November18 ,2020andNovember20 ,2020 THEMAXIMUMMODULUSPRINCIPLE&THEMEANVALUEPROPERTY ComplexVariables

MAT334H1-F – LEC0101

Complex Variables

THE MAXIMUM MODULUS PRINCIPLE &

THE MEAN VALUE PROPERTY

The open mapping theorem

Theorem – The open mapping theorem

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

Then its image𝑓 (𝑈 )is a domain (i.e. it is path-connected and open).

Proof.Since𝑓 is holomorphic, it is continuous. Hence𝑓 (𝑈 )is path-connected as the image of a path-connected set by a continuous function.

Let’s prove that𝑓 (𝑈 )is open. Let𝑏 ∈ 𝑓 (𝑈 )then𝑏 = 𝑓 (𝑎)for some𝑎 ∈ 𝑈.

Note that𝑎is a zero of𝑧 ↦ 𝑓 (𝑧) − 𝑓 (𝑎)but since this function is non-constant, by the isolated zero theorem there exists𝑟 > 0such that𝐷_𝑟(𝑎) ⊂ 𝑈 and∀𝑧 ∈ 𝐷_𝑟(𝑎) ⧵ {𝑎}, 𝑓 (𝑧) ≠ 𝑓 (𝑎). Set𝑚 =min_{|𝑧−𝑎|=𝑟}|𝑓 (𝑧) − 𝑓 (𝑎)| > 0. Let’s prove that𝐷_𝑚(𝑏) ⊂ 𝑓 (𝑈 ).

Let𝑤 ∈ 𝐷_𝑚(𝑏). Let𝑧be such that|𝑧 − 𝑎| = 𝑟then

|(𝑓 (𝑧) − 𝑤) − (𝑓 (𝑧) − 𝑏)| = |𝑤 − 𝑏| < 𝑚 ≤ |𝑓 (𝑧) − 𝑏|

By Rouché’s theorem,𝑓 (𝑧) − 𝑤and𝑓 (𝑧) − 𝑏have exactly the same number of zeroes on𝐷_𝑟(𝑎) (counted with multiplicities) so𝑓 (𝑧) − 𝑤 = 0for some𝑧 ∈ 𝐷_𝑟(𝑎)and𝑤 ∈ 𝑓 (𝐷_𝑟(𝑎)) ⊂ 𝑓 (𝑈 ). ■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 2 / 9

The open mapping theorem

Theorem – The open mapping theorem

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

Then its image𝑓 (𝑈 )is a domain (i.e. it is path-connected and open).

Proof.

Since𝑓 is holomorphic, it is continuous. Hence𝑓 (𝑈 )is path-connected as the image of a path-connected set by a continuous function.

Let’s prove that𝑓 (𝑈 )is open. Let𝑏 ∈ 𝑓 (𝑈 )then𝑏 = 𝑓 (𝑎)for some𝑎 ∈ 𝑈.

Note that𝑎is a zero of𝑧 ↦ 𝑓 (𝑧) − 𝑓 (𝑎)but since this function is non-constant, by the isolated zero theorem there exists𝑟 > 0such that𝐷_𝑟(𝑎) ⊂ 𝑈 and∀𝑧 ∈ 𝐷_𝑟(𝑎) ⧵ {𝑎}, 𝑓 (𝑧) ≠ 𝑓 (𝑎). Set𝑚 =min_{|𝑧−𝑎|=𝑟}|𝑓 (𝑧) − 𝑓 (𝑎)| > 0. Let’s prove that𝐷_𝑚(𝑏) ⊂ 𝑓 (𝑈 ).

