# Partial Differential Equations/Test function spaces

Partial Differential Equations
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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 $[0, 1), [1, 2), [2, 3), [3, 4), [4, 5)$ and zero elsewhere; like, for example, these two:

Let's call the left function $f_1$, and the right function $f_2$.

Of course we can easily see that the two functions are different; they differ on the interval $[4, 5)$; however, let's pretend that we are blind and our only way of finding out something about either function is evaluating the integrals

$\int_{\mathbb R} \varphi(x) f_1(x) dx$ and $\int_{\mathbb R} \varphi(x) f_2(x) dx$

for functions $\varphi$ in a given set of functions $\mathcal X$.

We proceed with choosing $\mathcal X$ sufficiently clever such that five evaluations of both integrals suffice to show that $f_1 \neq f_2$. To do so, we first introduce the characteristic function. Let $A \subseteq \mathbb R$ be any set. The characteristic function of $A$ is defined as

$\chi_A(x) := \begin{cases} 1 & x \in A \\ 0 & x \notin A \end{cases}$

With this definition, we choose the set of functions $\mathcal X$ as

$\mathcal X := \{\chi_{[0, 1)}, \chi_{[1, 2)}, \chi_{[2, 3)}, \chi_{[3, 4)}, \chi_{[4, 5)}\}$

It is easy to see (see exercise 1), that for $n \in \{1, 2, 3, 4, 5\}$, the expression

$\int_{\mathbb R} \chi_{[n-1, n)} (x) f_1 (x) dx$

equals the value of $f_1$ on the interval $[n - 1, n)$, and the same is true for $f_2$. But as both functions are uniquely determined by their values on the intervals $[n - 1, n), n \in \{1, 2, 3, 4, 5\}$ (since they are zero everywhere else), we can implement the following equality test:

$f_1 = f_2 \Leftrightarrow \forall \varphi \in \mathcal X : \int_{\mathbb R} \varphi(x) f_1(x) dx = \int_{\mathbb R} \varphi(x) f_2(x) dx$

This obviously needs five evaluations of each integral, as $\# \mathcal X = 5$.

Since we used the functions in $\mathcal X$ to test $f_1$ and $f_2$, we call them test functions. What we ask ourselves now is if this notion generalises from functions like $f_1$ and $f_2$, which are piecewise constant on certain intervals and zero everywhere else, to continuous functions. The following chapter shows that this is true.

## Bump functions

In order to write down the definition of a bump function more shortly, we need the following two definitions:

Definition 3.1:

Let $B \subseteq \mathbb R^d$, and let $f: B \to \mathbb R$. We say that $f$ is smooth iff all the partial derivatives

$\partial_\alpha f, \alpha \in \mathbb N_0^d$

exist in all points of $B$ and are continuous. We write $f \in \mathcal C^\infty(B)$.

Definition 3.2:

Let $f: \mathbb R^d \to \mathbb R$. We define the support of $f$, $\text{supp } f$, as follows:

$\text{supp } f := \overline{\{x \in \R^d | f(x) \neq 0\}}$

Now we are ready to define a bump function:

Definition 3.3:

$\varphi: \mathbb R^d \to \R$ is called a bump function iff $\varphi \in \mathcal C^\infty (\mathbb R^d)$ and $\text{supp } \varphi$ is compact.

These two properties make the function really look like a bump, as the following example shows:

The standard mollifier $\eta$ in dimension $d=1$

Example 3.4: The standard mollifier $\eta$, given by

$\eta: \mathbb R^d \to \mathbb R, \eta(x) = \frac{1}{c}\begin{cases} e^{-\frac{1}{1-\|x\|^2}}& \text{ if } \|x\|_2 < 1\\ 0& \text{ if } \|x\|_2 \geq 1 \end{cases}$

, where $c := \int_{B_1(0)} e^{-\frac{1}{1-\|x\|^2}} dx$, is a bump function.

## Schwartz functions

As for the bump functions, in order to write down the definition of Schwartz functions shortly, we first need two helpful definitions.

