To get solutions to the first more difficult partial differential equations (like, for example, Poisson's equation, the heat equation and Helmholtz' equation), we will now take a look at distributions. Distributions are continuous and linear functions which map a test function to a real number.
Definition and characterisationEdit
For each topological vector space, we can define a dual space.
Let be a topological vector space. The set
is called the dual space of .
Having defined the dual space of a topological vector space, we can define distributions as follows:
Let be open. We call the elements of distributions.
Furthermore, we also give a name to the elements of the dual space of the Schwartz space, but we give a slightly different name:
We call the elements of tempered distributions.
We will now show that distributions are exactly the linear and sequentially continuous functions from to for a domain .
Let be open and let be a function. Then the following two properties are equivalent:
- is continuous
- is sequentially continuous ()
The same holds true for tempered distributions:
Let be a function. Then the following two properties are equivalent:
- is continuous
- is sequentially continuous ()
Proof of theorems 4.4 and 4.5:
Let be continuous, and let be a sequence such that . We choose
for an arbitrary . As is continuous, is open. Therefore, is closed. Assume by contradiction that infinitely many elements of the sequence are not contained in . We choose as the subsequence of not contained in . Then by the second property of the notion of convergence, , and since is closed, which contradicts . Thus, for large enough , we have and therefore . Since was arbitrary, this shows that and thus is sequentially continuous.
Let be sequentially continuous. Let be a 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 it further follows (see exercise 1) that
and thus, again by definition of the preimage,
By the definitions of closed sets in and , which were given in the last chapter to be exactly the sets which are closed with respect to convergence, is closed.
Next, we will show that every tempered distribution is a distribution. But to do so, we need the following lemma:
Let be an arbitrary sequence of bump functions. If with respect to the notion of convergence for bump functions, then also with respect to the notion of convergence for Schwartz functions.
Let be a sequence in such that with respect to the notion of convergence of . Let thus be the compact set in which all the are contained. From this also follows that , since otherwise , where is any nonzero value takes outside ; this would contradict with respect to our notion of convergence.
In , ‘compact’ is equivalent to ‘bounded and closed’. Therefore, for an . Therefore, we have for all multiindices :
Therefore the sequence converges with respect to the notion of convergence for Schwartz functions.
Every tempered distribution is contained in every , where is an arbitrary domain.
Or to be more precise, the restriction of every tempered distribution to is contained in for every domain .
Let be a tempered distribution, and let be an arbitrary domain.
We show that has a well-defined value for .
Due to theorem 3.9, every bump function is a Schwartz function, which is why the expression
makes sense for every .
We show that the restriction is linear.
Let and . Since due to theorem 3.9 and are Schwartz functions as well, we have
due to the linearity of for all Schwartz functions. Thus is also linear for bump functions.
We show that the restriction of to is sequentially continuous.
Let with respect to the notion convergence of . Due to lemma 4.6, we have that with respect to the notion convergence of and therefore, since is sequentially continuous with respect to the notion convergence of (by theorem 4.5), . This shows sequential continuity and thus continuity (by theorem 4.4).
In this section, we shortly study a class of distributions which we call regular distributions. In particular, we will see that for certain kinds of functions there exist corresponding regular distributions. But let's first define what regular distributions are.
Let be a domain and let . If for all can be written as
for a function which is independent of , then we call a regular distribution.
Let . If for all can be written as
for a function which is independent of , then we call a regular tempered distribution.
Two questions related to this definition could be asked: Given a function , is given by
a distribution? Or is given by
a tempered distribution? In general, the answer to these two questions is no, but both questions can be answered with yes if the respective function has the respectively right properties, as the following two theorems show. But before we state the first theorem, we have to define what local integrability means, because in the case of bump functions, local integrability will be exactly the property which needs in order to define a corresponding regular distribution:
Let be a function. We say that is locally integrable iff for all compact subsets of
Now we are ready to give some sufficient conditions on to define a corresponding regular distribution or regular tempered distribution by the way of
Let be a locally integrable function, and let be a domain. Then
is a regular distribution.
Well-definedness follows from the monotony of the integral:
In order to be less or equal than something, the first integral must have a well-defined value in the first place. Since we excluded infinity among the possible values, really maps to and well-definedness is proven.
Continuity follows similarly due to
Linearity follows due to the linearity of the integral.
Let , i. e.
is a regular tempered distribution
Well-definedness follows from Hölder's inequality:
Due to the triangle inequality for integrals and Hölder's inequality, we have
If in the notion of convergence of the Schwartz function space, then this expression goes to zero. Therefore, continuity is verified.
Linearity follows from the linearity of the integral.
Operations on DistributionsEdit
For there are operations such as the differentiation of , the convolution of and and the multiplication of and . In the following section, we want to define these three operations (differentiation, convolution with and multiplication with ) for a distribution instead of .
Let be open sets and let be a linear function. If there is a linear and continuous function such that
, then for every distribution , the function is a distribution. Therefore, we may define a function
This function has the property
Noticing that differentiation, convolution and multiplication are linear, we will define these operations for distributions by taking as the respective of these three operations.
We have to prove two claims: First, that the function is a distribution, and second that as defined above has the property
We show that the function is a distribution.
has a well-defined value in as maps to , which is exactly the preimage of . The function is continuous since it is the composition of two continuous functions, and it is linear for the same reason (see exercise 2).
We show that has the property
For every , we have
Since equality of two functions is equivalent to equality of these two functions evaluated at every point, this shows the desired property.
Using the last lemma 4.13, we define operations like multiplication with a smooth function, differentiation and convolution with a smooth function for distributions.
For the bump functions and the Schwartz functions, we also may define the differentiation of distributions. Let and . Let's now define
Then, for the spaces or , the requirements for the above lemma 1.4 are met and we may define the differentiation of distribution in the following way:
This definition also satisfies .
Proof: By integration by parts, we obtain:
, where is the i-th component of the outward normal vector and is the boundary of . For bump functions, the boundary integral vanishes anyway, because the functions in are zero there. For Schwartz functions, we may use the identity
and the decreasing property of the Schwartz functions to see that the boundary integral goes to zero and therefore
To derive the equation
, we may apply the formula from above several times. This finishes the proof, because this equation was the only non-trivial property of , which we need for applying lemma 1.5.
Let , and let . Let's define
This function () is linear, because the integral is linear. It is called the convolution of and .
We can also define: , and:
By the theorem of Fubini, we can calculate as follows:
Therefore, the first assumption for Lemma 1.5 holds.
Due to the Leibniz integral rule, we obtain that for (i. e. is integrable) and (i. e. the partial derivatives of exist up to order and are also continuous):
With this formula, we can see (due to the monotony of the integral) that
From this follows sequential continuity for Schwartz and bump functions by defining and . Thus, with the help of lemma 1.5, we can define the convolution with a distribution of or as follows:
- Show that endowed with the usual topology is a topological vector space.