Distributions

Let $f\in L_{loc}^1({\mathbb{R}}^n)$, then for every $\phi \in C_c^{\infty }({\mathbb{R}}^n)$ it’s well defined the action of $f$ on $\phi$ by

$$⟨f,\phi ⟩=T_f(\phi ):={\int }_{{\mathbb{R}}^n}f\phi dx$$

Call $K=\text{supp }\phi$ since using Holder’s inequality

$$\Vert f\phi {\Vert }_{L^1({\mathbb{R}}^n)}=\Vert f\phi {\Vert }_{L^1(K)}\le \Vert f{\Vert }_{L^1(K)}\Vert \phi {\Vert }_{L^{\infty }(K)}$$

Let $\alpha =({\alpha }_1,\dots ,{\alpha }_n)$ with ${\alpha }_i\in \mathbb{N}$, $\alpha$ is known as a multi-index of order $|\alpha |:={\sum }_i{\alpha }_i$.

Let $\phi \in C^{\infty }(\Omega )$ with $\Omega \subseteq {\mathbb{R}}^n$ an open subset, we define

$$D^{\alpha }\phi :=\frac{{\partial }^{{\alpha }_1}}{\partial x_1^{{\alpha }_1}}\cdots \frac{{\partial }^{{\alpha }_n}}{\partial x_1^{{\alpha }_n}}\phi$$

The space of test functions §

Note

The space of test function $\mathcal{D}(\Omega )$ it’s the space $C_c^{\infty }(\Omega )$ together with the following convergence criterion: $\{{\phi }_n\}\subset C_c^{\infty }(\Omega )$ converges to $\phi \in C_c^{\infty }(\Omega )$ if:

1. there exists a compact set $K\subset {\mathbb{R}}^n$ such that $\text{supp }{\phi }_n\subset K$ for all $n$;
2. for every multi-index $\alpha$ we have $D^{\alpha }{\phi }_n\rightrightarrows D^{\alpha }\phi$ on $\Omega$ (on $K$).

We use the notation ${\phi }_n\stackrel{D}{\to }\phi$.

Note that $\text{supp }\phi \subseteq K$ is implied my uniform convergence. The motivation for condition $1.$ is clear, we don’t want limits with support wandering to infinity. The second rather strong conditions is usefull to make the continuous the linear functionals of the type:

$$F[x]:={\int }_{{\mathbb{R}}^n}x(t)\phi (t)dt$$

since with uniform convergence we can carlessly exchange the limit with the integral.

This notion of convergence can be obtained as a LC-topology from a family of seminorms ($\mathcal{D}(\Omega )$ is a Topological vector space).

Distributions §

Note

A linear map $T:\mathcal{D}({\mathbb{R}}^n)\to \mathbb{R}$ (or $\mathbb{C}$) is called distribution if:

1. $T$ is linear: $T(\lambda \phi +\psi )=\lambda T(\phi )+T(\psi )$
2. $T$ is continuous: if ${\phi }_n\stackrel{D}{\to }\phi$ then $T({\phi }_n)\to T(\phi )$

Since this space is the topological dual we use the notation ${\mathcal{D}}^{\prime }({\mathbb{R}}^n)$.

Example

1. Every $f\in L_{loc}^1({\mathbb{R}}^n)$ induces a distribution

$$T_f(\phi ):={\int }_{{\mathbb{R}}^n}f(t)\phi (t)dt$$

Distribution which can be represented by a local function are called regular.

1. The most famous not regular distribution is the Dirac’s delta:

$${\delta }_x(\phi ):=\phi (x)$$

it’s easy to see why is not regular. Suppose that there exists a function $f\in L_{loc}^1$ that represents the delta, then given any test function $\phi$. By the absolute continuity of the Lebesgue integral we can find $E\subseteq {\mathbb{R}}^n$ such that

$$\phi (x)={\delta }_x[\phi ]={\int }_{E}f(t)\phi (t)dt<ϵ$$

A clear contradiction if $\phi (x)\ne 0$.

Every Radon measure induces a distribution in a natural way:

$$T_{\mu }(\phi ):=\int \phi d\mu$$

in fact:

• it’s well defined since $\forall \phi \in C_c^{\infty }(\Omega )$

$${\int }_{\Omega }\phi d\mu \le \Vert \phi {\Vert }_{\infty }\mu (\text{supp}(\phi ))<\infty$$

• it’s linear and continuous by the properties of the Lebesgue’s integral

Note

Let $\mu$ be a Radon measure, then $T_{\mu }$ is representable by an $L_{loc}^1$ function iff $\mu$ is absolutely continuous w.r.t. the Lebesgue measure.

