Let $\Omega\subset \mathbb{R}^N$ open and $u\in C^2(\Omega)$. Show that the proprierties below are equivalent:

a) $\Delta u(x) \ge 0$ for all $x\in \Omega$

b) For all $x_0\in\Omega$ and all $r>0$ such that $\overline{B_r(x_0)}\subset \Omega$ it's true that $$u(x_0) \le \frac{1}{\omega_N}\int_{S_1(0)}u(x_0+ry)d\sigma(y)$$


Suggestion: for $1\to 2$, study the derivative of $$g(t) = \frac{1}{\omega_N}\int_{S_1(0)}u(x_0+ty)d\sigma(y), 0\le t\le r$$

For $2\to 1$, write the taylor expansion of rder $2$ of $u$ around $x_0$ and use:

Let $\Omega$ be a bounded open of $\mathbb{R}^N$ and consider a sequence >$\{u_j\}$ of harmonic functions in $\Omega$, each one of them continuous >in $\overline{\Omega}$. Suppose that $$\max_{y\in \partial\Omega}|u_j(y)-u_k(y)|\le \frac{1}{j}+\frac{1}{k}, \ k,k=1,2,\cdots$$ Then $\{u_j\}$ converges uniformly in $\overline{\Omega}$ for a function $u\in C(\overline{\Omega})$, which is still harmonic in $\Omega$

So to solve $a)\implies b)$ I must study

$$g'(t) = \frac{1}{\omega_N}\int_{S_1(0)}\sum_i\partial u_i(x_0+ty)y_i d\sigma(y)$$

can I do that? Take the derivative inside the integral. Well, I can see that the sum of the first derivatives appear inside the integral. Maybe if I take the derivative again I arrive at the laplacian inside the integral but there's nowhere to go from there.

To solve $b)\implies a)$ I did:

$$u(x_0+ry) = u(x_0) + r\nabla u(x_0)^Ty + \frac{1}{2}r^2y^t\nabla^2 u(x_0+try)y$$

for some $t\in (0,1)$

I think I must consider two versions of this expansion and thnk about the max of the difference of their absolute sum. I know that taylor expansions have bounds for the error terms, but I don't think the difference of two expansions would have these $\frac{1}{j}+\frac{1}{k}$ bounds.



Let's try the hint gave below:

$$g'(t)=\frac{1}{\omega_N}\int_{\partial \mathbb{S}_1(0)}\nabla u(x_0+ty)\cdot yd\sigma(y) = \frac{1}{\omega_N}\int_{B_1(0)}\Delta u(x_0+ty)\ d\sigma(y)\ge 0$$

So $g'$ is crescent. $g(0) =\frac{1}{\omega_N}\int_{S_1(0)}u(x_0)\ d\sigma (y) = u(x_0)$. So for $r>0$ we have $g'(0)\le g'(r)$ which implies the result.


Now, for the converse, by hypothesis and by the taylor expansion we have:

$$u(x_0) \le \frac{1}{\omega_N}\int_{S_1(0)} u(x_0+ry) = \\ \frac{1}{\omega_N}\int_{S_1(0)} \left(u(x_0) + r\nabla u(x_0)^Ty + \frac{1}{2}r^2y^t\nabla^2 u(x_0+try)y\right)\ d\sigma(y)$$

for some $t\in (0,1)$

I must somehow find $u_j$ and $u_k$ both harmonic in order to use that result and find another harmonic function

  • $\begingroup$ If you show that the $g'$ is positive (using $\Delta u(x) \ge 0$), you have that $g$ is increasing. Then just compute $g(0)$ $\endgroup$
    – Uskebasi
    Sep 14, 2018 at 20:43
  • $\begingroup$ @Uskebasi I can see how to do it for $g''$, I guess. Because $\Delta u$ appears inside the integrand. However, for $g'$ I don't see how $\endgroup$
    – Paprika
    Sep 14, 2018 at 21:19
  • $\begingroup$ A small hint: try with divergence theorem. Maybe it is clearer if you write $$g'(t)=\frac{1}{\omega_N}\int_{\partial \mathbb{S}_1(0)}\nabla u(x_0+ty)\cdot yd\sigma(y)$$ $\endgroup$
    – Uskebasi
    Sep 14, 2018 at 22:55
  • $\begingroup$ This is entirely standard, see for instance "Elliptic partial differential equations" (lecture notes) by F. Lin and Q. Han , the part (in the beginning of the book) where they discuss the mean value properties of harmonic functions. Although you have subharmonic here, the approach from the book works verbatim in this case as well. You do not need to use Taylor expansion, it's just divergence theorem and change of variables for (1) => (2), and convolution with smooth test function for (2) => (1). $\endgroup$
    – Hayk
    Sep 19, 2018 at 19:23

1 Answer 1


Since it looks like you've worked out the $(a) \implies (b)$ part, I'll focus on $(b) \implies (a)$. Suppose $u \in C^2(\Omega)$ satisfies condition (b). Then, by Taylor's theorem, $$u(x_0+ry) = u(x_0) + r\nabla u(x_0)^Ty + \frac{1}{2}r^2 y^T \nabla^2u(x_0)y + o(r^2).$$ Now, integrating gives $$u(x_0) \le \frac{1}{\Omega_n} \int_{S^1} u(x_0) + r\nabla u(x_0)^Ty + \frac{1}{2}r^2 y^T \nabla^2u(x_0)y + o(r^2)\;d\sigma(y)$$ I claim that $$\frac{1}{\Omega_n} \int_{S^1} u(x_0) + r\nabla u(x_0)^Ty + \frac{1}{2}r^2 y^T \nabla^2u(x_0)y + o(r^2)\;d\sigma(y) = u(x_0) + C\frac{N}{2}r^2 \Delta u(x_0) + o(r^2)$$ to see why, notice that by symmetry, $$\int_{S^1} \nabla u(x_0)^Ty d\sigma(y) = \int_{S^1} \nabla u(x_0)^T(-y) d\sigma(y)$$ since the sphere is symmetric under inversions. Similarly, by considering reflections of the coordinate axes, we can see that $$\int_{S^1} \partial_{x_ix_j} u(x_0) y_iy_j d\sigma(y) = 0 \text{ for } i \neq j$$ so $$\int_{S^1} y^T \nabla^2u(x_0) y \;d\sigma(y) = \Delta u(x_0) \int_{S^1} y_i^2\;d\sigma(y) = C\Delta u(x_0)$$

Thus, $$u(x_0) \leq u(x_0) + Cr^2\Delta u(x_0) + o(r^2).$$

Now, suppose for contradiction that $\Delta u(x_0) < 0$. Then, for $r$ sufficiently small, the $o(r^2)$ term is less than $-Cr^2 \Delta u(x_0)$, contradicting condition (b). It follows that $\Delta u(x_0) \ge 0$ for every $x_0$, which is precisely condition (a).

  • $\begingroup$ How do you know $o(r^2)$ is positive? What is $o(r^2)$ exactly? Is it an error? $\endgroup$
    – Paprika
    Sep 25, 2018 at 2:39
  • $\begingroup$ I mean, why the integral of $o(r^2)$ is still $o(r^2)$? $\endgroup$
    – Paprika
    Sep 25, 2018 at 2:39
  • $\begingroup$ is $C$ positive? $\endgroup$
    – Paprika
    Sep 25, 2018 at 2:41

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