Definition 3.5:

Let $X$ be an arbitrary set, and let $f : X \to \mathbb R$ be a function. Then we define the supremum norm of $f$ as follows:

$\|f\|_\infty := \sup\limits_{x \in X} |f(x)|$

Definition 3.6:

For a vector $x = (x_1, \ldots, x_d) \in \R^d$ and a $d$-dimensional multiindex $\alpha \in \N_0^d$ we define $x^\alpha$, $x$ to the power of $\alpha$, as follows:

$x^\alpha := x_1^{\alpha_1} \cdots x_d^{\alpha_d}$

Now we are ready to define a Schwartz function.

Definition 3.7:

We call $\phi: \R^d \to \mathbb R$ a Schwartz function iff the following two conditions are satisfied:

1. $\phi \in \mathcal C^\infty (\mathbb R^d)$
2. $\forall \alpha, \beta \in \mathbb{N}_0^d: \|x^\alpha \partial_\beta \phi \|_\infty < \infty$

By $x^\alpha \partial_\beta \phi$ we mean the function $x \mapsto x^\alpha \partial_\beta \phi(x)$.

$f(x, y) = e^{-x^2-y^2}$

Example 3.8: The function

$f: \R^2 \to \R, f(x, y) = e^{-x^2-y^2}$

is a Schwartz function.

Theorem 3.9:

Every bump function is also a Schwartz function.

This means that for example, the standard mollifier is also a Schwartz function.

Proof:

Let $\varphi$ be a bump function. Then, by definition of a bump function, $\varphi \in \mathcal C^\infty(\mathbb R^d)$. By the definition of bump functions, we choose $R > 0$ such that

$\text{supp } \varphi \subseteq \overline{B_R(0)}$

, as in $\mathbb R^d$, a set is compact iff it is closed & bounded. Further, for $\alpha, \beta \in \mathbb N_0^d$ arbitrary,

\begin{align} \|x^\alpha \partial_\beta \varphi(x)\|_\infty & := \sup_{x \in \mathbb R^d} |x^\alpha \partial_\beta \varphi(x)| & \\ & = \sup_{x \in \overline{B_R(0)}} |x^\alpha \partial_\beta \varphi(x)| & \text{supp } \varphi \subseteq \overline{B_R(0)} \\ & = \sup_{x \in \overline{B_R(0)}} \left( |x^\alpha| |\partial_\beta \varphi(x)| \right) & \text{rules for absolute value} \\ & \le \sup_{x \in \overline{B_R(0)}} \left( R^{|\alpha|} |\partial_\beta \varphi(x)| \right) & \forall i \in \{1, \ldots, d\}, (x_1, \ldots, x_d) \in \overline{B_R(0)} : |x_i| \le R \\ & < \infty & \text{Extreme value theorem} \end{align}
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## 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.

Definitions 3.10:

Let $X$ be a completely arbitrary set. Let $\tau \subseteq \mathcal P(X)$, where $\mathcal P(X)$ is the power set of $X$. We say that $\tau$ is a topology if and only if it satisfies the following three properties:

1. The whole set $X$ and the empty set are in $\tau$ ($X \in \tau, \emptyset \in \tau$)
2. Arbitrary unions of sets in $\tau$ are again in $\tau$ (for a family of sets $O_\alpha, \alpha \in A$ such that $\forall \alpha \in A : O_\alpha \in \tau$ holds that $\bigcup_{\alpha \in A}^{~} O_\alpha \in \tau$)
3. Finite intersections of sets in $\tau$ are again in $\tau$ ($O_1, \ldots, O_k \in \tau \Rightarrow \bigcap_{i = 1}^k O_k \in \tau$)

The elements of $\tau$ are called open sets. A set $F \in X$ is called closed set iff it's complement $X \setminus F$ is an open set, i. e. $X \setminus F \in \tau$.

Theorem 3.11:

Let $\tau \subseteq \mathcal P(X)$ be a topology, and let $X$ equipped with $\tau$ be an arbitrary topological space. That then arbitrary intersections of closed sets are closed again.

