Inequalities are an essential tool for proving central statements in function theory. Since
C
{\displaystyle \mathbb {C} }
complete/total order , one must rely on the magnitude of functions for estimations.
Inequality for the Sum of Real and Imaginary Parts - IRI
edit
Let
f
:
[
a
,
b
]
→
C
{\displaystyle f:[a,b]\rightarrow \mathbb {C} }
f
1
:
[
a
,
b
]
→
R
{\displaystyle f_{1}:[a,b]\rightarrow \mathbb {R} }
f
2
:
[
a
,
b
]
→
R
{\displaystyle f_{2}:[a,b]\rightarrow \mathbb {R} }
f
=
f
1
+
i
⋅
f
2
{\displaystyle f=f_{1}+i\cdot f_{2}}
|
∫
a
b
f
(
t
)
d
t
|
≤
∫
a
b
|
f
1
(
t
)
|
d
t
+
∫
a
b
|
f
2
(
t
)
|
d
t
{\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.
Inequality for the Absolute Value in the Integrand - AVI
edit
Let
f
:
[
a
,
b
]
→
C
{\displaystyle f:[a,b]\rightarrow \mathbb {C} }
|
∫
a
b
f
(
t
)
d
t
|
≤
∫
a
b
|
f
(
t
)
|
d
t
{\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)
∫
a
b
f
(
t
)
d
t
=
0
{\displaystyle \int _{a}^{b}f(t)\,dt=0}
(AVI-2)
∫
a
b
f
(
t
)
d
t
≠
0
{\displaystyle \int _{a}^{b}f(t)\,dt\not =0}
Since
∫
a
b
f
(
t
)
d
t
=
0
{\displaystyle \int _{a}^{b}f(t)\,dt=0}
|
∫
a
b
f
(
t
)
d
t
|
=
0
{\displaystyle \left|\int _{a}^{b}f(t)\,dt\right|=0}
|
f
(
t
)
|
≥
0
{\displaystyle |f(t)|\geq 0}
∫
a
b
|
f
(
t
)
|
d
t
≥
0
{\displaystyle \int _{a}^{b}|f(t)|\,dt\geq 0}
|
∫
a
b
f
(
t
)
d
t
|
=
0
≤
∫
a
b
|
f
(
t
)
|
d
t
{\displaystyle \left|\int _{a}^{b}f(t)\,dt\right|=0\leq \int _{a}^{b}|f(t)|\,dt}
The integral
β
=
∫
a
b
f
(
t
)
d
t
∈
C
{\displaystyle \beta =\int _{a}^{b}f(t)\,dt\in \mathbb {C} }
β
≠
0
{\displaystyle \beta \not =0}
|
β
|
=
β
⋅
β
¯
{\displaystyle |\beta |={\sqrt {\beta \cdot {\overline {\beta }}}}}
|
β
|
=
|
β
|
2
|
β
|
=
β
⋅
β
¯
|
β
|
=
β
¯
|
β
|
⏟
α
:=
⋅
β
=
α
⋅
β
{\displaystyle |\beta |={\frac {|\beta |^{2}}{|\beta |}}={\frac {\beta \cdot {\overline {\beta }}}{|\beta |}}=\underbrace {\frac {\overline {\beta }}{|\beta |}} _{\alpha :=}\cdot \beta =\alpha \cdot \beta }
edit
Since
β
≠
0
{\displaystyle \beta \not =0}
|
β
|
=
α
⋅
β
=
α
⋅
∫
a
b
f
(
t
)
d
t
=
∫
a
b
α
⋅
f
(
t
)
d
t
{\displaystyle |\beta |=\alpha \cdot \beta =\alpha \cdot \int _{a}^{b}f(t)\,dt=\int _{a}^{b}\alpha \cdot f(t)\,dt}
edit
Let
g
:=
α
⋅
f
{\displaystyle g:=\alpha \cdot f}
g
:
[
a
,
b
]
→
C
{\displaystyle g:[a,b]\rightarrow \mathbb {C} }
g
1
:
[
a
,
b
]
→
R
{\displaystyle g_{1}:[a,b]\rightarrow \mathbb {R} }
g
2
:
[
a
,
b
]
→
R
{\displaystyle g_{2}:[a,b]\rightarrow \mathbb {R} }
g
=
g
1
+
i
⋅
g
2
{\displaystyle g=g_{1}+i\cdot g_{2}}
R
e
(
∫
a
b
g
(
t
)
d
t
)
=
R
e
(
∫
a
b
g
1
(
t
)
d
t
⏟
∈
R
+
i
⋅
∫
a
b
g
2
(
t
)
d
t
⏟
∈
R
)
=
R
e
(
∫
a
b
g
1
(
t
)
d
t
⏟
∈
R
)
=
∫
a
b
g
1
(
t
)
d
t
=
∫
a
b
R
e
(
g
1
(
t
)
)
d
t
{\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}}}
edit
Since
|
β
|
=
∫
a
b
α
⋅
f
(
t
)
d
t
∈
R
{\displaystyle |\beta |=\int _{a}^{b}\alpha \cdot f(t)\,dt\in \mathbb {R} }
|
β
|
=
R
e
(
∫
a
b
α
⋅
f
(
t
)
d
t
⏟
∈
R
)
=
∫
a
b
R
e
(
α
⋅
f
(
t
)
)
⏟
∈
R
d
t
{\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}
edit
The following real part estimate against the absolute value of a complex number
z
{\displaystyle z}
R
e
(
z
)
=
z
1
≤
|
z
1
|
=
z
1
2
≤
z
1
2
+
z
2
2
=
|
z
|
{\displaystyle {\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
⋅
z
2
{\displaystyle z=z_{1}+i\cdot z_{2}}
R
e
(
α
⋅
f
(
t
)
)
⏟
∈
R
{\displaystyle \underbrace {{\mathfrak {Re}}\left(\alpha \cdot f(t)\right)} _{\in \mathbb {R} }}
edit
The following estimate is obtained analogously to Step 5 by the linearity of the integral
|
β
|
=
∫
a
b
