# Partial Differential Equations/Distributions

Partial Differential Equations
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## Distributions and tempered distributionsEdit

Definition 4.1:

Let $O \subseteq \mathbb R^d$ be open, and let $\mathcal T: \mathcal D(O) \to \mathbb R$ be a function. We call $\mathcal T$ a distribution iff

• $\mathcal T$ is linear ($\forall \varphi, \vartheta \in \mathcal D(O), b, c \in \mathbb R : \mathcal T(b \varphi + c \vartheta) = b \mathcal T(\varphi) + c \mathcal T(\vartheta)$)
• $\mathcal T$ is sequentially continuous (if $\varphi_l \to \varphi$ in the notion of convergence of bump functions, then $\mathcal T(\varphi_l) \to \mathcal T(\varphi)$ in the reals)

The set of all distributions for $\mathcal D(O)$ we denote by $\mathcal D(O)^*$

Definition 4.2:

Let $\mathcal T: \mathcal S(\mathbb R^d) \to \mathbb R$ be a function. We call $\mathcal T$ a tempered distribution iff

• $\mathcal T$ is linear ($\forall \varphi, \vartheta \in \mathcal S(\mathbb R^d), b, c \in \mathbb R : \mathcal T(b \varphi + c \vartheta) = b \mathcal T(\varphi) + c \mathcal T(\vartheta)$)
• $\mathcal T$ is sequentially continuous (if $\varphi_l \to \varphi$ in the notion of convergence of Schwartz functions, then $\mathcal T(\varphi_l) \to \mathcal T(\varphi)$ in the reals)

The set of all tempered distributions we denote by $\mathcal S(\mathbb R^d)$.

Theorem 4.3:

Let $\mathcal T$ be a tempered distribution. Then the restriction of $\mathcal T$ to bump functions is a distribution.

Proof:

Let $\mathcal T$ be a tempered distribution, and let $O \subseteq \mathbb R^d$ be open.

1.

We show that $\mathcal T(\varphi)$ has a well-defined value for $\varphi \in \mathcal D(O)$.

Due to theorem 3.9, every bump function is a Schwartz function, which is why the expression

$\mathcal T (\varphi)$

makes sense for every $\varphi \in \mathcal D(O)$.

2.

We show that the restriction is linear.

Let $a, b \in \mathbb R$ and $\varphi, \vartheta \in \mathcal D(O)$. Since due to theorem 3.9 $\varphi$ and $\vartheta$ are Schwartz functions as well, we have

$\forall a, b \in \mathbb R, \varphi, \vartheta \in \mathcal D(O) : \mathcal T (a \varphi + b \vartheta) = a \mathcal T (\varphi) + b \mathcal T (\vartheta)$

due to the linearity of $\mathcal T$ for all Schwartz functions. Thus $\mathcal T$ is also linear for bump functions.

3.

We show that the restriction of $\mathcal T$ to $\mathcal D(O)$ is sequentially continuous. Let $\varphi_l \to \varphi$ in the notion of convergence of bump functions. Due to theorem 3.11, $\varphi_l \to \varphi$ in the notion of convergence of Schwartz functions. Since $\mathcal T$ as a tempered distribution is sequentially continuous, $\mathcal T(\varphi_l) \to \mathcal T(\varphi)$.

$////$

## The convolutionEdit

Definition 4.4:

Let $f, g: \mathbb R^d \to \mathbb R$. The integral

$f * g : \mathbb R^d \to \mathbb R, (f * g)(y) := \int_{\mathbb R^d} f(x) g(y - x) dx$

is called convolution of $f$ and $g$ and denoted by $f * g$ if it exists.

The convolution of two functions may not always exist, but there are sufficient conditions for it to exist:

Theorem 4.5:

Let $p, q \in [1, \infty]$ such that $\frac{1}{p} + \frac{1}{q} = 1$ and let $f \in L^p(\mathbb R^d)$ and $g \in L^q(\mathbb R^d)$. Then for all $y \in O$, the integral

$\int_{\mathbb R^d} f(x) g(y - x) dx$

has a well-defined real value.

Proof:

Due to Hölder's inequality,

$\int_{\mathbb R^d} |f(x) g(y - x)| dx \le \left( \int_{\mathbb R^d} |f(x)|^p dx \right)^{1/p} \left( \int_{\mathbb R^d} |g(y - x)|^q dx \right)^{1/q} < \infty$.
$////$

We shall now prove that the convolution is commutative, i. e. $f * g = g * f$.

Theorem 4.6:

Let $p, q \in [1, \infty]$ such that $\frac{1}{p} + \frac{1}{q} = 1$ (where $\frac{1}{\infty} = 0$) and let $f \in L^p(\mathbb R^d)$ and $g \in L^q(\mathbb R^d)$. Then for all $y \in \mathbb R^d$:

$\forall y \in \mathbb R^d : (f * g)(y) = (g * f)(y)$

Proof:

We apply multi-dimensional integration by substitution using the diffeomorphism $x \mapsto y - x$ to obtain

$(f * g)(y) = \int_{\mathbb R^d} f(x) g(y - x) dx = \int_{\mathbb R^d} f(y - x) g(x) dx = (g * f)(y)$.
$////$

Lemma 4.7:

Let $O \subseteq \mathbb R^d$ be open and let $f \in L^1(\mathbb R^d)$. Then $f * \eta_\delta \in \mathcal C^\infty(\mathbb R^d)$.

Proof:

Let $\alpha \in \mathbb N_0^d$ be arbitrary. Then, since for all $y \in \mathbb R^d$

$\int_{\mathbb R^d} |f(x) \partial_\alpha \eta_\delta(y - x)| dx \le \|\partial_\alpha \eta_\delta\|_\infty \int_{\mathbb R^d} |f(x)| dx$

and further

$|f(x) \partial_\alpha \eta_\delta(y - x)| \le |f(x)|$,

Leibniz' integral rule (theorem 2.2) is applicable, and by repeated application of Leibniz' integral rule we obtain

$\partial_\alpha f * \eta_\delta= f * \partial_\alpha \eta_\delta$.
$////$

## Regular distributionsEdit

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 distributions.

Definition 4.8:

Let $O \subseteq \mathbb R^d$ be an open set and let $\mathcal T \in \mathcal D(O)^*$. If for all $\varphi \in \mathcal D(O)$ $\mathcal T(\varphi)$ can be written as

$\mathcal T(\varphi) = \int_O f(x) \varphi(x) dx$

for a function $f: O \to \mathbb R$ which is independent of $\varphi$, then we call $\mathcal T$ a regular distribution.

