I am studying theorem 6.26 (page 152) in Rudin's "Functional Analysis" that presents distributions as derivatives of continuous functions.
Right at the beginning of the proof, if $\Omega$ is the usual open subset, $Q = [0,1] ^n \subset \Omega$ is the unit cube and $\psi \in \mathcal D _Q (\Omega)$ (the space of test functions on $\Omega$ with support in $Q$), he applies the mean value theorem and states that $| \psi | \le \max \limits _{x \in Q} | ( \partial _i \psi ) (x) |$.
What precise statement of the MVT does he use, and how?
What I can think of is that the differential form of the MVT looks like $| \psi (x) - \psi (0) | \le \| \Bbb d \psi (\xi) \| \cdot \|x - 0\|$. Since $\text{supp} \ \psi \subset Q$ and $0 \in \partial Q$, then $\psi (0) = 0$, so the best I can get is $| \psi (x) | \le \| \Bbb d \psi (\xi) \| \cdot \|x\|$. Why is there no $\| x \|$ in Rudin's right-hand side? Does he work with $\| \cdot \| _\infty$, to make it $1$ on $Q$? But what then if $x \in \Omega \setminus Q$? And, in general, what is he doing here?
If $\psi=\psi(x_1\ldots x_n)$ is supported in the unit cube $Q$, then we may write $$ \begin{split} \psi(x_1\ldots x_n)&=\psi(x_1, x_2\ldots x_n) - \psi(0, x_2\ldots x_n) \\ &=\int_0^{x_1}\partial_{x_1}\psi(\xi_1, x_2\ldots x_n)\, d\xi_1 \end{split} $$ and so obtain the bound $$ \lvert\psi(x_1\ldots x_n)\rvert \le \lvert x_1\rvert \|\partial_{x_1}\psi\|_\infty\le\|\partial_{x_1} \psi\|_\infty,$$ where we used that $\lvert x_1\rvert \le 1$ because $(x_1\ldots x_n)$ lies in the unit cube.