# Using a cutoff to show an interior gradient estimate for a harmonic function

I am trying to follow a calculation shown in Chapter 1.4 of Han and Lin's Elliptic PDE textbook.

Let $$u$$ be a harmonic function on the unit ball $$B_1$$ in $$\mathbb{R}^n$$, and let $$\eta \in C^\infty_c (B_1)$$ with $$\eta = 1$$ on $$B_{\frac{1}{2}}$$.

First, one can show that $$\Delta(|Du|^2) = 2|D^2 u|^2 + 2\langle Du, D(\Delta u) \rangle = 2|D^2 u|^2 \geq 0$$. In other words, $$|Du|^2$$ is subharmonic.

Next, observe that

\begin{align*} \Delta(\eta^2|Du|^2) &= \Delta(\eta^2)|Du|^2 + 2 \langle D\eta^2, D(|Du|^2) \rangle + \eta^2 \Delta(|Du|^2) \\\\ &= 2(|D\eta|^2 + \eta \Delta \eta)|Du|^2 + 8 \langle \eta D\eta, D^2u \cdot Du \rangle + 2\eta^2|D^2u|^2. \end{align*}

In the text, the authors claim we can then "use Hölder's inequality" to bound the above quantity from below by $$(2\eta \Delta \eta - 6|D\eta|^2)|Du|^2 \geq -C|Du|^2)$$, where $$C>0$$ depends only on $$\eta$$. I've tried to apply various inequalities (I think the authors were referring to the Cauchy-Schwarz inequality $$|D\eta|^2|Du|^2 \geq |\langle D\eta, Du \rangle|^2$$) but I still haven't gotten the desired bound. I think the method involves completing a square I'm not seeing.

• When you write $|\mathrm D^2 u|$ which matrix norm are you using? May 15 at 15:50
• Frobenius norm. May 15 at 16:26
• I found a copy of the original text and I think you are missing a prefactor of $\eta^2$ in front of the last term. May 15 at 19:25
• Also, it should be noted that the Cauchy-Schwartz inequality is a special case of Holder's inequality, so they are technically not incorrect. May 15 at 19:30
• I just fixed that. May 16 at 18:25