Definition 4.9:

Let $\mathcal T \in \mathcal S(\mathbb R^d)^*$. If for all $\phi \in \mathcal S(\mathbb R^d)$ $\mathcal T(\phi)$ can be written as

$\mathcal T(\phi) = \int_{\mathbb R^d} f(x) \phi(x) dx$

for a function $f: \mathbb R^d \to \mathbb R$ which is independent of $\phi$, then we call $\mathcal T$ a regular tempered distribution.

Two questions related to this definition could be asked: Given a function $f: \mathbb R^d \to \mathbb R$, is $\mathcal T_f: \mathcal D(O) \to \mathbb R$ for $O \subseteq \mathbb R^d$ open given by

$\mathcal T_f(\varphi) := \int_O f(x) \varphi(x) dx$

well-defined and a distribution? Or is $\mathcal T_f: \mathcal S(\mathbb R^d) \to \mathbb R$ given by

$\mathcal T_f(\phi) := \int_{\mathbb R^d} f(x) \phi(x) dx$

well-defined and 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 $f$ 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 $f$ needs in order to define a corresponding regular distribution:

Definition 4.10:

Let $O \subseteq \mathbb R^d$ be open, $f: O \to \mathbb R$ be a function. We say that $f$ is locally integrable iff for all compact subsets $K$ of $O$

$-\infty < \int_K f(x) dx < \infty$

We write $f \in L^1_\text{loc}(O)$.

Now we are ready to give some sufficient conditions on $f$ to define a corresponding regular distribution or regular tempered distribution by the way of

$\mathcal T_f : \mathcal D(O) \to \mathbb R, \mathcal T_f(\varphi) := \int_O f(x) \varphi(x) dx$

or

$\mathcal T_f : \mathcal S(\mathbb R^d) \to \mathbb R, \mathcal T_f(\phi) := \int_{\mathbb R^d} f(x) \phi(x) dx$:

Theorem 4.11:

Let $O \subseteq \mathbb R^d$ be open, and let $f: O \to \mathbb R$ be a function. Then

$\mathcal T_f : \mathcal D(O) \to \mathbb R, \mathcal T_f(\varphi) := \int_O f(x) \varphi(x) dx$

is a regular distribution iff $f \in L^1_\text{loc}(O)$.

Proof:

1.

We show that if $f \in L^1_\text{loc}(O)$, then $\mathcal T_f : \mathcal D(O) \to \mathbb R$ is a distribution.

Well-definedness follows from the triangle inequality of the integral and the monotony of the integral:

\begin{align} \left| \int_U \varphi(x) f(x) dx \right| \le \int_U |\varphi(x) f(x)| dx = \int_{\text{supp } \varphi} |\varphi(x) f(x)| dx\\ \le \int_{\text{supp } \varphi} \|\varphi\|_\infty |f(x)| dx = \|\varphi\|_\infty \int_{\text{supp } \varphi} |f(x)| dx < \infty \end{align}

In order to have an absolute value strictly less than infinity, the first integral must have a well-defined value in the first place. Therefore, $\mathcal T_f$ really maps to $\mathbb R$ and well-definedness is proven.

Continuity follows similarly due to

$|T_f \varphi_l - T_f \varphi| = \left| \int_K (\varphi_l - \varphi)(x) f(x) dx \right| \le \|\varphi_l - \varphi\|_\infty \underbrace{\int_K |f(x)| dx}_{\text{independent of } l} \to 0, l \to \infty$

, where $K$ is the compact set in which all the supports of $\varphi_l, l \in \mathbb N$ and $\varphi$ are contained (remember: The existence of a compact set such that all the supports of $\varphi_l, l \in \mathbb N$ are contained in it is a part of the definition of convergence in $\mathcal D(O)$, see the last chapter. As in the proof of theorem 3.11, we also conclude that the support of $\varphi$ is also contained in $K$).

Linearity follows due to the linearity of the integral.

2.

We show that $\mathcal T_f$ is a distribution, then $f \in L^1_\text{loc}(O)$ (in fact, we even show that if $\mathcal T_f(\varphi)$ has a well-defined real value for every $\varphi \in \mathcal D(O)$, then $f \in L^1_\text{loc}(O)$. Therefore, by part 1 of this proof, which showed that if $f \in L^1_\text{loc}(O)$ it follows that $\mathcal T_f$ is a distribution in $\mathcal D^*(O)$, we have that if $\mathcal T_f(\varphi)$ is a well-defined real number for every $\varphi \in \mathcal D(O)$, $\mathcal T_f$ is a distribution in $\mathcal D(O)$.

Let $K \subset U$ be an arbitrary compact set. We define

$\mu: K \to \mathbb R, \mu(\xi) := \inf_{x \in \mathbb R^d \setminus O} \|\xi - x\|$

$\mu$ is continuous, even Lipschitz continuous with Lipschitz constant $1$: Let $\xi, \iota \in \mathbb R^d$. Due to the triangle inequality, both

$\forall (x, y) \in \mathbb R^2 : \|\xi - x\| \le \|\xi - \iota\| + \|\iota - y\| + \|y - x\| ~~~~~(*)$

and

$\forall (x, y) \in \mathbb R^2 : \|\iota - y\| \le \|\iota - \xi\| + \|\xi - x\| + \|x - y\| ~~~~~(**)$

, which can be seen by applying the triangle inequality twice.

We choose sequences $(x_l)_{l \in \mathbb N}$ and $(y_m)_{m \in \mathbb N}$ in $\mathbb R^d \setminus O$ such that $\lim_{l \to \infty} \|\xi - x_l\| = \mu(\xi)$ and $\lim_{m \to \infty} \|\iota - y_m\| = \mu(\iota)$ and consider two cases. First, we consider what happens if $\mu(\xi) \ge \mu(\iota)$. Then we have

\begin{align} |\mu(\xi) - \mu(\iota)| & = \mu(\xi) - \mu(\iota) & \\ & = \inf_{x \in \mathbb R^d \setminus O} \|\xi - x\| - \inf_{y \in \mathbb R^d \setminus O} \|\iota - y\| & \\ & = \inf_{x \in \mathbb R^d \setminus O} \|\xi - x\| - \lim_{m \to \infty} \|\iota - y_m\| & \\ & = \lim_{m \to \infty} \inf_{x \in \mathbb R^d \setminus O} \left( \|\xi - x\| - \|\iota - y_m\| \right) & \\ & \le \lim_{m \to \infty} \inf_{x \in \mathbb R^d \setminus O} \left( \|\xi - \iota\| + \|x - y_m\| \right) & (*) \text{ with } y = y_m \\ & = \|\xi - \iota\| & \end{align}.