Let𝑤 ∈ 𝐷_𝑚(𝑏). Let𝑧be such that|𝑧 − 𝑎| = 𝑟then

|(𝑓 (𝑧) − 𝑤) − (𝑓 (𝑧) − 𝑏)| = |𝑤 − 𝑏| < 𝑚 ≤ |𝑓 (𝑧) − 𝑏|

By Rouché’s theorem,𝑓 (𝑧) − 𝑤and𝑓 (𝑧) − 𝑏have exactly the same number of zeroes on𝐷_𝑟(𝑎) (counted with multiplicities) so𝑓 (𝑧) − 𝑤 = 0for some𝑧 ∈ 𝐷_𝑟(𝑎)and𝑤 ∈ 𝑓 (𝐷_𝑟(𝑎)) ⊂ 𝑓 (𝑈 ). ■

The open mapping theorem

Theorem – The open mapping theorem

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

Then its image𝑓 (𝑈 )is a domain (i.e. it is path-connected and open).

Proof.Since𝑓 is holomorphic, it is continuous. Hence𝑓 (𝑈 )is path-connected as the image of a path-connected set by a continuous function.

Let’s prove that𝑓 (𝑈 )is open. Let𝑏 ∈ 𝑓 (𝑈 )then𝑏 = 𝑓 (𝑎)for some𝑎 ∈ 𝑈.

Note that𝑎is a zero of𝑧 ↦ 𝑓 (𝑧) − 𝑓 (𝑎)but since this function is non-constant, by the isolated zero theorem there exists𝑟 > 0such that𝐷_𝑟(𝑎) ⊂ 𝑈 and∀𝑧 ∈ 𝐷_𝑟(𝑎) ⧵ {𝑎}, 𝑓 (𝑧) ≠ 𝑓 (𝑎). Set𝑚 =min_{|𝑧−𝑎|=𝑟}|𝑓 (𝑧) − 𝑓 (𝑎)| > 0. Let’s prove that𝐷_𝑚(𝑏) ⊂ 𝑓 (𝑈 ).

Let𝑤 ∈ 𝐷_𝑚(𝑏). Let𝑧be such that|𝑧 − 𝑎| = 𝑟then

|(𝑓 (𝑧) − 𝑤) − (𝑓 (𝑧) − 𝑏)| = |𝑤 − 𝑏| < 𝑚 ≤ |𝑓 (𝑧) − 𝑏|

By Rouché’s theorem,𝑓 (𝑧) − 𝑤and𝑓 (𝑧) − 𝑏have exactly the same number of zeroes on𝐷_𝑟(𝑎) (counted with multiplicities) so𝑓 (𝑧) − 𝑤 = 0for some𝑧 ∈ 𝐷_𝑟(𝑎)and𝑤 ∈ 𝑓 (𝐷_𝑟(𝑎)) ⊂ 𝑓 (𝑈 ). ■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 2 / 9

The open mapping theorem

Theorem – The open mapping theorem

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

Then its image𝑓 (𝑈 )is a domain (i.e. it is path-connected and open).

Proof.Since𝑓 is holomorphic, it is continuous. Hence𝑓 (𝑈 )is path-connected as the image of a path-connected set by a continuous function.

Let’s prove that𝑓 (𝑈 )is open. Let𝑏 ∈ 𝑓 (𝑈 )then𝑏 = 𝑓 (𝑎)for some𝑎 ∈ 𝑈.

Note that𝑎is a zero of𝑧 ↦ 𝑓 (𝑧) − 𝑓 (𝑎)but since this function is non-constant, by the isolated zero theorem there exists𝑟 > 0such that𝐷_𝑟(𝑎) ⊂ 𝑈and∀𝑧 ∈ 𝐷_𝑟(𝑎) ⧵ {𝑎}, 𝑓 (𝑧) ≠ 𝑓 (𝑎).

Set𝑚 =min_{|𝑧−𝑎|=𝑟}|𝑓 (𝑧) − 𝑓 (𝑎)| > 0. Let’s prove that𝐷_𝑚(𝑏) ⊂ 𝑓 (𝑈 ).

