# Translation operator on tempered distributions converges to derivative operator

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?

• as you write, $(U_aT)(\varphi) = T(U_{-a}\varphi)$; what is $U_{-a}\varphi$? Jun 18 at 8:25
• @user8268 Sorry I should have defined that in my post, it means translation by $-a$. Jun 18 at 8:30
• Isn't $U_aT(f)=T(U_{-a}f)=f(x-a)$? Jun 18 at 8:36
• @geetha290krm Yes you are right, thank you for catching that. Jun 18 at 8:42

## 1 Answer

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.

• Thank you. How did you use the weak* topology here? Jun 18 at 17:48
• Note that by definition of the weak* topology you have $(U_a -1)a^{-1}T\to T'$ if and only if $(U_a -1)a^{-1}T(f)\to T'(f)$ for all $f\in\mathscr S(\Bbb R)$. The answer shows that $(U_{-a}-1)^{-1}f\to -f'$ in the Schwartz topology, which implies (by continuity of the $T$): $$(U_a -1)a^{-1}T(f)=T(U_{-a} -1)a^{-1}f\to T(-f')=T'(f)$$ Jun 18 at 17:54
• I see now, thank you! Jun 18 at 17:57