Problem 3
Part a. Using the proof of the maximum principle, prove the maximum principle for subharmonic functions, and the minimum principle for superharmonic functions.
Answer:
Maximum principle for subharmonic functions:
Let: $u\left(\mathbf{x}\right)$ be subharmonic, $ \Delta u\left(\mathbf{x}\right) \ge 0$, on a bounded domain $\Omega$ with boundary $\partial\Omega = \Sigma$, $\mathbf{x} \in \Omega$. Let: $v\left(\mathbf{x}\right) = u\left(\mathbf{x}\right) + \epsilon |\mathbf{x}|^2$ where $|\mathbf{x}|^2 = \sum_{i=1}^{n} x_i^2 $, $\epsilon > 0$. Let $\delta$ be the diameter of the set $\bar\Omega$, namely the largest value of $| \mathbf{x}|^2 = \delta^2$.
At any interior maximum point of a function $f$ we require $f_{x_i x_i} \le 0$, $\Delta f = \sum_{i=1}^{n}f_{x_i x_i} \le 0$ by the second derivative test. Notice that for $\mathbf{x} \in \Omega$:
$$ \Delta v\left(\mathbf{x}\right) = \Delta u\left(\mathbf{x}\right) + \Delta \epsilon |\mathbf{x}|^2 \ge 0 + 2 n \epsilon > 0 $$
So $v$ has no interior maximum on $\Omega$. But $\Omega$ is bounded so $v$ has a maximum somewhere on $\bar\Omega$. As there is no interior maximum of $v$ it must have its maximum on $\partial\Omega = \Sigma$. Let: $\mathbf{x_o}$ be the maximum of $v$ on $\bar\Omega$. Then, for $\mathbf{x} \in \bar\Omega$:
$$ u\left(\mathbf{x}\right) \le v\left(\mathbf{x}\right) \le v\left(\mathbf{x_0}\right) = u\left(\mathbf{x_0}\right) + \epsilon |\mathbf{x_0}|^2 \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) + \epsilon \delta^2 $$
As $\delta$ is some constant and we have this inequality for all $\epsilon > 0$, we take the limit as $\epsilon \rightarrow 0$ to yield:
$$ \lim_{\epsilon \rightarrow 0} \{u\left(\mathbf{x}\right) \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) + \epsilon \delta^2 \} \rightarrow \{ u\left(\mathbf{x}\right) \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) \} $$
We have the maximum of $u\left(\mathbf{x}\right)$ is attained on $\mathbf{x} \in \Sigma = \partial\Omega$, as needed. $\square$
Minimum principle for subharmonic functions:
Let: $u\left(\mathbf{x}\right)$ be superharmonic, $ \Delta u\left(\mathbf{x}\right) \le 0$, on a bounded domain $\Omega$ with boundary $\partial\Omega = \Sigma$, $\mathbf{x} \in \Omega$. Let: $v\left(\mathbf{x}\right) = - u\left(\mathbf{x}\right)$.
Then $v\left(\mathbf{x}\right)$ is subharmonic and attains its maximum on $\Sigma$, by the maximum principle for subharmonic functions. This is equivalent to a minimum of $u$ and so $u\left(\mathbf{x}\right)$ attains its minimum on $\Sigma = \partial\Omega$. We are done. $\blacksquare$
Part b. Show that the minimum principle for subharmonic functions, and maximum principle for superharmonic functions do not hold.
Answer:
Let: $u\left(\mathbf{x}\right) = |\mathbf{x}|^2 = \sum_{i=1}^{n} x_i^2$ defined on some ball in $\mathbb{R}^n$ about the origin with radius larger than $0$.
Now: $\Delta u = \Delta |\mathbf{x}|^2 = 2 n > 0$ so $u$ is subharmonic. But clearly at $ \mathbf{x} = \mathbf{0}$, $u\left(\mathbf{x}\right) = |\mathbf{x}|^2 = 0$ but at any $\mathbf{x} \ne \mathbf{0}$, $ |\mathbf{x}|^2 > 0$. Then $u$ has an interior minimum and the minimum principle for subharmonic functions does not hold. $\square$
Let: $u\left(\mathbf{x}\right)$ be subharmonic, and $v\left(\mathbf{x}\right) = - u\left(\mathbf{x}\right)$. Then $v$ is superharmonic.
Because the minimum principle does not hold for subharmonic functions as shown, $u$ may have an interior minimum. This is equivalent to an interior maximum for $v$, so the maximum principle does not hold for superharmonic functions. We are done. $\blacksquare$
Part c. Prove that if $u, v, w$ are respectively harmonic, subharmonic, and superharmonic functions in the bounded domain $ \Omega$, coinciding on its boundary $ \{u \bigr|_\Sigma= v\bigr|_\Sigma= w\bigr|_\Sigma\}$, then $w \ge u \ge v$ in $\Omega$.
Answer:
Let: $\Delta u = 0$, $\Delta v \ge 0$, $ \Delta w \le 0$ on $ \Omega$, coinciding on its boundary $\partial\Omega = \Sigma$, $ \{u \bigr|_\Sigma= v\bigr|_\Sigma, w\bigr|_\Sigma\}$.
Define: $f = v - u$, $g = w - u$.
Then: $\Delta f = \Delta v - \Delta u \ge 0$, $\Delta g = \Delta w - \Delta u \le 0$, and: $ \{u \bigr|_\Sigma= v\bigr|_\Sigma= w\bigr|_\Sigma\}$ so:
$$ \{ f \bigr|_\Sigma= \left(v-u\right)\bigr|_\Sigma = 0, \, g \bigr|_\Sigma= \left(w-u\right)\bigr|_\Sigma = 0 \} $$
$\Delta f \ge 0$, so $f$ is subharmonic and by the maximum principle attains its maximum on $\Sigma$. Then in $\Omega$, $f \le f \bigr|_\Sigma = 0$ so $f = v - u \le 0 \implies v \le u$ in $\Omega$.
Similarly, $\Delta g \le 0$ so $g$ is superharmonic and by the minimum principle attains its minimum on $\Sigma$. So in $\Omega$, $g \ge g \bigr|_\Sigma = 0$ so $g = w - u \ge 0 \implies w \ge u$ in $\Omega$.
We have in $\Omega$, $v \le u$, and $ w \ge u$. So $w \ge u \ge v$ inside of $\Omega$, as needed. $\blacksquare$
