avewright/proofwiki-math huggingface dataset:
Theorem: Minkowski Functional of Open Ball with respect to Seminorm is Seminorm

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let: 
:$B = \set {x \in X : \map p x < 1}$ 
be the open ball centred at ${\mathbf 0}_X$ with radius $1$ in the pseudometric induced by $p$. 
Let $\mu_B$ the Minkowski functional of $B$. 
Then $p = \mu_B$.

Proof:
From Open Ball with respect to Seminorm is Convex, Balanced and Absorbing, $B$ is convex and absorbing.
Hence the definition is valid.
Let $x \in X$.
From {{SeminormAxiom|2}}:
:$\forall s > \map p x: \map p {\dfrac x s} = \dfrac 1 s \map p x < 1$ 
Then we have: 
:$s \in \set {t > 0 : t^{-1} x \in B}$
so that: 
:$\map {\mu_B} x \le s$ 
Taking the infimum over $s > \map p x$, we have: 
:$\map {\mu_B} x \le \map p x$ 
Let $\map p x = 0$.
Then:
:$\map p x \le \map {\mu_B} x$ 
Let $\map p x \ne 0$. 
Let $0 < t \le \map p x$.
Then: 
:$\map p {\dfrac x t} = \dfrac 1 t \map p x \ge 1$
So $t^{-1} x \notin B$.
So:
:$\forall x \in X: \map {\mu_B} x \ge \map p x$
We therefore obtain: 
:$\forall x \in X: \map {\mu_B} x = \map p x$
{{qed}}
Theorem: Continuous Linear Operator over Finite Dimensional Vector Space is Invertible

Let $\struct {X, \norm {\, \cdot\,}_X}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Let $S, T \in \map {CL} X$.
Suppose the dimension of $X$ is finite:
:$d = \dim X < \infty$
Suppose $T \circ S = I$ where $\circ$ denotes the composition of mappings.
Then $T$ and $S$ are invertible.

Proof:
Let $x \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {T \circ S} } x 
      | r = \map I x
}}
{{eqn | ll = \leadsto
      | l = \map T {\map {S} x}
      | r = x
}}
{{end-eqn}}
Let $\mathbf 0 \in X$ be the zero vector of $X$.
Suppose $\map S x = \mathbf 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map T {\map S x}
      | r = \map T {\mathbf 0}
}}
{{eqn | r = \mathbf 0
      | c = Linear Transformation Maps Zero Vector to Zero Vector
}}
{{eqn | r = x
}}
{{end-eqn}}
Hence:
:$\paren {\map S x = \mathbf 0} \implies \paren{x = \mathbf 0} \paren \star$
In other words:
:$\map \ker S = \set {\mathbf 0}$
where $\ker$ denotes the kernel.
Let $\set {v_1, \ldots v_d}$ be the basis for $X$.
Suppose $\forall k \in \N_{> 0} : k \le d : \alpha_k \in \mathbb F$ are scalars such that:
:$\ds \sum_{k \mathop = 1}^d \alpha_k \map S {v_k} = \mathbf 0$
where $\mathbb F$ is a number field.
By definition of linear transformation on vector space:
:$\ds \map S {\sum_{k \mathop = 1}^d \alpha_k v_k} = \mathbf 0$
By Eq. $\paren \star$:
:$\ds \sum_{k \mathop = 1}^d \alpha_k v_k = \mathbf 0$
Since $\set {v_1, \ldots v_d}$ was arbitrary, we have that:
:$\forall k \in \N_{> 0} : k \le d : \alpha_k = 0$
By definition of linearly independent sequence of vectors, $\set {\map S {v_1}, \ldots \map S {v_d} }$ is a basis for $X$.
Therefore:
:$\ds \forall x \in X : \forall k \in \N_{> 0} : k \le d : \exists \beta_k \in \mathbb F : x = \sum_{k \mathop = 1}^d \beta_k \map S {v_k}$
Thus:
{{begin-eqn}}
{{eqn | q = \forall x \in X
      | l = \map S {\map T x}
      | r = \map S {\map T {\sum_{k \mathop = 1}^d \beta_k \map S {v_k} } }
}}
{{eqn | r = \map S {\map T {\map S {\sum_{k \mathop = 1}^d \beta_k v_k} } }
      | c = {{Defof|Linear Transformation on Vector Space}}
}}
{{eqn | r = \map S {\map I {\sum_{k \mathop = 1}^d \beta_k v_k} }
      | c = $\map {\paren {T \circ S} } x = \map I x$
}}
{{eqn | r = \map S {\sum_{k \mathop = 1}^d \beta_k v_k}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = x
}}
{{end-eqn}}
All in all:
:$\paren {T \circ S = I} \implies \paren {T \circ S = S \circ T = I}$
{{qed}}
Theorem: Equivalence of Definitions of Riemann and Darboux Integrals

Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a real function.
Then:
:the Riemann integral of $f$ over $\closedint a b$ exists and is equal to $L$
{{iff}}:
:the Darboux integral of $f$ over $\closedint a b$ exists and is equal to $L$.

