§3.6 Alternative Integral Formulas

3.6 Alternative Integral Formulas

As in the Cauchy Integral Formula (Theorem 3.1.10 $Theorem 3.1.10 Theorem 3.1.10$ ), we can write holomorphic functions in terms of an integral representation. We define the Cauchy kernel to be

𝐻 (𝜁, 𝑧) = \frac{1}{2 π i (𝜁 - 𝑧)} .

Then Theorem 3.1.10 $Theorem 3.1.10 Theorem 3.1.10$ can be written as

𝑓 (𝑧) = \oint_{𝜕 𝑈} 𝑓 (𝜁) 𝐻 (𝜁, 𝑧) d 𝜁 .

There also exist other integral formulas for functions, varying in the kernel of the expression.

Let $Φ : 𝔻 \to ℝ$ be harmonic such that $Φ$ is continuous on $\overline{𝔻}$ . By the global mean-value property introduced in Lemma 3.4.1 $Lemma 3.4.1 Lemma 3.4.1$ , we have

Φ (0) = \frac{1}{2 π} \int_{0}^{2 π} Φ (𝜌 e^{i 𝑡}) d 𝑡,

where $0 < 𝜌 < 1$ . By the uniform continuity of $Φ$ on $\overline{𝔻}$ (Theorem 1.2.26 $Theorem 1.2.26 Theorem 1.2.26$ ), $\forall 𝜀 > 0$ , $\exists 𝛿 > 0$ such that for all $𝜌 \in (\frac{1}{2}, 1)$ satisfying $1 - 𝜌 < 𝛿$ and all $𝑡 \in [0, 2 π]$ ,

| Φ (e^{i 𝑡}) - Φ (𝜌 e^{i 𝑡}) | < 𝜀 .

It then follows that

| \frac{1}{2 π} \int_{0}^{2 π} Φ (e^{i 𝑡}) d 𝑡 - \frac{1}{2 π} \int_{0}^{2 π} Φ (𝜌 e^{i 𝑡}) d 𝑡 | < 𝜀 .

Hence,

lim_{𝜌 \to 1^{-}} \frac{1}{2 π} \int_{0}^{2 π} Φ (𝜌 e^{i 𝑡}) d 𝑡 = \frac{1}{2 π} \int_{0}^{2 π} Φ (e^{i 𝑡}) d 𝑡 = Φ (0) . (3.6 . 1)

Let $𝑧 \in 𝔻$ and notice that

𝜑_{𝑧} (𝜁) = \frac{𝜁 - 𝑧}{1 - \overline{𝑧} 𝜁} \in Aut (𝔻)

maps $𝜕 𝔻$ to $𝜕 𝔻$ bijectively. Let $𝑢$ be harmonic on $𝔻$ and continuous on $\overline{𝔻}$ . Then $𝑢 \circ 𝜑_{- 𝑧}$ is also harmonic on $𝔻$ , and by (3.6.1) $(3.6.1) (3.6.1)$ ,

𝑢 (𝑧) = 𝑢 \circ 𝜑_{- 𝑧} (0) = \frac{1}{2 π} \int_{0}^{2 π} 𝑢 \circ 𝜑_{- 𝑧} (e^{i 𝜓}) d 𝜓 .

By the univalence of $𝜑_{𝑧}$ , let $e^{i 𝜓} = 𝜑_{𝑧} (e^{i 𝑡})$ . It follows that

\begin{aligned} i e^{i 𝜓} d 𝜓 & = i \frac{1 - \overline{𝑧} 𝑧}{{(1 - \overline{𝑧} e^{i 𝑡})}^{2}} e^{i 𝑡} d 𝑡 \\ d 𝜓 & = \frac{1 - \overline{𝑧} 𝑧}{{(1 - \overline{𝑧} e^{i 𝑡})}^{2}} \frac{1 - \overline{𝑧} e^{i 𝑡}}{e^{i 𝑡} - 𝑧} e^{i 𝑡} d 𝑡 \\ = \frac{1 - {| 𝑧 |}^{2}}{{| 1 - \overline{𝑧} e^{i 𝑡} |}^{2}} d 𝑡 . (3.6 . 2) \end{aligned}

Then from (3.6.1) $(3.6.1) (3.6.1)$ ,

𝑢 (𝑧) = \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (e^{i 𝑡}) \frac{1 - {| 𝑧 |}^{2}}{{| 1 - \overline{𝑧} e^{i 𝑡} |}^{2}} d 𝑡 .

