Complex Analysis/Inequalities

Introduction

Inequalities are an essential tool for proving central statements in function theory. Since $ℂ$ does not have a complete/total order, one must rely on the magnitude of functions for estimations.

Inequality for the Sum of Real and Imaginary Parts - IRI

Let $f:[a,b]\to ℂ$ be a piecewise continuous function with $f_1:[a,b]\to ℝ$, $f_2:[a,b]\to ℝ$, and $f=f_1+i\cdot f_2$, then we have:

$|\underset{a}{\overset{b}{\int }}f(t)dt|\le \underset{a}{\overset{b}{\int }}|f_1(t)|dt+\underset{a}{\overset{b}{\int }}|f_2(t)|dt$

Learning Task - IRI

Prove the IRI inequality. The proof is done by decomposing into real part function and imaginary part function, linearity of the integral, and applying the triangle inequality.

Inequality for the Absolute Value in the Integrand - AVI

Let $f:[a,b]\to ℂ$ be a piecewise continuous function, then we have:

$|\underset{a}{\overset{b}{\int }}f(t)dt|\le \underset{a}{\overset{b}{\int }}|f(t)|dt$

Proof - AVI

The proof is done by a case distinction with:

• (AVI-1) $\underset{a}{\overset{b}{\int }}f(t)dt=0$
• (AVI-2)$\underset{a}{\overset{b}{\int }}f(t)dt=0$

Case - (AVI-1)

Since $\underset{a}{\overset{b}{\int }}f(t)dt=0$, we have $|\underset{a}{\overset{b}{\int }}f(t)dt|=0$. Since $|f(t)|\ge 0$, we have $\underset{a}{\overset{b}{\int }}|f(t)|dt\ge 0$ and we obtain:

$|\underset{a}{\overset{b}{\int }}f(t)dt|=0\le \underset{a}{\overset{b}{\int }}|f(t)|dt$

Case - (AVI-2)

The integral $\beta =\underset{a}{\overset{b}{\int }}f(t)dt\in ℂ$ is a complex number with $\beta =0$, for which we have with $|\beta |=\sqrt{\beta \cdot \bar{\beta }}$:

$|\beta |=\frac{|\beta {|}^2}{|\beta |}=\frac{\beta \cdot \bar{\beta }}{|\beta |}=\underset{\alpha :=}{\underbrace{\frac{\bar{\beta }}{|\beta |}}}\cdot \beta =\alpha \cdot \beta$

Case - (AVI-2) - Step 1

Since $\beta =0$, we have by the linearity of the integral:

$|\beta |=\alpha \cdot \beta =\alpha \cdot \underset{a}{\overset{b}{\int }}f(t)dt=\underset{a}{\overset{b}{\int }}\alpha \cdot f(t)dt$

Case - (AVI-2) - Step 3

Let $g:=\alpha \cdot f$ and $g:[a,b]\to ℂ$ be a piecewise continuous function with $g_1:[a,b]\to ℝ$, $g_2:[a,b]\to ℝ$, and $g=g_1+i\cdot g_2$, then we have by the linearity of the integral:

$\begin{array}{ccc}\Re 𝔢(\underset{a}{\overset{b}{\int }}g(t)dt)& =& \Re 𝔢(\underset{\in ℝ}{\underbrace{\underset{a}{\overset{b}{\int }}{g}_1(t)dt}}+i\cdot \underset{\in ℝ}{\underbrace{\underset{a}{\overset{b}{\int }}{g}_2(t)dt}})\\ & =& \Re 𝔢\left(\underset{\in ℝ}{\underbrace{\underset{a}{\overset{b}{\int }}{g}_1(t)dt}}\right)=\underset{a}{\overset{b}{\int }}{g}_1(t)dt\\ & =& \underset{a}{\overset{b}{\int }}\Re 𝔢(g_1(t))dt\end{array}$

Case - (AVI-2) - Step 4

Since $|\beta |=\underset{a}{\overset{b}{\int }}\alpha \cdot f(t)dt\in ℝ$ holds, we have by the above calculation from Step 3 for the real part:

