Continuous Subharmonic U C Log z
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Everything but the last definition and lemma 1
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The last definition and lemma 2
Recall from Definition 3.6 that \(u \in C^2(\Omega)\) is called subharmonic if \(\Delta u \ge 0\) in \(\Omega\text{.}\) Inspired by the comparison principle (Proposition 3.5), we now introduce a more general notion of subharmonic which makes sense when \(u\) is merely continuous. Such functions will be the main characters in Section 4.5.
Definition 4.10 .
We say \(u \in C^0(\overline \Omega)\) is subharmonic if for every ball with \(\overline B \subseteq \Omega\) and every function \(h \in C^0(\overline B) \cap C^2(B)\) which is harmonic \(B\) and satisfies \(u\le h\) on \(\partial B\text{,}\) we have \(u \le h\) in \(B\text{.}\) Superharmonic functions are defined similarly but with all of the inequalities reversed.
We now check that our two definitions of subharmonic are consistent.
Lemma 4.11 .
Let \(u \in C^2(\Omega) \cap C^0(\overline\Omega)\text{.}\) Then \(u\) is subharmonic according to Definition 3.6 if and only if it is subharmonic according to Definition 4.10.
Proof.
First suppose that \(u\) is subharmonic according to Definition 3.6, i.e. that \(\Delta u \ge 0\) on \(\Omega\text{.}\) Let \(B\subseteq \Omega\) be a ball and \(h \in C^0(\overline B) \cap C^2(B)\) be a harmonic function on \(B\) with \(u\le h\) on \(\partial B\text{.}\) Applying the comparison principle (Proposition 3.5) we find that that \(u\le h\) on \(B\text{.}\) Thus \(u\) is also subharmonic according to Definition 4.10.
Next suppose that \(u\) is subharmonic according to Definition 4.10, and assume for the sake of contradiction that \(\Delta u(x) \lt 0\) for some \(x \in \Omega\text{.}\) Then we can find a ball \(B\subseteq \Omega\) centred at \(x\) on which \(\Delta u\lt 0\text{.}\) By Theorem 4.6, there is a harmonic function \(U\in C^2(B) \cap C^0(\overline B)\) which agrees with \(u\) on \(\partial B\text{.}\) Applying the Strong maximum principle to \(U-u\text{,}\) we deduce that either \(u \gt U\) in \(B\text{,}\) which would contradict Definition 4.10, or else \(u \equiv U\) in \(B\text{.}\) But \(\Delta(U-u) = \Delta u \gt 0\) in \(B\text{,}\) and so \(U-u\) cannot be constant there.
Proposition 4.12 .
Let \(\Omega\) be bounded and connected, and let \(u \in C^0(\overline\Omega)\) be subharmonic and \(v \in C^0(\overline\Omega)\) be superharmonic. If \(v \ge u\) on \(\partial \Omega\text{,}\) then either \(v\gt u\) in \(\Omega\) or \(v \equiv u\text{.}\)
Proof.
Let \(w = u-v\text{,}\) and note that \(w \le 0\) on \(\partial\Omega\text{.}\) Assume for contradiction that there exists \(x_0\in \Omega\) with
\begin{equation*} w(x_0)=\sup_\Omega w =M \ge 0. \end{equation*}
If \(w \equiv M\text{,}\) then \(w \le 0\) on \(\partial \Omega\) forces \(M = 0\) and hence \(u \equiv v\text{.}\) So assume that \(w \not \equiv M\) and hence that there exists some \(x_1\in \Omega\) with \(w(x_1) \lt M\text{.}\) As in the proof of Theorem 3.8, we can choose \(x_0,x_1\) so that \(x_1 \in \partial B_r(x_0)\) and \(\overline{B_r(x_0)} \subseteq \Omega\text{.}\)
Applying Theorem 4.6, we can find harmonic functions \(\bar u\) and \(\bar v\) in \(C^2(B_r(x_0)) \cap C^0(\overline{B_r(x_0)})\) which are equal to \(u\) and \(v\) on \(\partial B_r(x_0)\text{.}\) By Definition 4.10, we have \(u \le \bar u\) and \(v \ge \bar v\) on \(B_r(x_0)\text{.}\) Taking differences, we find that the harmonic function
\begin{equation} \bar w = \bar u - \bar v \ge u - v = w\tag{4.12} \end{equation}
on \(B_r(x_0)\text{.}\) Since \(\bar w = w \le M\) on \(\partial B_r(x_0)\text{,}\) the strong maximum principle implies that with \(\bar w \lt M\) in \(B_r(x_0)\) or else \(\bar w \equiv M\text{.}\) The latter cannot happen since \(\bar w(x_1) = w(x_1) \lt M\text{,}\) and so \(\bar w \lt M\) in \(B\text{.}\) But then at \(x_0\) this and (4.12) imply
\begin{equation*} w(x_0) \le \bar w(x_0) \lt M, \end{equation*}
contradicting the fact that \(w(x_0)=M\text{.}\)
Definition 4.13 .
