# Use of Holder inequality in gradient estimate for harmonic function.

While reading the book "Elliptic Partial Differential Equations" by Han and Lin, I failed to understand the proof of the interior gradient estimate for harmonic functions. The theorem says that if $$u$$ is harmonic in $$B_1 (\subset \mathbb{R}^n)$$, then $$\sup_{B_{1/2}}{|Du|} \le c\sup_{\partial B_1}{|u|}$$ for some positive constant $$c=c(n).$$ The proof illustrated in the book begins with choosing a cut-off function $$\phi = \eta^2$$ for some $$\eta \in C_{0}^{1}(B_1)$$ with $$\eta = 1$$ in $$B_{1/2}.$$ Directly calculating, one can show $$\Delta(\eta^2|Du|^2) = 2\eta\Delta\eta|Du|^2 + 2|D\eta|^2|Du|^2 + 8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u} + 2\eta^2\sum_{i,j=1}^{n}(D_{ij}u)^2.$$ What I do not know is the following inequality: $$\Delta(\eta^2|Du|^2) \ge (2\eta\Delta\eta-6|D\eta|^2)|Du|^2.$$ The book explains that "Holder inequality" is used here, but the only thing that I know regarding to Holder inequality is what appears in the usual real analysis book in the chapter of $$L^p$$ spaces (inequality for integral). I guess by the form of the equation, we need to say something about the third term: $$8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u}.$$ To be specific, obtaining the inequality $$8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u} \ge -8|D\eta|^2|Du|^2$$ will end the proof. Can anyone help me to see how Holder inequality could be used to prove the claimed inequality? Thanks in advance.

$$8 \eta \sum_{ij} D_i \eta D_j u D_{ij} u + 2\eta^2 \sum_{ij} (D_{ij}u)^2 \ge - 8 |D\eta |^2 |Du|^2$$
Note that in each term, $$\eta$$ and $$D_{ij}u$$ appear in equal powers. So define $$X_{ij} = \eta D_{ij} u ,\quad Y_{ij} = D_i \eta D_j u$$ Then the above inequality is equivalent to $$\sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2 \ge 0$$ but of course this LHS is nothing but $$\sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = \sum_{ij} (2Y_{ij}+X_{ij})^2$$ which is clearly non-negative, so the original inequality that we wanted to prove is also true.
For vectors, the inequality $$|a+b|^2\ge 0$$ can be used to prove Cauchy-Schwarz ($$\sum_i a_ib_i \le |a||b|$$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.