# Complex Analysis/Inequalities
## Introduction

Inequalities are an essential tool for proving central statements in function theory. Since $\mathbb {C}$ 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 $f:[a,b]\rightarrow \mathbb {C}$ be a piecewise continuous function with $f_{1}:[a,b]\rightarrow \mathbb {R}$, $f_{2}:[a,b]\rightarrow \mathbb {R}$, and $f=f_{1}+i\cdot f_{2}$, then we have:
    $\left|\int _{a}^{b}f(t)\,dt\right|\leq \int _{a}^{b}|f_{1}(t)|\,dt+\int _{a}^{b}|f_{2}(t)|\,dt$

### 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 $f:[a,b]\rightarrow \mathbb {C}$ be a piecewise continuous function, then we have:
    $\left|\int _{a}^{b}f(t)\,dt\right|\leq \int _{a}^{b}|f(t)|\,dt$

## Proof - AVI

The proof is done by a case distinction with:
* (AVI-1) $\int _{a}^{b}f(t)\,dt=0$
* (AVI-2)$\int _{a}^{b}f(t)\,dt\not =0$

### Case - (AVI-1)

Since $\int _{a}^{b}f(t)\,dt=0$, we have $\left|\int _{a}^{b}f(t)\,dt\right|=0$. Since $|f(t)|\geq 0$, we have $\int _{a}^{b}|f(t)|\,dt\geq 0$ and we obtain:
    $\left|\int _{a}^{b}f(t)\,dt\right|=0\leq \int _{a}^{b}|f(t)|\,dt$

### Case - (AVI-2)

The integral $\beta =\int _{a}^{b}f(t)\,dt\in \mathbb {C}$ is a complex number with $\beta \not =0$, for which we have with $|\beta |={\sqrt {\beta \cdot {\overline {\beta }}}}$:
    $|\beta |={\frac {|\beta |^{2}}{|\beta |}}={\frac {\beta \cdot {\overline {\beta }}}{|\beta |}}=\underbrace {\frac {\overline {\beta }}{|\beta |}} _{\alpha :=}\cdot \beta =\alpha \cdot \beta$

#### Case - (AVI-2) - Step 1

Since $\beta \not =0$, we have by the linearity of the integral:
    $|\beta |=\alpha \cdot \beta =\alpha \cdot \int _{a}^{b}f(t)\,dt=\int _{a}^{b}\alpha \cdot f(t)\,dt$

#### Case - (AVI-2) - Step 3

Let $g:=\alpha \cdot f$ and $g:[a,b]\rightarrow \mathbb {C}$ be a piecewise continuous function with $g_{1}:[a,b]\rightarrow \mathbb {R}$, $g_{2}:[a,b]\rightarrow \mathbb {R}$, and $g=g_{1}+i\cdot g_{2}$, then we have by the linearity of the integral:
    ${\begin{array}{rcl}{\mathfrak {Re}}{\bigg\int _{a}^{b}g(t)\,dt{\bigg )}&=&{\mathfrak {Re}}{\bigg\underbrace {\int _{a}^{b}g_{1}(t)\,dt} _{\in \mathbb {R} }+i\cdot \underbrace {\int _{a}^{b}g_{2}(t)\,dt} _{\in \mathbb {R} }{\bigg )}\\&=&{\mathfrak {Re}}{\bigg\underbrace {\int _{a}^{b}g_{1}(t)\,dt} _{\in \mathbb {R} }{\bigg )}=\int _{a}^{b}g_{1}(t)\,dt\\&=&\int _{a}^{b}{\mathfrak {Re}}(g_{1}(t))\,dt\\\end{array}}$

#### Case - (AVI-2) - Step 4

Since $|\beta |=\int _{a}^{b}\alpha \cdot f(t)\,dt\in \mathbb {R}$ holds, we have by the above calculation from Step 3 for the real part:
    $|\beta |={\mathfrak {Re}}{\bigg\underbrace {\int _{a}^{b}\alpha \cdot f(t)\,dt} _{\in \mathbb {R} }{\bigg )}=\int _{a}^{b}\underbrace {{\mathfrak {Re}}\left(\alpha \cdot f(t)\right)} _{\in \mathbb {R} }\,dt$