Proof:

Let

$F := \bigcap_{\upsilon \in \Upsilon} F_\upsilon$

where $\Upsilon$ is an arbitrary set and $F_\upsilon, \upsilon \in \Upsilon$ are arbitrary closed sets. We have

\begin{align} F &:= \bigcap_{\upsilon \in \Upsilon} F_\upsilon & \\ &= X \setminus \left( \bigcup_{\upsilon \in \Upsilon} (X \setminus F_\upsilon) \right) & \text{exercise ?} \end{align}

Since $\bigcup_{\upsilon \in \Upsilon} (X \setminus F_\upsilon)$ is open as the union of open sets, $F$ is closed.

$////$

Definition 3.12:

Let $\mathbb F$ be a field. A set $V$ is called a vector space over $\mathbb F$ iff it is endowed with a vector addition

$+ : V \times V \to V$

and a scalar-vector multiplication

$\cdot : \mathbb F \times V \to V$

such that the following conditions are fulfilled:

1. $\forall \mathbf u, \mathbf v, \mathbf w \in V : (\mathbf u + \mathbf v) + \mathbf w = \mathbf u + (\mathbf v + \mathbf w)$
2. $\forall \mathbf u, \mathbf v \in V : \mathbf u + \mathbf v = \mathbf v + \mathbf u$
3. $\forall a, b \in \mathbb F, \mathbf v \in V : (ab)\mathbf v = a(b\mathbf v)$
4. $\forall a \in \mathbb F, \mathbf u, \mathbf v \in V : a(\mathbf u + \mathbf v) = a\mathbf u + a\mathbf v$
5. $\forall a, b \in \mathbb F, \mathbf v \in V : (a + b)\mathbf v = a\mathbf v + b\mathbf v$
6. $\exists \mathbf 0 \in V : \forall \mathbf v \in V : \mathbf v + \mathbf 0 = \mathbf v$ (we call $\mathbf 0$ the zero element)
7. $\forall \mathbf v \in V : \exists -\mathbf v \in V : \mathbf v + -\mathbf v = 0$ (we call $-\mathbf v$ ‘minus $\mathbf v$’)
8. If $1$ is the multiplicative identity in $\mathbb F$, then $\forall \mathbf v \in V : 1\mathbf v = \mathbf v$

Definition 3.13:

Let $\mathcal X$, $\mathcal Y$ be two sets with topology, and let $f : \mathcal X \to \mathcal Y$ be a function. We say that $f$ is continuous iff for all open sets $O \subseteq \mathcal Y$, $f^{-1}(O)$ is open in $\mathcal X$.

Theorem 3.14:

Let $\mathcal X$, $\mathcal Y$ be two sets with topology, and let $f : \mathcal X \to \mathcal Y$ be a function. $f$ is continuous iff for all closed sets $F \subseteq \mathcal Y$, $f^{-1}(F)$ is closed in $\mathcal X$

Proof:

Let $A \subseteq \mathcal Y$ be any set. Then by definition of closed sets, $A$ is closed iff $\mathcal Y \setminus A$ is open. Therefore, $f$ is continuous iff for all closed sets $F \subseteq \mathcal Y$, we have that $f^{-1}(\mathcal Y \setminus F)$ is open. But

$f^{-1}(\mathcal Y \setminus F) = \mathcal X \setminus f^{-1}(F)$

, and therefore $f$ is continuous iff for all closed sets $F \subseteq \mathcal Y$, we have that $f^{-1}(F)$ is closed.

$////$

Definition 3.15:

Let $\mathcal X$, $\mathcal Y$ be sets with a topology. We consider the cartesian product $\mathcal X \times \mathcal Y$. The topology on $\mathcal X \times \mathcal Y$ with the fewest open sets such that both projections

$\text{pr}_{\mathcal X}: \mathcal X \times \mathcal Y \to \mathcal X, \text{pr}_{\mathcal X}((x, y)) := x$
$\text{pr}_{\mathcal Y}: \mathcal X \times \mathcal Y \to \mathcal Y, \text{pr}_{\mathcal Y}((x, y)) := y$

are continuous is called the product topology on $\mathcal X \times \mathcal Y$.