Let𝑤 ∈ 𝐷_𝑚(𝑏). Let𝑧be such that|𝑧 − 𝑎| = 𝑟then

|(𝑓 (𝑧) − 𝑤) − (𝑓 (𝑧) − 𝑏)| = |𝑤 − 𝑏| < 𝑚 ≤ |𝑓 (𝑧) − 𝑏|

By Rouché’s theorem,𝑓 (𝑧) − 𝑤and𝑓 (𝑧) − 𝑏have exactly the same number of zeroes on𝐷_𝑟(𝑎) (counted with multiplicities) so𝑓 (𝑧) − 𝑤 = 0for some𝑧 ∈ 𝐷_𝑟(𝑎)and𝑤 ∈ 𝑓 (𝐷_𝑟(𝑎)) ⊂ 𝑓 (𝑈 ). ■

The open mapping theorem

Theorem – The open mapping theorem

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

Then its image𝑓 (𝑈 )is a domain (i.e. it is path-connected and open).

Proof.Since𝑓 is holomorphic, it is continuous. Hence𝑓 (𝑈 )is path-connected as the image of a path-connected set by a continuous function.

Let’s prove that𝑓 (𝑈 )is open. Let𝑏 ∈ 𝑓 (𝑈 )then𝑏 = 𝑓 (𝑎)for some𝑎 ∈ 𝑈.

Note that𝑎is a zero of𝑧 ↦ 𝑓 (𝑧) − 𝑓 (𝑎)but since this function is non-constant, by the isolated zero theorem there exists𝑟 > 0such that𝐷_𝑟(𝑎) ⊂ 𝑈and∀𝑧 ∈ 𝐷_𝑟(𝑎) ⧵ {𝑎}, 𝑓 (𝑧) ≠ 𝑓 (𝑎).

Set𝑚 =min_{|𝑧−𝑎|=𝑟}|𝑓 (𝑧) − 𝑓 (𝑎)| > 0. Let’s prove that𝐷_𝑚(𝑏) ⊂ 𝑓 (𝑈 ).

Let𝑤 ∈ 𝐷_𝑚(𝑏). Let𝑧be such that|𝑧 − 𝑎| = 𝑟then

|(𝑓 (𝑧) − 𝑤) − (𝑓 (𝑧) − 𝑏)| = |𝑤 − 𝑏| < 𝑚 ≤ |𝑓 (𝑧) − 𝑏|

By Rouché’s theorem,𝑓 (𝑧) − 𝑤and𝑓 (𝑧) − 𝑏have exactly the same number of zeroes on𝐷_𝑟(𝑎) (counted with multiplicities) so𝑓 (𝑧) − 𝑤 = 0for some𝑧 ∈ 𝐷_𝑟(𝑎)and𝑤 ∈ 𝑓 (𝐷_𝑟(𝑎)) ⊂ 𝑓 (𝑈 ). ■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 2 / 9

The maximum modulus principle – 1

We saw on October 7 that if the range of a holomorphic function defined on a domain lies on a horizontal line, or on a vertical line, or on a circle, then the function is constant. We are now going to give a stronger version of this result.

Theorem – The maximum modulus principle

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic. If|𝑓 |has a local maximum on𝑈then𝑓 is constant on𝑈.

Proof.Assume that𝑧₀is a local maximum of|𝑓 |then there exists𝑟 > 0such that𝐷_𝑟(𝑧₀) ⊂ 𝑈 and

∀𝑧 ∈ 𝐷_𝑟(𝑧₀), |𝑓 (𝑧)| ≤ |𝑓 (𝑧₀)|.

Assume by contradiction that𝑓 is not constant, then, by the open mapping theorem,𝑓 (𝐷_𝑟(𝑧₀))is open so∃𝛿 > 0, 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀)).

Set𝑤 = (1 +2|𝑓 (𝑧^𝛿₀)|) 𝑓 (𝑧⁰)then𝑤 ∈ 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀))but|𝑤| > |𝑓 (𝑧₀)|.

Contradiction. ■

Once again this phenomenon is specific to complex calculus: it is possible for a real differentiable function𝑓 to be non-constant whereas|𝑓 |has local maxima.

The maximum modulus principle – 1

We saw on October 7 that if the range of a holomorphic function defined on a domain lies on a horizontal line, or on a vertical line, or on a circle, then the function is constant. We are now going to give a stronger version of this result.