R
e
(
α
⋅
f
(
t
)
)
d
t
≤
∫
a
b
|
α
⋅
f
(
t
)
|
d
t
=
|
α
|
⋅
∫
a
b
|
f
(
t
)
|
d
t
{\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}
edit
Since
|
α
|
=
|
β
¯
|
β
|
|
=
|
β
¯
|
|
β
|
=
1
{\displaystyle |\alpha |=\left|{\frac {\overline {\beta }}{|\beta |}}\right|={\frac {|{\overline {\beta }}|}{|\beta |}}=1}
|
∫
a
b
f
(
t
)
d
t
|
=
α
⋅
∫
a
b
|
f
(
t
)
|
d
t
≤
|
α
|
⋅
∫
a
b
|
f
(
t
)
|
d
t
=
∫
a
b
|
f
(
t
)
|
d
t
{\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}
Inequality - Length of Integration Path - LIP
edit
Let
γ
:
[
a
,
b
]
→
C
{\displaystyle \gamma :[a,b]\rightarrow \mathbb {C} }
f
:
U
→
C
{\displaystyle f:U\rightarrow \mathbb {C} }
γ
{\displaystyle \gamma }
S
p
u
r
(
γ
)
:=
{
γ
(
t
)
:
t
∈
[
a
,
b
]
}
⊂
U
{\displaystyle Spur(\gamma ):=\{\gamma (t)\,:\,t\in [a,b]\}\subset U}
|
∫
γ
f
(
z
)
d
z
|
≤
max
z
∈
S
p
u
r
(
γ
)
|
f
(
z
)
|
⋅
L
(
γ
)
{\displaystyle \left|\int _{\gamma }f(z)\,dz\right|\leq \max _{z\in Spur(\gamma )}|f(z)|\cdot {\mathcal {L}}(\gamma )}
where
L
(
γ
)
=
∫
a
b
|
γ
′
(
t
)
|
d
t
{\displaystyle {\mathcal {L}}(\gamma )=\int _{a}^{b}|\gamma '(t)|\,dt}
By using the above estimate for the absolute value of the integrand
|
f
(
z
)
d
z
|
≤
max
z
∈
S
p
u
r
(
γ
)
|
f
(
z
)
|
{\displaystyle \left|f(z)\,dz\right|\leq \max _{z\in Spur(\gamma )}|f(z)|}
|
∫
γ
f
(
z
)
d
z
|
≤
A
V
I
∫
a
b
|
f
(
γ
(
t
)
)
⋅
γ
′
(
t
)
|
d
t
=
∫
a
b
|
f
(
γ
(
t
)
)
|
⋅
|
γ
′
(
t
)
|
d
t
≤
∫
a
b
max
z
∈
S
p
u
r
(
γ
)
|
f
(
z
)
|
⏟
M
:=
⋅
|
γ
′
(
t
)
|
d
z
=
M
⋅
∫
γ
|
γ
′
(
t
)
|
d
z
=
M
⋅
L
(
γ
)
{\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 Spur(\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
edit
Let
γ
:
[
a
,
b
]
→
C
{\displaystyle \gamma :[a,b]\rightarrow \mathbb {C} }
Integration path and
f
:
U
→
C
{\displaystyle f:U\rightarrow \mathbb {C} }
γ
{\displaystyle \gamma }
T
r
a
c
e
(
γ
)
:=
γ
(
t
)
,
:
,
t
∈
[
a
,
b
]
⊂
U
{\displaystyle Trace(\gamma ):={\gamma (t),:,t\in [a,b]}\subset U}
|
∫
γ
f
(
z
)
,
d
z
|
≤
max
z
∈
T
r
a
c
e
(
γ
)
|
f
(
z
)
|
⋅
L
(
γ
)
{\displaystyle \left|\int _{\gamma }f(z),dz\right|\leq \max _{z\in Trace(\gamma )}|f(z)|\cdot L(\gamma )}
Here,
L
(
γ
)
=
∫
a
b
|
γ
′
(
t
)
|
,
d
t
{\displaystyle L(\gamma )=\int _{a}^{b}|\gamma '(t)|,dt}
![{\displaystyle f:[a,b]\rightarrow \mathbb {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/808e07bf306bb74c974003e5e32cd5dc306fd093)
![{\displaystyle f_{1}:[a,b]\rightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/df0b18b6ad6edaf3ecd6b3b7535c236f05a1b849)
![{\displaystyle f_{2}:[a,b]\rightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/195234869081ea96612e5c856d95934b8d7454e6)
![{\displaystyle g:[a,b]\rightarrow \mathbb {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d83fdf44c36d1d1b29c40fcb8099b2ef2958be3e)
![{\displaystyle g_{1}:[a,b]\rightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd83219d7e4faefa8fce23877bcc8f05bc29d902)
![{\displaystyle g_{2}:[a,b]\rightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f7ae99967030ebd812fe113ed68a87b801fed23)
![{\displaystyle \gamma :[a,b]\rightarrow \mathbb {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/90453f67f4a2b31d84e5f1a2a705a23d2049d4f4)
![{\displaystyle Spur(\gamma ):=\{\gamma (t)\,:\,t\in [a,b]\}\subset U}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6200c6cf0c599b1deeac094332c9d53622a2fd14)
![{\displaystyle Trace(\gamma ):={\gamma (t),:,t\in [a,b]}\subset U}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9fb0d2ab0a79749f8388ec41930c15b75b71c6b)