We can also describe the product topology with respect to the closed sets of the topological spaces $\mathcal X$ and $\mathcal Y$:

Theorem 3.16:

Let $\mathcal X$, $\mathcal Y$ be sets with a topology. We consider the cartesian product $\mathcal X \times \mathcal Y$. The topology on $\mathcal X \times \mathcal Y$ which induces the fewest closed sets such that both projections

$\text{pr}_{\mathcal X}: \mathcal X \times \mathcal Y \to \mathcal X, \text{pr}_{\mathcal X}((x, y)) := x$
$\text{pr}_{\mathcal Y}: \mathcal X \times \mathcal Y \to \mathcal Y, \text{pr}_{\mathcal Y}((x, y)) := y$

are continuous is the product topology on $\mathcal X \times \mathcal Y$.

Proof:

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.

$////$

Definition 3.17:

A field $\mathbb F$ is called a topological field iff it is endowed with a topology in which addition $+ : \mathbb F \times \mathbb F \to \mathbb F$, multiplication $\cdot : \mathbb F \times \mathbb F \to \mathbb F$ and inversion $~^{-1} : \mathbb F \setminus \{0\} \to \mathbb F \setminus \{0\}$ are continuous, where we endow $\mathbb F \times \mathbb F$ 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.

Definition 3.18:

A vector space $V$ 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 $V \times V$ or $\mathbb F \times V$ 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 $\mathbb R$ with the usual topology, i. e. a set $U \subseteq \mathbb R$ is open iff

$\forall x \in U : \exists \epsilon > 0 : (x - \epsilon, x + \epsilon) \subseteq U$

Then $\mathbb R$ is a topological field. Further, if we equipp $\mathbb R^d$, which is a vector space over $\mathbb R$, with the usual topology, which means that a set $O \subseteq \mathbb R^d$ is open iff

$\forall x \in O : \exists \epsilon > 0 : B_\epsilon(x) \subseteq O$

, then $\mathbb R^d$ is a topological vector space over $\mathbb R$, given that $\mathbb R$ is equipped with this usual topology.

## Sequential spaces

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.

Definition 3.20:

Let $X$ be any set. We say that $X$ has a notion of convergence iff for every sequence $(x_l)_{l \in \mathbb N}$ with elements in $X$ and every $x \in X$, we have either $x_l \to x, l \to \infty$ or $x_l \not\to x, l \to \infty$.

If we have two sets $X$ and $Y$ which both have a notion of convergence, we may define sequential continuity for functions $f: X \to Y$.

Definition 3.21:

Let $X$, $Y$ be two sets with a notion of convergence and let $f: X \to Y$ be a function. $f$ is defined to be sequentially continuous iff

$x_l \to x, l \to \infty \text{ in } X \Rightarrow f(x_l) \to f(x), l \to \infty \text{ in } Y$

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.

Definition 3.22:

Let $\mathcal X$ be a topological space which has a notion of convergence. We call $\mathcal X$ a sequential space iff for all sets $B \subseteq \mathcal X$

$B \text{ is closed } \Leftrightarrow \forall \{x_l | l \in \mathbb N\} \subseteq B, x_l \to x : x \in B$

Now let's show that for functions from a sequential space into a sequential space, sequential continuity implies continuity.

Theorem 3.23:

Let $\mathcal X, \mathcal Y$ be sequential spaces, and let $f: \mathcal X \to \mathcal Y$ be a function. Then we have

$f \text{ sequentially continuous } \Rightarrow f \text{ continuous}$

Proof:

Let $f: \mathcal X \to \mathcal Y$ be sequentially continuous. Let $F \subseteq \mathcal Y$ be an arbitrary closed set. We show that $\mathcal T^{-1}(F)$ 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 $(x_l)_{l \in \mathbb N}$ be a sequence such that $\forall l \in \mathbb N : x_l \in f^{-1}(F)$ and $x_l \to x, l \to \infty$. We have

$\forall l \in \mathbb N : f(x_l) \in F$

, because else $x_l$ would by definition of the preimage not be in $f^{-1}(F)$. Since $f$ is sequentially continuous, we have

$\lim_{l \to \infty} f(x_l) = f(x)$

Due to the closedness of $F$ and due to the fact that $\mathcal Y$ is a sequential space it further follows that

$f(x) \in F$

and thus, again by definition of the preimage,

$x \in f^{-1}(F)$

Since $\mathcal X$ is a sequential space, $f^{-1}(F)$ thus is closed (as in sequential spaces a set is closed if and only if it contains all the limit points).

