I'm having difficulty proving Leibniz' formula for weak derivatives as presented in Evans PDE. I was hoping that someone could maybe shed some light on it for me? To be honest I'm not entirely comfortable with weak derivative concept. I'm actually an economist by trade hoping to improve my maths. My question is very closely related to the following one posted on this website: Evans's proof of the Leibniz's formula for the weak derivatives in Sobolev spaces however my undrstanding isn't as great as the original poster. I hope this exercise will help others struggling with weak derivatives.
Disclaimer. I have copied the latex code from @Cookie's question as it's exactly the same as Evans' theorem in the book in a bid to save time. Since I'm relatively new to this site I don't know if this is acceptable or not. If not please tell me and I'll rewrite the code.
The theorem and proof is as follows:
- $W^{k,p}(U)$ 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}
Just to make sure I am understanding this correctly, the first part of the proof follows because
\begin{align}
\int_U u D^\alpha (\zeta \phi) – u(D^\alpha \zeta) \phi \, dx &=\int_U u \zeta D^\alpha \phi\, dx + \int_U u \phi D^\alpha \zeta\,dx – \int_U u(D^\alpha \zeta) \phi \, dx \\ &=\int_U \zeta u D^\alpha \phi \, dx
\end{align}
and we have since $\zeta \phi \in C^\infty_c(U)$
\begin{align*}
\int_U \zeta u D^\alpha \phi \, dx=\int_U u D^\alpha(\zeta \phi)\, dx – \int_U u(D^\alpha \zeta)\phi \,dx&=-\int_U D^\alpha u (\zeta \phi)\,dx-\int_U u(D^\alpha \zeta)\phi \,dx \\ &= -\int(\zeta D^\alpha u + u D^\alpha \zeta)\phi\,dx
\end{align*}
I also understand how to get (A). My main problem is that I don't see how to get from (A) to (B), (B) to (C) and (C) to (D). Many thanks.
Best Answer
(A) to (B): Usual integration by parts formula ($|\gamma|=1$).
(B) to (C): $|\alpha|=|\beta|+|\gamma|,$ and $\rho=\sigma+\gamma,$ so $\beta-\sigma=(\beta+\gamma)-(\sigma+\gamma)=\alpha-\rho$.
(C) to (D): Let us call the terms in (C) with $\rho$ "sum 1" and the terms without "sum 2." Observe that $\binom{\alpha}{\sigma}=\binom{\alpha_{1}}{\sigma_{1}}\binom{\alpha_{2}}{\sigma_{2}}\cdots\binom{\alpha_{n}}{\sigma_{n}},$ so if $\gamma=e_{j},$ we can see that (when $\sigma\geq \gamma$): $$\binom{\beta}{\sigma}+\binom{\beta}{\sigma-\gamma}=\left(\binom{\beta_{j}}{\sigma_{j}}+\binom{\beta_{j}}{\sigma_{j}-1}\right)\prod_{i=1,i\neq j}^{n}\binom{\beta_{i}}{\sigma_{i}}=\binom{\beta_{j}+1}{\sigma_{j}}\prod_{i=1,i\neq j}^{n}\binom{\beta_{i}}{\sigma_{i}}=\binom{\alpha}{\sigma}.$$ Similarly, when $\sigma+\gamma\leq\beta,$ $\binom{\beta}{\sigma}+\binom{\beta}{\sigma+\gamma}=\binom{\alpha}{\sigma+\gamma}$ (just let $\sigma'=\sigma+\gamma$ and use the above).
We have three cases to consider. Either $\rho\leq\beta,$ so there is a match between a term in sum 1 and a term in sum 2; $\rho\not\leq\beta,$ so there is a term in sum 1 with no match; or $\sigma\leq\beta$ is such that there is no corresponding $\rho=\sigma,$ so there is a term in sum 2 with no match. For case 1, if $\rho\leq \beta,$ applying the equality above for $\binom{\beta}{\sigma}D^{\rho}\zeta D^{\alpha-\rho}u+\binom{\beta}{\sigma+\gamma}D^{\sigma+\gamma}\zeta D^{\alpha-\sigma-\gamma}u$ (where the first term comes from sum 1, and the second term comes from sum 2), we get $\binom{\alpha}{\sigma+\gamma}D^{\sigma+\gamma}\zeta D^{\alpha-\sigma-\gamma}u$ (note that the derivatives are exactly the same, so we're just using the formula for these binomial coefficients). For case 2, if $\rho\not\leq\beta,$ then $\sigma_{j}=\beta_{j},$ so $\binom{\beta_{j}}{\sigma_{j}}=1=\binom{\alpha_{j}}{\sigma_{j}+1},$ which gives us $\binom{\beta}{\sigma}=\binom{\alpha}{\sigma+\gamma},$ since none of the other factors are affected by adding $\gamma.$ This leaves case 3, where $\sigma\leq\beta$ is not of the form $\sigma'+\gamma$ for some $\sigma'\leq\beta.$ But this means that $\sigma_{j}=0,$ so again $\binom{\beta_{j}}{\sigma_{j}}=1=\binom{\alpha_{j}}{\sigma_{j}},$ which implies that $\binom{\beta}{\sigma}=\binom{\alpha}{\sigma}$ for such $\sigma.$ It should be clear that each $\sigma\leq \alpha$ appears as one of the three cases above, so this completes the proof.