Second, we consider what happens if $\mu(\xi) \le \mu(\iota)$:

\begin{align} |\mu(\xi) - \mu(\iota)| & = \mu(\iota) - \mu(\xi) & \\ & = \inf_{y \in \mathbb R^d \setminus O} \|\iota - y\| - \inf_{x \in \mathbb R^d \setminus O} \|\xi - x\| & \\ & = \inf_{y \in \mathbb R^d \setminus O} \|\iota - y\| - \lim_{l \to \infty} \|\xi - x_l\| & \\ & = \lim_{l \to \infty} \inf_{y \in \mathbb R^d \setminus O} \left( \|\iota - y\| - \|\xi - x_l\| \right) & \\ & \le \lim_{l \to \infty} \inf_{y \in \mathbb R^d \setminus O} \left( \|\xi - \iota\| + \|y - x_l\| \right) & (**) \text{ with } x = x_l \\ & = \|\xi - \iota\| & \end{align}

Since always either $\mu(\xi) \ge \mu(\iota)$ or $\mu(\xi) \le \mu(\iota)$, we have proven Lipschitz continuity and thus continuity. By the extreme value theorem, $\mu$ therefore has a minimum $\kappa \in \mathbb R^d$. Since $\mu(\kappa) = 0$ would mean that $\|\xi - x_l\| \to 0, l \to \infty$ for a sequence $(x_l)_{l \in \mathbb N}$ in $\mathbb R^d \setminus O$ which is a contradiction as $\mathbb R^d \setminus O$ is closed and $\kappa \in K \subset O$, we have $\mu(\kappa) > 0$.

Hence, if we define $\delta := \mu(\kappa)$, then $\delta > 0$. Further, the function

$\vartheta: \mathbb R^d \to \mathbb R, \vartheta(x) := (\chi_{K + B_{\delta/4}(0)} * \eta_{\delta/4})(x) = \int_{\mathbb R^d} \eta_{\delta/4}(y) \chi_{K + B_{\delta/4}(0)}(x - y) dy = \int_{B_{\delta/4}(0)} \eta_{\delta/4}(y) \chi_{K + B_{\delta/4}(0)}(x - y) dy$

has support contained in $O$, is equal to $1$ within $K$ and further is contained in $\mathcal C^\infty(\mathbb R^d)$ due to lemma 4.7. Hence, it is also contained in $\mathcal D(O)$. Since therefore, by the monotonicity of the integral

$\int_K |f(x)| dx = \int_O |f(x)| \chi_K(x) dx \le \int_{\mathbb R^d} |f(x)| \vartheta(x) dx$

, $f$ is indeed locally integrable.

$////$

Theorem 4.12:

Let $f \in L^2(\mathbb R^d)$, i. e.

$\int_{\mathbb R^d} |f(x)|^2 dx < \infty$

Then

$\mathcal T_f : \mathcal S(\mathbb R^d) \to \mathbb R, \mathcal T_f(\phi) := \int_{\mathbb R^d} f(x) \phi(x) dx$

is a regular tempered distribution.

Proof:

From Hölder's inequality we obtain

$\int_{\R^d} |\phi(x)| |f(x)| dx \le \|\phi\|_{L^2} \|f\|_{L^2} < \infty$.

Hence, $\mathcal T_f$ is well-defined.

Due to the triangle inequality for integrals and Hölder's inequality, we have

$|T_f(\phi_l) - T_f(\phi)| \le \int_{\R^d} |(\phi_l - \phi)(x)| |f(x)| dx \le \|\phi_l - \phi\|_{L^2} \|f\|_{L^2}$

Furthermore

\begin{align} \|\phi_l - \phi\|_{L^2}^2 & \le \|\phi_l - \phi\|_\infty \int_{\R^d} |(\phi_l - \phi)(x)| dx \\ & = \|\phi_l - \phi\|_\infty \int_{\R^d} \prod_{j=1}^d (1 + x_j^2) |(\phi_l - \phi)(x)| \frac{1}{\prod_{j=1}^d (1 + x_j^2)} dx \\ & \le \|\phi_l - \phi\|_\infty \left\|\prod_{j=1}^d (1 + x_j^2) (\phi_l - \phi)\right\|_\infty \underbrace{\int_{\R^d} \frac{1}{\prod_{j=1}^d (1 + x_j^2)} dx}_{= \pi^d} \end{align}.

If $\phi_l \to \phi$ 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.

$////$

## EquicontinuityEdit

We now introduce the concept of equicontinuity.

Definition 4.13:

Let $M$ be a metric space equipped with a metric which we shall denote by $d$ here, let $X \subseteq M$ be a set in $M$, and let $\mathcal Q$ be a set of continuous functions mapping from $X$ to the real numbers $\mathbb R$. We call this set $\mathcal Q$ equicontinuous if and only if

$\forall x \in X : \exists \delta \in \mathbb R_{>0} : \forall y \in X : d(x, y) < \delta \Rightarrow \forall f \in \mathcal Q : |f(x) - f(y)| < \epsilon$.

So equicontinuity is in fact defined for sets of continuous functions mapping from $X$ (a set in a metric space) to the real numbers $\mathbb R$.

Theorem 4.14:

Let $M$ be a metric space equipped with a metric which we shall denote by $d$, let $Q \subseteq M$ be a sequentially compact set in $M$, and let $\mathcal Q$ be an equicontinuous set of continuous functions from $Q$ to the real numbers $\mathbb R$. Then follows: If $(f_l)_{l \in \mathbb N}$ is a sequence in $\mathcal Q$ such that $f_l(x)$ has a limit for each $x \in Q$, then for the function $f(x) := \lim_{l \to \infty} f_l(x)$, which maps from $Q$ to $\mathbb R$, it follows $f_l \to f$ uniformly.

Proof:

In order to prove uniform convergence, by definition we must prove that for all $\epsilon > 0$, there exists an $N \in \mathbb N$ such that for all $l \ge N : \forall x \in Q : |f_l(x) - f(x)| < \epsilon$.

