nyuuzyou/wikis huggingface dataset:
Residue at Multiple Pole

== Theorem ==
Let $f: \C \to \C$ be a function meromorphic on some region, $D$, containing $a$. 
Let $f$ have a single pole in $D$, of order $N$, at $a$.
Then the residue of $f$ at $a$ is given by:
:$\ds \Res f a = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac {\d^{N - 1} } {\d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$
== Proof == 
By Existence of Laurent Series, there exists a Laurent series: 
:$\ds \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$ 
convergent on $D \setminus \set a$. 
As $f$ has a pole of order $N$ at $a$, we have $c_n = 0$ for $n < -N$. 
So: 
:$\ds \paren {z - a}^N \map f z = \sum_{n \mathop = -N}^\infty c_n \paren {z - a}^{n + N}$ 
Which can be rewritten: 
:$\ds \paren {z - a}^N \map f z = \sum_{n \mathop = 0}^\infty c_{n - N} \paren {z - a}^n$
Note that this is a Taylor series with centre $a$. 
By the definition of a residue: 
:$\ds \Res f a = c_{-1}$
This corresponds to the $\paren {N - 1}$th in the Taylor series of $\paren {z - a}^N \map f z$ about $a$. 
We therefore have by Taylor Series of Holomorphic Function: 
:$\ds c_{-1} = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac {\d^{N - 1} } {\d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$
Hence the result. 
Category:Complex Analysis
Residue at Simple Pole

== Theorem ==
Let $f: \C \to \C$ be a function meromorphic on some region, $D$, containing $a$. 
Let $f$ have a simple pole at $a$. 
Then the residue of $f$ at $a$ is given by:
:$\ds \Res f a = \lim_{z \mathop \to a} \paren {z - a} \map f z$
== Proof ==
By Existence of Laurent Series, there exists a Laurent series:
:$\ds \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$
which is convergent in $D \setminus \set a$, where $\sequence {c_n}$ is a doubly infinite sequence of complex coefficients. 
We are given that $f$ has only a simple pole at $a$.
Thus $c_n = 0$ for $n < -1$.
So we can write: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n + \frac {c_{-1} } {z - a}$
Then:
Category:Complex Analysis
Residue of Fibonacci Number Modulo Fibonacci Number

== Theorem ==
Let $F_n$ denote the $n$th Fibonacci number.
Let $m, r$ be non-negative integers.
Then:
:$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\
\paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\
\paren {-1}^n F_r & : m \bmod 4 = 2 \\
\paren {-1}^{r + 1 + n} F_{n - r} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
== Proof ==
=== Lemma ===
We prove the result for all $r$ by induction on $r$:
For all $r \in \N$, let $\map P r$ be the proposition:
:$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\
\paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\
\paren {-1}^n F_r & : m \bmod 4 = 2 \\
\paren {-1}^{r + 1 + n} F_{n - r} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
This fact follows from Divisibility of Fibonacci Number.
$\map P 1$ is the case:
:$F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\
F_{n - 1} & : m \bmod 4 = 1 \\
\paren {-1}^n F_1 & : m \bmod 4 = 2 \\
\paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
This is our lemma.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P {k - 1}$ and $\map P {k - 2}$ are true, where $k \ge 2$, then it logically follows that $\map P k$ is true.
So this is our induction hypotheses:
:$F_{m n + k - 1} \equiv \paren {\begin{cases} F_{k - 1} & : m \bmod 4 = 0 \\
\paren {-1}^k F_{n - k + 1} & : m \bmod 4 = 1 \\
\paren {-1}^n F_{k - 1} & : m \bmod 4 = 2 \\
\paren {-1}^{k + n} F_{n - k + 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ 
:$F_{m n + k - 2} \equiv \paren {\begin{cases} F_{k - 2} & : m \bmod 4 = 0 \\
\paren {-1}^{k + 1} F_{n - k + 2} & : m \bmod 4 = 1 \\
\paren {-1}^n F_{k - 2} & : m \bmod 4 = 2 \\
\paren {-1}^{k + 1 + n} F_{n - k + 2} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ 
Then we need to show:
:$F_{m n + k} \equiv \paren {\begin{cases} F_k & : m \bmod 4 = 0 \\
\paren {-1}^{k + 1} F_{n - k} & : m \bmod 4 = 1 \\
\paren {-1}^n F_k & : m \bmod 4 = 2 \\
\paren {-1}^{k + 1 + n} F_{n - k} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ 
=== Induction Step ===
This is our induction step:
So $\map P {k - 2} \land \map P {k - 1} \implies \map P k$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\
\paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\
\paren {-1}^n F_r & : m \bmod 4 = 2 \\
\paren {-1}^{r + 1 + n} F_{n - r} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $32$
Category:Fibonacci Numbers
Category:Proofs by Induction
Category:Residue of Fibonacci Number Modulo Fibonacci Number
Residue of Fibonacci Number Modulo Fibonacci Number/Lemma

