# Finding proof of representation theorem for distributions

So, we have a theorem, we call it representation theorem for distributions:

Theorem: Let $f$ be a distribution with support $\{0\}$. Then, there is a $k\in N$ and $c_{\alpha} \in \mathbb{R}$ for all $\alpha \in Z_{+}^n$ multi index such that $|\alpha| \leq k$ (where $\alpha = (\alpha_1,\alpha_2,...,\alpha_n), |\alpha| = \alpha_1+\alpha_2+...+\alpha_n$) such that $f = \sum_{|\alpha| \leq k} c_{\alpha} \partial^{\alpha} \delta$.

Proof which professor gave I can not consider correct. Please help I have an oral exam and no ideas where to find this proof.

• And what was the proof given by your professor? Why do you consider it incorrect? – TZakrevskiy Feb 13 '17 at 9:13
• @TZakrevskiy would you accept as a proof of theorem, which begins with some diagrams with no sense at all and now, it goes linear algebra? Do I hear no? – nikola Feb 18 '17 at 11:42
• It is all hearsay, your word against his word. Nothing eliminates the case that you did not understand something, unless you provide the details and point out which part of the proof does not seem plausible to you. – TZakrevskiy Feb 18 '17 at 12:10
• Since you did not accept my answer, I realize that it was not helpful to you so I deleted it. – NeedForHelp Feb 21 '17 at 3:06
• @NeedForHelp I am sorry, I did not pass the written exam, so I have to listen the same subject again, and when professor comes to that proof, I will ask him to do that in detail. Thanks anyway! – nikola Feb 21 '17 at 7:28

Here's the proof given by my professor. Hopefully it meets your standards.

Recall the following definitions concerning distributions with compact support:

Definitions:

1. If $X \subset \mathbb{R}^d$ is open then $$\mathcal{E}'(X) := \{u \in \mathcal{D}'(X) : \operatorname{supp} u \text{ is a compact subset of } X\}$$
2. If $u \in \mathcal{E}'(X)$ and $\psi \in C^{\infty}(X)$ then $$\langle u, \psi \rangle := \langle u, \rho \psi\rangle$$ where $\rho \in C^{\infty}_C(X)$ with $\rho = 1$ on a neighborhood of $\operatorname{supp} u$.

Remark: Definition 2. is independent of the choice of $\rho$ and coincides with the usual definition when $\psi \in C^{\infty}_C(X)$.

Lemma: If $u \in \mathcal{E}'(X)$ and $V$ is a neighborhood of $\operatorname{supp} u$ then there exists constants $C$, $k$ such that $$|\langle u,\psi \rangle |\leq C\sum_{|\alpha|\leq k}\sup_V |\partial^{\alpha}\psi|\quad\quad\forall\psi\in C^{\infty}(X)$$

Proof: Fix $\rho \in C^{\infty}_C(X)$ such that $\rho = 1$ on a neighborhood of $\operatorname{supp} u$ and $\operatorname{supp} \rho \subset V$. Let $K := \operatorname{supp} \rho$. Then there are constants $C$, $k$ such that $$|\langle u,\varphi \rangle |\leq C\sum_{|\alpha|\leq k}\sup |\partial^{\alpha}\varphi|\quad\quad\forall\varphi\in C^{\infty}_C(X), \operatorname{supp} \varphi \subset K$$

Let's take an arbitrary $\psi \in C^{\infty}(X)$ and apply this last inequality with $\varphi = \rho \psi$. We obtain \begin{align} |\langle u,\psi \rangle| &= |\langle u,\rho\psi \rangle| \\ &\leq C\sum_{|\alpha|\leq k}\sup |\partial^{\alpha}(\rho\psi)| \\ &= C\sum_{|\alpha|\leq k}\sup \left|\sum_{\substack{\beta,\gamma \\ \beta+\gamma=\alpha}}\frac{\alpha!}{\beta!\, \gamma!}\partial^{\beta}\rho\,\partial^{\gamma}\psi\right| \\ &\leq C\sum_{|\alpha|\leq k} \sum_{\substack{\beta,\gamma \\ \beta+\gamma=\alpha}} \frac{\alpha!}{\beta!\, \gamma!} \sup_V |\partial^{\beta}\rho|\sup_V|\partial^{\gamma}\psi| \\ &\leq C' \sum_{|\alpha|\leq k}\sup_V|\partial^{\gamma}\psi| \end{align}

Theorem: Let $u \in \mathcal{D}'(X)$ with $\operatorname{supp}u=\{0\}$. Then $\displaystyle u = \sum_{|\alpha|\leq k} c_{\alpha} \delta^{(\alpha)}$ where $k \in \Bbb{Z}^+$, $c_{\alpha} \in \Bbb{C}$ and $\delta^{(\alpha)}=\partial^{\alpha} \delta$.