$////$

Theorem 3.24:

Let $X$ be a set with a notion of convergence with the properties

1. $X$ is closed with respect to the notion of convergence (if $x_n \to x$, then $x \in X$)
2. Infinite subsequences converge, and to the same limit (if $x_n \to x$ and $(x_{n_k})$ is an infinite subsequence of $(x_n)$, then $x_{n_k} \to x$)

If we define a set $F$ as closed iff $\left( x_n \to x \wedge \{x_n : n \in \mathbb N\} \subseteq F \right) \Rightarrow x \in F$ and define $\tau := \{O \subset X : X \setminus O \text{ is closed}\}$ , then $\tau$ is a topology, and $X$ equipped with $\tau$ is a sequential space.

Proof:

1.

We verify that $\tau$ has the defining properties of a topology.

1.1.

We first show that $X \in \tau$ and $\emptyset \in \tau$.

$\emptyset$ is closed with respect to our definition, because there simply are no sequences in the empty set and therefore

$\neg x \in \emptyset \Rightarrow \neg \left( x_n \to x \wedge \{x_n : n \in \mathbb N\} \subseteq \emptyset \right)$

Therefore, we have $X \setminus \emptyset = X \in \tau$.

Furthermore, the whole set $X$ is closed in $X$. This is because of the first property we required for the notion of convergence. Therefore, $X \setminus X = \emptyset \in \tau$.

1.2.

We show that arbitrary unions of open sets (i. e. elements of $\tau$) are open again (i. e. elements of $\tau$ again).

Let $O_\alpha, \alpha \in A$ be a family of open sets, and let

$O = \bigcup_{\alpha \in A} O_\alpha$

To show that $O$ is open, we need, by the definition of $\tau$, to show that $X \setminus O$ is closed with respect to the definition of closedness we have introduced. Let now $x_n \to x$ such that $\{x_n : n \in \mathbb N\} \in X \setminus O$. Assume that $x \notin X \setminus O$. Due to the second de Morgan formula, we have

$X \setminus O = X \setminus \bigcup_{\alpha \in A} O_\alpha = \bigcap_{\alpha \in A} (X \setminus O_\alpha)$

, which is why $x \notin X \setminus O$ is equivalent to $\exists \alpha \in A : x \notin X \setminus O_\alpha$. But since $\{x_n : n \in \mathbb N\} \in X \setminus O \subseteq X \setminus O_\alpha$, $X \setminus O_\alpha$ is not closed, which is why $O_\alpha$ is not open. Contradiction! Therefore, it must $x \in X \setminus O$, and therefore $X \setminus O$ is closed, and $O$ is open.

1.3.

We show that the intersection of finitely many open sets is open.

Let $O_j, j \in \{1, \ldots, n\}$ be open, and let

$U := \bigcap_{j=1}^n O_j$

Again, by definition of $\tau$, in order to show that $U$ is open, we need to show that $X \setminus U$ is closed. Let $(x_l)_{l \in \mathbb N}$ be a sequence such that $\{x_l : l \in \mathbb N\} \subseteq X \setminus U$ and $x_l \to x$ and assume that $x \notin X \setminus U$. Again, we use one of the two de Morgan formulas, this time the first one:

$X \setminus U = X \setminus \bigcap_{j=1}^n O_j = \bigcup_{j=1}^n (X \setminus O_j)$