== Theorem ==
Let $F_n$ denote the $n$th Fibonacci number.
Let $m$ be a non-negative integer.
Then:
:$F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\
F_{n - 1} & : m \bmod 4 = 1 \\
\paren {-1}^n F_1 & : m \bmod 4 = 2 \\
\paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
== Proof ==
We prove this by induction on $m$.
For all $m \in \N$, let $\map P m$ be the proposition:
:$F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\
F_{n - 1} & : m \bmod 4 = 1 \\
\paren {-1}^n F_1 & : m \bmod 4 = 2 \\
\paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$F_1 \equiv F_1 \pmod {F_n}$
So $\map P 0$ is true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P m$ is true then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
:$F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\
F_{n - 1} & : m \bmod 4 = 1 \\
\paren {-1}^n F_1 & : m \bmod 4 = 2 \\
\paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
Then we need to show:
=== Induction Step ===
This is our induction step:
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Category:Proofs by Induction
Category:Residue of Fibonacci Number Modulo Fibonacci Number
Residue of Gamma Function

== Theorem == 
Let $\Gamma$ be the Definition:Gamma Function. 
Let $n$ be a non-negative integer.
Then: 
:$\Res \Gamma {-n} = \dfrac {\paren {-1}^n} {n!}$
== Proof == 
By Poles of Gamma Function, $\Gamma$ has simple poles at the non-positive integers, so $-n$ is a simple pole of $\Gamma$. 
Then:
Category:Gamma Function
Residue of Quotient

== Theorem == 
Let $f$ and $g$ be functions holomorphic on some region containing $a$. 
Let $g$ have a zero of multiplicity $1$ at $a$. 
Then:
:$\Res {\dfrac f g} a = \dfrac {\map f a} {\map {g'} a}$
== Proof == 
As $g$ has a zero of multiplicity $1$ at $a$, $\dfrac f g$ has a simple pole at $a$ by definition.
So: 
 
Category:Complex Analysis
Resolvent Mapping Converges to 0 at Infinity

== Theorem ==
Let $B$ be a Banach space.
Let $\map \LL {B, B}$ be the set of bounded linear operators from $B$ to itself.
Let $T \in \map \LL {B, B}$.
Let $\map \rho T$ be the resolvent set of $T$ in the complex plane.
Then the resolvent mapping $f : \map \rho T \to \map \LL {B, B}$ given by $\map f z = \paren {T - z I}^{-1}$ is such that $\ds \lim_{z \mathop \to \infty} \norm {\map f z}_* = 0$.
== Proof ==
Pick $z \in \Bbb C$ such that $\size z > 2 \norm T_*$.
By Operator Norm is Norm:
:$\norm {\dfrac T z}_* = \dfrac {\norm T_*} {\size z} < \dfrac 1 2$
Hence:
Taking limits of both sides as $\size z \to \infty$:
:$\norm {\map f z}_* \to 0$
Category:Functional Analysis
Resolvent Mapping is Analytic/Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$. 
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Let $x \in A$.
Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.
Define $R : \map {\rho_A} x \to A$ by:
:$\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$
Then $R$ is analytic with derivative:
:$\map {R'} \lambda = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$
== Proof ==
Let $\lambda, \mu \in \map {\rho_A} x$ be such that $\lambda \ne \mu$.
Then, we have:
From Resolvent Mapping is Continuous: Banach Algebra, we have:
:$\paren {\mu {\mathbf 1}_A - x}^{-1} \to \paren {\lambda {\mathbf 1}_A - x}^{-1}$ as $\mu \to \lambda$ in $\map {\rho_A} x$. 
So, from Product Rule for Sequence in Normed Algebra, we have:
:$-\paren {\mu {\mathbf 1}_A - x}^{-1} \paren {\lambda {\mathbf 1}_A - x}^{-1} \to -\paren {\lambda {\mathbf 1}_A - x}^{-2}$ as $\mu \to \lambda$.
So, we have that:
:$\ds \lim_{\mu \mathop \to \lambda} \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} - \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda}$ exists
and:
:$\ds \lim_{\mu \mathop \to \lambda} \frac {\map R \mu - \map R \lambda} {\mu - \lambda} = \lim_{\mu \mathop \to \lambda} \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} - \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda} = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$
So, we obtain:
:$\map {R'} \lambda = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$
Category:Resolvent Mapping is Analytic
Category:Banach Algebras
Resolvent Mapping is Analytic/Bounded Linear Operator