So let's assume the contrary, which equals by negating the logical statement

$\exists \epsilon > 0 : \forall N \in \mathbb N : \exists l \ge N : \exists x \in Q : |f_l(x) - f(x)| \ge \epsilon$.

We choose a sequence $(x_m)_{m \in \mathbb N}$ in $Q$. We take $x_1$ in $Q$ such that $|f_{l_1}(x_1) - f(x_1)| \ge \epsilon$ for an arbitrarily chosen $l_1 \in \mathbb N$ and if we have already chosen $x_k$ and $l_k$ for all $k \in \{1, \ldots, m\}$, we choose $x_{m+1}$ such that $|f_{l_{m+1}}(x_{m+1}) - f(x_{m+1})| \ge \epsilon$, where $l_{m+1}$ is greater than $l_m$.

As $Q$ is sequentially compact, there is a convergent subsequence $(x_{m_j})_{j \in \mathbb N}$ of $(x_m)_{m \in \mathbb N}$. Let us call the limit of that subsequence sequence $x$.

As $\mathcal Q$ is equicontinuous, we can choose $\delta \in \mathbb R_{>0}$ such that

$\|x - y\| < \delta \Rightarrow \forall f \in \mathcal Q : |f(x) - f(y)| < \frac{\epsilon}{4}$.

Further, since $x_{m_j} \to x$ (if $j \to \infty$ of course), we may choose $J \in \mathbb N$ such that

$\forall j \ge J : \|x_{m_j} - x\| < \delta$.

But then follows for $j \ge J$ and the reverse triangle inequality:

$|f_{l_{m_j}}(x) - f(x)| \ge \left| |f_{l_{m_j}}(x) - f(x_{m_j})| - |f(x_{m_j}) - f(x)| \right|$

Since we had $|f(x_{m_j}) - f(x)| < \frac{\epsilon}{4}$, the reverse triangle inequality and the definition of t

$|f_{l_{m_j}}(x) - f(x_{m_j})| \ge \left| |f_{l_{m_j}}(x_{m_j}) - f(x_{m_j})| - |f_{l_{m_j}}(x) - f_{l_{m_j}}(x_{m_j})| \right| \ge \epsilon - \frac{\epsilon}{4}$

, we obtain:

\begin{align} |f_{l_{m_j}}(x) - f(x)| & \ge \left| |f_{l_{m_j}}(x) - f(x_{m_j})| - |f(x_{m_j}) - f(x)| \right| \\ & = |f_{l_{m_j}}(x) - f(x_{m_j})| - |f(x_{m_j}) - f(x)| \\ & \ge \epsilon - \frac{\epsilon}{4} - \frac{\epsilon}{4} \\ & \ge \frac{\epsilon}{2} \end{align}

Thus we have a contradiction to $f_l(x) \to f(x)$.

$////$

Theorem 4.15:

Let $\mathcal Q$ be a set of differentiable functions, mapping from the convex set $X \subseteq \mathbb R^d$ to $\mathbb R$. If we have, that there exists a constant $b \in \mathbb R_{>0}$ such that for all functions in $\mathcal Q$, $\forall x \in X : \| \nabla f(x) \| \le b$ (the $\nabla f$ exists for each function in $\mathcal Q$ because all functions there were required to be differentiable), then $\mathcal Q$ is equicontinuous.

Proof: We have to prove equicontinuity, so we have to prove

$\forall x \in X : \exists \delta \in \mathbb R_{>0} : \forall y \in X: \|x - y\| < \delta \Rightarrow \forall f \in \mathcal Q : |f(x) - f(y)| < \epsilon$.

Let $x \in X$ be arbitrary.

We choose $\delta := \frac{\epsilon}{b}$.

Let $y \in X$ such that $\|x - y\| < \delta$, and let $f \in \mathcal Q$ be arbitrary. By the mean-value theorem in multiple dimensions, we obtain that there exists a $\lambda \in [0, 1]$ such that:

$f(x) - f(y) = \nabla f(\lambda x + (1 - \lambda) y) \cdot (x - y)$

The element $\lambda x + (1 - \lambda) y$ is inside $X$, because $X$ is convex. From the Cauchy-Schwarz inequality then follows:

$|f(x) - f(y)| = | \nabla f(\lambda x + (1 - \lambda) y) \cdot (x - y) | \le \|\nabla f(\lambda x + (1 - \lambda) y)\| \|x - y\| < b \delta = \frac{b}{b} \epsilon = \epsilon$
$////$

## The generalised product ruleEdit

Definition 4.16:

If $\alpha = (\alpha_1, \ldots, \alpha_d), \beta = (\beta_1, \ldots, \beta_d) \in \N_0^d$ are two $d$-dimensional multiindices, we define the binomial coefficient of $\alpha$ over $\beta$ as

$\binom{\alpha}{\beta} := \binom{\alpha_1}{\beta_1} \binom{\alpha_2}{\beta_2} \cdots \binom{\alpha_d}{\beta_d}$.

We also define less or equal relation on the set of multi-indices.

Definition 4.17:

Let $\alpha = (\alpha_1, \ldots, \alpha_d), \beta = (\beta_1, \ldots, \beta_d) \in \N_0^d$ be two $d$-dimensional multiindices. We define $\beta$ to be less or equal than $\alpha$ if and only if

$\beta \le \alpha :\Leftrightarrow \forall n \in \{1, \ldots, d\} : \beta_n \le \alpha_n$.

For $d \ge 2$, there are vectors $\alpha, \beta \in \mathbb N_0^d$ such that neither $\alpha \le \beta$ nor $\beta \le \alpha$. For $d = 2$, the following two vectors are examples for this:

$\alpha = (1, 0), \beta = (0, 1)$

This example can be generalised to higher dimensions (see exercise 6).

With these multiindex definitions, we are able to write down a more general version of the product rule. But in order to prove it, we need another lemma.

Lemma 4.18:

If $n \in \{1, \ldots, d\}$ and $e_n := (0, \ldots, 0, 1, 0, \ldots, 0)$, where the $1$ is at the $n$-th place, we have

$\binom{\alpha - e_n}{\beta -e_n} + \binom{\alpha - e_n}{\beta} = \binom{\alpha}{\beta}$

for arbitrary multiindices $\alpha, \beta \in \mathbb N_0^d$.

Proof:

For the ordinary binomial coefficients for natural numbers, we had the formula

$\binom{n - 1}{k - 1} + \binom{n - 1}{k} = \binom{n}{k}$.

