I know that a function $u \in C^2(\Omega)$ is subharmonic on $\Omega$ if $\Delta u(x) \ge 0$ for all $x \in \Omega$. But I have just acquainted with another definition of subharmonic functions which is given as follows $:$

"A function $u \in C^0 (\Omega)$ is subharmonic on $\Omega$ if for any $h \in C^2(\Omega) \cup C^0 (\bar \Omega)$ with $\Delta h = 0$ on $\Omega$ and for every ball $B \subset \subset \Omega$, $u \le h$ on $\partial B \implies u \le h$ on $B$."

To see that previous definition implies the later is immediate. For this we just need to apply Strong Maximum Principle on the function $w:=u-h$ by observing that $w \le 0$ on $\partial B$ and $h$ is superharmonic on $\Omega$ since it is harmonic on $\Omega$. But how can I do the converse part? Please give me some suggestions.

Thank you in advance.


The "converse part", as you say, is false for regularity reasons. The function $u(x)=|x|$ is subharmonic on $\mathbb R$ in the sense of the second definition, but it is not in the sense of the first, because it is not differentiable.

  • $\begingroup$ I think in order to hold the converse we need to restrict $u$ in $C^2 (\Omega)$ in the second definition. $\endgroup$ – Arnab Chattopadhyay. Apr 23 '18 at 8:48
  • 1
    $\begingroup$ In that case I am sure the two definitions are equivalent. There is a third approach to (sub-)harmonic functions, using the mean value property. The equivalence of the various definitions becomes apparent in that context, I think. Try looking here (not a good source, but links to a great book). $\endgroup$ – Giuseppe Negro Apr 23 '18 at 8:58
  • $\begingroup$ @GiuseppeNegro If $u$ is subharmonic then $u(x) \leq \fint u$. But the converse is obviously not true. $\endgroup$ – user586752 Aug 29 '18 at 9:45
  • $\begingroup$ @OhDaeSu: Does \fint u mean $\frac{1}{|B|}\int_B u$? In that case, it is true that $u(x)\le \frac{1}{|B|}\int_B u$ for all balls centered at $x$ if and only if $u$ is subharmonic in some sense. I am not fresh in these things but they are in the linked book, I'm sure $\endgroup$ – Giuseppe Negro Aug 29 '18 at 9:57
  • 1
    $\begingroup$ @OhDaeSu: Agreed, but the inequality must be satisfied for all points and for all balls centered at the point. In that example, this fails at $|x|\ge 5$. $\endgroup$ – Giuseppe Negro Aug 29 '18 at 10:18

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.