Cfrgtkky

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 $barpsi in D(Bbb R)$ with a good property that $exists a>0$ such that $bar psibig|_{[-a,a]}equiv Bbb 1_{[-a,a]}$. The final step is to consider the product $xbar psi(x)$.

For the second part we can write by definition

$$0=fdelta_0'iff forall phi in Dquad langle fdelta_0',phirangle=0 iff$$

$$forall phi in Dquad 0=langledelta_0',fphirangle iff forall phi in Dquad 0=f'(0)phi(0)+f(0)phi'(0).$$

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 $gin C^infty$ such that $f(x)=x^2 g(x)$.