Therefore,

\begin{align} \binom{\alpha - e_n}{\beta -e_n} + \binom{\alpha - e_n}{\beta} &= \binom{\alpha_1}{\beta_1} \cdots \binom{\alpha_n - 1}{\beta_n - 1} \cdots \binom{\alpha_d}{\beta_d} + \binom{\alpha_1}{\beta_1} \cdots \binom{\alpha_n - 1}{\beta_n} \cdots \binom{\alpha_d}{\beta_d} \\ &= \binom{\alpha_1}{\beta_1} \cdots \left( \binom{\alpha_n - 1}{\beta_n - 1} + \binom{\alpha_n - 1}{\beta_n} \right) \cdots \binom{\alpha_d}{\beta_d} \\ &= \binom{\alpha_1}{\beta_1} \cdots \binom{\alpha_n}{\beta_n} \cdots \binom{\alpha_d}{\beta_d} = \binom{\alpha}{\beta} \end{align}
$////$

This is the general product rule:

Theorem 4.19:

Let $f \in \mathcal C^n(\mathbb R^d)$ and let $|\alpha| \le n$. Then

$\partial_\alpha (f g) = \sum_{\beta \le \alpha} \binom{\alpha}{\beta} \partial_\beta f \partial_{\alpha - \beta} g$

Proof:

We prove the claim by induction over $|\alpha|$.

1.

We start with the induction base $|\alpha| = 0$. Then the formula just reads

$f(x)g(x) = f(x)g(x)$

, and this is true. Therefore, we have completed the induction base.

2.

Next, we do the induction step. Let's assume the claim is true for all $\alpha \in \N_0^d$ such that $|\alpha| = n$. Let now $\alpha \in \N_0^d$ such that $|\alpha| = n+1$. Let's choose $k \in \{1, \ldots, d\}$ such that $\alpha_k > 0$ (we may do this because $|\alpha| = k+1 > 0$). We define again $e_k = (0, \ldots, 0, 1, 0, \ldots, 0)$, where the $1$ is at the $k$-th place. Due to Schwarz' theorem and the ordinary product rule, we have

$\partial_\alpha f g = \partial_{\alpha - e_k} \left( \partial_{x_k} f g \right) = \partial_{\alpha - e_k} \left( \partial_{x_k} f g + f \partial_{x_k} g\right)$.

By linearity of derivatives and induction hypothesis, we have

\begin{align} \partial_{\alpha - e_k} \left(\partial_{x_k} f g + f \partial_{x_k} g \right) & = \partial_{\alpha - e_k} \left( \partial_{x_k} f g \right) + \partial_{\alpha - e_k} \left( f \partial_{x_k} g \right) \\ & = \sum_{\varsigma \le \alpha - e_k} \binom{\alpha - e_k}{\varsigma} \partial_\varsigma \partial_{x_k} f \partial_{\alpha - e_k - \varsigma} g + \sum_{\varsigma \le \alpha - e_k} \binom{\alpha - e_k}{\varsigma} \partial_\varsigma f \partial_{\alpha - e_k - \varsigma} \partial_{x_k} g \end{align}.

Since

$\partial_{\alpha - e_k - \varsigma} = \partial_{\alpha - (\varsigma + e_k)}$

and

$\{\varsigma \in \N_0^d | 0 \le \varsigma \le \alpha - e_k\} = \{\varsigma - e_k \in \N_0^d | e_k \le \varsigma \le \alpha\}$,

we are allowed to shift indices in the first of the two above sums, and furthermore we have by definition

$\partial_\varsigma \partial_{x_k} = \partial_{\varsigma + e_k}$.

With this, we obtain

$\sum_{\varsigma \le \alpha - e_k} \binom{\alpha - e_k}{\varsigma} \partial_\varsigma \partial_{x_k} f \partial_{\alpha - e_k - \varsigma} g + \sum_{\varsigma \le \alpha - e_k} \binom{\alpha - e_k}{\varsigma} \partial_\varsigma f \partial_{\alpha - e_k - \varsigma} \partial_{x_k} g = \sum_{e_k \le \varsigma \le \alpha} \binom{\alpha - e_k}{\varsigma - e_k} \partial_\varsigma f \cdot \partial_{\alpha - \varsigma} g + \sum_{\varsigma \le \alpha - e_k} \binom{\alpha - e_k}{\varsigma} \partial_\varsigma f \partial_{\alpha - \varsigma} g$

Due to lemma 4.18,

$\binom{\alpha - e_k}{\beta -e_i} + \binom{\alpha - e_k}{\beta} = \binom{\alpha}{\beta}$.

Further, we have

$\binom{\alpha - e_i}{0} = \binom{\alpha}{0} = 1$ where $0 = (0, \ldots, 0)$ in $\N_0^d$,

and

$\binom{\alpha - e_k}{\alpha - e_k} = \binom{\alpha}{\alpha} = 1$

(these two rules may be checked from the definition of $\binom{\alpha}{\beta}$). It follows

\begin{align} \partial_\alpha (f g) & = \sum_{e_k \le \varsigma \le \alpha} \binom{\alpha - e_k}{\varsigma - e_k} \partial_\varsigma f \cdot \partial_{\alpha - \varsigma} g + \sum_{\varsigma \le \alpha - e_k} \binom{\alpha - e_k}{\varsigma} \partial_\varsigma f \partial_{\alpha - \varsigma} g \\ & = \binom{\alpha - e_k}{0} f \partial_\alpha g + \sum_{e_k \le \varsigma \le \alpha - e_k} \left[ \binom{\alpha - e_k}{\varsigma - e_k} + \binom{\alpha - e_k}{\varsigma} \right] \partial_\varsigma f \partial_{\alpha - \varsigma} g + \binom{\alpha - e_k}{\alpha - e_k} f \partial_\alpha g \\ & = \sum_{\varsigma \le \alpha} \binom{\alpha}{\varsigma} \partial_\varsigma f \partial_{\alpha - \varsigma} \end{align}.
$////$

## Operations on DistributionsEdit

For $\varphi, \vartheta \in \mathcal D(\mathbb R^d)$ there are operations such as the differentiation of $\varphi$, the convolution of $\varphi$ and $\vartheta$ and the multiplication of $\varphi$ and $\vartheta$. In the following section, we want to define these three operations (differentiation, convolution with $\vartheta$ and multiplication with $\vartheta$) for a distribution $\mathcal T$ instead of $\varphi$.