Therefore, $x \notin X \setminus U \Leftrightarrow \forall j \in \{1, \ldots, n\} : x \notin X \setminus O_j$. We now choose $j \in \{1, \ldots, n\}$ such that infinitely many entries of the sequence $(x_l)_{l \in \mathbb N}$ are in $X \setminus O_j$. This is possible because if each $(X \setminus O_j), j \in \{1, \ldots, n\}$ would contain only finitely many entries of $(x_l)_{l \in \mathbb N}$, then since $\{x_l : l \in \mathbb N \} \subseteq \bigcup_{j=1}^n (X \setminus O_j)$, the natural numbers would have finite cardinality. Furthermore, we choose $x_{l_m}$ as the subsequence of $x_l$ whose entries are in $X \setminus O_j$. Due to our second requirement on the notion of convergence, $x_{l_m} \to x$, and since $X \setminus O_j$ is closed, $x \in X \setminus O_j \subseteq X \setminus U$, contradiction! Therefore, $x \in X \setminus U$, and $X \setminus U$ is closed, which is why $U$ is open.

Therefore $\tau$ has all three defining properties of a topology.

2.

From the definition of sequential spaces, it is clear that the resulting space is sequential.

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## 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 $d \in \mathbb N$ and $O \subseteq \R^d$ an open set. The set of bump functions for the set $O$, written $\mathcal D(O)$, is defined as

$\mathcal D(O) := \left\{ \varphi \in \mathcal C^{\infty}(\mathbb R^d) | \text{supp } \varphi \text{ is a compact subset of } O \right\}$

A notion of convergence for this space we define as follows: A sequence of bump functions $(\varphi_i)_{i \in \N}$ is said to converge to another bump function $\varphi$ iff the following two conditions are satisfied:

1. There is a compact set $K \subset \Omega$ such that $\forall i \in \N: \text{supp } \varphi_i \subseteq K$
2. $\forall \alpha \in \mathbb N_0^d : \lim_{i \rightarrow \infty} \|\partial_\alpha \varphi_i - \partial_\alpha \varphi\|_\infty = 0$

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 $\mathcal D(O)$ a topological vector space. The vectors in this space will be the functions themselves.

Theorem 3.26:

Let $d \in \mathbb N$ and let $O \subseteq \R^d$ be an open set. If we endow $\mathcal D(O)$ with pointwise definitions of addition and scalar-vector multiplication, i. e.

$\forall \varphi, \vartheta \in \mathcal D(O) : \forall x \in \mathbb R^d : (\varphi + \vartheta)(x) := \varphi(x) + \vartheta(x)$
$\forall c \in \mathbb R, \varphi \in \mathcal D(O) : \forall x \in \mathbb R^d : (c \varphi)(x) := c \varphi(x)$

and endow $\mathcal D(O)$ with the topology which is defined by the definition of closed sets in theorem 2.16 and the notion of convergence of $\mathcal D(O)$, then $\mathcal D(O)$ is a topological vector space.

Proof:

We have to show three things:

1. $\mathcal D(O)$ is a vector space with respect to the given definitions of vector-vector addition and scalar-vector multiplication
2. $\mathcal D(O)$ is a topological space with respect to the topology defined by the notion of convergence and theorem 3.24
3. 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.

3.1.

We show that the product topology in the space $\mathcal D(O) \times \mathcal D(O)$ is given by the topology generated by theorem 3.24 and the notion of convergence in $\mathcal D(O) \times \mathcal D(O)$ defined by:

$(\varphi_l, \vartheta_l) \to (\varphi, \vartheta) :\Leftrightarrow \varphi_l \to \varphi \text{ in } \mathcal D(O) \wedge \vartheta_l \to \vartheta \text{ in } \mathcal D(O)$

For every $n \in \mathbb N$ and every $\varrho \in \mathcal D(O)$, we define

$U_n(\varrho) := \left\{ \varphi \in \mathcal D(O) \big| \forall \alpha \in \mathbb N_0^d \text{ such that } |\alpha| \le n : \|\partial_\alpha(\varphi - \varrho)\|_\infty < \frac{1}{n} \right\}$

We show that $U_n(\varrho)$ is open in $\mathcal D(O)$ for arbitrary $n \in \mathbb N$ and $\vartheta \in \mathcal D(O)$. We recall that the closed sets of $\mathcal D(O)$ 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 $U_n(\varrho)$ contains all the limits of the sequences which are completely contained in it.