Let \(u\) be subharmonic in \(\Omega\text{,}\) and \(B\) a ball with \(\overline B \subseteq \Omega\text{.}\) Letting \(\bar u\) be the harmonic function on \(B\) satisfying \(\bar u =u\) on \(\partial B\text{,}\) the function
\begin{equation} U(x) = \begin{cases} \bar u(x) \amp x \in B \\ u(x) \amp x \in \Omega \setminus B \end{cases}\tag{4.13} \end{equation}
is called the harmonic lifting of \(u\text{.}\)
Lemma 4.14 . Operations on subharmonic functions.
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The harmonic lifting \(U\) of a subharmonic function \(u\) is also subharmonic. In particular, \(u\le U\) in \(\Omega\text{.}\)
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If \(u_1,u_2\) are subharmonic, then \(\max(u_1,u_2)\) is subharmonic.
Proof.
Exercise 4.4.3 and Exercise 4.4.2.
Exercises Exercises
1. *Directly checking \(\abs x\) is subharmonic.
Let \(\Omega=\R\) so that \(N=1\text{,}\) and consider the function \(u \maps \R \to \R\) given by \(u(x)=\abs x\text{.}\) We will show directly that \(u\) is subharmonic in the sense of Definition 4.10. Fix an arbitrary ball \(B = (a,b) \subset \R\) and a harmonic function \(h \in C^2(B) \cap C^0(\overline B)\) with \(u \le h\) on \(\partial B\text{.}\)
(a)
Argue that the graph of \(h\) is a straight line, and deduce that for any \(t \in (0,1)\text{,}\) we have
\begin{gather} h(ta+(1-t)b) = th(a)+(1-t)h(b)\tag{✶} \end{gather}
Solution .
The function \(h\) being harmonic means that \(h''=0\text{,}\) i.e. that \(h\) is an affine function whose graph is a straight line. The identity (✶) is one of the many 'standard forms' for affine functions, and can easily be derived from your preferred form by algebraic manipulations, or by appealing to basic facts about lines in the plane.
(b)
Use the triangle inequality to prove that, for any \(t \in (0,1)\text{,}\)
\begin{equation*} u(ta+(1-t)b) \le tu(a)+(1-t)u(b)\text{.} \end{equation*}
Solution .
For any \(t \in (0,1)\) we have \(t,1-t \ge 0\text{,}\) and so by the triangle inequality
\begin{align*} u(ta+(1-t)b) \amp = \abs{ta+(1-t)b}\\ \amp \le \abs{ta}+\abs{(1-t)b}\\ \amp = t\abs{a}+(1-t)\abs{b}\\ \amp = tu(a)+(1-t)u(b) \end{align*}
as desired.
(c)
Conclude that \(u \le h\) on \(B\) as desired, and hence that \(u\) is subharmonic.
Hint .
Note that an arbitrary point \(x \in (a,b)\) may be expressed as \(x=ta+(1-t)b\) for some \(t \in (0,1)\text{.}\)
Solution .
Combining the previous two parts, we have, for \(t \in (0,1)\)
\begin{align*} u(ta+(1-t)b) \amp \le tu(a)+(1-t)u(b)\\ \amp \le th(a)+(1-t)h(b)\\ \amp = h(ta+(1-t)b)\text{,} \end{align*}
where in the second step we have used that \(u \le h\) on \(\partial B = \{a,b\}\text{.}\) Thus \(u \le h\) on \(B\text{.}\) By the arbitrariness of \(h\) and \(B\text{,}\) we conclude that \(u\) is subharmonic in the sense of Definition 4.10.
2. *Pointwise maximum of subharmonic functions.
