Translation operator on tempered distributions converges to derivative operator

Question

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?

Answer 1

Let $f\in \mathscr S(\Bbb R)$, to keep expressions from looking terrible write $f_a:= (U_{-a}-1)a^{-1}f$. Then for all $a$ there is some $y_a\in[x-a, x]$ $$f_a(x)=\frac{f(x-a)-f(x)}{a}=-f'(y_a)$$ which is the mean value theorem. Then, since $f'$ is Lipschitz, you get $$\sup_{x\in \Bbb R}|f_a(x)+f'(x)| ≤ L(f')\, a$$ Here $L$ is the Lipschitz constant.

You can adapt this to show that $$\sup_{x\in\Bbb R}\left|(x+1)^n (\frac{d^m}{dx^n}f_a + \frac{d^m}{dx^m}f')(x)\right|$$ goes to $0$ as $a\to0$ for all $n, m$. This implies that $f_a\to -f'$ in the topology of $\mathscr S(\Bbb R)$. Then for all $T\in\mathscr S'(\Bbb R)$ you get $T(f_a)\to T(-f')$, which is where you want to go.