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Let $u$ be a distribution on $\mathbb{R}^n$ with support = $\left\{0\right\}$. Then there exists $N$ such that $u$ has order $N$. Let $\chi\in C_0^{\infty}(\mathbb{R}^n)$ a smooth function with

$\chi(x) = 1$ for $0\leq |x|\leq 1 $,

$\chi(x)\in [0,1]$ for $1\leq |x|\leq 2$,

$\chi(x)=0$ for $|x|\geq 2$

Denote $ \chi(x/r) = k_r(x)$ for $r\in (0,1]$. For the case $N=0$ I want to show that there exists $c_1$ such that $$|\left\langle u,\phi \right\rangle|\leq c_1 |\phi(0)| $$ for all $\phi \in C_0^{\infty}(\mathbb{R}^n)$. The idea would be to apply $u$ on $\phi = k_r\phi+(1-k_r)\phi$ and letting $r\to 0$.

So then I guess we can write $$ |\left\langle u,\phi \right\rangle| \leq |\left\langle u,k_r\phi \right\rangle| + |\left\langle u,(1-k_r)\phi \right\rangle |$$ But not really sure how this follows...

Moreover how can i show that $\left\langle u,\phi \right\rangle = \left\langle u,\phi \chi\right\rangle = \left\langle u,\chi\right\rangle\phi(0) $?

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up vote 1 down vote accepted

Assume for the moment $\phi(0)=0$. In this case, by Bunder's answer and Taylor's theorem, $$ |\langle u,k_r\phi\rangle| \leq C ||k_r\phi||_\infty \leq C'r, $$ hence $\langle u,\phi\rangle = 0$ by letting $r\to 0$. Now, for general $\phi$, set $$ \psi := \phi - \chi\phi(0). $$ As $\psi\in C_c^\infty$ and $\psi(0)=0$, by the above $\langle u,\psi\rangle = 0$, so $$ \langle u,\phi\rangle = \phi(0) \langle u,\chi\rangle.$$

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Sorry, Id love to understand how you used Taylor's Theorem...and where you used $\phi(0)=0$.Cause the way I see it $C ||k_r\phi||_\infty = C\sup_{x\in B(0,r)}|\phi(x)k_r(x)| \leq C\sup_{x\in B(0,r)}|\phi(x)|$ – DinkyDoe Mar 4 '13 at 12:24
Wait..I Guess thats the point, so when u let $r\to 0$ we get $C|\phi(0)|$ – DinkyDoe Mar 4 '13 at 12:35
thank you both. I think I understand it now. – DinkyDoe Mar 4 '13 at 12:38
@DinkyDoe You may use Hadamard's lemma for a rigorous treatment of $C||k_r\phi||\leq C'r$. – Vobo Mar 4 '13 at 13:23
How to show that support$=\{0\}$ implies $u$ is of finite order (i.e. there exists $N$ such that $u$ has order $N$)? – Xiang Yu Feb 29 at 15:31

The precise and quite clear proof of the statement above can be found in Hoermander "Linear PDE's" vol. 1. p. 12-13, theorems 1.5.3 and 1.5.4.


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The second term is zero since $(1-k_r)\phi$ has support outside of $\{0\}$. For the first term we have from the "Cauchy-Shwartz" property of distributions that $$ |\left <u, k_r \phi\right>| \leq C \| k_r \phi \|_\infty$$ since the distribution is of order $0$. Since the $k_r$ converges to a dirac delta as $r \to \infty$ the supremum norm above will converge to the value of $\phi$ at $0$

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Someone call the $\langle$angle bracket$\rangle$ police: \langle and \rangle are what they should be. – user53153 Mar 4 '13 at 3:13
this isn't a paper, it is just a quick answer. I think i can relax a little bit with the typesetting rules. – Bunder Mar 4 '13 at 7:17
@Bunder But this justifies only the "$\leq$". It doesn't answer the "$=$". – Vobo Mar 4 '13 at 11:49
@Bunder In addition, $k_r/r$ would converge to Dirac delta, $k_r$ itself does not. – Vobo Mar 4 '13 at 14:55

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