Let therefore $(\varphi_l)_{l \in \mathbb N}$ be a sequence in $\mathcal D(O) \setminus U_n(\varrho)$ such that $\varphi_l$ converges to a $\varphi \in \mathcal D(O)$. Let's assume that its limit, instead of being contained in $\mathcal D(O) \setminus U_n(\varrho)$, was contained in $U_n(\varrho)$. Due to our notion of convergence on the set of bump functions from $\varphi_l \to \varphi, l \to \infty$ follows that

$\forall \alpha \in \mathbb N_0^d : \|\partial_\alpha (\varphi_l - \varphi)\|_\infty \to 0, l \to \infty$

or, inserting the epsilon definition of convergence,

$\forall \alpha \in \mathbb N_0^d : \forall \epsilon \in \mathbb R_{>0} : \exists l_{\epsilon, \alpha} \in \mathbb N : \forall l > l_{\epsilon, \alpha} : \|\partial_\alpha (\varphi_l - \varphi)\|_\infty < \epsilon (*)$

We now choose $\epsilon$ so small that

$\forall \alpha \in \mathbb N_0^d \text{ such that } |\alpha| \le n : \|\partial_\alpha(\varphi - \varrho)\| + \epsilon < \frac{1}{n} (**)$

and

$l_\epsilon := \max_{\alpha \in \mathbb N_0^d \text{ such that } |\alpha| < n} l_{\epsilon, \alpha} (***)$

, where the maximum exists because finite subsets of the real numbers always have a maximum. From $(*), (**) \text{ and } (***)$ then follows

\begin{align} \forall l > l_\epsilon : \forall \alpha \in \mathbb N_0^d \text{ such that } |\alpha| \le n : \|\partial_\alpha (\varrho - \varphi_l)\|_\infty & \le \|\partial_\alpha (\varrho - \varphi)\|_\infty + \|\partial_\alpha (\varphi_l - \varphi)\|_\infty & \text{triangle ineq.} \\ & < \|\partial_\alpha(\varphi - \varrho)\|_\infty + \epsilon & (*) \text{ and } (***) \\ & < \frac{1}{n} & (**) \end{align}

This shows that for $l > l_\epsilon$, $\varphi_l \in U_n(\varrho)$, which contradicts the assumption that $(\varphi_l)_{l \in \mathbb N}$ was a sequence entirely contained in $\mathcal D(O) \setminus U_n(\varrho)$. Therefore, we must have $\varphi \in \mathcal D(O) \setminus U_n(\varrho)$ and since $(\varphi_l)_{l \in \mathbb N}$ was an arbitrary convergent sequence, this shows that $\mathcal D(O) \setminus U_n(\varrho)$ is closed, which shows that $U_n(\varrho)$ is open.

As in the product topology of $\mathcal D(O) \times \mathcal D(O)$ both projections are continuous, we have that for arbitrary $n \in \mathbb N$ and $\varrho_1, \varrho_2 \in \mathcal D(O)$

$\text{pr}_1^{-1}(U_n(\varrho_1))$

and

$\text{pr}_2^{-1}(U_n(\varrho_2))$

are open, and as the product topology is a topology,

$\text{pr}_1^{-1}(U_n(\varrho_1)) \cap \text{pr}_2^{-1}(U_n(\varrho_2))$

is also open, which is why

$\mathcal D(O) \times \mathcal D(O) \setminus \left( \text{pr}_1^{-1}(U_n(\varrho_1)) \cap \text{pr}_2^{-1}(U_n(\varrho_2)) \right)$

is closed with respect to the product topology.

