Equations in D' (the space of distributions)

Question

i have the following exercice:

1. prouve the existence of $\psi_1 \in \mathcal{D}(\mathbb{R})$ such as $\psi_1(0)=0$ and $\psi_1'(0)=0$.

2. Let $\psi_0 \in \mathcal{D}(\mathbb{R})$ such as $\psi_0=1$ in $V(0)$, and let $\varphi \in \mathcal{D}(\mathbb{R})$ We note $f(x)= \varphi(x)-\varphi(0)\psi_0(x)- \varphi'(0)\psi_1(x)$ such as $f(0)=f'(0)=0$.

   • prouve that there exist $g \in \mathcal{D}(\mathbb{R})$ such $f(X)= x^2 g(x)$.

3. resolve the equation $x^2 T=0$.

4. resolve the equation $x^2 T=\delta$.

I try to do this. for question 1, it's OK. for question 2. i try to methods, but i have questions

Method 1. for $x \neq 0$, we have $f(x)=\dfrac{g(x)}{x^2}$. My difficultie is how we define $g(0)$?

b. With the Taylor developement integral remainder of $f$, in the neighbourhood of zero, in order 2

$$ f(x)= f(0)+ xf'(0)+ \dfrac{x^2}{2!} \displaystyle\int_0^1 (1-t)^2 f''(tx) dt. $$ beacause of $f(0)= f'(0)=0$, then we have $g(x)= \dfrac{1}{2!} \displaystyle\int_0^1(1-t)^2 f''(tx) dt.$ it's clear that $g \in C^\infty(\Bbb R)$ and $\mathrm{supp\,} g $ is compact beacause $\mathrm{supp\,} f'' \subset \mathrm{supp\,} f' \subset \mathrm{supp\,} f$ et $\mathrm{supp\,} f$ est compact car $\mathrm{supp\,} f \subset \{\mathrm{supp\,} \varphi\} \cup \{\mathrm{supp\,} \psi_0\} \cup \{\mathrm{supp\,} \psi_1\}$.

the method 2 is it true?

for question 3 and 4 i don't understand how we can resolve the two equations in $\mathcal{D}'(\mathbb{R})$.

Thank's to help me.

Answer 1

In order to solve $x^2T=0$, you need to define $T$ on all test functions. Let $\phi$ be a test function, then take as in (1) and (2) the decomposition $$\phi = \phi(0)\psi_0(x) + \phi'(0)\psi_1(x) + x^2g(x)$$ and then apply $T$: $$\langle T,\phi \rangle = \phi(0)\langle T,\psi_0 \rangle+ \phi'(0)\langle T,\psi_1 \rangle + \langle T,x^2g \rangle$$ $$=\phi(0)\langle T,\psi_0 \rangle+ \phi'(0)\langle T,\psi_1 \rangle + \langle x^2T,g \rangle$$ $$=\phi(0)\langle T,\psi_0 \rangle+ \phi'(0)\langle T,\psi_1 \rangle $$ We do not know the values $\langle T,\psi_0 \rangle$ and $\langle T,\psi_1 \rangle$, and we put them as arbitrary constants $c_0$ and $c_1$,respectively.

Therefore, $$T = c_0\delta_0 - c_1\delta_0'.$$

Can you apply the same method to solve $x^2T=\delta_0$?