# How to prove that $(D_e h)(x)=(D_e L \tau_x\phi)(x)-(L\tau_xD_e\phi)(x)$ ($h(x)=L\tau_x\phi(x)$ and $\phi$ smooth with compact support)?

This title may look confusing but that's exactly why this question is posted. Let me elaborate these symbols.

1. $$L$$ is a continous linear mapping of $$C^\infty_c(R^n)$$ (smooth function with compact support) into $$C^\infty(R^n)$$ (smooth function).

2. For any function $$u$$ in $$R^n$$, we define $$\tau_x u(y)=u(y-x)$$.

3. $$D_e$$ is the directional derivative in the direction $$e$$. To be precise, Let $$e$$ be a unit vector in $$R^n$$, put $$\eta_r = r^{-1}(\tau_0-\tau_{re})$$, then $$D_e$$ is the limit as $$r \to 0$$.

Fix $$\phi \in C_c^\infty(R^n)$$. Put $$h(x)=\tau_{-x}L\tau_x\phi(0)=L\tau_x\phi(x)$$. Show that

$$(D_e h)(x)=(D_e L \tau_x\phi)(x)-(L\tau_xD_e\phi)(x)$$

How do I even start? I guess this indicates that $$D_e$$ is anti-derivation of some sort. Is it feasible to consider the limit

$$\lim_{r \to 0} \eta_r (L\tau_x \phi (x)) = \lim_{r \to 0} \frac{1}{r}(\tau_0-\tau_{re})L\tau_x\phi(x)=\lim_{r \to 0} L \tau_x \frac{\phi(x)-\phi(x-re)}{r}?$$

Hope I didn't write anything incorrect. I'm wondering if this is even close to the equation in title...

Let me give the context (can be found by W. Rudin's Functional Analysis exercise 6.25) . I'm asked to show that if $$L$$ commutes with differential operator of arbitrary order $$\alpha$$, then $$\tau_xL=L\tau_x$$ for all $$x$$. The commutativity is given as follows. Suppose $$\alpha=(\alpha_1,\alpha_2,\dots,\alpha_n) \in \mathbb{N}^n$$, then

$$D^\alpha f = D_1^{\alpha_1}D_2^{\alpha_2}\cdots D_n^{\alpha_n} f$$

where $$D_i$$ is the partial derivative on the $$i$$-th variable. So the $$L$$ I'm studying is a linear mappings satisfying $$D^\alpha L\phi = LD^\alpha \phi$$ for all $$\phi \in C_c^\infty (R^n)$$.

I am suggested to study this $$h(x)$$, whence the equation in title is given. If $$L$$ commutes with differential operators, then it follows that $$D_e h(x)=0$$ for all $$x$$, hence $$h(x)=h(0)$$ and I can deduce that $$\tau_x L = L \tau_x$$. I'm OK with these following but the equation in title is really weird to me. Basically it is saying $$D_e h(x) \ne (D_e L\tau_x\phi)(x)$$ even though $$h(x)=L\tau_x\phi(x)$$.

• What is $h$? ${}$ Aug 14, 2021 at 18:48
• @WillM. the definition of $h$ is in the title.
– user614535
Aug 15, 2021 at 1:11
• Don't do that or if you do, repeat the definition in the main question, tablets do not compile the mathjax of the title. (At least mine doesn't.) Aug 15, 2021 at 14:57
• @WillM. It should be OK now.
– user614535
Aug 15, 2021 at 15:47

Let $$h(x) = L \tau_x \varphi(x).$$ Since $$L$$ is just multiplication for a constant, $$h'(x) = L \partial_x (\tau_x \varphi(x)).$$ We focus on the latter. Let $$u(\varphi, x) = \varphi(x)$$ defined on $$\mathscr{C}_c^\infty \times \mathbf{R}$$ with values in $$\mathbf{R}.$$ Then, for a given point $$(\varphi, x),$$ $$\partial_\varphi u \cdot \psi = \psi(x), \quad \partial_x u \cdot t = \varphi'(x) t.$$ Let me check these formulas, $$u(\varphi + \epsilon, x) = \varphi(x) + \epsilon(x) = u(x) + \partial_\varphi u \cdot \epsilon + 0,$$ in other words, we expanded in the form $$f(x + h) = f(x) + f'(x) h + o(h)$$ (the error term being zero here). The second case is similarly easy, $$u(\varphi, x + e) = \varphi(x + e) = \varphi(x) + \varphi'(x) e + o(e)$$ since $$\varphi$$ is differentiable. Notice that $$\partial_\varphi u$$ and $$\partial_x u$$ are continuous functions (since $$\varphi$$ is continuously differentiable). A fortiori, $$u$$ is differentiable and $$u'(\varphi, x) \cdot (\psi, t) = \partial_\varphi \cdot \psi + \partial_x u \cdot t = \psi(x) + \varphi'(t).$$

Let $$v:\mathscr{C}_c^\infty \times \mathbf{R} \to \mathscr{C}_c^\infty$$ by $$v(\psi, x) = \tau_x \psi.$$ It should be clear $$\partial_\psi v \cdot \xi = \tau_x \xi.$$ Now, $$v(\psi, x+e) = \tau_{x+e} \psi = \tau_e \tau_x \psi = \tau_x \psi + (\tau_e \tau_x \psi - \tau_x \psi).$$

Lemma. For every $$\psi \in \mathscr{C}_c^\infty,$$ $$\tau_e \psi - \psi = -\psi'(\cdot) e + o(e).$$

Proof of lemma. By definition, $$\tau_e \psi(t) - \psi(t) = - \int\limits_{t-e}^t \psi' = -\psi'(t) e - \int\limits_{t-e}^t \big( \psi' - \psi'(t) \big) \leq -\psi'(t) + \sup \big| \psi'(s) - \psi'(s - t) \big| e,$$ where $$s$$ lies between $$t$$ and $$t - e.$$ Clearly, $$\sup \big| \psi'(s) - \psi'(s - t) \big| \leq \sup_{|t| \leq |e|} \| \tau_t \psi - \psi\|_\infty$$ and since $$\psi'$$ is uniformly continuous, $$\sup\limits_{|t| \leq |e|} \| \tau_t \psi - \psi\|_\infty = o(1)$$ as $$e \to 0$$ in $$\mathbf{R}$$ and this shows that $$\sup \big| \psi'(s) - \psi'(s - t) \big| e = o(e), \text{ as }e \to 0\text{ in } \mathscr{C}_c^\infty.$$ The proof of the lemma is complete.

We may now apply the lemma, $$v(\psi, x + e) = \tau_x \psi - (\tau_x \psi)'(\cdot) e + o(e).$$ It is very easy to check $$(\tau_x \psi)' = \tau_x \psi'.$$ A fortiori, $$\partial_x v = -\tau_x \psi'.$$

Finally, notice that $$\tau_x \varphi(x)$$ means $$(\tau_x \varphi)(x),$$ so that $$\tau_x \varphi(x) = u(v(\varphi, x), x).$$ By the chain rule, $$\partial_x \tau_x \varphi(x) = \partial_\varphi u\Big|_{(v(\varphi,x), x)} \partial_x v\Big|_{(\varphi, x)} + \partial_x u \Big|_{(v(\varphi,x), x)} = -\tau_x \varphi'(x) + \tau_x \varphi'(x) = 0.$$

Of course this surprised me first, but the way you wrote the function is either wrong or misleading since $$\tau_x \varphi(x) = \varphi(x - x) = \varphi(0)$$ is a constant (its derivative is of course zero). I hope the ideas in along the way help you, though.