Math 618
Theory of Functions of a Complex Variable II
Spring 2015
Exercise on three-circles theorems
In this assignment, you will apply subharmonic functions to prove Hadamard’s
three-circles theorem (which is one of the topics on the official syllabus for the
qualifying examination in complex analysis).
1. Suppose π’ is a subharmonic function in an annulus with inner radius π and
outer radius π: namely, { π§ ∈ β βΆ π < |π§| < π }. Let π(π) denote the
maximum of π’(π§) when |π§| = π. Prove that if π < π1 < π < π2 < π, then
π(π) ≤
log π − log π1
log π2 − log π
π(π1 ) +
π(π2 ).
log π2 − log π1
log π2 − log π1
Sometimes this inequality is described by saying that π(π) is “a convex function of log π.” An equivalent formulation of the inequality is that
log
π(π) ≤
log
π2
π
π2
π1
π(π1 ) +
log ππ
1
log
π2
π1
π(π2 ).
Suggestion: Use a harmonic comparison function of the form π΄ + π΅ log |π§|.
2. Deduce that if π is a holomorphic function in the annulus, and π(π) denotes
the maximum of |π (π§)| when |π§| = π, then
π
1−π
π(π) ≤ π(π1 ) π(π2 )
log
,
where π < π1 < π < π2 < π, and π =
log
π2
π
π2
π1
.
This statement is Hadamard’s three-circles theorem (Theorem 23.0.3 on
page 387 of the textbook).
3. Suppose in Part 1 that π’ is bounded above, and π = 0. Let π1 approach 0 to
deduce that π(π) is an increasing function of π. Extend π’ to the whole disk
{ π§ ∈ β βΆ |π§| < π } by declaring π’(0) to be limπ→0+ π(π). Prove that the
extended function π’ is subharmonic on the whole disk.
This statement is a removable singularity theorem for subharmonic functions
that are bounded above on a punctured disk.
April 2, 2015
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Dr. Boas