localization for distributions

DefinitionSuppose $U$ is an open set in ${ℝ}^n$ and $T$ is adistribution $T\in {𝒟}^{\prime }(U)$. Then we say that $T$ vanishes onan open set $V\subset U$, if the restriction of $T$ to $V$ isthe zero distribution on $V$. In other words, $T$ vanishes on $V$, if$T(v)=0$ for all $v\in C_0^{\mathrm{\infty }}(V)$. (Here $C_0^{\mathrm{\infty }}(V)$ is the setof smooth function with compact support in $V$.) Similarly, we say thattwo distributions $S,T\in {𝒟}^{\prime }(U)$ are equal, orcoincide on $V$, if $S-T$ vanishes on $V$. We thenwrite: $S=T$ on $V$.

Theorem[1, 3]Suppose $U$ is an open set in ${ℝ}^n$ and${\{U_i\}}_{i\in I}$ is an open cover of $U$, i.e.,

$$U=\bigcup _{i\in I}{U}_i.$$

Here, $I$ is an arbitrary index set. If $S,T$ are distributions on $U$,such that $S=T$ on each $U_i$, then $S=T$ (on U).

Proof. Suppose $u\in 𝒟(U)$. Our aim is to show that $S(u)=T(u)$.First, we have $\mathrm{supp}u\subset K$ for some compact $K\subset U$.It follows (http://planetmath.org/YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover) that there exist a finite collection of $U_i$:s from the open cover,say $U_1,\mathrm{\dots },U_N$, such that $K\subset {\cup }_{i=1}^N{U}_i$.By a smooth partition of unity, thereare smooth functions ${\varphi }_1,\mathrm{\dots },{\varphi }_N:U\to ℝ$ such that

1. $\mathrm{supp}{\varphi }_i\subset U_i$ for all $i$.

2. ${\varphi }_i(x)\in [0,1]$ for all $x\in U$ and all $i$,

3. ${\sum }_{i=1}^N{\varphi }_i(x)=1$ for all $x\in K$.

From the first property, and from a property for the support of a function (http://planetmath.org/SupportOfFunction),it follows that$\mathrm{supp}{\varphi }_{i}u\subset \mathrm{supp}{\varphi }_i\cap \mathrm{supp}u\subset U_i$.Therefore, for each $i$,$S({\varphi }_{i}u)=T({\varphi }_{i}u)$ since $S$ and $T$ conicideon $U_i$.Then

$S(u)={\displaystyle \sum _{i=1}^N}S({\varphi }_{i}u)={\displaystyle \sum _{i=1}^N}T({\varphi }_{i}u)=T(u),$

and the theorem follows. $\mathrm{\square }$