Lemma 4.20:

Let $O, U \subseteq \mathbb R^d$ be open sets and let $L : \mathcal D(O) \to L^1_\text{loc}(U)$ be a linear function. If there is a linear and sequentially continuous (in the sense of definition 4.1) function $\mathcal L : \mathcal D(U) \to \mathcal D(O)$ such that

$\forall \varphi \in \mathcal D(O), \vartheta \in \mathcal D(U) : \int_O \varphi(x) \mathcal L(\vartheta)(x) dx = \int_U L(\varphi)(x) \vartheta(x) dx$

, then for every distribution $\mathcal T \in \mathcal D(O)^*$, the function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is a distribution. Therefore, the function

$\Lambda : \mathcal D(O)^* \to \mathcal D(U)^*, \Lambda(\mathcal T) := \mathcal T \circ \mathcal L$

really maps to $\mathcal D(U)^*$. This function has the property

$\forall \varphi \in \mathcal D(O) : \Lambda(\mathcal T_\varphi) = \mathcal T_{L (\varphi)}$

Proof:

We have to prove two claims: First, that the function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is a distribution, and second that $\Lambda$ as defined above has the property

$\forall \varphi \in \mathcal D(O) : \Lambda(\mathcal T_\varphi) = \mathcal T_{L (\varphi)}$

1.

We show that the function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is a distribution.

$\mathcal T(\mathcal L(\varphi))$ has a well-defined value in $\mathbb R$ as $\mathcal L$ maps to $\mathcal D(O)$, which is exactly the preimage of $\mathcal T$. The function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is continuous since it is the composition of two continuous functions, and it is linear for the same reason (see exercise 2).

2.

We show that $\Lambda$ has the property

$\forall \varphi \in \mathcal D(O) : \Lambda(\mathcal T_\varphi) = \mathcal T_{L (\varphi)}$

For every $\vartheta \in \mathcal D(U)$, we have

$\Lambda(\mathcal T_\varphi)(\vartheta) := (\mathcal T_\varphi \circ \mathcal L)(\vartheta) := \int_O \varphi(x) \mathcal L(\vartheta)(x) dx \overset{\text{by assumption}}{=} \int_U L(\varphi)(x) \vartheta(x) dx =: \mathcal T_{L (\varphi)}(\vartheta)$

Since equality of two functions is equivalent to equality of these two functions evaluated at every point, this shows the desired property.

$////$

We also have a similar lemma for Schwartz distributions:

Lemma 4.21:

Let $L : \mathcal S(\mathbb R^d) \to L^1_\text{loc}(\mathbb R^d)$ be a linear function. If there is a linear and sequentially continuous (in the sense of definition 4.2) function $\mathcal L : \mathcal S(\mathbb R^d) \to \mathcal S(\mathbb R^d)$ such that

$\forall \phi, \theta \in \mathcal S(\mathbb R^d) : \int_{\mathbb R^d} \phi(x) \mathcal L(\theta)(x) dx = \int_{\mathbb R^d} L(\phi)(x) \theta(x) dx$

, then for every distribution $\mathcal T \in S(\mathbb R^d)^*$, the function $\phi \mapsto \mathcal T(\mathcal L(\phi))$ is a distribution. Therefore, we may define a function

$\Lambda : \mathcal S(\mathbb R^d)^* \to \mathcal S(\mathbb R^d)^*, \Lambda(\mathcal T) := \mathcal T \circ \mathcal L$

This function has the property

$\forall \phi \in \mathcal S(\mathbb R^d) : \Lambda(\mathcal T_\phi) = \mathcal T_{L(\phi)}$

The proof is exactly word-for-word the same as the one for lemma 4.20.

Noting that multiplication, differentiation and convolution are linear, we will define these operations for distributions by taking $L$ in the two above lemmas as the respective of these three operations.

Theorem and definitions 4.22:

Let $f \in \mathcal C^\infty(\mathbb R^d)$, and let $O \subseteq \mathbb R^d$ be open. Then for all $\varphi \in \mathcal D(O)$, the pointwise product $f \varphi$ is contained in $\mathcal D(O)$, and if further $f$ and all of it's derivatives are bounded by polynomials, then for all $\phi \in \mathcal S(\mathbb R^d)$ the pointwise product $f \phi$ is contained in $\mathcal S(\mathbb R^d)$. Also, if $\varphi_l \to \varphi$ in the sense of bump functions, then $f \varphi_l \to f \varphi$ in the sense of bump functions, and if $f$ and all of it's derivatives are bounded by polynomials, then $\phi_l \to \phi$ in the sense of Schwartz functions implies $f \phi_l \to f \phi$ in the sense of Schwartz functions. Further:

• Let $\mathcal T: \mathcal D(O) \to \mathbb R$ be a distribution. If we define

$f \mathcal T: \mathcal D(O) \to \mathbb R, f \mathcal T(\varphi) := \mathcal T(f \varphi)$,

then the expression on the right hand side is well-defined and for all $\vartheta \in \mathcal D(O)$ we have

$f \mathcal T_\vartheta = \mathcal T_{f \vartheta}$,

and $f \mathcal T$ is a distribution.

• Assume that $f$ and all of it's derivatives are bounded by polynomials. Let $\mathcal T: \mathcal S(\mathbb R^d) \to \mathbb R$ be a tempered distribution. If we define

$f \mathcal T: \mathcal S(\mathbb R^d) \to \mathbb R, f \mathcal T(\phi) := \mathcal T(f \phi)$,

then the expression on the right hand side is well-defined and for all $\theta \in \mathcal S(\mathbb R^d)$ we have

$f \mathcal T_\theta = \mathcal T_{f \theta}$,

and $f \mathcal T$ is a tempered distribution.

Proof:

The product of two $\mathcal C^\infty$ functions is again $\mathcal C^\infty$, and further, if $\varphi(x) = 0$, then also $(f \varphi)(x) = f(x) \varphi(x) = 0$. Hence, $f \varphi \in \mathcal D(O)$.

Also, if $\varphi_l \to \varphi$ in the sense of bump functions, then, if $K \subset \mathbb R^d$ is a compact set such that $\text{supp } \varphi_n \subseteq K$ for all $n \in \mathbb N$,

\begin{align} \|\partial_\alpha (f (\varphi_l - \varphi))\|_\infty & = \left\| \sum_{\varsigma \le \alpha} \binom{\alpha}{\varsigma} \partial_\varsigma f \partial_{\alpha - \varsigma} (\varphi_l - \varphi) \right\|_\infty \\ & \le \sum_{\varsigma \le \alpha} \|\partial_\varsigma f \partial_{\alpha - \varsigma} (\varphi_l - \varphi)\|_\infty \\ & \le \sum_{\varsigma \le \alpha} \max_{x \in K} |\partial_\varsigma f| \|\partial_{\alpha - \varsigma} (\varphi_l - \varphi)\|_\infty \to 0, l \to \infty \end{align}.