Theorem – The maximum modulus principle

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

If|𝑓 |has a local maximum on𝑈then𝑓 is constant on𝑈.

Proof.Assume that𝑧₀is a local maximum of|𝑓 |then there exists𝑟 > 0such that𝐷_𝑟(𝑧₀) ⊂ 𝑈 and

∀𝑧 ∈ 𝐷_𝑟(𝑧₀), |𝑓 (𝑧)| ≤ |𝑓 (𝑧₀)|.

Assume by contradiction that𝑓 is not constant, then, by the open mapping theorem,𝑓 (𝐷_𝑟(𝑧₀))is open so∃𝛿 > 0, 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀)).

Set𝑤 = (1 +2|𝑓 (𝑧^𝛿₀)|) 𝑓 (𝑧⁰)then𝑤 ∈ 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀))but|𝑤| > |𝑓 (𝑧₀)|.

Contradiction. ■

Once again this phenomenon is specific to complex calculus: it is possible for a real differentiable function𝑓 to be non-constant whereas|𝑓 |has local maxima.

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 3 / 9

The maximum modulus principle – 1

We saw on October 7 that if the range of a holomorphic function defined on a domain lies on a horizontal line, or on a vertical line, or on a circle, then the function is constant. We are now going to give a stronger version of this result.

Theorem – The maximum modulus principle

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

If|𝑓 |has a local maximum on𝑈then𝑓 is constant on𝑈.

Proof.Assume that𝑧₀is a local maximum of|𝑓 |then there exists𝑟 > 0such that𝐷_𝑟(𝑧₀) ⊂ 𝑈 and

∀𝑧 ∈ 𝐷_𝑟(𝑧₀), |𝑓 (𝑧)| ≤ |𝑓 (𝑧₀)|.

Assume by contradiction that𝑓 is not constant, then, by the open mapping theorem,𝑓 (𝐷_𝑟(𝑧₀))is open so∃𝛿 > 0, 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀)).

Set𝑤 = (1 +2|𝑓 (𝑧^𝛿₀)|) 𝑓 (𝑧⁰)then𝑤 ∈ 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀))but|𝑤| > |𝑓 (𝑧₀)|.

Contradiction. ■

Once again this phenomenon is specific to complex calculus: it is possible for a real differentiable function𝑓 to be non-constant whereas|𝑓 |has local maxima.

The maximum modulus principle – 1

We saw on October 7 that if the range of a holomorphic function defined on a domain lies on a horizontal line, or on a vertical line, or on a circle, then the function is constant. We are now going to give a stronger version of this result.

Theorem – The maximum modulus principle

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

If|𝑓 |has a local maximum on𝑈then𝑓 is constant on𝑈.

Proof.Assume that𝑧₀is a local maximum of|𝑓 |then there exists𝑟 > 0such that𝐷_𝑟(𝑧₀) ⊂ 𝑈 and

∀𝑧 ∈ 𝐷_𝑟(𝑧₀), |𝑓 (𝑧)| ≤ |𝑓 (𝑧₀)|.

Assume by contradiction that𝑓 is not constant, then, by the open mapping theorem,𝑓 (𝐷_𝑟(𝑧₀))is open so∃𝛿 > 0, 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀)).

Set𝑤 = (1 +2|𝑓 (𝑧^𝛿₀)|) 𝑓 (𝑧⁰)then𝑤 ∈ 𝐷_𝛿(𝑓 (𝑧₀)) ⊂ 𝑓 (𝐷_𝑟(𝑧₀))but|𝑤| > |𝑓 (𝑧₀)|.

Contradiction. ■

Once again this phenomenon is specific to complex calculus: it is possible for a real differentiable function𝑓 to be non-constant whereas|𝑓 |has local maxima.

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 3 / 9

The maximum modulus principle – 2

Corollary

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

1 ℜ(𝑓 )has no local extremum on𝑈.

2 ℑ(𝑓 )has no local extremum on𝑈.