Proof:
=== Riemann Integral $\implies$ Darboux Integral ===
Let $L$ be the Riemann integral of $f$ over $\closedint a b$.
Then $\forall \epsilon > 0: \exists \delta > 0$ such that for every finite subdivision with norm $< \delta$
and every sequence of corresponding sample points, the Riemann sum of the subdivision is in $\openint {L - \epsilon} {L + \epsilon}$.
{{AimForCont}} the lower Darboux integral of $f$ over $\closedint a b$ is less than $L$.
Let:
:$\ds \epsilon = L - \underline {\int_a^b} \map f x \rd x$
Because $\ds \underline {\int_a^b} \map f x \rd x < L$ it follows that $\epsilon > 0$.
Therefore, there is a $\delta$ that satisfies the conditions above.
Let $n = \ceiling {\dfrac {b - a} \delta}$.
Define:
:$x_i = a + \dfrac i n \paren {b - a}$
Then the sequence $\sequence {x_i}_{0 \mathop \le i \mathop \le n}$ is a finite subdivision of $\closedint a b$.
Additionally, for every $i$:
{{begin-eqn}}
{{eqn | l = x_i - x_{i - 1}
      | r = \paren {a + \frac i n \paren {b - a} } - \paren {a + \frac {i - 1} n \paren {b - a} }
}}
{{eqn | r = \frac {b - a} n
}}
{{eqn | o = <
      | r = \paren {b - a} \frac \delta {b - a}
}}
{{eqn | r = \delta
}}
{{end-eqn}}
So the norm of $\sequence {x_i} < \delta$.
Let $K$ be the lower Darboux sum of of $f$ over $\sequence {x_i}$.
The lower Darboux sum is a Riemann sum, so by the definition of $\delta$:
:$\ds K > L - \epsilon = \underline {\int_a^b} \map f x \rd x$
But by the definition of lower Darboux integral:
:$\ds K \le \underline {\int_a^b} \map f x \rd x$
These inequalities are contradictory, so our supposition that the lower Darboux integral is less than $L$ was false.
Therefore:
:$\ds \underline {\int_a^b} \map f x \rd x \ge L$
A similar argument shows that:
:$\ds \overline {\int_a^b} \map f x \rd x \le L$
But by Upper Darboux Integral Never Smaller than Lower Darboux Integral:
:$\ds \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x$
Thus, the following is true:
:$\ds L \le \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x \le L$
and we can conclude that:
:$\ds L = \underline {\int_a^b} \map f x \rd x = \overline {\int_a^b} \map f x \rd x$
{{qed|lemma}}
Theorem: Wave Equation is Linear