Note \mu be a Radon measure, following Lebesgue decomposition theorem we can decompose it as \mu = \mu_{as} + \mu_s. A measure defines a distribution via integration:

Let

$$T_{\mu }(\phi )=\int \phi d\mu =\int \phi d{\mu }_{as}+\int \phi d{\mu }_s$$

Using the Radon-Nikodym theorem we can write the integral with the absolutely continuous measure as

$$\int g\phi d\lambda$$

w.r.t. the Lebesgue measure. Since $\mu$ is Radon, it gives finite measure to every compact set, so $g\in L_{loc}^1$. Clearly equality holds iff ${\mu }_s\equiv 0$, i.e. $\mu$ is absolutely continuous w.r.t. the Lebesgue measure. $\square$

The following proposition is usefull to check if a given linear functional is a distribution.

Note

Let $T:\mathcal{D}({\mathbb{R}}^n)\to \mathbb{R}$ (or $\mathbb{C}$) be linear. The following are equivalent:

1. $T$ is a distribution
2. $\forall K\subset {\mathbb{R}}^n$ compat $\exists m\in \mathbb{N}$ $\exists C>0\forall \phi \in \mathcal{D}({\mathbb{R}}^n)$ $\text{supp }\phi \subseteq K$ s.t.

$$|T(\phi )|\le C\sum _{|\alpha |\le m}\Vert D^{\alpha }\phi {\Vert }_{\infty }$$

Note T is continous. (\impliedby) Let \varphi_n, \varphi \in \mathcal{D}(\mathbb{R}^n) with \varphi_n \xrightarrow{D}\varphi. Then by the definition of convergence there exists K compact such that \text{supp }\varphi_n \subseteq K and for all \alpha \in \mathbb{N} we have \Vert D^\alpha \varphi_n - D^\alpha \varphi\Vert_\infty \to 0. To prove that T is continuous we need to check that

We only need to check that

$$|T({\phi }_n)-T(\phi )|=|T({\phi }_n-\phi )|$$

goes to zero. We can use the bound $2.$

$$\le C\sum _{|\alpha |\le m}\Vert D^{\alpha }{\phi }_n-D^{\alpha }\phi {\Vert }_{\infty }\stackrel{n\to \infty }{\to }0$$

since on the right we have a finite sum. $(⟹)$ We prove the contrapositive statement: if $2.$ doesn not hold, then $T$ is not continuous (hence not a distribution). $\exists K$ compact $\forall m\in \mathbb{N}$ and $\forall C>0$ s.t. $\forall \phi \in \mathcal{D}$ with $\text{supp }\phi \subseteq K$

$$|T(\phi )|>C\sum _{|\alpha |\le m}\Vert D^{\alpha }\phi {\Vert }_{\infty }$$

Choose $C=m=k\in \mathbb{N}$, we can construct a sequence ${\phi }_k\in \mathcal{D}$ with

$$|T({\phi }_k)|>k\sum _{|\alpha |\le k}\Vert D^{\alpha }{\phi }_k{\Vert }_{\infty }$$

The rescaled sequence ${\psi }_k:=\frac{{\phi }_k}{|T({\phi }_k)|}$ is well defined, also

$${\psi }_k\stackrel{D}{\to }0$$

since

$$|T({\phi }_k)|>k\Vert {\phi }_k{\Vert }_{\infty }>k|{\phi }_k(x)|⟹\frac{|{\phi }_k(x)|}{|T({\phi }_k)|}<\frac{1}{k}$$

this works for any multi-index $\alpha$. But if we compute

$$|T({\psi }_k)|=\frac{1}{|{\phi }_k|}|T({\phi }_k)|=1\ne 0$$

hence $T$ is not continous. $\square$

Given a distribution, the $m$ above is called its order. For example any regular distribution has order $m=0$, since:

$$|T_f(\phi )|\le \int |f(t)\phi (t)|dt\le C\Vert \phi {\Vert }_{\infty }$$

where $C$ is just the integral (which is finite since $f\in L_{loc}^1$). In general any Radon measure induces a distribution of order zero:

$$|T_{\mu }(\phi )|\le \mu (\text{supp }\phi )\Vert \phi {\Vert }_{\infty }$$

The converse is also true by the Riesz representation theorem: Let $T$ be an order zero distributuion, then there exists $4$ unique Radon measures such that

$$T(\phi )=\int \phi d{\mu }_1-\int \phi d{\mu }_2+i\int \phi d{\mu }_2-i\int \phi d{\mu }_4$$

In general for a distribution of order $m$ there exists a set of Radon measures for any multi-index of oder \leq m$$: \mu_\alpha$,$\alpha \in \mathbb{N}^n$,$|\alpha| \leq m$ such that

$$T(\phi )=\sum _{|\alpha |\le m}\int D^{\alpha }\phi d{\mu }_{\alpha }$$

Tip

We can think of a distribution of order $m$ to depend only on derivatives of order less or equal to $m$

Distributional derivatives §

Since $C^1({\mathbb{R}}^n)\subset L_{loc}^1({\mathbb{R}}^n)$ it makes sense to define $T_f$ for any differentiable function $f$. In $n=1$ we would like:

$$T_f^{\prime }=T_{f^{\prime }}$$

so our notion of derivatives of distribution needs to be compatible with the equality above. If we apply $T_{f^{\prime }}$ on a test function and use integration by parts

$$T_{f^{\prime }}(\phi )=\int f^{\prime }(t)\phi (t)dt=-\int f(t){\phi }^{\prime }(t)dt=-T_f({\phi }^{\prime })$$

(boundary terms are zero) the formula on the right makes sense for any distribution $T$, and ca be generalized for all partial derivatives (note that it also holds more generally for Absolutely continuous functions).

