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Visit Stack Exchange$\newcommand{\singsupp}{\operatorname{sing supp}}$Let$u, v$ be two distributions with disjoint singular support. Take$\psi_u, \psi_v \in C^\infty$ such that$\psi_u = 1$ on a neighborhood of$\singsupp u$ and$\phi_u = 0$$\psi_u = 0$ on a neighborhood of$\singsupp v$, reversely for$\psi_v$. Then we decompose$u$ and$v$ according to$u = \psi_u u + (1-\psi_u) u$ and$v = \psi_v v + (1-\psi_v) v$.Further we have$\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally,$$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\= \left(\psi_u u\right) \left(\psi_v v\right)+ \left(\psi_u u\right) \left((1-\psi_v) v\right)+ \left((1-\psi_u) u\right) \left(\psi_v v\right)+ \left((1-\psi_u) u\right) \left((1-\psi_v) v\right)$$
The first term,$\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as$\left(\psi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution$\psi_u u$ times a$C^\infty$ function$(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with$u$ and$v$ swapped, so it is also defined.
The forth term is a product of two$C^\infty$ functions and thus defined.
Therefore we can take the above formula (edit: with$\left(\psi_u u\right) \left(\psi_v v\right)$ replaced with$\left(\psi_v u\right) \left(\psi_u v\right)$) as a definition of$uv$.
$\newcommand{\singsupp}{\operatorname{sing supp}}$Let$u, v$ be two distributions with disjoint singular support. Take$\psi_u, \psi_v \in C^\infty$ such that$\psi_u = 1$ on a neighborhood of$\singsupp u$ and$\phi_u = 0$ on a neighborhood of$\singsupp v$, reversely for$\psi_v$. Then we decompose$u$ and$v$ according to$u = \psi_u u + (1-\psi_u) u$ and$v = \psi_v v + (1-\psi_v) v$.Further we have$\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally,$$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\= \left(\psi_u u\right) \left(\psi_v v\right)+ \left(\psi_u u\right) \left((1-\psi_v) v\right)+ \left((1-\psi_u) u\right) \left(\psi_v v\right)+ \left((1-\psi_u) u\right) \left((1-\psi_v) v\right)$$
The first term,$\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as$\left(\psi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution$\psi_u u$ times a$C^\infty$ function$(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with$u$ and$v$ swapped, so it is also defined.
The forth term is a product of two$C^\infty$ functions and thus defined.
Therefore we can take the above formula (edit: with$\left(\psi_u u\right) \left(\psi_v v\right)$ replaced with$\left(\psi_v u\right) \left(\psi_u v\right)$) as a definition of$uv$.
$\newcommand{\singsupp}{\operatorname{sing supp}}$Let$u, v$ be two distributions with disjoint singular support. Take$\psi_u, \psi_v \in C^\infty$ such that$\psi_u = 1$ on a neighborhood of$\singsupp u$ and$\psi_u = 0$ on a neighborhood of$\singsupp v$, reversely for$\psi_v$. Then we decompose$u$ and$v$ according to$u = \psi_u u + (1-\psi_u) u$ and$v = \psi_v v + (1-\psi_v) v$.Further we have$\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally,$$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\= \left(\psi_u u\right) \left(\psi_v v\right)+ \left(\psi_u u\right) \left((1-\psi_v) v\right)+ \left((1-\psi_u) u\right) \left(\psi_v v\right)+ \left((1-\psi_u) u\right) \left((1-\psi_v) v\right)$$
The first term,$\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as$\left(\psi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution$\psi_u u$ times a$C^\infty$ function$(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with$u$ and$v$ swapped, so it is also defined.
The forth term is a product of two$C^\infty$ functions and thus defined.
Therefore we can take the above formula (edit: with$\left(\psi_u u\right) \left(\psi_v v\right)$ replaced with$\left(\psi_v u\right) \left(\psi_u v\right)$) as a definition of$uv$.
