Complex Analysis/Inequalities

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

Let ${\displaystyle f:[a,b]\rightarrow \mathbb {C} }$ be a piecewise continuous function with ${\displaystyle f_{1}:[a,b]\rightarrow \mathbb {R} }$, ${\displaystyle f_{2}:[a,b]\rightarrow \mathbb {R} }$, and ${\displaystyle f=f_{1}+i\cdot f_{2}}$, then we have:

${\displaystyle \left|\int _{a}^{b}f(t)\,dt\right|\leq \int _{a}^{b}|f_{1}(t)|\,dt+\int _{a}^{b}|f_{2}(t)|\,dt}$

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.

Let ${\displaystyle f:[a,b]\rightarrow \mathbb {C} }$ be a piecewise continuous function, then we have:

${\displaystyle \left|\int _{a}^{b}f(t)\,dt\right|\leq \int _{a}^{b}|f(t)|\,dt}$

The proof is done by a case distinction with:

• (AVI-1) ${\displaystyle \int _{a}^{b}f(t)\,dt=0}$
• (AVI-2)${\displaystyle \int _{a}^{b}f(t)\,dt\not =0}$

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

${\displaystyle \left|\int _{a}^{b}f(t)\,dt\right|=0\leq \int _{a}^{b}|f(t)|\,dt}$

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

${\displaystyle |\beta |={\frac {|\beta |^{2}}{|\beta |}}={\frac {\beta \cdot {\overline {\beta }}}{|\beta |}}=\underbrace {\frac {\overline {\beta }}{|\beta |}} _{\alpha :=}\cdot \beta =\alpha \cdot \beta }$

Since ${\displaystyle \beta \not =0}$, we have by the linearity of the integral:

${\displaystyle |\beta |=\alpha \cdot \beta =\alpha \cdot \int _{a}^{b}f(t)\,dt=\int _{a}^{b}\alpha \cdot f(t)\,dt}$

Let ${\displaystyle g:=\alpha \cdot f}$ and ${\displaystyle g:[a,b]\rightarrow \mathbb {C} }$ be a piecewise continuous function with ${\displaystyle g_{1}:[a,b]\rightarrow \mathbb {R} }$, ${\displaystyle g_{2}:[a,b]\rightarrow \mathbb {R} }$, and ${\displaystyle g=g_{1}+i\cdot g_{2}}$, then we have by the linearity of the integral:

${\displaystyle {\begin{array}{rcl}{\mathfrak {Re}}{\bigg (}\int _{a}^{b}g(t)\,dt{\bigg )}&=&{\mathfrak {Re}}{\bigg (}\underbrace {\int _{a}^{b}g_{1}(t)\,dt} _{\in \mathbb {R} }+i\cdot \underbrace {\int _{a}^{b}g_{2}(t)\,dt} _{\in \mathbb {R} }{\bigg )}\\&=&{\mathfrak {Re}}{\bigg (}\underbrace {\int _{a}^{b}g_{1}(t)\,dt} _{\in \mathbb {R} }{\bigg )}=\int _{a}^{b}g_{1}(t)\,dt\\&=&\int _{a}^{b}{\mathfrak {Re}}(g_{1}(t))\,dt\\\end{array}}}$

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

${\displaystyle |\beta |={\mathfrak {Re}}{\bigg (}\underbrace {\int _{a}^{b}\alpha \cdot f(t)\,dt} _{\in \mathbb {R} }{\bigg )}=\int _{a}^{b}\underbrace {{\mathfrak {Re}}\left(\alpha \cdot f(t)\right)} _{\in \mathbb {R} }\,dt}$

The following real part estimate against the absolute value of a complex number ${\displaystyle z}$

${\displaystyle {\mathfrak {Re}}(z)=z_{1}\leq |z_{1}|={\sqrt {z_{1}^{2}}}\leq {\sqrt {z_{1}^{2}+z_{2}^{2}}}=|z|}$

for ${\displaystyle z=z_{1}+i\cdot z_{2}}$ is now applied to the integrand of the above integral ${\displaystyle \underbrace {{\mathfrak {Re}}\left(\alpha \cdot f(t)\right)} _{\in \mathbb {R} }}$.

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

${\displaystyle |\beta |=\int _{a}^{b}{\mathfrak {Re}}\left(\alpha \cdot f(t)\right)\,dt\leq \int _{a}^{b}\left|\alpha \cdot f(t)\right|\,dt=|\alpha |\cdot \int _{a}^{b}\left|f(t)\right|\,dt}$

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

${\displaystyle \left|\int _{a}^{b}f(t)\,dt\right|=\alpha \cdot \int _{a}^{b}|f(t)|\,dt\leq |\alpha |\cdot \int _{a}^{b}|f(t)|\,dt=\int _{a}^{b}|f(t)|\,dt}$

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

${\displaystyle \left|\int _{\gamma }f(z)\,dz\right|\leq \max _{z\in trace(\gamma )}|f(z)|\cdot {\mathcal {L}}(\gamma )}$

where ${\displaystyle {\mathcal {L}}(\gamma )=\int _{a}^{b}|\gamma '(t)|\,dt}$ is the length of the integral.

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

${\displaystyle {\begin{array}{rcl}\displaystyle \left|\int _{\gamma }f(z)\,dz\right|&{\stackrel {AVI}{\leq }}&\displaystyle \int _{a}^{b}\left|f(\gamma (t))\cdot \gamma {\,}'(t)\right|\,dt=\int _{a}^{b}\left|f(\gamma (t))\right|\cdot \left|\gamma {\,}'(t)\right|\,dt\\&\leq &\displaystyle \int _{a}^{b}\underbrace {\max _{z\in trace(\gamma )}|f(z)|} _{M:=}\cdot |\gamma {\,}'(t)|\,dz=M\cdot \int _{\gamma }|\gamma {\,}'(t)|\,dz\\&=&\displaystyle M\cdot {\mathcal {L}}(\gamma )\\\end{array}}}$

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

${\displaystyle \left|\int _{\gamma }f(z),dz\right|\leq \max _{z\in Trace(\gamma )}|f(z)|\cdot L(\gamma )}$

Here, ${\displaystyle L(\gamma )=\int _{a}^{b}|\gamma '(t)|,dt}$ is the length of the integral.

• rectifiable curve

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• Date: 12/17/2024