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 $0$.

I have a few questions about this proof.

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.

  • $\begingroup$ If $\phi(x)=x^n/n!$ then $\partial^n\phi(0)\ne0$. $\endgroup$
    – Did
    Jan 7, 2019 at 7:12
  • $\begingroup$ @Did Ah, I see. Can you provide any info for the second question I asked? $\endgroup$ Jan 7, 2019 at 18:14
  • $\begingroup$ Simply by differentiating $\alpha$ times the identity $$ \psi(x)=\phi(x) - \sum_{|\beta| \leq k}\frac{\partial^{\beta}\phi(0)(x)^{\beta}}{\beta!} $$ $\endgroup$
    – Did
    Jan 7, 2019 at 18:16
  • $\begingroup$ 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. $\endgroup$ Jan 7, 2019 at 18:48
  • $\begingroup$ 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. $\endgroup$
    – Did
    Jan 7, 2019 at 22:12


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