#### Case - (AVI-2) - Step 5

The following real part estimate against the absolute value of a complex number $z$
    ${\mathfrak {Re}}(z)=z_{1}\leq |z_{1}|={\sqrt {z_{1}^{2}}}\leq {\sqrt {z_{1}^{2}+z_{2}^{2}}}=|z|$
for $z=z_{1}+i\cdot z_{2}$ is now applied to the integrand of the above integral $\underbrace {{\mathfrak {Re}}\left(\alpha \cdot f(t)\right)} _{\in \mathbb {R} }$.

#### Case - (AVI-2) - Step 6

The following estimate is obtained analogously to Step 5 by the linearity of the integral
    $|\beta |=\int _{a}^{b}{\mathfrak {Re}}\left(\alpha \cdot f(t)\right)\,dt\leq \int _{a}^{b}\left|\alpha \cdot f(t)\right|\,dt=|\alpha |\cdot \int _{a}^{b}\left|f(t)\right|\,dt$

#### Case - (AVI-2) - Step 7

Since $|\alpha |=\left|{\frac {\overline {\beta }}{|\beta |}}\right|={\frac {|{\overline {\beta }}|}{|\beta |}}=1$ holds, we have in total the desired estimate:
    $\left|\int _{a}^{b}f(t)\,dt\right|=\alpha \cdot \int _{a}^{b}|f(t)|\,dt\leq |\alpha |\cdot \int _{a}^{b}|f(t)|\,dt=\int _{a}^{b}|f(t)|\,dt$

## Inequality - Length of Integration Path - LIP

Let $\gamma :[a,b]\rightarrow \mathbb {C}$ be an integration path and $f:U\rightarrow \mathbb {C}$ be a function on the trace of $\gamma$ (i.e. ${trace}(\gamma ):=\{\gamma (t)\,:\,t\in [a,b]\}\subset U$). Then we have:
    $\left|\int _{\gamma }f(z)\,dz\right|\leq \max _{z\in trace(\gamma )}|f(z)|\cdot {\mathcal {L}}(\gamma )$
where ${\mathcal {L}}(\gamma )=\int _{a}^{b}|\gamma '(t)|\,dt$ is the length of the integral.

### Proof - LIP

By using the above estimate for the absolute value of the integrand $\left|f(z)\,dz\right|\leq \max _{z\in trace(\gamma )}|f(z)|$ and the UG-BI inequality, we obtain:
    ${\begin{array}{rcl}\displaystyle \left|\int _{\gamma }f(z)\,dz\right|&{\stackrel {AVI}{\leq }}&\displaystyle \int _{a}^{b}\left|f(\gamma (t))\cdot \gamma {\,}'(t)\right|\,dt=\int _{a}^{b}\left|f(\gamma (t))\right|\cdot \left|\gamma {\,}'(t)\right|\,dt\\&\leq &\displaystyle \int _{a}^{b}\underbrace {\max _{z\in trace(\gamma )}|f(z)|} _{M:=}\cdot |\gamma {\,}'(t)|\,dz=M\cdot \int _{\gamma }|\gamma {\,}'(t)|\,dz\\&=&\displaystyle M\cdot {\mathcal {L}}(\gamma )\\\end{array}}$

## Inequality for Estimation Over Integration Paths

Let $\gamma :[a,b]\rightarrow \mathbb {C}$ be an Integration path and $f:U\rightarrow \mathbb {C}$ a continuous function on the trace of $\gamma$ ($Trace(\gamma ):={\gamma (t),:,t\in [a,b]}\subset U$). Then, the following holds:
    $\left|\int _{\gamma }f(z),dz\right|\leq \max _{z\in Trace(\gamma )}|f(z)|\cdot L(\gamma )$
Here, $L(\gamma )=\int _{a}^{b}|\gamma '(t)|,dt$ is the length of the integral.