Proof.

1 Define𝑔 = 𝑒^𝑓, then𝑔and1/𝑔are holomorphic and non-constant.

Hence, by the maximum modulus principle, neither|𝑔| = 𝑒^{ℜ(𝑓 )}nor|𝑔¹| = 𝑒^{−ℜ(𝑓 )}have a local maximum.

Since the real exponential is increasing, we get that neitherℜ(𝑓 )nor−ℜ(𝑓 )have a local maximum.

Soℜ(𝑓 )has no local maximum and no local minimum.

2 Apply ¹ to−𝑖𝑓. ■

The maximum modulus principle – 2

Corollary

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

1 ℜ(𝑓 )has no local extremum on𝑈.

2 ℑ(𝑓 )has no local extremum on𝑈. Proof.

1 Define𝑔 = 𝑒^𝑓, then𝑔and1/𝑔are holomorphic and non-constant.

Hence, by the maximum modulus principle, neither|𝑔| = 𝑒^{ℜ(𝑓 )}nor|¹𝑔| = 𝑒^{−ℜ(𝑓 )}have a local maximum.

Since the real exponential is increasing, we get that neitherℜ(𝑓 )nor−ℜ(𝑓 )have a local maximum.

Soℜ(𝑓 )has no local maximum and no local minimum.

2 Apply ¹ to−𝑖𝑓. ■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 4 / 9

The maximum modulus principle – 2

Corollary

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a non-constant holomorphic/analytic function.

1 ℜ(𝑓 )has no local extremum on𝑈.

2 ℑ(𝑓 )has no local extremum on𝑈. Proof.

1 Define𝑔 = 𝑒^𝑓, then𝑔and1/𝑔are holomorphic and non-constant.

Hence, by the maximum modulus principle, neither|𝑔| = 𝑒^{ℜ(𝑓 )}nor|¹𝑔| = 𝑒^{−ℜ(𝑓 )}have a local maximum.

Since the real exponential is increasing, we get that neitherℜ(𝑓 )nor−ℜ(𝑓 )have a local maximum.

Soℜ(𝑓 )has no local maximum and no local minimum.

Apply to−𝑖𝑓. ■

Schwarz’s lemma

Schwarz’s lemma

Assume that𝑓 ∶ 𝐷₁(0) → ℂis holomorphic/analytic.

If ¹ 𝑓 (0) = 0

2 ∀𝑧 ∈ 𝐷₁(0), |𝑓 (𝑧)| ≤ 1 then ¹ ∀𝑧 ∈ 𝐷₁(0), |𝑓 (𝑧)| ≤ |𝑧|.

2 |𝑓^′(0)| ≤ 1

Moreover, if there exists𝑧₀≠ 0s.t. |𝑓 (𝑧₀)| = |𝑧₀|or if|𝑓^′(0)| = 1then𝑓 (𝑧) = 𝜆𝑧where|𝜆| = 1.

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 5 / 9

Schwarz’s lemma

Schwarz’s lemma

Assume that𝑓 ∶ 𝐷₁(0) → ℂis holomorphic/analytic.

If ¹ 𝑓 (0) = 0

2 ∀𝑧 ∈ 𝐷₁(0), |𝑓 (𝑧)| ≤ 1 then ¹ ∀𝑧 ∈ 𝐷₁(0), |𝑓 (𝑧)| ≤ |𝑧|.

2 |𝑓^′(0)| ≤ 1

Moreover, if there exists𝑧₀≠ 0s.t. |𝑓 (𝑧₀)| = |𝑧₀|or if|𝑓^′(0)| = 1then𝑓 (𝑧) = 𝜆𝑧where|𝜆| = 1.

Remark

Once again, Schwarz’s lemma doesn’t hold for real differentiable functions.

For instance, define𝑢(𝑥) = 2𝑥

𝑥²+ 1 on[−1, 1]:

it is𝐶¹,𝑢(0) = 0,|𝑢(𝑥)| ≤ 1but|𝑢(𝑥)| > |𝑥|on[−1, 1] ⧵ {0}.