== Theorem ==
Let $B$ be a Banach space.
Let $\map \LL {B, B}$ be the set of bounded linear operators from $B$ to itself.
Let $T \in \map \LL {B, B}$.
Let $\map \rho T$ be the resolvent set of $T$ in the complex plane.
Then the resolvent mapping $f : \map \rho T \to \map \LL {B, B}$ given by $\map f z = \paren {T - z I}^{-1}$ is analytic, and:
:$\map {f'} z = \paren {T - z I}^{-2}$
where $f'$ denotes the derivative of $f$ $z$.
== Proof 1 ==
== Proof 2 ==
Category:Resolvent Mapping is Analytic
Category:Functional Analysis
Resolvent Mapping is Analytic/Bounded Linear Operator/Proof 1

== Theorem ==
== Proof ==
For $a \in \map \rho T$, define:
:$R_a = \paren {T - a I}^{-1}$
Then we have:
From Resolvent Mapping is Continuous we have:
:$R_{z + h} \to R_z$ as $h \to 0$
Taking limits of both sides and using Norm is Continuous, we get:
:$\ds \lim_{h \mathop \to 0} \dfrac {\norm {\map f {z + h} - \map f z - \paren {T - z I}^{-2} h }_*} {\size h} = \norm {R_z^2 - R_z^2}_* = 0$
which is the result.
Category:Resolvent Mapping is Analytic
Category:Functional Analysis
Resolvent Mapping is Analytic/Bounded Linear Operator/Proof 2

== Theorem ==
== Proof ==
If $B=\set {\mathbf 0_B}$, the statement is trivial, since $\map f z = \mathbf 0_{\map \LL {B, B}}$ for all $z \in \C$.
We assume that $B \ne \set {\mathbf 0_B}$.
Especially, for all $z \in \map \rho T$:
:$\map f z \ne \mathbf 0_{\map \LL {B, B}}$
since:
:$\paren {T - z I} \map f z = I \ne \mathbf 0_{\map \LL {B, B}}$
Let $a \in \map \rho T$.
Then for each $z \in \C$:
Recall that $\map B {c, r}$ denotes the open disc of center $c \in \C$ and radius $r>0$.
For all $z \in \map B {a, \frac{1}{\norm {\map f a} } }$, we have:
Therefore:
That is, in a neighborhood of each $a \in \map \rho T$, $f$ can be written as:
:$\ds \map f z = \sum_{n \mathop \ge 0} \paren {\map f a}^{n+1} \paren {z - a}^n$
This means that $f$ is analytic on $\map \rho T$.
Finally, by Derivative of Power Series, in the neighborhood of each $a \in \map \rho T$:
:$\ds \map {f'} z = \sum_{n \mathop \ge 1} \paren {\map f a}^{n+1} n \paren {z - a}^{n-1}$
Choosing $z = a$, we obtain:
:$\map {f'} a = \paren {\map f a}^2 = \paren {T - aI}^{-2}$
Category:Resolvent Mapping is Analytic
Category:Functional Analysis
Resolvent Mapping is Continuous/Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$. 
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Let $x \in A$.
Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.
Define $R : \map {\rho_A} x \to A$ by:
:$\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$
Then $R$ is continuous. 
== Proof ==
=== Lemma ===
From the Lemma, we have:
:the mapping $S : \C \to A$ defined by:
::$\map S \lambda = \lambda {\mathbf 1}_A - x$
:for each $\lambda \in \C$, is continuous.
From Restriction of Continuous Mapping is Continuous, $S \restriction_{\map {\rho_A} x}$ is continuous.
From the definition of the resolvent set, we have $S \sqbrk {\map {\rho_A} x} \subseteq \map G A$.
From Inverse Mapping on Group of Units in Unital Banach Algebra is Continuous, the mapping $\phi : \map G A \to \map G A$ defined by:
:$\map \phi x = x^{-1}$ for each $x \in \map G A$
is continuous.
From Composite of Continuous Mappings is Continuous, $\phi \circ \paren {S \restriction_{\map {\rho_A} x} } : \map {\rho_A} x \to A$ is continuous.
Since $\phi \circ S = R$, we have that $R$ is continuous.
Category:Spectral Theory of Banach Algebras
Category:Resolvent Mapping is Continuous
Resolvent Mapping is Continuous/Banach Algebra/Lemma