Let $\phi_1$ and $\phi_2$ be particular solutions to the wave equation.
Then:
:$a_1 \phi_1 + a_2 \phi_2$ is also a particular solution to the wave equation.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac {\partial^2} {\partial t^2} } {a_1 \phi_1 + a_2 \phi_2}
      | r = a_1 \map {\dfrac {\partial^2} {\partial t^2} } {\phi_1} + a_2 \map {\dfrac {\partial^2} {\partial t^2} } {\phi_2}
      | c = Linear Combination of Derivatives
}}
{{eqn | l = \map {\dfrac {\partial^2} {\partial x^2} } {a_1 \phi_1 + a_2 \phi_2}
      | r = a_1 \map {\dfrac {\partial^2} {\partial x^2} } {\phi_1} + a_2 \map {\dfrac {\partial^2} {\partial x^2} } {\phi_2}
      | c = Linear Combination of Derivatives
}}
{{eqn | l = \map {\dfrac {\partial^2} {\partial y^2} } {a_1 \phi_1 + a_2 \phi_2}
      | r = a_1 \map {\dfrac {\partial^2} {\partial y^2} } {\phi_1} + a_2 \map {\dfrac {\partial^2} {\partial x^2} } {\phi_2}
      | c = Linear Combination of Derivatives
}}
{{eqn | l = \map {\dfrac {\partial^2} {\partial z^2} } {a_1 \phi_1 + a_2 \phi_2}
      | r = a_1 \map {\dfrac {\partial^2} {\partial z^2} } {\phi_1} + a_2 \map {\dfrac {\partial^2} {\partial x^2} } {\phi_2}
      | c = Linear Combination of Derivatives
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map {\dfrac {\partial^2} {\partial t^2} } {a_1 \phi_1 + a_2 \phi_2}
      | r = a_1 \map {\dfrac {\partial^2} {\partial t^2} } {\phi_1} + a_2 \map {\dfrac {\partial^2} {\partial t^2} } {\phi_2}
      | c = from above
}}
{{eqn | r = a_1 c^2 \paren {\map {\dfrac {\partial^2} {\partial x^2} } {\phi_1} + \map {\dfrac {\partial^2} {\partial y^2} } {\phi_1} + \map {\dfrac {\partial^2} {\partial z^2} } {\phi_1} } + a_2 c^2 \paren {\map {\dfrac {\partial^2} {\partial x^2} } {\phi_2} + \map {\dfrac {\partial^2} {\partial y^2} } {\phi_2} + \map {\dfrac {\partial^2} {\partial z^2} } {\phi_2} }
      | c = as $\phi_1$ and $\phi_2$ satisfy the wave equation
}}
{{eqn | r = c^2 \paren {\map {\dfrac {\partial^2} {\partial x^2} } {a_1 \phi_1 + a_2 \phi_2} + \map {\dfrac {\partial^2} {\partial y^2} } {a_1 \phi_1 + a_2 \phi_2} + \map {\dfrac {\partial^2} {\partial z^2} } {a_1 \phi_1 + a_2 \phi_2} }
      | c = from above
}}
{{end-eqn}}
and it is seen that $a_1 \phi_1 + a_2 \phi_2$ also satisfies the wave equation.
{{qed}}
Theorem: Well-Defined Jordan Content Equals Content

Let $M$ be a bounded subspace of Euclidean space.
Let the Jordan content of $M$ be $\map m M$.
Then the content $\map V M = \map m M$.

Proof:
Let $C$ be a finite covering of $M$.
By {{EuclidCommonNotionLink|5}}, $\map V C \ge \map V M$.
Therefore, $\map V M$ is a lower bound of all $\map V C$.
So by the definition of greatest lower bound:
:$\map V M \le \map {m^*} M$
Let $D$ be a finite covering of $S \setminus M$.
By the same reasoning:
:$\map V {S \setminus M} \le \map {m^*} {S \setminus M}$
But:
{{begin-eqn}}
{{eqn | l = \map V M + \map V {S \setminus M}
      | r = \map V S
}}
{{eqn | l = \map V M + \map {m^*} {S \setminus M}
      | o = \ge
      | r = \map V S
}}
{{eqn | l = \map V M
      | o = \ge
      | r = \map V S - \map {m^*} {S \setminus M}
}}
{{end-eqn}}
Therefore:
:$\map V S - \map {m^*} {S \setminus M} \le \map V M \le \map {m^*} M$
But by hypothesis:
:$\map V S - \map {m^*} {S \setminus M} = \map {m^*} M = \map m M$
So it follows that:
:$\map V M = \map m M$
{{qed}}
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Theorem: Outer Jordan Content Never Smaller than Inner Jordan Content

Let $M$ be a bounded subspace of Euclidean space.
Then:
:$\map {m^*} M \ge \map {m_*} M$
where:
:$m^*$ denotes the outer Jordan content
:$m_*$ denotes the inner Jordan content

Proof:
Let $\RR$ be a closed rectangle that contains $M$.
By definition of inner Jordan content:
:$\map {m_*} M = \map V \RR - \map {m^*} {\RR \setminus M}$
But since:
:$\RR = M \cup \paren {\RR \setminus M}$
we have by Outer Jordan Content is Subadditive that:
:$\map {m^*} \RR \le \map {m^*} M + \map {m^*} {\RR \setminus M}$
Finally, by Outer Jordan Content of Closed Rectangle:
:$\map {m^*} \RR = \map V \RR$
and hence:
:$\map V \RR - \map {m^*} {\RR \setminus M} \le \map {m^*} M$
The result follows.
{{qed}}
Category:Outer Jordan Content
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Theorem: Sum of Equal and Opposite Harmonic Waves form Stationary Wave

Let $\phi_1$ be a harmonic wave which is propagated along the $x$-axis in the positive direction with constant velocity $c$.
Let $\phi_2$ be a harmonic wave travelling with constant velocity $-c$, that is, at the same speed as $\phi_1$ but in the opposite direction.
Then their sum $\phi_1 + \phi_2$ describes a stationary wave.