Schwarz’s lemma

Schwarz’s lemma

Assume that𝑓 ∶ 𝐷₁(0) → ℂis holomorphic/analytic.

If ¹ 𝑓 (0) = 0

2 ∀𝑧 ∈ 𝐷₁(0), |𝑓 (𝑧)| ≤ 1 then ¹ ∀𝑧 ∈ 𝐷₁(0), |𝑓 (𝑧)| ≤ |𝑧|.

2 |𝑓^′(0)| ≤ 1

Moreover, if there exists𝑧₀≠ 0s.t. |𝑓 (𝑧₀)| = |𝑧₀|or if|𝑓^′(0)| = 1then𝑓 (𝑧) = 𝜆𝑧where|𝜆| = 1.

Proof.

We define𝑔 ∶ 𝐷₁(0) → ℂby𝑔(𝑧) = {

𝑓 (𝑧)

𝑧 if𝑧 ≠ 0

𝑓^′(0) otherwise . Then𝑔is holomorphic/analytic.

By the maximum modulus principle, for𝑟 ∈ (0, 1),max

|𝑧|≤𝑟 |𝑔| =max

|𝑧|=𝑟 |𝑔| ≤1 𝑟. As𝑟 → 1, we see that|𝑔(𝑧)| ≤ 1, i.e. |𝑓 (𝑧)| ≤ |𝑧|. It follows that|𝑓^′(0)| ≤ 1.

Assume that there exists𝑧₀≠ 0such that|𝑓 (𝑧₀)| = |𝑧₀|, then𝑧₀is local max of𝑔. Hence, by the maximum modulus principle,𝑔is constant. Besides it is of modulus1, i.e.𝑔(𝑧) = 𝜆where|𝜆| = 1. Hence𝑓 (𝑧) = 𝜆𝑧.

Assume that|𝑓^′(0)| = 1then|𝑔(0)| = |𝑓^′(0)| = 1|and0is a local max of𝑔. Hence we may conclude as in

the above case. ■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 5 / 9

Maximum modulus and boundary – 1

In the above proof we used the fact that if𝑓 is defined on𝑈and holomorphic on𝑈, then the local maximum of|𝑓 |, it there are some, are located on𝜕𝑈.

Corollary

Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂa function.

Assume that𝑓 is holomorphic/analytic and non-constant on𝑈. Then the possible

1 local extrema ofℜ(𝑓 ),

2 local extrema ofℑ(𝑓 ), and,

3 local maxima of|𝑓 | are located on𝜕𝑈.

Maximum modulus and boundary – 2

Corollary

Let𝑈 ⊂ ℂbe a bounded domain and𝑓 ∶ 𝑈 → ℂa continuous function.

Assume that𝑓 is holomorphic/analytic.

If𝑓_|𝜕𝑈 = 0then𝑓 = 0.

Proof.|𝑓 |is continuous on the compact set𝑈(closed and bounded), hence it admits a maximum.

By the above corollary, either the function is constant equal to0on𝑈or the local maxima of|𝑓 | are located on𝜕𝑈(and so are the global maxima).

In either case the maximum of|𝑓 |is reached on the boundary of𝑈 so that it has to be0.

Hence∀𝑧 ∈ 𝑈 , 𝑓 (𝑧) = 0. ■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 7 / 9

The mean value property for holomorphic functions

Theorem – The mean value property for holomorphic functions

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

Let𝑧₀∈ 𝑈 and𝑟 > 0be such that𝐷_𝑟(𝑧₀) ⊂ 𝑈. Then

𝑓 (𝑧₀) = 1 2𝜋 ∫

2𝜋 0

𝑓 (𝑧₀+ 𝑟𝑒^𝑖𝑡)d𝑡

Proof.Define𝛾 ∶ [0, 2𝜋] → ℂby𝛾(𝑡) = 𝑧₀+ 𝑟𝑒^𝑖𝑡then, by Cauchy’s theorem,

𝑓 (𝑧₀) = 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤)