$|\beta |=\Re 𝔢\left(\underset{\in ℝ}{\underbrace{\underset{a}{\overset{b}{\int }}\alpha \cdot f(t)dt}}\right)=\underset{a}{\overset{b}{\int }}\underset{\in ℝ}{\underbrace{\Re 𝔢(\alpha \cdot f(t))}}dt$

Case - (AVI-2) - Step 5

The following real part estimate against the absolute value of a complex number $z$

$\Re 𝔢(z)=z_1\le |z_1|=\sqrt{z_1^2}\le \sqrt{z_1^2+z_2^2}=|z|$

for $z=z_1+i\cdot z_2$ is now applied to the integrand of the above integral $\underset{\in ℝ}{\underbrace{\Re 𝔢(\alpha \cdot f(t))}}$.

Case - (AVI-2) - Step 6

The following estimate is obtained analogously to Step 5 by the linearity of the integral

$|\beta |=\underset{a}{\overset{b}{\int }}\Re 𝔢(\alpha \cdot f(t))dt\le \underset{a}{\overset{b}{\int }}|\alpha \cdot f(t)|dt=|\alpha |\cdot \underset{a}{\overset{b}{\int }}|f(t)|dt$

Case - (AVI-2) - Step 7

Since $|\alpha |=\left|\frac{\bar{\beta }}{|\beta |}\right|=\frac{|\bar{\beta }|}{|\beta |}=1$ holds, we have in total the desired estimate:

$|\underset{a}{\overset{b}{\int }}f(t)dt|=\alpha \cdot \underset{a}{\overset{b}{\int }}|f(t)|dt\le |\alpha |\cdot \underset{a}{\overset{b}{\int }}|f(t)|dt=\underset{a}{\overset{b}{\int }}|f(t)|dt$

Inequality - Length of Integration Path - LIP

Let $\gamma :[a,b]\to ℂ$ be an integration path and $f:U\to ℂ$ be a function on the trace of $\gamma$ (i.e. $trace(\gamma ):=\{\gamma (t):t\in [a,b]\}\subset U$). Then we have:

$|\underset{\gamma }{\int }f(z)dz|\le \underset{z\in trace(\gamma )}{\max }|f(z)|\cdot ℒ(\gamma )$

where $ℒ(\gamma )=\underset{a}{\overset{b}{\int }}|{\gamma }^{\prime }(t)|dt$ is the length of the integral.

Proof - LIP

By using the above estimate for the absolute value of the integrand $|f(z)dz|\le \underset{z\in trace(\gamma )}{\max }|f(z)|$ and the UG-BI inequality, we obtain:

$\begin{array}{ccc}|\underset{\gamma }{\int }f(z)dz|& \stackrel{AVI}{\le }& \underset{a}{\overset{b}{\int }}|f(\gamma (t))\cdot \gamma {}^{\prime }(t)|dt=\underset{a}{\overset{b}{\int }}|f(\gamma (t))|\cdot |\gamma {}^{\prime }(t)|dt\\ & \le & \underset{a}{\overset{b}{\int }}\underset{M:=}{\underbrace{\underset{z\in trace(\gamma )}{\max }|f(z)|}}\cdot |\gamma {}^{\prime }(t)|dz=M\cdot \underset{\gamma }{\int }|\gamma {}^{\prime }(t)|dz\\ & =& M\cdot ℒ(\gamma )\end{array}$

Inequality for Estimation Over Integration Paths

Let $\gamma :[a,b]\to ℂ$ be an Integration path and $f:U\to ℂ$ a continuous function on the trace of $\gamma$ ($Trace(\gamma ):=\gamma (t),:,t\in [a,b]\subset U$). Then, the following holds:

$|\underset{\gamma }{\int }f(z),dz|\le \underset{z\in Trace(\gamma )}{\max }|f(z)|\cdot L(\gamma )$

Here, $L(\gamma )=\underset{a}{\overset{b}{\int }}|{\gamma }^{\prime }(t)|,dt$ is the length of the integral.

See also

• rectifiable curve

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