Let $\varphi \in D(\mathbb{R}) $ and $f(x)=C \varphi(x)$ , C constant with $\int_{-\infty}^{+\infty}f(x)dx=1$. Let $0 < \alpha < \frac{1}{2}$ and $\psi_{\alpha}(x)=\frac{1}{\alpha}f(\frac{x}{\alpha})$
I proved with a change of variables that $\int_{-\infty}^{+\infty}\psi(x)dx=1$.
Now let $\ell>0$ and $$E(x)=\begin{cases} 1 \quad \text{if} \quad 0\le x\le \ell, \\ 0 \quad \text{else.}\end{cases} $$
My question is how prove that the convolution $h=\psi*E \in D(\mathbb{R}) $ and $h \equiv 1$ in $[\alpha; \ell-\alpha]$
You are just willing to say that if $\phi \in L^1$ is supported on $[-r,r]$ then for $\ell > 2r$ $$\phi \ast 1_{[0,\ell]}$$ is constant $=\int_{-r}^r\phi(x)dx$ on $[r,\ell-r]$,
and $0$ on $(-\infty,-r]\cup [\ell+r,\infty)$.
If $\phi$ was $C^\infty$ then $\phi \ast 1_{[0,\ell]}$ is $C^\infty$ because its derivative is $\phi(x)-\phi(x+\ell)$.