# On the proof that a distribution with $\{0\}$ as support can be written as a sum of point masses

We have the following Theorem and its accompanying proof from pages 46-47 of Hormander's The Analysis of Linear Partial Differential Operators.

Theorem 2.3.4 If $$F$$ is a distribution of order $$k$$ with support equal to $$\{0\}$$, then for $$\phi \in C^k$$, $$F$$ is of the form:

$$F(\phi) = \sum_{|\alpha| \leq k}c_{\alpha}\partial^{\alpha}\phi(0)$$

Proof: Expanding $$\phi$$ in a taylor series gives us:

$$\phi(x) = \sum_{|\alpha| \leq k}\frac{\partial^{\alpha}\phi(0)(x)^{\alpha}}{\alpha!} + \psi(x)$$

We have that $$\partial^{\alpha}\psi(0) = 0$$ when $$|\alpha| \leq k$$, so $$F(\psi) = 0$$ by theorem 2.3.3. Hence, the result follows with $$c_{\alpha} = F\left(\frac{(x)^{\alpha}}{\alpha!}\right) \space \space \blacksquare$$

As a note, theorem 2.3.3 mentioned in the proof is the statement that if all partials of a $$C^k$$ function vanish on a point in the support of $$F$$, then $$F$$ acting on that function is $$0$$.

1) It seems as though $$\psi$$ is the remainder term of the taylor expansion. Is this true?

2) Why does it follow that the partials of $$\psi$$ vanish at 0? Does this have to do with Taylor's theorem? It is not immediately clear.

3) Why does theorem 2.3.3 not apply to the function $$\frac{x^{\alpha}}{\alpha!}$$? It seems to me that all partials of this function evaluated at 0, should be 0. So by theorem 2.3.3, $$F\left(\frac{x^{\alpha}}{\alpha!}\right)$$ should equal 0.

• If $\phi(x)=x^n/n!$ then $\partial^n\phi(0)\ne0$.
– Did
Jan 7, 2019 at 7:12
• @Did Ah, I see. Can you provide any info for the second question I asked? Jan 7, 2019 at 18:14
• Simply by differentiating $\alpha$ times the identity $$\psi(x)=\phi(x) - \sum_{|\beta| \leq k}\frac{\partial^{\beta}\phi(0)(x)^{\beta}}{\beta!}$$
– Did
Jan 7, 2019 at 18:16
• Thank you @Did I just have one more little question. Suppose that the distribution acted on $C^{\infty}$ functions with compact support. In the context of this question, would the expression $F(\frac{x^{\alpha}}{\alpha!})$ even make sense? Since the input does not have compact support. Jan 7, 2019 at 18:48
• The answer is already in the question: if the distribution is only defined on functions with compact support then it cannot act on functions with noncompact support. Tautological.
– Did
Jan 7, 2019 at 22:12