test function - borderintegral

09.12.2012, 16:21

clancy

test function - borderintegral

Meine Frage:
hello

if $\begin{eqnarray*} \phi\in C_0^{\infty}(B_1(0)), B_1(0)\subset\mathbb{R}^2 \end{eqnarray*}$ , i.e. $\begin{eqnarray*} \phi \end{eqnarray*}$ is a test function on the two dimensional unit ball,

then the borderintegral

$\begin{eqnarray*} \int\limits_{\partial (B_1(0)\setminus B_{\varepsilon}(0))}\psi(x)\, dx \end{eqnarray*}$ is the same as the borderintegral

$\begin{eqnarray*} \int\limits_{\partial B_{\varepsilon}(0)}\psi(x)\, dx \end{eqnarray*}$

but why?

Meine Ideen:
i guess it has something to do with the feature that $\begin{eqnarray*} \phi \end{eqnarray*}$ as a test function has compact support within $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ .. but i dont know how to make this clear

is there somebody to help?

greetings

09.12.2012, 16:26

Che Netzer

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RE: test function - borderintegral
If $\begin{eqnarray*} \phi \end{eqnarray*}$ (or $\begin{eqnarray*} \psi \end{eqnarray*}$ ?) is a test function on $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ , it has to vanish on $\begin{eqnarray*} \partial B_1(0) \end{eqnarray*}$ (i.e. on a neighborhood of it).
Now think about how to write $\begin{eqnarray*} \partial(B_1(0)\setminus B_\varepsilon(0)) \end{eqnarray*}$ as the union of two sets.

09.12.2012, 16:33

clancy

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RE: test function - borderintegral
hi ché netzer - thank you for answering that quickly!

sorry for my disturbing mixture of notation
i mean the same function $\begin{eqnarray*} \phi \end{eqnarray*}$

Zitat:

Original von Che Netzer
If $\begin{eqnarray*} \phi \end{eqnarray*}$ (or $\begin{eqnarray*} \psi \end{eqnarray*}$ ?) is a test function on $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ , it has to vanish on $\begin{eqnarray*} \partial B_1(0) \end{eqnarray*}$ (i.e. on a neighborhood of it).

oh - why?? i cannnot see it

Zitat:

Now think about how to write $\begin{eqnarray*} \partial(B_1(0)\setminus B_\varepsilon(0)) \end{eqnarray*}$ as the union of two sets.

i guess

$\begin{eqnarray*} \partial(B_1(0)\setminus B_{\varepsilon}(0))=\partial B_1(0)\cup\partial B_{\varepsilon}(0) \end{eqnarray*}$ .

if $\begin{eqnarray*} \phi \end{eqnarray*}$ vanishes on $\begin{eqnarray*} \partial B_1(0) \end{eqnarray*}$ as you say then it is clear that it is enough to integrate over $\begin{eqnarray*} \partial B_{\varepsilon}(0) \end{eqnarray*}$ .

09.12.2012, 16:39

Che Netzer

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RE: test function - borderintegral

Zitat:

Original von clancy
oh - why?? i cannnot see it

We assume that the support of $\begin{eqnarray*} \phi \end{eqnarray*}$ is a compact subset of $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ . Thus it must have a positive distance to the border of the open (!) set $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ : If it would "touch" that border, it could not be completely contained in $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ .
If we could not find any open neighborhood of some $\begin{eqnarray*} x\in\partial B_1(0) \end{eqnarray*}$ which does not intersect the support, then $\begin{eqnarray*} x \end{eqnarray*}$ would be a limiting point of $\begin{eqnarray*} \operatorname{supp}\phi \end{eqnarray*}$ .

Zitat:

i guess
$\begin{eqnarray*} \partial(B_1(0)\setminus B_{\varepsilon}(0))=\partial B_1(0)\cup\partial B_{\varepsilon}(0) \end{eqnarray*}$ .

Exactly

09.12.2012, 16:47

clancy

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ah!

1. $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ is the OPEN unit ball

2. $\begin{eqnarray*} \operatorname{supp}(\phi)\subsetneqq B_1(0) \end{eqnarray*}$ and is compact - and therefore especially closed

?

in general the support of a testfunction is a REAL subset of the concerning set?

09.12.2012, 16:52

Che Netzer

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Zitat:

Original von clancy
1. $\begin{eqnarray*} B_1(0) \end{eqnarray*}$ is the OPEN unit ball

I guess so (and at least we always defined it to be open – you surely did so as well). If it were closed, it would not make much sense.

Zitat:

2. $\begin{eqnarray*} \operatorname{supp}(\phi)\subsetneqq B_1(0) \end{eqnarray*}$ and is compact - and therefore especially closed

Well, first we have $\begin{eqnarray*} \operatorname{supp}\phi\subset B_1(0) \end{eqnarray*}$ . But as a compact set cannot be open, " $\begin{eqnarray*} \subsetneq \end{eqnarray*}$ " follows immediately.

Zitat:

in general the support of a testfunction is a REAL subset of the concerning set?

In most cases, one considers testfunction on an open set. Then surely the support has to be a real subset. But for a compact set $\begin{eqnarray*} A \end{eqnarray*}$ , we simply have $\begin{eqnarray*} C_0^\infty(A)=C^\infty(A) \end{eqnarray*}$ . That would be quite boring Augenzwinkern

09.12.2012, 16:55

clancy

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now i understood

thank you very much!

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