== Lemma ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$. 
Let $x \in A$.
Define $S : \C \to A$ by:
:$\map S \lambda = \lambda {\mathbf 1}_A - x$
Then $S$ is continuous.
== Proof ==
We have, for $\lambda, \mu \in \C$:
Let $\epsilon > 0$ and $\lambda, \mu \in \C$ be such that:
:$\cmod {\lambda - \mu} < \epsilon$
Then, we have:
:$\norm {\map S \lambda - \map S \mu} < \epsilon$
Hence $S$ is continuous.
Category:Resolvent Mapping is Continuous
Resolvent Mapping is Continuous/Bounded Linear Operator

== Theorem ==
Let $B$ be a Banach space.
Let $\mathfrak{L}(B, B)$ be the set of bounded linear operators from $B$ to itself.
Let $T \in \mathfrak{L}(B, B)$.
Let $\rho(T)$ be the resolvent set of $T$ in the complex plane.
Then the resolvent mapping $f : \rho(T) \to \mathfrak{L}(B,B)$ given by:
:$f(z) = (T - zI)^{-1}$
is continuous in the operator norm $\|\cdot\|_*$.
== Proof ==
Pick $z\in\rho(T)$. Since $z\in\rho(T)$, the operator $R_z = (T - zI)^{-1}$ exists and has finite norm $C \geq 0$.
Since Resolvent Set is Open, $z+h\in\rho(T)$ for any $h\in \Bbb C$ smaller than some $\delta > 0$. For such $h$,
Consider:
:$(I - hR_z)^{-1}$
Either:
:$\|R_z\|_* = 0$ 
so that:
:$\forall h \in \Bbb C: \|hR_z\|_* = |h|\|R_z\|_* = 0$ (by Operator Norm is Norm)
or else:
:$\|hR_z\|_* < 1$ for $h < 1/\|R_z\|_*$
In either case:
:$\|hR_z\|_* < 1$ 
for all sufficiently small $h$.
Therefore, by Invertibility of Identity Minus Operator, for sufficiently small $h$ we have $(I - hR_z)^{-1} = I + hR_z + h^2R_z^2 + \ldots$. Substituting this into $(1)$, we get
as long as:
:$|h| \leq \frac{1}{2\norm{R_z}}$
The expression on the right-hand side goes to zero as $h$ gets small. Therefore, taking limits, we get
$$
\lim_{h\to0} \norm{ f(z+h) - f(z) }_* = 0.
$$
This establishes continuity of $f$ at arbitrary $z\in\rho(T)$.
Category:Bounded Linear Operators
Category:Resolvent Mapping is Continuous
Resolvent Set of Bounded Linear Operator equal to Resolvent Set as Densely-Defined Linear Operator

== Theorem ==
Let $\HH$ be a Hilbert space over $\C$. 
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\map {\rho_1} T$ be the resolvent set of $T$ as a bounded linear operator.
Let $\map {\rho_2} T$ be the resolvent set of $T$ as a densely-defined linear operator $\struct {\HH, T}$.
Then: 
:$\map {\rho_1} T = \map {\rho_2} T$
 