Proof:
From Equation of Harmonic Wave: Wave Number and Frequency:
:$(1): \quad \map {\phi_1} {x, t} = a \map \cos {2 \pi \paren {k x - \nu t} }$
where:
:$k$ denotes the wave number of $\phi_1$
:$\nu$ denotes the frequency of $\phi_1$.
From Equation of Wave with Constant Velocity: Corollary, the equation of $\phi_2$ is:
:$(2): \quad \map {\phi_2} {x, t} = a \map \cos {2 \pi \paren {k x + \nu t} }$
Then we have:
{{begin-eqn}}
{{eqn | l = \map {\phi_1} {x, t} + \map {\phi_2} {x, t}
      | r = a \map \cos {2 \pi \paren {k x - \nu t} } + a \map \cos {2 \pi \paren {k x + \nu t} }
      | c = from $(1)$ and $(2)$
}}
{{eqn | r = 2 a \map \cos {\dfrac {2 \pi \paren {k x - \nu t} + 2 \pi \paren {k x + \nu t} } 2} \map \cos {\dfrac {2 \pi \paren {k x - \nu t} - 2 \pi \paren {k x + \nu t} } 2}
      | c = Cosine plus Cosine
}}
{{eqn | r = 2 a \map \cos {2 \pi k x} \map \cos {2 \pi \paren {-\nu t} }
      | c = simplification
}}
{{eqn | r = 2 a \map \cos {2 \pi k x} \map \cos {2 \pi \nu t}
      | c = Cosine Function is Even
}}
{{end-eqn}}
{{finish|It remains to be shown that this is the equation of a stationary wave.}}
Theorem: Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube/Lemma 2

:the family of continuous mappings $\family{f_{U,V}}_{\tuple{U,V} \in \AA}$ separates points from closed sets

Proof:
Let $F$ be a closed subset of $T$.
Let $x \in S \setminus F$.
By definition of closed subset:
:$S \setminus F$ is open in $S$
Let $V = S \setminus F$.
By definition of regular space:
:$\exists U \in \tau: x \subseteq U, U^- \subseteq V$
Hence $\tuple{U, V} \in \AA$.
Consider the Urysohn function $f_{U, V}$ for $U^-$ and $S \setminus V$:
By definition of Urysohn function:
:$f_{U, V} \sqbrk {U^-} = \set 0, f_{U, V} \sqbrk {S \setminus V} = \set 1$ 
Hence:
:$\map {f_{U, V}} x = 0, f_{U, V} \sqbrk F = \set 1$
Since $\set 1^- = \set 1$, then:
:$\map {f_{U, V}} x = 0 \notin \set 1 = f_{U, V} \sqbrk F^-$
By definition, the family of continuous mappings $\family{f_{U,V}}_{\tuple{U,V} \in \AA}$ separates points from closed sets
{{qed}}
Category:Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube
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Theorem: Minkowski Functional of Convex Absorbing Set is Sublinear

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let $A$ be a convex absorbing set. 
Let $\mu_A$ be the Minkowski functional of $A$. 
Then for each $x, y \in X$ we have: 
:$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$

Proof:
Let $\epsilon > 0$. 
By the definition of infimum, we can pick $t > 0$ such that:
{{explain|domain of $t$ -- this is not completely obvious, as the underlying set of $X$ includes the case where it's $\C$.}}
:$\map {\mu_A} x \le t \le \map {\mu_A} x + \epsilon$
and $t^{-1} x \in A$.
We can also pick $s > 0$ such that: 
:$\map {\mu_A} y \le s \le \map {\mu_A} y + \epsilon$
and $s^{-1} y \in A$. 
We have: 
:$\ds \frac t {s + t} + \frac s {s + t} = 1$
So, since $A$ is convex, we have: 
:$\ds \frac t {s + t} \paren {\frac x t} + \frac s {s + t} \paren {\frac y s} = \frac {x + y} {s + t} \in A$ 
So, we have: 
:$\map {\mu_A} {x + y} \le s + t \le \map {\mu_A} x + \map {\mu_A} y + 2 \epsilon$
Since this inequality holds for all $\epsilon > 0$, we have: 
:$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$ 
{{qed}}
Theorem: Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\struct {\map D T, T}$ be a symmetric densely-defined linear operator.
Let $\struct {\map D {T^\ast}, T^\ast}$ be the adjoint of $T$.
Then $\map D T \subseteq \map D {T^\ast}$ and: 
:$T x = T^\ast x$ for each $x \in \map D T$.