Hence, $f \varphi_l \to f \varphi$ in the sense of bump functions.

Further, also $f \phi \in \mathcal C^\infty(\mathbb R^d)$. Let $\alpha, \beta \in \mathbb N_0^d$ be arbitrary. Then

$\partial_\beta f \phi = \sum_{\varsigma \le \beta} \binom{\beta}{\varsigma} \partial_\varsigma f \partial_{\beta - \varsigma} \phi$.

Since all the derivatives of $f$ are bounded by polynomials, by the definition of that we obtain

$\forall x \in \mathbb R^d: |\partial_\varsigma f(x)| \le |p_\varsigma(x)|$

, where $p_\varsigma, \varsigma \in \mathbb N_0^d$ are polynomials. Hence,

$\|x^\alpha \partial_\beta f \phi\|_\infty \le \sum_{\varsigma \le \beta} \|x^\alpha p_\varsigma \partial_{\beta - \varsigma} \phi\|_\infty < \infty$.

Similarly, if $\phi_l \to \phi$ in the sense of Schwartz functions, then by exercise 3.6

$\|x^\alpha \partial_\beta f (\phi - \phi_l)\|_\infty \le \sum_{\varsigma \le \beta} \|x^\alpha p_\varsigma \partial_{\beta - \varsigma} (\phi - \phi_l)\|_\infty \to 0, l \to \infty$

and hence $f \phi_l \to f \phi$ in the sense of Schwartz functions.

If we define $L(\varphi) := \mathcal L(\varphi) := f\varphi$, from lemmas 4.20 and 4.21 follow the other claims.

$////$

Theorem and definitions 4.23:

Let $O \subseteq \mathbb R^d$ be open. We define

$L: \mathcal S(\mathbb R^d) \to \mathcal C^\infty(\mathbb R^d), L(\phi) := \sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \phi$

, where $a_\alpha \in \mathcal C^\infty(\mathbb R^d)$ such that only finitely many of the $a_\alpha$ are different from the zero function (such a function is also called a linear partial differential operator), and further we define

$\mathcal L : \mathcal S(\mathbb R^d) \to \mathcal C^\infty(\mathbb R^d), \mathcal L(\phi) := \sum_{|\alpha| \le k} (-1)^{|\alpha|} \partial_\alpha (a_\alpha \phi)$.
• Let $\mathcal T: \mathcal D(O) \to \mathbb R$ be a distribution. If we define

$\sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T: \mathcal D(O) \to \mathbb R, \sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T(\varphi) := \mathcal T(\mathcal L(\varphi))$,

then the expression on the right hand side is well-defined, for all $\vartheta \in \mathcal D(O)$ we have

$\sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T_\vartheta = \mathcal T_{L(\vartheta)}$,

and $\sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T$ is a distribution.

• Assume that all $a_\alpha$s and all their derivatives are bounded by polynomials. Let $\mathcal T: \mathcal S(\mathbb R^d) \to \mathbb R$ be a tempered distribution. If we define

$\sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T: \mathcal D(O) \to \mathbb R, \sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T(\varphi) := \mathcal T(\mathcal L(\varphi))$,

then the expression on the right hand side is well-defined, for all $\vartheta \in \mathcal D(O)$ we have

$\sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T_\vartheta = \mathcal T_{L(\vartheta)}$,

and $\sum_{\alpha \in \mathbb N_0^d} a_\alpha \partial_\alpha \mathcal T$ is a tempered distribution.

Proof:

We want to apply lemmas 4.20 and 4.21. Hence, we prove that the requirements of these lemmas are met.

Since the derivatives of bump functions are again bump functions, the derivatives of Schwartz functions are again Schwartz functions (see exercise 3.3 for both), and because of theorem 4.22, we have that $L$ and $\mathcal L$ map $\mathcal D(O)$ to $\mathcal D(O)$, and if further all $a_\alpha$ and all their derivatives are bounded by polynomials, then $L$ and $\mathcal L$ map $\mathcal S(\mathbb R^d)$ to $\mathcal S(\mathbb R^d)$.

The sequential continuity of $\mathcal L$ follows from theorem 4.22.

Further, for all $\phi, \theta \in \mathcal S(\mathbb R^d)$,

$\int_{\mathbb R^d} \phi(x) \mathcal L(\theta)(x) dx = \sum_{\alpha \in \mathbb N_0^d} (-1)^{|\alpha|} \int_{\mathbb R^d} \phi(x) \partial_\alpha(a_\alpha \theta)(x) dx$.

Further, if we single out an $\alpha \in \mathbb N_0^d$, by Fubini's theorem and integration by parts we obtain

\begin{align} \int_{\mathbb R^d} \phi(x) \partial_\alpha(a_\alpha \theta)(x) dx & = \int_{\mathbb R^{d-1}} \int_{\mathbb R} \phi(x) \partial_\alpha(a_\alpha \theta)(x) dx_1 d(x_2, \ldots, x_d) \\ & = \int_{\mathbb R^{d-1}} \int_{\mathbb R} \phi(x) \partial_\alpha(a_\alpha \theta)(x) dx_1 d(x_2, \ldots, x_d) \\ & = \int_{\mathbb R^{d-1}} (-1)^{\alpha_1} \int_{\mathbb R} \partial_{(\alpha_1, 0, \ldots, 0)} \phi(x) \partial_{\alpha - (\alpha_1, 0, \ldots, 0)} (a_\alpha \theta)(x) dx_1 d(x_2, \ldots, x_d) \\ & = \cdots = (-1)^{|\alpha|} \int_{\mathbb R^d} \partial_\alpha \phi(x) a_\alpha(x) \theta(x) dx \end{align}.

Hence,

$\int_{\mathbb R^d} \phi(x) \mathcal L(\theta)(x) dx = \int_{\mathbb R^d} L(\phi)(x) \theta(x) dx$

and the lemmas are applicable.