$\newcommand{\singsupp}{\operatorname{sing supp}}$Let$u, v$ be two distributions with disjoint singular support. Take$\psi_u, \psi_v \in C^\infty$ such that$\psi_u = 1$ on a neighborhood of$\singsupp u$ and$\phi_u = 0$ on a neighborhood of$\singsupp v$, reversely for$\psi_v$. Then we decompose$u$ and$v$ according to$u = \psi_u u + (1-\psi_u) u$ and$v = \psi_v v + (1-\psi_v) v$.Further we have$\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally,$$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\= \left(\psi_u u\right) \left(\psi_v v\right)+ \left(\psi_u u\right) \left((1-\psi_v) v\right)+ \left((1-\psi_u) u\right) \left(\psi_v v\right)+ \left((1-\psi_u) u\right) \left((1-\psi_v) v\right)$$
The first term,$\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as$\left(\phi_v u\right) \left(\psi_u v\right) \in C^\infty$$\left(\psi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution$\psi_u u$ times a$C^\infty$ function$(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with$u$ and$v$ swapped, so it is also defined.
The forth term is a product of two$C^\infty$ functions and thus defined.
Therefore we can take the above formula (edit: with$\left(\psi_u u\right) \left(\psi_v v\right)$ replaced with$\left(\psi_v u\right) \left(\psi_u v\right)$) as a definition of$uv$.
$\newcommand{\singsupp}{\operatorname{sing supp}}$Let$u, v$ be two distributions with disjoint singular support. Take$\psi_u, \psi_v \in C^\infty$ such that$\psi_u = 1$ on a neighborhood of$\singsupp u$ and$\phi_u = 0$ on a neighborhood of$\singsupp v$, reversely for$\psi_v$. Then we decompose$u$ and$v$ according to$u = \psi_u u + (1-\psi_u) u$ and$v = \psi_v v + (1-\psi_v) v$.Further we have$\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally,$$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\= \left(\psi_u u\right) \left(\psi_v v\right)+ \left(\psi_u u\right) \left((1-\psi_v) v\right)+ \left((1-\psi_u) u\right) \left(\psi_v v\right)+ \left((1-\psi_u) u\right) \left((1-\psi_v) v\right)$$
The first term,$\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as$\left(\phi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution$\psi_u u$ times a$C^\infty$ function$(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with$u$ and$v$ swapped, so it is also defined.
The forth term is a product of two$C^\infty$ functions and thus defined.
Therefore we can take the above formula as a definition of$uv$.
$\newcommand{\singsupp}{\operatorname{sing supp}}$Let$u, v$ be two distributions with disjoint singular support. Take$\psi_u, \psi_v \in C^\infty$ such that$\psi_u = 1$ on a neighborhood of$\singsupp u$ and$\phi_u = 0$ on a neighborhood of$\singsupp v$, reversely for$\psi_v$. Then we decompose$u$ and$v$ according to$u = \psi_u u + (1-\psi_u) u$ and$v = \psi_v v + (1-\psi_v) v$.Further we have$\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally,$$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\= \left(\psi_u u\right) \left(\psi_v v\right)+ \left(\psi_u u\right) \left((1-\psi_v) v\right)+ \left((1-\psi_u) u\right) \left(\psi_v v\right)+ \left((1-\psi_u) u\right) \left((1-\psi_v) v\right)$$
The first term,$\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as$\left(\psi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution$\psi_u u$ times a$C^\infty$ function$(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with$u$ and$v$ swapped, so it is also defined.
The forth term is a product of two$C^\infty$ functions and thus defined.
Therefore we can take the above formula (edit: with$\left(\psi_u u\right) \left(\psi_v v\right)$ replaced with$\left(\psi_v u\right) \left(\psi_u v\right)$) as a definition of$uv$.
$\newcommand{\singsupp}{\operatorname{sing supp}}$Let$u, v$ be two distributions with disjoint singular support. Take$\psi_u, \psi_v \in C^\infty$ such that$\psi_u = 1$ on a neighborhood of$\singsupp u$ and$\phi_u = 0$ on a neighborhood of$\singsupp v$, reversely for$\psi_v$. Then we decompose$u$ and$v$ according to$u = \psi_u u + (1-\psi_u) u$ and$v = \psi_v v + (1-\psi_v) v$.Further we have$\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally,$$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\= \left(\psi_u u\right) \left(\psi_v v\right)+ \left(\psi_u u\right) \left((1-\psi_v) v\right)+ \left((1-\psi_u) u\right) \left(\psi_v v\right)+ \left((1-\psi_u) u\right) \left((1-\psi_v) v\right)$$
The first term,$\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as$\left(\phi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution$\psi_u u$ times a$C^\infty$ function$(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with$u$ and$v$ swapped, so it is also defined.
The forth term is a product of two$C^\infty$ functions and thus defined.
Therefore we can take the above formula as a definition of$uv$.