𝑤 − 𝑧₀d𝑤 = 1 2𝑖𝜋 ∫

2𝜋 0

𝑓 (𝑧₀+ 𝑟𝑒^𝑖𝑡)

𝑟𝑒^𝑖𝑡 𝑟𝑖𝑒^𝑖𝑡d𝑡 = 1 2𝜋 ∫

2𝜋 0

𝑓 (𝑧₀+ 𝑟𝑒^𝑖𝑡)d𝑡

■

The mean value property for holomorphic functions

Theorem – The mean value property for holomorphic functions

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

Let𝑧₀∈ 𝑈 and𝑟 > 0be such that𝐷_𝑟(𝑧₀) ⊂ 𝑈. Then

𝑓 (𝑧₀) = 1 2𝜋 ∫

2𝜋 0

𝑓 (𝑧₀+ 𝑟𝑒^𝑖𝑡)d𝑡

Proof.Define𝛾 ∶ [0, 2𝜋] → ℂby𝛾(𝑡) = 𝑧₀+ 𝑟𝑒^𝑖𝑡then, by Cauchy’s theorem,

𝑓 (𝑧₀) = 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤)

𝑤 − 𝑧₀d𝑤 = 1 2𝑖𝜋 ∫

2𝜋 0

𝑓 (𝑧₀+ 𝑟𝑒^𝑖𝑡)

𝑟𝑒^𝑖𝑡 𝑟𝑖𝑒^𝑖𝑡d𝑡 = 1 2𝜋 ∫

2𝜋 0

𝑓 (𝑧₀+ 𝑟𝑒^𝑖𝑡)d𝑡

■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 8 / 9

The mean value property for harmonic functions

Theorem – The mean value property for harmonic functions

Let𝒰 ⊂ ℝ²be open and simply connected. Let𝑢 ∶ 𝒰 → ℝbe harmonic.

Let𝑝₀= (𝑥₀, 𝑦₀) ∈ 𝒰 and𝑟 > 0be such that𝐷_𝑟(𝑝₀) ⊂ 𝒰. Then

𝑢(𝑝₀) = 1 2𝜋 ∫

2𝜋

0 𝑢 (𝑥0+ 𝑟cos𝑡, 𝑦₀+ 𝑟sin𝑡)d𝑡

Proof.Since𝑢is harmonic on𝒰simply connected, we know that𝑢is the real part of a holomorphic function𝑓 = 𝑢 + 𝑖𝑣.

By the previous theorem

𝑓 (𝑥₀+ 𝑖𝑦₀) = 1 2𝜋 ∫

2𝜋 0

𝑓 (𝑥₀+ 𝑖𝑦₀+ 𝑟𝑒^𝑖𝑡)d𝑡

We conclude by taking the real part of both sides. ■

The mean value property for harmonic functions

Theorem – The mean value property for harmonic functions

Let𝒰 ⊂ ℝ²be open and simply connected. Let𝑢 ∶ 𝒰 → ℝbe harmonic.

Let𝑝₀= (𝑥₀, 𝑦₀) ∈ 𝒰 and𝑟 > 0be such that𝐷_𝑟(𝑝₀) ⊂ 𝒰. Then

𝑢(𝑝₀) = 1 2𝜋 ∫

2𝜋

0 𝑢 (𝑥0+ 𝑟cos𝑡, 𝑦₀+ 𝑟sin𝑡)d𝑡

Proof.Since𝑢is harmonic on𝒰simply connected, we know that𝑢is the real part of a holomorphic function𝑓 = 𝑢 + 𝑖𝑣.

By the previous theorem

𝑓 (𝑥₀+ 𝑖𝑦₀) = 1 2𝜋 ∫

2𝜋 0

𝑓 (𝑥₀+ 𝑖𝑦₀+ 𝑟𝑒^𝑖𝑡)d𝑡

We conclude by taking the real part of both sides. ■

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Nov 18, 2020 and Nov 20, 2020 9 / 9