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.



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

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.

  • $\begingroup$ Thank you very much. Your answer is clear and great! I think I have a mistake part from my class. $\endgroup$ – jakeoung Mar 16 '14 at 13:49
  • $\begingroup$ I have a question. If I replace $D_iu(y)$ with $D_u(y)$ (gradient of u), is it okay? $\endgroup$ – jakeoung Mar 16 '14 at 13:50
  • 1
    $\begingroup$ 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 $$\lVert Du(y)\rVert \leqslant \frac{1}{w_nR^n} \int_{\partial B_R(y)} \lvert u\rvert\,dS \leqslant \frac{n}{R}\sup_{\partial B_R(y)} \lvert u\rvert.$$ $\endgroup$ – Daniel Fischer Mar 16 '14 at 14:00
  • $\begingroup$ 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. $\endgroup$ – jakeoung Mar 16 '14 at 14:06
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Mar 16 '14 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.