Let $\mathscr{S}(\mathbb{R})$ be the Schwartz space over $\mathbb{R}$ and $\mathscr{S}'(\mathbb{R})$ the space of tempered distributions. Define an operator $U_a$ on the space of tempered distributions as translation by $a$: $$(U_aT)(\varphi) = T(U_{-a}\varphi)$$ for all $T \in \mathscr{S}'(\mathbb{R})$ and $\varphi \in \mathscr{S}(\mathbb{R})$.
If $d/dx$ is the derivative operator on $\mathscr{S}'(\mathbb{R})$, I would like to prove that $$(U_a - 1)a^{-1} \rightarrow \frac{d}{dx}$$ where the convergence is pointwise and in the $\sigma(\mathscr{S}'(\mathbb{R}), \mathscr{S}(\mathbb{R}))$ topology, that is the weak* topology on $\mathscr{S}'(\mathbb{R})$.
I see that $$(U_a - 1)a^{-1}T(f) = T\Big(\frac{f(x - a) - f(a)}{a}\Big) \rightarrow T\big(f'(x)\big). \tag{1}$$
It seems that I am missing a negative sign on the right hand side of (1) to agree with the definition of the derivative in the sense of distributions: $$\Big(\frac{d}{dx}T\Big)(f) = -T\Big(\frac{d}{dx}f\Big).$$
Aside from this issue, I would also like to make (1) rigorous. To do this I will have to show that $(U_a - 1)a^{-1} \rightarrow \frac{d}{dx}$ in the weak* topology. I am having some trouble with this part. Usually weak* convergence means showing that $T_*(f) \rightarrow T(f)$ for all $f \in \mathscr{S}(\mathbb{R})$. However here the situation is a little different since the image is also in $\mathscr{S}'(\mathbb{R})$. How would one show that an operator converges in the weak* topology on tempered distributions?