In the proof Theorem 5.2.1 (equality (5.2.5)), Hörmander claims that for any smooth function $\psi \in C^{\infty} (\mathbb{R}^n)$ the following equality is true:
$ \frac{\partial}{\partial \epsilon}(\epsilon^{-n} \psi (\frac{x}{\epsilon})) = \sum_{j=1}^n \frac{\partial}{\partial x_j}(\epsilon^{-n}\psi_j(\frac{x}{\epsilon})) $
where $\psi_j(x) = -x_j\psi(x)$.
But with product and chain rule I get that for the left side
$\frac{\partial}{\partial \epsilon}(\epsilon^{-n} \psi (\frac{x}{\epsilon})) = -n\epsilon^{-n-1}\psi(\frac{x}{\epsilon})-\epsilon^{-2}\sum_{j=1}^n \frac{\partial}{\partial x_j}(\epsilon^{-n}\psi(\frac{x}{\epsilon}))$
and for the right side
$\sum_{j=1}^n \frac{\partial}{\partial x_j}(\epsilon^{-n}\psi_j(\frac{x}{\epsilon})) = -n\epsilon^{-n-1}\psi(\frac{x}{\epsilon})-\epsilon^{-1}\sum_{j=1}^n \frac{\partial}{\partial x_j}(\epsilon^{-n}\psi(\frac{x}{\epsilon})) $
so in the 2nd equality the factor $1/\epsilon$ is missing. I have gone through these calculations multiple times and always came to the same results. Is it right that
$\frac{\partial}{\partial \epsilon} \psi (\frac{x}{\epsilon}) = -\frac{x_1}{\epsilon^2}\frac{\partial}{\partial x_1}\psi(\frac{x}{\epsilon})-...-\frac{x_n}{\epsilon^2}\frac{\partial}{\partial x_n}\psi(\frac{x}{\epsilon})$ ?
And I am not stupid saying $\psi_j(\frac{x}{\epsilon}) = -\frac{x_j}{\epsilon}\psi(\frac{x}{\epsilon})$, am I?
Thanks for you help..
is not correct, it should be
$\frac{\partial}{\partial \varepsilon}(\varepsilon^{-n} \psi (\frac{x}{\varepsilon})) = -n\varepsilon^{-n-1}\psi(\frac{x}{\varepsilon})-\varepsilon^{-1}\sum_{j=1}^n \frac{\partial}{\partial x_j}(\varepsilon^{-n}\psi(\frac{x}{\varepsilon}))$, since $\frac{\partial}{\partial x_j}(\varepsilon^{-n}\psi(\frac{x}{\varepsilon})) = \varepsilon^{-n}\cdot \frac{1}{\varepsilon}\frac{\partial \psi}{\partial x_j}(x/\varepsilon)$ and not $\frac{\partial}{\partial x_j}(\varepsilon^{-n}\psi(\frac{x}{\varepsilon})) = \varepsilon^{-n}\frac{\partial \psi}{\partial x_j}(x/\varepsilon)$.
You can caluclate like this:
First, it is $$\varepsilon \frac{\partial}{\partial \varepsilon}(\varepsilon^{-n} \psi(x/\varepsilon)) + \sum_{j = 1}^n x_j \frac{\partial}{\partial x_j}(\varepsilon^{-n} \psi(x/\varepsilon)) = -n\varepsilon^{-n}\psi(x/\varepsilon).$$
Now $$-x_j \frac{\partial}{\partial x_j}(\varepsilon^{-n} \psi(x/\varepsilon)) = \frac{\partial}{\partial x_j}(\varepsilon^{-n} (-x_j)\psi(x/\varepsilon)) +\varepsilon^{-n}\psi(x/\varepsilon) = \varepsilon\frac{\partial}{\partial x_j}(\varepsilon^{-n} \psi_j(x/\varepsilon)) +\varepsilon^{-n}\psi(x/\varepsilon).$$
Therefore $$\varepsilon \frac{\partial}{\partial \varepsilon}(\varepsilon^{-n} \psi(x/\varepsilon)) = -\sum_{j = 1}^n x_j \frac{\partial}{\partial x_j}(\varepsilon^{-n} \psi(x/\varepsilon))-n\varepsilon^{-n}\psi(x/\varepsilon) \\ = \sum_{j = 1}^n (-x_j \frac{\partial}{\partial x_j}(\varepsilon^{-n} \psi(x/\varepsilon))-\varepsilon^{-n}\psi(x/\varepsilon)) = \varepsilon\sum_{j=1}^n \frac{\partial}{\partial x_j}(\varepsilon^{-n} \psi_j(x/\varepsilon)).$$
Devide by $\varepsilon$ and get $\frac{\partial}{\partial \varepsilon}(\varepsilon^{-n} \psi(x/\varepsilon)) = \sum_{j=1}^n \frac{\partial}{\partial x_j}(\varepsilon^{-n} \psi_j(x/\varepsilon))$.