== Proof ==
Let $\lambda \in \map {\rho_1} T$.
Then $T - \lambda I$ is invertible in the sense of a bounded linear transformation.
That is, $T - \lambda I$ is bijective and $\paren {T - \lambda I}^{-1}$ is bounded.
From Underlying Set of Topological Space is Everywhere Dense, we have that $\HH$ is everywhere dense in $\HH$. 
So, $T - \lambda I$ is injective, $\map {\paren {T - \lambda I} } {\map D T}$ is everywhere dense in $\HH$, and $\paren {T - \lambda I}^{-1}$ is bounded.
So $\lambda \in \map {\rho_2} T$.
We therefore have: 
:$\map {\rho_1} T \subseteq \map {\rho_2} T$
by the definition of set inclusion.
Now let $\lambda \in \map {\rho_2} T$. 
Then:
:$T - \lambda I$ is injective, $\map {\paren {T - \lambda I} } {\map D T}$ is everywhere dense in $\HH$, and $\paren {T - \lambda I}^{-1}$ is bounded.
To show that $\lambda \in \map {\rho_1} T$, we just need to show that: 
:$\map {\paren {T - \lambda I} } {\map D T} = \HH$.
Let $y \in \HH$.
For brevity, let $S_\lambda = \paren {T - \lambda I}^{-1}$. 
From Point in Closure of Subset of Metric Space iff Limit of Sequence, there exists a sequence in $\map {\paren {T - \lambda I} } {\map D T}$ with: 
:$y_n \to y$
Define:
:$x_n = \map {S_\lambda^{-1} } {y_n}$
for each $n \in \N$.
We show that $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Since $\HH$ is a Hilbert space, we will then have that $\sequence {x_n}_{n \mathop \in \N}$ converges.
We have: 
From Convergent Sequence in Normed Vector Space is Cauchy Sequence, we have: 
:$\sequence {y_n}_{n \mathop \in \N}$ is Cauchy.
So, for each $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\ds \norm {y_n - y_m} < \frac \epsilon {\norm {S_\lambda^{-1} } }$ for all $n \ge N$.
Then: 
:$\norm {x_n - x_m} < \epsilon$ for all $n \ge N$. 
So $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy, and so converges to a limit $x$.
We then have: 
That is: 
:$y \in \map {\paren {T - \lambda I} } {\map D T}$
Since $y \in \HH$ was arbitrary, we have: 
:$\HH \subseteq \map {\paren {T - \lambda I} } {\map D T}$
from the definition of set inclusion. 
We also have: 
:$\map {\paren {T - \lambda I} } {\map D T} \subseteq \HH$
by definition and so:
:$\map {\paren {T - \lambda I} } {\map D T} = \HH$
as required.
So $\lambda \in \map {\rho_1} T$.
We therefore have: 
:$\map {\rho_2} T \subseteq \map {\rho_1} T$
and can therefore conclude:
:$\map {\rho_1} T = \map {\rho_2} T$
== Sources ==
* $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators
Category:Resolvent Sets (Bounded Linear Operators)
Category:Resolvent Sets (Densely-Defined Linear Operators)
Resolvent Set of Bounded Linear Operator is Open

== Theorem ==
Let $X$ be a Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator.
Let $\map \rho T$ be the resolvent set of $T$. 
Then $\map \rho T$ is open. 
== Proof ==
Let $\lambda \in \map \rho T$. 
Then $T - \lambda I$ is invertible as a bounded linear operator.
Let $\delta > 0$ be such that: 
:$\cmod \delta < \norm {\paren {T - \lambda I}^{-1} }_{\map \BB X}^{-1}$ 
Then, we have: 
:$\norm {\delta I}_{\map \BB X} \norm {\paren {T - \lambda I}^{-1} } < 1$ 
From Invertibility of Identity Minus Operator: Corollary, we have: 
:$T - \paren {\lambda + \delta} I$ is invertible as a bounded linear operator.
That is, if: 
:$\ds \cmod {\mu - \lambda} < \norm {\paren {T - \lambda I}^{-1} }_{\map \BB X}^{-1}$
we have: 
:$T - \mu I$ is invertible as a bounded linear operator
so that $\mu \in \map \rho T$. 
So $\map \rho T$ contains an open neighborhood of each of its points. 
So $\map \rho T$ is open.
== Sources ==
* $14.1$: The Resolvent and Spectrum
Category:Resolvent Sets (Bounded Linear Operators)
Resolvent Set of Element in Banach Algebra has same Unique Unbounded Component as Resolvent Set of Element in Closed Subalgebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.
Let $B$ be a closed unital subalgebra of $A$. 
Let $x \in B$.
Let $\map {\rho_A} x$ and $\map {\rho_B} x$ be the resolvent sets of $x$ in $A$ and $B$ respectively.
Then $\map {\rho_A} x$ and $\map {\rho_B} x$ both have unique unbounded component, and further these components coincide. 
== Proof ==
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively. 
From Spectrum of Element of Banach Algebra is Bounded, we have:
:$\map {\sigma_A} x$ and $\map {\sigma_B} x$ are bounded.
Hence, from Complement of Bounded Set in Complex Plane has at most One Unbounded Component:
:$\map {\rho_A} x$ and $\map {\rho_B} x$ both have a unique unbounded component.
Let $U$ be the unbounded component of $\map {\rho_A} x$.
From Complement of Closed Disk in Complex Plane is Path-Connected, we have that:
:$\set {z \in \C : \cmod z > \norm x}$ is path-connected.
From Path-Connected Space is Connected:
:$\set {z \in \C : \cmod z > \norm x}$ is connected.
From Spectrum of Element of Banach Algebra is Bounded, we have:
:$\map {\rho_A} x \supseteq \set {z \in \C : \cmod z > \norm x}$
and:
:$\map {\rho_B} x \supseteq \set {z \in \C : \cmod z > \norm x}$
and hence:
:$U \cap \map {\rho_B} x \supseteq \set {z \in \C : \cmod z > \norm x} \ne \O$
From Component of Resolvent Set of Element in Unital Banach Algebra is Disjoint from or Component of Resolvent Set in Closed Subalgebra, we therefore have:
:$U$ is a component of $\map {\rho_B} x$.
Since $U$ is unbounded, we conclude that $U$ is also the unique unbounded component of $\map {\rho_B} x$. 
Category:Resolvent Sets
Category:Connected Spaces
Category:Unital Banach Algebras
Resolvent Set of Element in Closed Unital Subalgebra of Banach Algebra is Clopen in Resolvent Set