Let

𝑃 (𝜁, 𝑧) = \frac{1 - {| 𝑧 |}^{2}}{2 π {| 1 - \overline{𝑧} 𝜁 |}^{2}} = \frac{1 - {| 𝑧 |}^{2}}{2 π {| 𝜁 - 𝑧 |}^{2}},

known as the Poisson kernel on $𝔻$ . Then,

𝑢 (𝑧) = \int_{0}^{2 π} 𝑢 (𝜁) 𝑃 (𝜁, 𝑧) d 𝑡, | 𝑧 | < 1, (3.6 . 3)

where $𝜁 = e^{i 𝑡}$ . (3.6.3) $(3.6.3) (3.6.3)$ is also known as the Poisson Integral Formula.

For all $𝑧 \in 𝐷 (0, 𝑅)$ , where $𝑅 > 0$ , we can apply the transformation

{\tilde{𝜑}}_{𝑧} (𝜁) = 𝑅 𝜑_{\frac{𝑧}{𝑅}} (\frac{𝜁}{𝑅})

to extend the automorphism to $𝐷 (0, 𝑅)$ . Let $𝑢$ instead be harmonic on $𝐷 (0, 𝑅)$ and continuous on $\overline{𝐷 (0, 𝑅)}$ . Then,

𝑢 (0) = \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑅 e^{i 𝜓}) d 𝜓 .

It follows that $𝑢 \circ {\tilde{𝜑}}_{- 𝑧}$ is also harmonic on $𝐷 (0, 𝑅)$ with

𝑢 (𝑧) = 𝑢 \circ {\tilde{𝜑}}_{- 𝑧} (0),

and from the bijectivity of $𝑅 e^{i 𝜓} = {\tilde{𝜑}}_{𝑧} (𝑅 e^{i 𝑡})$ ,

\begin{aligned} d 𝜓 & = \frac{1 - \frac{{| 𝑧 |}^{2}}{𝑅^{2}}}{{(1 - \frac{\overline{𝑧}}{𝑅} e^{i 𝑡})}^{2}} e^{i 𝑡} e^{- i 𝜓} d 𝑡 \\ = \frac{1 - \frac{{| 𝑧 |}^{2}}{𝑅^{2}}}{{(1 - \frac{\overline{𝑧}}{𝑅} e^{i 𝑡})}^{2}} \frac{1 - \frac{\overline{𝑧}}{𝑅} e^{i 𝑡}}{1 - \frac{𝑧}{𝑅} e^{- i 𝑡}} d 𝑡 \\ = \frac{𝑅^{2} - {| 𝑧 |}^{2}}{{| 𝑅 e^{i 𝑡} - 𝑧 |}^{2}} d 𝑡 . (3.6 . 4) \end{aligned}

Then because ${\tilde{𝜑}}_{𝑧}^{- 1} = {\tilde{𝜑}}_{- 𝑧}$ ,

𝑢 (𝑧) = \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑅 e^{i 𝑡}) \frac{𝑅^{2} - {| 𝑧 |}^{2}}{{| 𝑅 e^{i 𝑡} - 𝑧 |}^{2}} d 𝑡 .

The expression

𝑃 (𝜁, 𝑧) = \frac{{| 𝜁 |}^{2} - {| 𝑧 |}^{2}}{2 π {| 𝜁 - 𝑧 |}^{2}}

is a general form of the Poisson kernel on a disk. Then with $𝜁 = 𝑅 e^{i 𝑡}$ ,

𝑢 (𝑧) = \int_{0}^{2 π} 𝑢 (𝜁) 𝑃 (𝜁, 𝑧) d 𝑡 . (3.6 . 5)

The Poisson kernel can also be rewritten as

\begin{aligned} 𝑃 (𝜁, 𝑧) & = \frac{{| 𝜁 |}^{2} - {| 𝑧 |}^{2}}{2 π (𝜁 - 𝑧) (\overline{𝜁} - \overline{𝑧})} \\ = \frac{1}{4 π} (\frac{𝜁 + 𝑧}{𝜁 - 𝑧} + \frac{\overline{𝜁} + \overline{𝑧}}{\overline{𝜁} - \overline{𝑧}}) \\ = \frac{1}{2 π} ℜ𝔢 (\frac{𝜁 + 𝑧}{𝜁 - 𝑧}) . (3.6 . 6) \end{aligned}

Thus, (3.6.5) $(3.6.5) (3.6.5)$ is equivalent to

𝑢 (𝑧) = \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝜁) ℜ𝔢 (\frac{𝜁 + 𝑧}{𝜁 - 𝑧}) d 𝑡 .

