I need prove the following extention doesn't defines a distribution on $\mathbb{R}$ $$\sum\limits^{\infty}_{k=0}\varphi^{k}(0)~~~,~~~\varphi\in D(\mathbb{R})$$ My attempts :
I will prove the above serie are not convergent so I use :
$$\varphi(x)=e^{x}\psi(x)$$
Where $\psi\in D(\mathbb{R})$ and $\psi(x)=1$ if $x\in [-a;a]$ and $\psi(x)=0$ otherwise
So we find : $\varphi^{k}(0)=1$ this mean that the sum are not convergence
Are my Steps correct ?
Is there another ideas ?
With $$\rho(x)= e^{-1/(1-x^2)} 1_{|x|<1}, \qquad C=\int_\Bbb{R} \rho(x)dx$$ $\rho$ is smooth on $\Bbb{R}$ and supported on $[-1,1]$. then $$\psi(x)= \frac1C \int_{-3}^3 \rho(x-y)dy$$ is smooth on $\Bbb{R}$, supported on $[-4,4]$, and constant $=1$ on $[-2,2]$.
So $$\varphi(x)= e^x \psi(x)$$ works, it is smooth on $\Bbb{R}$, supported on $[-4,4]$, and $$\varphi^{(k)}(0)=1$$ which proves that $$\phi \mapsto \sum\limits^{\infty}_{k=0}\phi^{k}(0)$$ is not a distribution.