$////$

Definition 4.24:

Let $\mathcal T \in \mathcal D(\mathbb R^d)^*$ and let $\varphi \in \mathcal D(\mathbb R^d)$. Then we define the function

$\mathcal T * \varphi(x) := \mathcal T(\varphi(x - \cdot))$.

This function is called the convolution of $\mathcal T$ and $\varphi$.

Theorem 4.25:

Let $\mathcal T \in \mathcal D(\mathbb R^d)^*$ and let $\varphi \in \mathcal D(\mathbb R^d)$. Then

1. $\mathcal T * \varphi$ is continuous,
2. $\forall \alpha \in \mathbb N_0^d : \partial_\alpha (\mathcal T * \varphi) = \mathcal T * (\partial_\alpha \varphi)$ and
3. $\mathcal T * \varphi \in \mathcal C^\infty(\mathbb R^d)$.

Proof:

1.

Let $x \in \mathbb R^d$ be arbitrary, and let $(x_l)_{l \in \mathbb N}$ be a sequence converging to $x$ and let $N \in \mathbb N$ such that $\forall n \ge N : \|x_n - x\| \le 1$. Then

$K := \overline{\bigcup_{n \ge N} \text{supp } \varphi(x_n - \cdot) \cup \bigcup_{n < N} \text{supp } \varphi(x_n - \cdot)}$

is compact. Hence, if $\beta \in \mathbb N_0^d$ is arbitrary, then $\partial_\beta \varphi(x_l - \cdot)|_K \to \partial_\beta \varphi(x - \cdot)|_K$ uniformly. But outside $K$, $\partial_\beta \varphi(x_l - \cdot) - \partial_\beta \varphi(x - \cdot) = 0$. Hence, $\partial_\beta \varphi(x_l - \cdot) \to \partial_\beta \varphi(x - \cdot)$ uniformly. Further, for all $n \in \mathbb N$ $\text{supp } \varphi(x_n - \cdot) \subseteq K$. Hence, $\varphi(x_l - \cdot) \to \varphi, l \to \infty$ in the sense of bump functions. Thus, by continuity of $\mathcal T$,

$(\mathcal T * \varphi)(x_l) = \mathcal T(\varphi(x_l - \cdot)) \to \mathcal T(\varphi(x - \cdot)) = (\mathcal T * \varphi)(x), l \to \infty$.

2.

We proceed by induction on $|\alpha|$.

The induction base $|\alpha| = 0$ is obvious, since $\partial_{(0, \ldots, 0)} f = f$ for all functions $f: \mathbb R^d \to \mathbb R$ by definition.

Let the statement be true for all $\alpha \in \mathbb N_0^d$ such that $|\alpha| = n$. Let $\beta \in \mathbb N_0^d$ such that $|\beta| = n+1$. We choose $k \in \{1, \ldots, d\}$ such that $\beta_k > 0$ (this is possible since otherwise $\beta = \mathbf 0$). Further, we define

$e_k := (0, \ldots, 0, \overbrace{1}^{k\text{th place}}, 0, \ldots, 0)$.

Then $|\beta - e_k| = n$, and hence $\partial_{\beta - e_k} (\mathcal T * \varphi) = \mathcal T * (\partial_{\beta - e_k} \varphi)$.

Furthermore, for all $\vartheta \in \mathcal D(\mathbb R^d)$,

$\lim_{\lambda \to 0} \frac{\mathcal T * \vartheta (x + \lambda e_k) - \mathcal T * \vartheta (x)}{\lambda} = \lim_{\lambda \to 0} \mathcal T \left( \frac{\vartheta(x + \lambda e_k- \cdot) - \vartheta(x- \cdot)}{\lambda} \right)$.

But due to Schwarz' theorem, $\frac{\vartheta(x + \lambda e_k- \cdot) - \vartheta(x- \cdot)}{\lambda} \to \partial_{x_k} \vartheta, \lambda \to 0$ in the sense of bump functions, and thus

$\lim_{\lambda \to 0} \mathcal T \left( \frac{\vartheta(x + \lambda e_k - \cdot) - \vartheta(x- \cdot)}{\lambda} \right) = \mathcal T(\vartheta(x- \cdot))$.

Hence, $\partial_\beta (\mathcal T * \varphi) = \partial_{e_k} \mathcal T * (\partial_{\beta - e_k} \varphi) = \mathcal T * (\partial_\beta \varphi)$, since $\partial_{\beta - e_k} \varphi$ is a bump function (see exercise 3.3).

3.

This follows from 1. and 2., since $\partial_\beta \varphi$ is a bump function for all $\beta \in \mathbb N_0^d$ (see exercise 3.3).

$////$

## ExercisesEdit

1. Let $\mathcal T_1, \ldots, \mathcal T_n$ be (tempered) distributions and let $c_1, \ldots, c_n \in \mathbb R$. Prove that also $\sum_{j=1}^n c_j \mathcal T_j$ is a (tempered) distribution.
2. Let $f : \mathbb R^d \to \mathbb R$ be essentially bounded. Prove that $\mathcal T_f$ is a tempered distribution.
3. Prove that if $\mathcal Q$ is a set of differentiable functions which go from $[0, 1]^d$ to $\mathbb R$, such that there exists a $c \in \mathbb R_{>0}$ such that for all $g \in \mathcal Q$ it holds $\forall x \in \mathbb R^d : \|\nabla g(x)\| < c$, and if $(f_l)_{l \in \mathbb N}$ is a sequence in $\mathcal Q$ for which the pointwise limit $\lim_{l \to \infty} f_l(x)$ exists for all $x \in \mathbb R^d$, then $f_l$ converges to a function uniformly on $[0, 1]^d$ (hint: $[0, 1]^d$ is sequentially compact; this follows from the Bolzano–Weierstrass theorem).
4. Let $f: \mathbb R^d \to \mathbb R$ such that $\mathcal T_f$ is a distribution. Prove that for all $\varphi \in \mathcal D(O)$ $\mathcal T_f * \varphi = f * \varphi$.
5. Prove that for $x \in \mathbb R^d$ the function $\delta_x: \mathcal S(\mathbb R^d) \to \mathbb R, \delta(\phi) := \phi(x)$ is a tempered distribution (this function is called the Dirac delta distribution after Paul Dirac).
6. For each $d \in \mathbb N$, find $\alpha_d, \beta_d \in \mathbb N_0^d$ such that neither $\alpha_d \le \beta_d$ nor $\beta_d \le \alpha_d$.

## SourcesEdit

Partial Differential Equations
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