Since $d 𝜁 = i 𝜁 d 𝑡$ , $d 𝑡 = \frac{d 𝜁}{i 𝜁}$ , and

𝑢 (𝑧) = \frac{1}{2 π i} \oint_{𝜕 𝐷 (0, 𝑅)} \frac{𝑢 (𝜁)}{𝜁} ℜ𝔢 (\frac{𝜁 + 𝑧}{𝜁 - 𝑧}) d 𝜁 = ℜ𝔢 (\frac{1}{2 π i} \oint_{𝜕 𝐷 (0, 𝑅)} \frac{𝑢 (𝜁)}{𝜁} \frac{𝜁 + 𝑧}{𝜁 - 𝑧} d 𝜁),

where $𝑧 \in 𝐷 (0, 𝑅)$ . Since $𝑅 > 0$ and $𝜁 - 𝑧 \neq 0$ for all $𝜁 \in 𝜕 𝐷 (0, 𝑅)$ and $𝑧 \in 𝐷 (0, 𝑅)$ , the function

𝐹 (𝑧) = \frac{1}{2 π i} \oint_{𝜕 𝐷 (0, 𝑅)} \frac{𝑢 (𝜁)}{𝜁} \frac{𝜁 + 𝑧}{𝜁 - 𝑧} d 𝜁

is holomorphic on $𝐷 (0, 𝑅)$ : for each fixed $𝜁 \in 𝜕 𝐷 (0, 𝑅)$ , the integrand is holomorphic in $𝑧$ , and on compact subsets of $𝐷 (0, 𝑅)$ we may differentiate under the integral sign. Therefore, $𝑢 (𝑧)$ is the real part of a holomorphic function

𝑓 (𝑧) = \frac{1}{2 π i} \oint_{𝜕 𝐷 (0, 𝑅)} \frac{𝑢 (𝜁)}{𝜁} \frac{𝜁 + 𝑧}{𝜁 - 𝑧} d 𝜁 + i 𝑐,

where $𝑐 \in ℝ$ . Since $𝑐 \in ℝ$ is holomorphic, by Proposition 2.2.5 $Proposition 2.2.5 Proposition 2.2.5$ , $𝑐$ is constant. For $𝑓 (𝑧) = 𝑢 (𝑧) + i 𝑣 (𝑧)$ ,

𝑣 (𝑧) = 𝑐 + \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝜁) ℑ𝔪 (\frac{𝜁 + 𝑧}{𝜁 - 𝑧}) d 𝑡 . (3.6 . 7)

Letting $𝑧 = 0$ , the integral vanishes, and we obtain $𝑐 = 𝑣 (0) = ℑ𝔪 (𝑓 (0))$ .

Define the Schwarz kernel to be

𝑆 (𝜁, 𝑧) = \frac{𝜁 + 𝑧}{2 π i (𝜁 - 𝑧) 𝜁} .

Then for a holomorphic function $𝑓$ on $𝐷 (0, 𝑅)$ that is continuous on $\overline{𝐷 (0, 𝑅)}$ , we obtain the Schwarz Integral Formula:

𝑓 (𝑧) = \oint_{𝜕 𝐷 (0, 𝑅)} ℜ𝔢 (𝑓 (𝜁)) 𝑆 (𝜁, 𝑧) d 𝜁 + i ℑ𝔪 (𝑓 (0)) . (3.6 . 8)

The significance of this alternative formula implies that a holomorphic function can be recovered from the real part on the boundary of a disk and the imaginary part at a single point.

From (3.6.7) $(3.6.7) (3.6.7)$ , we can rewrite

\begin{aligned} ℑ𝔪 (\frac{𝜁 + 𝑧}{𝜁 - 𝑧}) & = ℑ𝔪 (1 + \frac{2 𝑧}{𝜁 - 𝑧}) \\ = ℑ𝔪 (\frac{2 𝑧 (\overline{𝜁} - \overline{𝑧})}{{| 𝜁 - 𝑧 |}^{2}}) \\ = \frac{2 ℑ𝔪 (𝑧 \overline{𝜁})}{{| 𝜁 - 𝑧 |}^{2}} . (3.6 . 9) \end{aligned}

Let

𝑄 (𝜁, 𝑧) = \frac{ℑ𝔪 (𝑧 \overline{𝜁})}{π {| 𝜁 - 𝑧 |}^{2}},

which is known as the conjugate Poisson kernel. Then (3.6.7) $(3.6.7) (3.6.7)$ yields yet another integral representation of harmonic functions:

𝑣 (𝑧) = 𝑣 (0) + \int_{0}^{2 π} 𝑢 (𝜁) 𝑄 (𝜁, 𝑧) d 𝑡 . (3.6 . 10)

where $𝜁 = 𝑅 e^{i 𝑡}$ . Two harmonic functions are said to be conjugate if they are the real and imaginary parts of a holomorphic function. As seen above, on open disks, any harmonic function will admit a unique conjugate, up to an additive constant $𝑣 (0)$ . For a harmonic function $𝑢$ , we can construct its harmonic conjugate from (3.6.9) $(3.6.9) (3.6.9)$ .

The Poisson kernel is important in many branches of mathematics. We will introduce two of the important uses below.