== Theorem ==
Let $A$ be a Banach algebra.
Let $B$ be a closed unital subalgebra of $A$.
Let $x \in B$.
Let $\map {\rho_A} x$ and $\map {\rho_B} x$ be the resolvent sets of $x$ in $A$ and $B$ respectively. 
Then $\map {\rho_B} x$ is clopen in $\map {\rho_A} x$. 
== Proof ==
Let $\map G A$ and $\map G B$ be the group of units of $A$ and $B$ respectively. 
First, from Resolvent Set of Element in Subalgebra we have:
:$\map {\rho_B} x \subseteq \map {\rho_A} x$ 
Define $R : \map {\rho_A} x \to A$ by: 
:$\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$ 
for each $\lambda \in \map {\rho_A} x$. 
From Resolvent Mapping is Continuous: Banach Algebra, $R$ is continuous.
Hence from the definition of a continuous mapping and since $B$ is closed, we have that:
:$R^{-1} \sqbrk B$ is closed in $\map {\rho_A} x$.
Note that $\lambda {\mathbf 1}_A - x \in \map G B$ $\lambda {\mathbf 1}_A - x \in \map G A$ and $\paren {\lambda {\mathbf 1}_A - x}^{-1} \in B$.
That is, $R^{-1} \sqbrk B = \map {\rho_B} x$.
So $\map {\rho_B} x$ is closed in $\map {\rho_A} x$.
From Resolvent Set of Element of Banach Algebra is Open, $\map {\rho_B} x$ is open in $A$.
Hence $\map {\rho_B} x$ is open in $\map {\rho_A} x$.
We conclude that $\map {\rho_B} x$ is clopen in $\map {\rho_A} x$.
Category:Resolvent Sets
Resolvent Set of Element in Subalgebra

== Theorem ==
Let $A$ be a Banach algebra.
Let $B$ be a closed unital subalgebra of $A$.
Let $x \in B$.
Let $\map {\rho_A} x$ and $\map {\rho_B} x$ be the resolvent sets of $x$ in $A$ and $B$ respectively. 
Then:
:$\map {\rho_B} x \subseteq \map {\rho_A} x$
== Proof ==
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively. 
From Spectrum of Element in Unital Subalgebra, we have:
:$\map {\sigma_A} x \subseteq \map {\sigma_B} x$
From Set Complement inverts Subsets, we have:
:$\C \setminus \map {\sigma_B} x \subseteq \C \setminus \map {\sigma_A} x$
From the definition of the resolvent set, we have:
:$\map {\rho_B} x \subseteq \map {\rho_A} x$
Category:Resolvent Sets
Resolvent Set of Element of Banach Algebra is Open

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x \in A$.
Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.
Then $\map {\rho_A} x$ is open.
== Proof ==
 suppose that $A$ is unital, swapping $A$ for its unitization if necessary.
Let $\map G A$ be the group of units of $A$. 
Define $S : \C \to A$ by:
:$\map S \lambda = \lambda {\mathbf 1}_A - x$
From Resolvent Mapping is Continuous: Continuous, $S$ is continuous.
From Group of Units in Unital Banach Algebra is Open, $\map G A$ is open. 
From the definition of the resolvent set, we have:
:$\map {\rho_A} x = \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in \map G A}$
That is:
:$\map {\rho_A} x = S^{-1} \sqbrk {\map G A}$ 
Since $S$ is continuous, and $\map G A$ is open, $S^{-1} \sqbrk {\map G A}$ is open.
We conclude that $\map {\rho_A} x$ is open.
Category:Resolvent Sets
Category:Spectral Theory of Banach Algebras