Proof:
For each $y \in \HH$, define the linear functional $f_x : \map D T \to \Bbb F$ by:
:$\map {f_y} x = \innerprod {T x} y$ for each $x \in \map D T$.
We show that for $y \in \map D T$, $f_y$ is bounded.
Let $y \in \map D T$, then: 
:$\innerprod {T x} y = \innerprod x {T y}$ for each $x \in \map D T$.
Then we have: 
{{begin-eqn}}
{{eqn	| l = \cmod {\map {f_y} x}
		| r = \cmod {\innerprod x {T y} }
}}
{{eqn	| o = \le
		| r = \norm {T y} \norm x
		| c = Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
}}
{{end-eqn}}
So $f_y$ is bounded and $\map D T \subseteq \map D {T^\ast}$. 
Now, for $x, y \in \map D T$, we have: 
:$\innerprod {T x} y = \innerprod x {T y}$ 
since $\struct {\map D T, T}$ is symmetric and: 
:$\innerprod {T x} y = \innerprod x {T^\ast y}$ 
from the definition of the adjoint.
So we have: 
:$\innerprod x {T y - T^\ast y} = 0$
for each $x \in \map D T$. 
Taking a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ such that $x_n \to T y - T^\ast y$. 
Since: 
:$\innerprod {x_n} {T y - T^\ast y} = 0$
for each $n \in \N$, we have: 
:$\innerprod {T y - T^\ast y} {T y - T^\ast y} = \norm {T y - T^\ast y}^2 = 0$ 
from Inner Product is Continuous.
So $T y = T^\ast y$ for each $y \in \map D T$.
{{qed}}
Category:Symmetric Densely-Defined Linear Operators
Category:Adjoints (Densely-Defined Linear Operators)
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Theorem: Spectrum of Bounded Linear Operator is Compact

Let $X$ be a Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator. 
Let $\map \sigma T$ be the spectrum of $T$. 
Then $\map \sigma T$ is compact, and: 
:$\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$

Proof:
From Spectrum of Bounded Linear Operator is Closed, $\map \sigma T$ is closed in $\C$.
It therefore suffices to show that $\map \sigma T$ is bounded.
Hence it will suffice to show that: 
:$\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
Let $\lambda \in \C$ be such that $\cmod \lambda > \norm T_{\map \BB X}$. 
Then, we have: 
:$\ds \norm {\frac 1 \lambda T}_{\map \BB X} = \frac 1 {\cmod \lambda} \norm T_{\map \BB X} < 1$
from {{NormAxiomVector|2}}.
From Invertibility of Identity Minus Operator:
:$\ds I - \frac 1 \lambda T$ is invertible as a bounded linear operator.
Since $\map \BB X$ is a vector space, we have: 
:$\ds -\frac 1 \lambda \paren {I - \frac 1 \lambda T} = T - \lambda I$ is invertible as a bounded linear operator.
That is $\lambda \in \map \rho T$, where $\map \rho T$ is the resolvent set of $T$. 
So if $\cmod \lambda > \norm T_{\map \BB X}$, we have $\lambda \in \C \setminus \map \sigma T$. 
So if $\lambda \in \map \sigma T$, we have $\cmod \lambda \le \norm T_{\map \BB X}$.
That is: 
:$\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
{{qed}}
Theorem: Resolvent Set of Bounded Linear Operator is Open

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.
{{qed}}
Theorem: Spectrum of Adjoint of Bounded Linear Operator

Let $X$ be a Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator.
Let $T^\ast : X \to X$ be the adjoint of $T$. 
Let $\map \sigma T$ and $\map \sigma {T^\ast}$ be the spectrum of $T$ and $T^\ast$ respectively. 
Then: 
:$\map \sigma {T^\ast} = \set {\overline \lambda : \lambda \in \map \sigma T}$ 
where $\overline \lambda$ denotes the complex conjugate of $\lambda$.