Prove the second part of Lemma 4.14, and use this to give a (indirect but much quicker) solution to Exercise 4.4.1.
Solution .
Let \(B\) be a ball whose closure is contained in \(\Omega\text{,}\) and let \(h \in C^2(B) \cap C^0(\overline B)\) be a harmonic function with
\begin{gather*} u = \max(u_1,u_2) \le h \ona \partial B. \end{gather*}
Then in particular \(u_1 \le h\) on \(\partial B\text{,}\) and so since \(u_1\) is subharmonic we have \(u_1 \le h\) in \(B\text{.}\) Similarly we get \(u_2 \le h\text{.}\) Thus \(u = \max(u_1,u_2) \le h\) as desired.
This can be applied to the function \(u(x)=\abs x\) in Exercise 4.4.1. Indeed, we have \(u(x) = \max(x,-x)\text{,}\) where the functions \(x \mapsto \pm x\) are harmonic and hence in particular subharmonic, and so \(u\) is subharmonic by the above argument.
Comment .
Several students this year never wrote something like \(\max(u_1,u_2) \le h\) on \(\partial B\text{,}\) but instead jumped straight to the pair of inequalities \(u_1, u_2 \le h\text{.}\) In the context of this problem, where the whole point is passing between the single function \(\max(u_1,u_2)\) and the pair of functions \(u_1,u_2\text{,}\) I think this is too vague. On an exam, depending on how clearly the rest of the problem was written out, I think this could easily end up costing a mark or two.
3. *Harmonic liftings.
Prove the first part of Lemma 4.14, namely that harmonic liftings of subharmonic functions are subharmonic.
Hint 1 .
This is much more subtle than the second part of Lemma 4.14. I highly recommend that you draw yourself a picture of the various sets involved. In particular, this is useful when you need to think about the boundary of an intersection.
Hint 2 .
First show that \(u \le U\) in \(\Omega\text{.}\) Then, consider an arbitrary ball \(B'\) with \(\overline{B'} \subseteq \Omega\) and a harmonic function \(h\) on \(B'\) with \(U \le h\) on \(\partial B'\text{.}\) Certainly \(U \le h\) on \(B' \without B\text{,}\) and so it remains to show \(U \le h\) on the set \(B' \cap B\text{.}\) Since \(U\) and \(h\) are both harmonic on this set, there is a chance you can apply the comparison principle.
Solution .
By construction, \(U\) is harmonic on \(B\) and agrees with \(u\) on \(\partial B\text{,}\) and so \(u \le U\) on \(B\) by the definition of \(u\) being subharmonic. Moreover, \(u \equiv U\) on \(\Omega\without B\text{,}\) and so we deduce that \(u \le U\) on all of \(\Omega\) as desired.
It remains to show that \(U\) is subharmonic. Let \(B'\) be a ball whose closure is contained in \(\Omega\text{,}\) and let \(h \in C^2(B') \cap C^0(\overline{B'})\) be a harmonic function in \(B'\) with \(U \le h\) on \(\partial B'\text{.}\) We need to show that \(U \le h\) in \(B'\text{.}\) Since \(u \le U\) on \(\Omega\text{,}\) we have \(u \le U \le h\) on \(\partial B'\text{,}\) and since \(u\) is subharmonic this implies \(u \le h\) on \(B'\text{.}\) Thus on \(B' \without B\) (which may of course be empty) we have \(u = U \le h\text{,}\) and so it remains to show that \(U \le h\) on the intersection \(B' \cap B\text{.}\)
But \(U\) and \(h\) are both harmonic on \(B' \cap B\text{,}\) and so by the comparison principle it suffices to show that \(U \le h\) on the boundary (draw a picture!!)
\begin{gather*} \partial(B' \cap B) = (\overline{\partial B' \cap B}) \cup (\overline{B' \cap \partial B}). \end{gather*}
On \(\partial B' \cap B \subseteq \partial B'\) we have \(U \le h\) by assumption. On \(B' \cap \partial B \subseteq \partial B\) we have \(u=U\text{,}\) and since \(B' \cap \partial B \subseteq B'\) this gives \(U=u\le h\) by the argument in the previous paragraph. Since \(U\) and \(h\) are continuous, we don't have to worry about the closures, and so we conclude that \(U \le h\) on \(\partial(B' \cap B)\) and hence in \(B' \cap B\) by the comparison principle.
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