# Estimates of derivatives of harmonic function

I have a PDE class as a undergraduate student. The professor said that we want to approximate functions by using derivatives. (title: Estimates of derivatives of harmonic function)

The proof is as follows in my memory. So there can be some errors.

$$\Delta u=0 \implies \Delta D_i u = 0$$ Let $B_R(y)\subset\subset\Omega$. Apply MVT to $D_iu$ : $$D_iu(y) = \frac 1 {w_nR^n}\int_{B_R(y)} D_i u \, dx$$ $D_iu = e_i \cdot Du = \operatorname{div}(e_i u)$

$(\because \operatorname{div}(\mathbf vf)=\mathrm{div} (\mathbf v) f + \mathbf v\cdot \nabla f)$

By the Divergence Theorem, $$D_iu(y) = \frac 1 {w_nR^n}\int_{B_R(y)} \mathrm{div}(e_iu) dx = \frac 1 {w_nR^n}\int_{\partial B_R(y)} e_iu \cdot \nu ds$$ So $$|D_iu(y)| \leq \frac 1 {w_nR^n} \sup_{\partial B_R(y)}|u| \cdot \underbrace{ nw_nR^{n-1}}_{\text{surface area}} = \frac n R \sup_{\partial B_R(y)} |u|$$

At this time, I have a question. I don't know how the following inequality is derived, particularly in the second, even if $R < d(y,\partial\Omega)$.

$$|D_u(y)| \leq \frac nR \sup_{\partial\Omega}|u| \leq \frac n {d(y,\partial\Omega)} \sup_\Omega |u|$$

A similar theorem appears on page 23 in Elliptic Partial Differential Eqns of Second Order 2001 by Gilbarg.

This

$$|D_iu(y)| \leq \frac nR \sup_{\partial\Omega}|u| \leq \frac n {d(y,\partial\Omega)} \sup_\Omega |u|\tag{\ast}$$

is wrong for $R < d(y,\partial\Omega)$ and harmonic $u \not\equiv 0$. The first of the two inequalities is okay if $u$ has decent boundary values on $\partial\Omega$ or we replace $\partial\Omega$ with $\partial B_R(y)$ or $\Omega$ there, but not the second. Since $u$ is harmonic, if it has nice (for example continuous) boundary values, we have

$$\sup_{\Omega} \lvert u\rvert = \sup_{\partial\Omega} \lvert u\rvert,$$

and since $R < d(y,\partial\Omega)$, the second inequality can only hold if the factor $\sup\limits_\Omega \lvert u\rvert$ is $0$. If we don't have good boundary values, the $\sup\limits_{\partial\Omega}$ would anyway have to be replaced with $\sup\limits_\Omega$ or $\sup\limits_{\partial B_R(y)}$. In either case, there are harmonic $u$ (constants, for example) for which the second inequality in $(\ast)$ doesn't hold.

However, the first inequality of $(\ast)$ [with the supremum taken over $\Omega$ to avoid demanding boundary regularity] holds for all $R < d(y,\partial\Omega)$, and hence

$$\lvert D_i u(y)\rvert \leqslant \inf_{R < d(y,\partial\Omega)} \frac{n}{R}\sup_{\Omega} \lvert u\rvert = \frac{n}{d(y,\partial\Omega)} \sup_{\Omega} \lvert u\rvert$$

gives us the estimate we are interested in.

• Thank you very much. Your answer is clear and great! I think I have a mistake part from my class. – jakeoung Mar 16 '14 at 13:49
• I have a question. If I replace $D_iu(y)$ with $D_u(y)$ (gradient of u), is it okay? – jakeoung Mar 16 '14 at 13:50
• Yes. Ignoring the normalisation factors to save space, we have $D_i u(y) = \int (e_i u)\cdot\nu\,dS = \int u\cdot\nu_i\,dS$. So $Du(y) = (D_1u(y),\dotsc,D_nu(y)) = \int u(\nu_1,\dotsc,\nu_n)\,dS$, and hence $\lVert Du(y)\rVert \leqslant \int \lvert u\rvert\cdot\lVert\nu\rVert\,dS$, but $\lVert\nu\rVert = 1$, so $$\lVert Du(y)\rVert \leqslant \frac{1}{w_nR^n} \int_{\partial B_R(y)} \lvert u\rvert\,dS \leqslant \frac{n}{R}\sup_{\partial B_R(y)} \lvert u\rvert.$$ – Daniel Fischer Mar 16 '14 at 14:00
• Thank you for your clear answer. There is a minor question. When is 'leqslant' symbol used? Actually I've saw it for the first time. – jakeoung Mar 16 '14 at 14:06
• Whether you use $\leq$ or $\leqslant$ is purely a matter of aesthetic preference. I find $\leqslant$ looks better, so I use that. It's the same thing. – Daniel Fischer Mar 16 '14 at 14:11