Note

Let $T\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$. The (distributional) partial derivative of $T$ w.r.t. $x_i$ is the distribution ${\partial }_{x_i}\phi \in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ defined as:

$$⟨{\partial }_{x_i}T,\phi ⟩=-⟨T,{\partial }_{x_i}\phi ⟩,\forall \phi \in \mathcal{D}({\mathbb{R}}^n)$$

Using induction see that for any multi-index $\alpha$

$$⟨D^{\alpha }T,\phi ⟩=(-1{)}^{|\alpha |}⟨T,D^{\alpha }\phi ⟩,\forall \phi \in \mathcal{D}({\mathbb{R}}^n)$$

Example

In general for differentiable a.e. functions the distributional derivatives doesn’t match the a.e. derivative. Consider the Haveside function $H(x)=1_{[0,+\infty )}(x)$ (a jump function) which has derivative $0$ a.e, but it’s distributional derivative is:

$$T_H^{\prime }(\phi )=-T_H({\phi }^{\prime })=-{\int }_{[0,+\infty )}{\phi }^{\prime }(t)dt=\phi (0)={\delta }_0(\phi )$$

Definizione §

Per definire la derivata debole, abbiamo integrato una funzione localmente integrabile $f\in L_{loc}^1({\mathbb{R}}^n)$ con una funzione test $\psi \in C_c^{\infty }(\Omega )$. Possiamo interpetare questa procedura come una funzionale definito dalla $f$ che manda $C_{cpt}^{\infty }\to \mathbb{R}$

$$\psi \mapsto {\int }_{\Omega }f\psi d^{n}x$$

Una distribuzione su $\Omega \subset {\mathbb{R}}^n$ è un funzionale generale, non necessariamente rappresentabile in forma integrale, es. Dirac’s delta ${\delta }_x$ :

$${\delta }_x[\psi ]=\psi (x)$$

ovviamente non vogliamo tutti i funzionali, ma quelli lineari, per avere uno spazio vettoriale e continui. Lo spazio delle distribuzioni è quindi il duale delle funzioni test, con somma e prodotto per scalare definiti nella maniera ovvia, Swartz indicava $C_{cpt}^{\infty }$ con il simbolo $\mathcal{D}$ , quindi il suo duale:

$${\mathcal{D}}^{\prime }(\Omega )$$

Siccome ogni funzione localmente integrabile definisce la relativa distribuzione mediante l’integrale, è evidente che vale l’inclusione:

$$L_{loc}^1(\Omega )\subset {\mathcal{D}}^{\prime }(\Omega )$$

dove abbiamo commesso un abuso di notazione (le funzioni ed i rispettivi funzionali integrali). In generale ad ogni funzione in $L^p$ si può associare una distribuzione (integrale).

Convergenza e continutà §

Prima dobbiamo definire la convergenza nello spazio di partenza, le funzioni test. Siano $\psi \in C_{cpt}^{\infty }(\Omega )$ , $\{{\psi }_k{\}}_{k\in \mathbb{N}}\subset C_{cpt}^{\infty }(\Omega )$, diciamo che la successione converge

$${\psi }_k\to \psi$$

se e solo se il supporto di tutte le ${\psi }_k$ è contenuto in un insieme compatto $K\subset \Omega$ e le funzioni e tutte le derivate parziali convergono uniformemente in $K$.

Definiamo quindi la continuità come al solito. La distribuzione $u$ è continua se data una successione tale che ${\psi }_k\to \psi$, vale:

$$\underset{k\to \infty }{\lim }(u,{\psi }_k)=(u,\psi )$$

abbiamo introdotto la notazione comune $(u,\psi ):=u[\psi ]$.

Derivate distribuzionali §

Definiamo in maniera praticamente uguale alle derivate debolil e derivate di una distribuzione:

$$(D^{\alpha }u,\psi ):=(-1{)}^{|\alpha |}(u,D^{\alpha }\psi )$$

ben definito perchè è una mappa lineare e continua.

Osservazioni §

Le derivate deboli non sempre esistono, mentre quelle distribuzionali esistono sempre, $\psi$ sono $C^{\infty }$!!!. Data la somiglianza spesso i termini derivata debole e distribuzionale vengono usati come sinonimi, ma sono cose diverse! (spazi diversi), infatti la derivata debole (quando esiste) è rappresentata da una funzione localmente integrabile, cosa non sempre vera per le distibuzioni.