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?

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

1 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.

  • $\begingroup$ Thank you. How did you use the weak* topology here? $\endgroup$
    – CBBAM
    Jun 18 at 17:48
  • 1
    $\begingroup$ 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 $T$): $$(U_a -1)a^{-1}T(f)=T(U_{-a} -1)a^{-1}f\to T(-f')=T'(f)$$ $\endgroup$
    – s.harp
    Jun 18 at 17:54
  • $\begingroup$ I see now, thank you! $\endgroup$
    – CBBAM
    Jun 18 at 17:57

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