Furthermore, for $B \subseteq \mathcal D(O)$ we can calculate $\text{pr}_i^{-1}(B)$ explicitly for $i \in \{1, 2\}$:

\begin{align} \text{pr}_1^{-1}(B) & = \{(\varphi, \vartheta) \in \mathcal D(O) \times \mathcal D(O) | \varphi \in B\} \\ & = B \times \mathcal D(O) \end{align}
\begin{align} \text{pr}_2^{-1}(B) & = \{(\varphi, \vartheta) \in \mathcal D(O) \times \mathcal D(O) | \vartheta \in B\} \\ & = \mathcal D(O) \times B \end{align}

Therefore,

\begin{align} \text{pr}_1^{-1}(U_n(\varrho_1)) \cap \text{pr}_2^{-1}(U_n(\varrho_2)) & = U_n(\varrho_1) \times \mathcal D(O) \cap \mathcal D(O) \times U_n(\varrho_2) \\ & = \{(\varphi, \vartheta) \in \mathcal D(O) \times \mathcal D(O) | \varphi \in U_n(\varrho_1) \text{ and } \vartheta \in U_n(\varrho_2) \} \\ & = U_n(\varrho_1) \times U_n(\varrho_2) \end{align}

and thus

$\mathcal D(O) \times \mathcal D(O) \setminus \left( \text{pr}_1^{-1}(U_n(\varrho_1)) \cap \text{pr}_2^{-1}(U_n(\varrho_2)) \right) = \mathcal D(O) \times \mathcal D(O) \setminus \left( U_n(\varrho_1) \times U_n(\varrho_2) \right)$

Let now $A \subseteq \mathcal D(O) \times \mathcal D(O)$ 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 $\mathcal D(O) \times \mathcal D(O) \setminus U_n(\varrho_1) \times U_n(\varrho_2)$ being closed follows that

$\bigcap_{n \in \mathbb N, \varrho_1, \varrho_2 \in \mathcal D(O) \text{ such that } U_n(\varrho_1) \times U_n(\varrho_2) \cap A = \emptyset} \left( \mathcal D(O) \times \mathcal D(O) \setminus U_n(\varrho_1) \times U_n(\varrho_2) \right)$

is closed with respect to the product topology.

3.2

3.3

We also define the space of the Schwartz functions.

Definition 3.27:

Let $d \in \mathbb N$. Then the $d$-dimensional Schwartz space, written $\mathcal S(\R^d)$, is defined as

$\mathcal S(\R^d) := \left\{ \phi \in \mathcal C^\infty(\R^d) | \forall \alpha, \beta \in \mathbb N_0^d: \|x^\alpha \partial_\beta \phi\|_\infty < \infty \right\}$

The space of Schwartz functions also has a notion of convergence: We say that the sequence of Schwartz functions $(\phi_i)_{i \in \N}$ converges to $\phi$ iff the following condition is satisfied:

$\forall \alpha, \beta \in \mathbb N_0^d: \|x^\alpha \partial_\beta \phi_i - x^\alpha \partial_\beta \phi\|_\infty \to 0, i \to \infty$

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.

Theorem 3.28:

Let $d \in \mathbb N$. If we endow $\mathcal S(\mathbb R^d)$ with pointwise definitions of addition and scalar-vector multiplication, i. e.

$\forall \phi, \theta \in \mathcal S(\mathbb R^d) : \forall x \in \mathbb R^d : (\phi + \theta)(x) := \phi(x) + \theta(x)$
$\forall c \in \mathbb R, \phi \in \mathcal S(\mathbb R^d) : \forall x \in \mathbb R^d : (c \phi)(x) := c \phi(x)$

and endow $\mathcal S(\mathbb R^d)$ with the topology which is defined by from the definition of closed sets in theorem 2.16 and the notion of convergence of $\mathcal S(\mathbb R^d)$, then $\mathcal S(\mathbb R^d)$ is a topological vector space.

Proof:

## Exercises

1. Let $b \in \mathbb R$ and $f : \mathbb R \to \mathbb R$ be constant on the interval $[b-1, b)$. Show that

$\forall y \in [b-1, b) : \int_{\mathbb R} \chi_{[b-1, b)} (x) f(x) dx = f(y)$

2. Prove that $\mathbb R$ endowed with the topology stated in example 3.17 is a topological field.
3. Prove that $\mathbb R^d$ 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).
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