Proof:
We show that for $\lambda \in \C$, we have $\lambda \not \in \map \sigma T$ {{iff}} $\overline \lambda \not \in \map \sigma {T^\ast}$.
Let $\lambda \in \C$ have $\lambda \not \in \map \sigma T$.
Then $T - \lambda I$ is invertible as a bounded linear operator.
Note that $I^\ast = I$ from Adjoint of Identity Transformation. 
From Adjoint is Conjugate Linear, we have that $T^\ast - \overline \lambda I = \paren {T - \lambda I}^\ast$.
From Adjoining Commutes with Inverting, we have: 
:$T^\ast - \overline \lambda I = \paren {T - \lambda I}^\ast$ is invertible
So we have $\overline \lambda \not \in \map \sigma {T^\ast}$. 
So if $\lambda \not \in \map \sigma T$ then $\overline \lambda \not \in \map \sigma {T^\ast}$. 
From Adjoint is Involutive, swapping $T$ for $T^\ast$ we have that $\lambda \not \in \map \sigma {T^\ast}$ implies $\overline \lambda \not \in \map \sigma T$. 
From Complex Conjugation is Involution, swapping $\lambda$ for $\overline \lambda$ we have that $\lambda \not \in \map \sigma {T^\ast}$ implies $\overline \lambda \not \in \map \sigma T$. 
So for $\lambda \in \C$, we have $\lambda \not \in \map \sigma T$ {{iff}} $\overline \lambda \not \in \map \sigma {T^\ast}$.
So we have: 
:$\map \sigma {T^\ast} = \set {\overline \lambda : \lambda \in \map \sigma T}$ 
{{qed}}
Theorem: Weak-* Topology is Hausdorff

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$. 
Let $X^\ast$ be the topological dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
Then $\struct {X^\ast, w^\ast}$ is Hausdorff space.

Proof:
From Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $w^\ast$ can be induced by: 
:$\PP = \set {p_{x^\wedge} : x \in X}$
where we define $p_{x^\wedge} : X^\ast \to \R_{\ge 0}$ by: 
:$\map {p_{x^\wedge} } f = \cmod {\map {x^\wedge} f} = \cmod {\map f x}$
From Locally Convex Space is Hausdorff iff induces Hausdorff Topology, it suffices to show that $\struct {X^\ast, w^\ast}$ is Hausdorff as a locally convex space.
That is, that $\PP$ is separating.
Let $f \in X^\ast$ be such that $f \ne 0$. 
Then there exists $x \in X$ such that $\map f x \ne 0$. 
Then $\map {p_{x^\wedge} } f = \map {x^\wedge} f \ne 0$. 
So $\PP$ is indeed separating.
{{qed}}
Category:Weak-* Topologies
Category:Hausdorff Spaces
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Theorem: Vector Addition is Continuous in Weak Topology

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$ with weak topology $w$. 
Define $s : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ by: 
:$\map s {x, y} = x + y$
for each $x, y \in X$. 
Then $s$ is continuous. 
That is, vector addition remains continuous when restricting to the weak topology.

Proof:
Let $X^\ast$ be the topological dual space of $X$.
From Continuity in Initial Topology, it suffices to show that for each $f \in X^\ast$ we have: 
:$f \circ s : \struct {X, w} \times \struct {X, w} \to K$ is continuous.
Define the projections $\pr_1 : \struct {X, w} \times {X, w} \to \struct {X, w}$ and $\pr_2 : \struct {X, w} \times {X, w} \to \struct {X, w}$ as the projection onto the first and second factors.
Then for each $x, y \in X$ we have: 
{{begin-eqn}}
{{eqn	| l = \map {\paren {f \circ s} } {x, y}
		| r = \map f {x + y}
}}
{{eqn	| r = \map f x + \map f y
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| r = \map f {\map {\pr_1} {\tuple {x, y} } } + \map f {\map {\pr_2} {\tuple {x, y} } }
}}
{{end-eqn}}
That is: 
:$f \circ s = f \circ \pr_1 + f \circ \pr_2$
From the definition of the product topology:
:$\pr_1 : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$
and:
:$\pr_2 : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ 
are continuous.
From the definition of the weak topology, $f : \struct {X, w} \to K$ is continuous.
From Composite of Continuous Mappings is Continuous, $f \circ \pr_1 : \struct {X, w} \times \struct {X, w} \to K$ and $f \circ \pr_2 : \struct {X, w} \times \struct {X, w} \to K$ are continuous.
From Sum of Continuous Functions on Topological Ring is Continuous, $f \circ \pr_1 + f \circ \pr_2 : \struct {X, w} \times \struct {X, w} \to K$ is continuous.
So we have that $f \circ s : \struct {X, w} \to K$ is continuous for each $f \in X^\ast$. 
So from Continuity in Initial Topology, $s : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ is continuous in the weak topology. 
{{qed}}
Category:Weak Topologies on Topological Vector Spaces
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