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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 $h(x)=L\tau_x\phi(x)$.

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  • $\begingroup$ What is $h$? ${}$ $\endgroup$
    – William M.
    Aug 14, 2021 at 18:48
  • $\begingroup$ @WillM. the definition of $h$ is in the title. $\endgroup$
    – user614535
    Aug 15, 2021 at 1:11
  • $\begingroup$ 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.) $\endgroup$
    – William M.
    Aug 15, 2021 at 14:57
  • $\begingroup$ @WillM. It should be OK now. $\endgroup$
    – user614535
    Aug 15, 2021 at 15:47

1 Answer 1

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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.

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