Proof: Let's fix $r>0$ such that $\{x : |x|\leq r\} \subset X$. By the lemma there exists constants $C$, $k$ such that $$|\langle u,\psi \rangle |\leq C\sum_{|\alpha|\leq k}\,\sup_{|x|<r} |\partial^{\alpha}\psi|\quad\quad\forall\psi\in C^{\infty}(X)$$

If $\chi \in C^{\infty}(X)$ and if $\partial^{\alpha}\chi(0)=0$ for all $|\alpha|\leq k$ then $\langle u, \chi \rangle = 0$

Let's fix $\rho \in C^{\infty}_C(\mathbb{R}^d)$ with $\operatorname{supp} \rho \subset \{x : |x| < r\}$ and $\rho = 1$ on $\{x : |x| \leq r/2\}$. Then for all $\epsilon > 0$, \begin{align} |\langle u,\chi \rangle| &= |\langle u, \rho(x/\epsilon) \chi(x)\rangle| \\ &\leq C\sum_{|\alpha|\leq k}\,\sup_{|x|<r} |\partial^{\alpha}(\rho(x/\epsilon) \chi(x))| \\ &= C\sum_{|\alpha|\leq k}\,\sup_{|x|<r} \left|\sum_{\substack{\beta,\gamma \\ \beta+\gamma=\alpha}}\frac{\alpha!}{\beta!\, \gamma!}\partial^{\beta}(\rho(x/\epsilon))\,\partial^{\gamma}\chi(x)\right| \\ &\leq C\sum_{|\alpha|\leq k}\sum_{\substack{\beta,\gamma \\ \beta+\gamma=\alpha}}\frac{\alpha!}{\beta!\, \gamma!} \sup_{|x|<r\epsilon}|\partial^{\beta}(\rho(x/\epsilon))\,\partial^{\gamma}\chi(x)| \\ &\leq C\sum_{|\alpha|\leq k}\sum_{\substack{\beta,\gamma \\ \beta+\gamma=\alpha}} \frac{\alpha!}{\beta!\, \gamma!} \sup_{|x|<r\epsilon} \frac{1}{\epsilon^{|\beta|}}|(\partial^{\beta}\rho)(x/\epsilon)| \sup_{|x|<r\epsilon} |\partial^{\gamma}\chi(x)|\tag{$\star$} \end{align}

But the function $\partial^{\gamma}\chi$ satisfies $\partial^{\eta}(\partial^{\gamma}\chi)(0)=0$ for all multi index $\eta$ with $|\eta|\leq k-|\gamma|$. Hence, by Taylor's theorem applied to $\partial^{\gamma}\chi$ at $0$, we have $$\partial^{\gamma}\chi (x) = o(|x|^{k-|\gamma|}) \quad \quad (x \to 0)$$

So $$\sup_{|x|<r\epsilon} |\partial^{\gamma}\chi(x)| = o(\epsilon^{k-|\gamma|})\quad\quad(\epsilon\to0)$$

Substituting this information in $(\star)$, it follows that $$|\langle u,\chi \rangle| = o(1)\quad\quad (\epsilon \to 0)$$

Since $|\langle u,\chi \rangle|$ does not depend on $\epsilon$, it follows that $\langle u,\chi \rangle = 0$.

$\displaystyle u = \sum_{|\alpha| \leq k} c_{\alpha} \delta^{(\alpha)}$ for some $c_{\alpha} \in \Bbb{C}$

Let's take $\varphi \in C^{\infty}_C(X)$. Let $\displaystyle \chi(x) := \varphi(x) - \sum_{|\alpha| \leq k}\frac{\partial^{\alpha}\varphi(0)}{\alpha!}x^{\alpha}$. Then $\chi \in C^{\infty}(X)$ and $\partial^{\alpha}\chi(0)=0$ for all $|\alpha| \leq k$. By the first part of the proof, $\langle u,\chi \rangle=0$. Hence \begin{align} \langle u,\varphi \rangle &= \left\langle u, \sum_{|\alpha| \leq k}\frac{\partial^{\alpha}\varphi(0)}{\alpha!}x^{\alpha}\right\rangle \\ &= \sum_{|\alpha| \leq k} \partial^{\alpha}\varphi(0) \langle u,x^{\alpha}/\alpha! \rangle \\ &= \sum_{|\alpha| \leq k} \langle u,x^{\alpha}/\alpha! \rangle (-1)^{|\alpha|} \langle \delta^{(\alpha)}, \varphi \rangle \end{align} and it follows that $$u = \sum_{|\alpha| \leq k} c_{\alpha} \delta^{(\alpha)}$$ where $c_{\alpha} = (-1)^{|\alpha|}\langle u,x^{\alpha}/\alpha! \rangle$.

• @TZakrevskiy I'm not the asker! lol – NeedForHelp Feb 16 '17 at 22:36
• Ooops, I'm sorry, my bad. Kudos for typing all that! – TZakrevskiy Feb 16 '17 at 23:06
• @TZakrevskiy Yeah, well, kudos to my professor for making the proof clear! I feel like I should make my answer Community Wiki since it's just a copy of someone else's work... :[ – NeedForHelp Feb 16 '17 at 23:14