since is a sequential space, thus is closed (as in sequential spaces a set is closed if and only if it contains all the limit points).
Let be a set with a notion of convergence with the properties
- is closed with respect to the notion of convergence (if , then )
- Infinite subsequences converge, and to the same limit (if and is an infinite subsequence of , then )
If we define a set as closed iff and define , then is a topology, and equipped with is a sequential space.
We verify that has the defining properties of a topology.
We first show that and .
is closed with respect to our definition, because there simply are no sequences in the empty set and therefore
Therefore, we have .
Furthermore, the whole set is closed in . This is because of the first property we required for the notion of convergence. Therefore, .
We show that arbitrary unions of open sets (i. e. elements of ) are open again (i. e. elements of again).
Let be a family of open sets, and let
To show that is open, we need, by the definition of , to show that is closed with respect to the definition of closedness we have introduced. Let now such that . Assume that . Due to the second de Morgan formula, we have
, which is why is equivalent to . But since , is not closed, which is why is not open. Contradiction! Therefore, it must , and therefore is closed, and is open.
We show that the intersection of finitely many open sets is open.
Let be open, and let
Again, by definition of , in order to show that is open, we need to show that is closed. Let be a sequence such that and and assume that . Again, we use one of the two de Morgan formulas, this time the first one:
Therefore, . We now choose such that infinitely many entries of the sequence are in . This is possible because if each would contain only finitely many entries of , then since , the natural numbers would have finite cardinality. Furthermore, we choose as the subsequence of whose entries are in . Due to our second requirement on the notion of convergence, , and since is closed, , contradiction! Therefore, , and is closed, which is why is open.
Therefore has all three defining properties of a topology.
From the definition of sequential spaces, it is clear that the resulting space is sequential.
The topological vector spaces of the bump and the Schwartz functions, respectively
We introduce the set of bump and Schwartz functions.
Definition 3.25: Let and an open set. The set of bump functions for the set , written , is defined as
A notion of convergence for this space we define as follows: A sequence of bump functions is said to converge to another bump function iff the following two conditions are satisfied:
- There is a compact set such that
By giving the right definitions of vector addition and scalar-vector multiplication, and constructing a topology from the notion of convergence due to theorem 2.16, we can make a topological vector space. The vectors in this space will be the functions themselves.
Let and let be an open set. If we endow with pointwise definitions of addition and scalar-vector multiplication, i. e.
and endow with the topology which is defined by the definition of closed sets in theorem 2.16 and the notion of convergence of , then is a topological vector space.
We have to show three things:
- is a vector space with respect to the given definitions of vector-vector addition and scalar-vector multiplication
- is a topological space with respect to the topology defined by the notion of convergence and theorem 3.24
- The vector-vector addition and scalar-vector multiplication are continuous with respect to this topology and the respective subspace topologies
1. is left for you as an exercise (see exercise ?), 2. follows directly from the statement of theorem 3.24 and 3. we shall now show in three steps.
We show that the product topology in the space is given by the topology generated by theorem 3.24 and the notion of convergence in defined by:
For every and every , we define
We show that is open in for arbitrary and . We recall that the closed sets of were defined due to theorem 3.24 by the way of defining the closed sets as the sets which contain all the limits of the sequences they contain, and the open sets were defined as the complements of the closed sets. Therefore, it suffices for us to show that the complement of contains all the limits of the sequences which are completely contained in it.
Let therefore be a sequence in such that converges to a . Let's assume that its limit, instead of being contained in , was contained in . Due to our notion of convergence on the set of bump functions from follows that
or, inserting the epsilon definition of convergence,
We now choose so small that
, where the maximum exists because finite subsets of the real numbers always have a maximum. From then follows
This shows that for , , which contradicts the assumption that was a sequence entirely contained in . Therefore, we must have and since was an arbitrary convergent sequence, this shows that is closed, which shows that is open.
As in the product topology of both projections are continuous, we have that for arbitrary and
are open, and as the product topology is a topology,
is also open, which is why
is closed with respect to the product topology.
Furthermore, for we can calculate explicitly for :
Let now be a set which is closed with respect to the topology defined by theorem 3.24 and the notion of convergence defined at the beginning of this section 3.1 of the proof. From theorem 3.11 and from the being closed follows that
is closed with respect to the product topology.
We also define the space of the Schwartz functions.
Let . Then the -dimensional Schwartz space, written , is defined as
The space of Schwartz functions also has a notion of convergence: We say that the sequence of Schwartz functions converges to iff the following condition is satisfied:
Like the space of the bump functions, the space of Schwartz functions is a topological vector space, if endowed with the topology from the notion of convergence and the right definitions for vector addition and scalar-vector multiplication.
Let . If we endow with pointwise definitions of addition and scalar-vector multiplication, i. e.
and endow with the topology which is defined by from the definition of closed sets in theorem 2.16 and the notion of convergence of , then is a topological vector space.
The ‘testing’ property of test functions
Let and be constant on the interval . Show that
- Prove that endowed with the topology stated in example 3.17 is a topological field.
- Prove that endowed with the topology stated in example 3.17 is a topological vector space (for the usual definitions of vector addition and scalar-vector multiplication).