I was solving an exercise and I faced these 2 parts 1) Prove that there exist a function $\gamma \in D(\mathbb{R})$ such that $\gamma(0)=0$ and $\gamma'(0)=1$.
I tried for this part the function $$\gamma(x)= \mathbb{1}_{[-a,a]}(x)xe^x$$ but then I thought that its not continuous over $\mathbb{R}$, so is there any other function that have these characteristics? And can I edit mine so that it belongs to $D(\mathbb{R})$
2)Let $f \in C^{\infty}(\mathbb{R})$ Prove the following equivalence :
$f \delta_0'=0$ in $D^*(\mathbb{R}) \iff$ $\exists g \in C^{\infty}(\mathbb{R})$ such that $f(x)=x^2g(x)$
I think that the sufficient condition is easy, but I didn't know how to prove the sufficient condition.
Can anyone help please? Thank you
For the first part, you should start with mollifiers. By a convolution of $\mathbb 1_{[-1,1]}$ with a well-chosen mollifier $\psi$ you will get a function $\bar\psi \in D(\Bbb R)$ with a good property that $\exists a>0$ such that $\bar \psi\big|_{[-a,a]}\equiv \Bbb 1_{[-a,a]}$. The final step is to consider the product $x\bar \psi(x)$.
For the second part we can write by definition
$$0=f\delta_0'\iff \forall \phi \in D\quad \langle f\delta_0',\phi\rangle=0 \iff$$
$$\forall \phi \in D\quad 0=\langle\delta_0',f\phi\rangle \iff \forall \phi \in D\quad 0=f'(0)\phi(0)+f(0)\phi'(0).$$
Now, using the first part of your question, put $\phi \in D$ such that $\phi'(0)=1$ and $\phi(0)=0$ to obtain $f(0)=0$; similarly, if $\phi'(0)=0$ and $\phi(0)=1$ you will get $f'(0)=0$.
By this time you definitely saw an exercise allowing you to show that $f(0)=0=f'(0)$ implies the existence of $g\in C^\infty$ such that $f(x)=x^2 g(x)$.