Let $g_j(x,t) := g(t/\epsilon_j)t^jf_i(x)/j!$ be given and $g_j$ is infinitely differentiable with compact support in $K\times I$, where $K$ is a compact subset of $\mathbb{R}^n$ and $I$ a compact subset of $\mathbb{R}$.

$g$ is the "solution" of $g'(t) = \phi(-t) - \phi(t)$, where $\phi$ is also infinitely differentiable with compact support in $(0,\epsilon)\subset \mathbb{R}$, and also $\int\phi(t)dt=1$. The book isn't so clear about this, so I'm guessing the author means by "solution" that $g(t) = \int_{-\infty}^t \phi(-x)-\phi(x)dx$, since $g$ is supposed to satisfy the condition $\text{d}^{j}(g(t)-1)/\text{d}t^j = 0$ for $t=0$ and $j=0,1,...$

According to the author it should be easy to see that if one takes $t/\epsilon_j$ as a new variable one gets $|\partial^\alpha g_j(x,t)|\leq C_{\alpha,j}\epsilon_j^{j-\alpha_t}$ where $\alpha_t$ is the order of differentiation with respect to $t$. $\alpha\leq j-1$ is also given.

$C_{\alpha,j}$ is just a constant that comes from estimating the functions with compact support. I don't understand the $\epsilon_j^{j-\alpha_t}$ part which should be due to the substitution when taking $t/\epsilon_j$ as a new variable.

Does anyone see this?

P.S. The book is Lars Hörmander's "Analysis of partial differential operators I" (Theorem 1.2.6)


If $\frac{dg}{dt}\leq C$ on a domain, then by changing the name of the variable $u=\frac{t}{\epsilon}$, you have $t=\epsilon u$ and $\frac{dt}{du}=\epsilon$, so

$$ \frac{d}{du}g(u)=\frac{dt}{du}\frac{d}{dt}g(u)=\epsilon g'\left(\frac{t}{\epsilon}\right)\leq C. $$

Thus $g'(t/\epsilon)\leq C\epsilon^{-1}$, by the chain rule. Take higher derivatives, get more factors of $\epsilon$ by repeated applications of the chain rule.

  • $\begingroup$ Yeah, that's what I'm thinking, too. But why is he talking about taking $t/\epsilon_j$ as a new variable when all he's doing is just differentiating with respect to t? $\endgroup$ – Miriam Nov 7 '17 at 20:20
  • $\begingroup$ People often think of the chain rule in terms of a change of variables. $\endgroup$ – ziggurism Nov 7 '17 at 20:22
  • $\begingroup$ Ah, I see. Thank you. $\endgroup$ – Miriam Nov 7 '17 at 20:23
  • $\begingroup$ it's also called $u$-substitution, especially when doing antiderivatives $\endgroup$ – ziggurism Nov 7 '17 at 20:24
  • $\begingroup$ Let me rewrite to use Leibniz notation, I think it's clearer $\endgroup$ – ziggurism Nov 7 '17 at 20:29

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