# Evans's proof of the Leibniz's formula for the weak derivatives in Sobolev spaces

In PDE Evans 2nd edition, pages 261-263, there is a theorem and its proof which concerns the four properties of weak derivatives. Unfortunately, I do not understand the fourth property, which I will type here accordingly. Note that, according to Evans' notation in his textbook,

• $W^{k,p}(U)$ denotes a Sobolev space that consists of all locally summable functions $u : U \rightarrow \mathbb{R}$ such that for each multiindex $\alpha$ with $|\alpha| \le k$, $D^\alpha u$ exists in the weak sense and belongs to $L^p(U)$.
• $C_c^\infty(U)$ denotes the space of infinitely differentiable functions $\phi : U \rightarrow \mathbb{R}$, with compact support in $U$, and the function $\phi$ is called a test function.

Theorem 1 (Properties of weak derivatives). Assume $u,v, \in W^{k,p}(U), |\alpha| \le k$. Then

$\quad$ (iv) If $\zeta \in C_c^\infty(U)$, then $\zeta u \in W^{k,p}(U)$ and $$D^\alpha (\zeta u)=\sum_{\beta \le \alpha} {\alpha \choose \beta} D^\beta \zeta D^{\alpha - \beta} u \qquad \textit{(Leibniz' formula)} \tag{7}$$ $\quad$ where ${\alpha \choose \beta} = \frac{\alpha!}{\beta!(\alpha-\beta)!}$.

Again, I didn't list properties $\text{(i)-(iii)}$ because I understood them already.

Proof (of property $\text{(iv)}$). We prove $\text{(7)}$ by induction on $|\alpha|$. Suppose first $|\alpha|=1$. Choose any $\phi \in C_c^\infty (U)$. Then \begin{align} \int_U \zeta u D^\alpha \phi \, dx &= \int_U u D^\alpha (\zeta \phi) - u(D^\alpha \zeta) \phi \, dx \\ &= - \int_U (\zeta D^\alpha u + u D^\alpha \zeta) \phi \, dx \end{align} Thus $D^\alpha (\zeta u)=\zeta D^\alpha u + u D^\alpha \zeta$, as required.

$\quad$ Next assume $l < k$ and formula $\text{(7)}$ is valid for all $|\alpha| \le l$ and all functions $\zeta$. Choose a multiindex $\alpha$ with $|\alpha| = l+1$. Then $\alpha = \beta + \gamma$ for some $|\beta|=l, |\gamma| = 1$. Then for $\phi$ as above,

\begin{align} \int_U \zeta u D^\alpha \phi \, dx &= \int_U \zeta u D^\beta (D^\gamma \phi) \, dx \\ &= (-1)^{|\beta|} \int_U \sum_{\sigma \le \beta} {\beta \choose \sigma} D^\sigma \zeta D^{\beta - \sigma} u D^\gamma \phi \, dx \tag{A} \\ &= (-1)^{|\beta|+|\gamma|} \int_U \sum_{\sigma \le \beta} {\beta \choose \sigma} D^\gamma(D^\sigma \zeta D^{\beta - \sigma} u ) \phi \, dx \tag{B} \\ &= (-1)^{|\alpha|} \int_U \sum_{\sigma\le \beta} {\beta \choose \sigma} [D^\rho \zeta D^{\alpha - \rho} u + D^\sigma \zeta D^{\alpha - \sigma} u] \phi \, dx \tag{C} \\ &= (-1)^{|\alpha|} \int_u \left[\sum_{\sigma \le \alpha} {\alpha \choose \sigma} D^\sigma \zeta D^{\alpha - \sigma} u \right] \phi \, dx. \tag{D} \end{align}

• $\text{(A)}$: by the induction assumption
• $\text{(B)}$: by the induction assumption again
• $\text{(C)}$: where $\rho = \sigma + \gamma$
• $\text{(D)}$: since $\displaystyle {\beta \choose \sigma-\gamma} + {\beta \choose \sigma} = { \alpha \choose \sigma}$

I especially would like to know how the last step of the work was derived, that is, the line with the tag $\text{(D)}$. I have trouble filling in the details there.

• It's easy to get lost in all those indices. It might be easier to just prove it yourself, since nothing fancy is going on here, it's just the base case iterated several times. It would be very helpful to compare to a proof of the Leibniz formula for 2 smooth functions - there the induction argument is identical. Jul 13, 2014 at 20:55
• "It might be easier to just prove it yourself," -- I always do that first before I ask here. I only ask if I am truly stuck proving this myself. Jul 13, 2014 at 22:19

Step (B) isn't really by the induction hypothesis, it's the definition of weak $D^\gamma$ (a.k.a., "formal integration by parts").

Step (C) is by the induction hypothesis, distributing $D^\gamma$ according to the Leibniz rule.

To understand (D), split the sum in (C) in two, express the first one in terms of $\rho$, and then rename the index $\rho$ as $\sigma$: $$\sum_{\sigma\le \beta} {\beta \choose \sigma} D^\rho \zeta D^{\alpha - \rho} u + \sum_{\sigma\le \beta} {\beta \choose \sigma}D^\sigma \zeta D^{\alpha - \sigma} u \\ = \sum_{\gamma \le \rho\le \alpha} {\beta \choose \rho-\gamma} D^\rho \zeta D^{\alpha - \rho} u + \sum_{\sigma\le \beta} {\beta \choose \sigma}D^\sigma \zeta D^{\alpha - \sigma} u \\ = \sum_{\gamma\le \sigma\le \alpha} {\beta \choose \sigma-\gamma} D^\sigma \zeta D^{\alpha - \sigma} u + \sum_{\sigma\le \beta} {\beta \choose \sigma}D^\sigma \zeta D^{\alpha - \sigma} u$$ It remains to use the aforementioned identity $\displaystyle {\beta \choose \sigma-\gamma} + {\beta \choose \sigma} = { \alpha \choose \sigma}$, which can be proved by recalling that multinomial coefficient ${ \alpha \choose \sigma}$ is the coefficient of $x^\sigma$ in $$(1+x)^\alpha = (1+x)^\beta(1+x)^\gamma = (1+x)^\beta + x^\gamma (1+x)^\beta$$

Here I am (ab)using notation, how it's customary with multiindices: e.g., $$(1+x)^\alpha = \prod_i (1+x_i)^{\alpha_i}$$ Since $|\gamma|=1$, the factor $(1+x)^\gamma$ is linear: it's simply $1+x_i$ where $i$ is whatever coordinate has $\gamma_i=1$.

• @glace See the added remark at the end.
– user147263
Jul 14, 2014 at 19:27
• "Step (B) isn't really by the induction hypothesis" -- then I'm confused as to why the textbook stated that it is. Admittedly, I still do not know where the $\sum_{\sigma \le \beta} {\beta \choose \sigma}$ came from, but perhaps that is due to this "formal integration by parts" you mentioned above? Jul 15, 2014 at 20:25
• @glace In the book, the text (by the induction assumption) appears between two lines (not next to a line as in your formatting). I think it sometimes refers to preceding line, and sometimes to the next. In any event, the book is not a holy script: it was written by a human.... The sum over $\sigma$ appears at step (A) by the induction hypothesis.
– user147263
Jul 15, 2014 at 21:55