### Author Topic: Solution of Question 3  (Read 2921 times)

#### Chaojie Li

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• Karma: 0 ##### Solution of Question 3
« on: March 19, 2015, 09:08:28 PM »
1.  Using the proof of maximum principle prove the maximum principle for
subharmonic functions and minimum principle for superharmonic
functions.

2.  Show that minimum principle for subharmonic functions and maximum
principle for superharmonic functions do not hold (*Hint*: construct
counterexamples with $f=f(r)$).

3.  Prove that if $u,v,w$ are respectively harmonic, subharmonic and
superharmonic functions in the bounded domain $\Omega$,
coinciding on its boundary ($u|_\Sigma=v|_\Sigma=w|_\Sigma$)
then in $w\ge u \ge v$ in $\Omega$.

SO i just post what i write on paper
« Last Edit: March 20, 2015, 08:08:11 AM by Victor Ivrii »

#### Chaojie Li

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• Karma: 0 ##### Re: Solution of Question 3
« Reply #1 on: March 19, 2015, 09:10:17 PM »
This is my solution of Q3 b

#### Victor Ivrii ##### Re: Solution of Question 3
« Reply #2 on: March 20, 2015, 08:12:54 AM »
c Is easy: if $u$ is subharmonic (or superharmonic)  and $v$ is harmonic with the same boundary value $(u-v)|_\Sigma =0$ then $u-v$ is subharmonic (or superharmonic) and then by maximum (or minimum) principle $(u-v)\le \max _\Sigma (u-v) =0$ (or $(u-v)\ge \min _\Sigma (u-v) =0$ and thus $u\le v$ ($u\ge v$ respectively)