nyuuzyou/wikis huggingface dataset:
Continuity of Mapping to Cartesian Product under Chebyshev Distance

== Theorem ==
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
For all $i \in \set {1, 2, \ldots, n}$, let $\pr_i: \AA \to A_i$ be the $i$th projection on $\AA$:
:$\forall a \in \AA: \map {\pr_i} a = a_i$
where $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
Let $M' = \struct {X, d'}$ be a metric space.
Let $f: X \to \AA$ be a mapping.
Then $f$ is continuous on $X$ each of $\pr_i \circ f: X \to A_i$ is continuous on $X$.
== Proof ==
, let $i \in \set {1, 2, \ldots, n}$ be arbitrary.
=== Necessary Condition ===
Let $f$ be continuous.
From Projection from Cartesian Product under Chebyshev Distance is Continuous, $\pr_i: \AA \to A_i$ is continuous on $\AA$.
From Composite of Continuous Mappings between Metric Spaces is Continuous it follows that $\pr_1 \circ f$ is continuous on $\AA$.
=== Sufficient Condition ===
Let $\pr_i \circ f$ be continuous on $X$.
Let $a \in X$.
By definition of continuity at a point:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \pr_i \circ f \sqbrk {\map {B_\delta} {a; d'} } \subseteq \map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$
where $\map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$ denotes the open $\epsilon$-ball of $\map {\pr_i \circ f} a$ with respect to the metric $d_i$, and similarly for $\map {B_\delta} {a; d'}$.
From Projection from Cartesian Product under Chebyshev Distance is Continuous, we have that:
:$\forall \epsilon \in \R_{>0}: \pr_i \sqbrk {\map {B_\epsilon} {\map f a; d_\infty} } \subseteq \map {B_\epsilon} {\map {\pr_i} {\map f a}; d_i}$
From Open Ball in Cartesian Product under Chebyshev Distance:
:$\ds \map {B_\epsilon} {\map f a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$
From Metric Space Continuity by Inverse of Mapping between Open Balls:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map {B_\delta} {a; d'} \subseteq \paren {\pr_i \circ f}^{-1} \sqbrk {\map {B_\epsilon} {\map {\pr_i \circ f} a; d_i} }$
where $\map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$ denotes the open $\epsilon$-ball of $\map {\pr_i \circ f} a$ with respect to the metric $d_i$, and similarly for $\map {B_\delta} {a; d'}$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $7$
Category:Chebyshev Distance
Category:Projections
Category:Continuous Mappings on Metric Spaces
Continuity of Root Function

== Theorem ==
Let $n \in \N_{>0}$ be a non-zero natural number.
Let $f: \hointr 0 \infty \to \R$ be the real function defined by $\map f x = x^{1/n}$.
Then $f$ is continuous at each $\xi > 0$ and continuous on the right at $\xi = 0$.
== Proof ==
First suppose that $\xi > 0$.
Let $X, Y \in \R$ such that $0 < X < \xi < Y$.
Let $x \in \R$ such that $X < x < Y$.
From Inequalities Concerning Roots:
:$X Y^{1/n} \ \size {x - \xi} \le n X Y \ \size {x^{1/n} - \xi^{1/n} } \le Y X^{1/n} \ \size {x - \xi}$
Thus:
:$\dfrac 1 {n Y} Y^{1/n} \ \size {x - \xi} \le \size {x^{1/n} - \xi^{1/n} } \le \dfrac 1 {n X} X^{1/n} \ \size {x - \xi}$
The result follows by applying the Squeeze Theorem.
Now we need to show that $\map f x \to 0$ as $x \to 0^+$.
We need to show that:
:$\forall \epsilon > 0: \exists \delta > 0: x^{1/n} = \size {x^{1/n} - 0} < \epsilon$
provided $0 < x < \delta$.
Clearly, for any given $\epsilon$, we can choose $\delta = \epsilon^n$.
Hence the result.
== Sources ==
* : $\S 8.15 \ (5)$
Category:Roots of Numbers
Category:Continuous Functions
Continuity under Integral Sign

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $U$ be a non-empty open set of a metric space.
Let $f: U \times X \to \R$ be a mapping satisfying:
:$(1): \quad$ For all $\lambda \in U$, the mapping $x \mapsto \map f {\lambda, x}$ is $\mu$-integrable
:$(2): \quad$ For $\mu$-almost all $x \in X$, the mapping $\lambda \mapsto \map f {\lambda, x}$ is continuous
:$(3): \quad$ There exists a $\mu$-integrable $g: X \to \R$ such that:
::::$\forall \tuple {\lambda, x} \in U \times X: \size {\map f {\lambda, x} } \le \map g x$
Then the mapping $h: U \to \R$ defined by:
:$\ds \map h \lambda := \int \map f {\lambda, x} \map {\rd \mu} x$
is continuous.
== Proof ==
Let $\lambda_0 \in U$ be arbitrary.
Let $\sequence {\lambda_n}_{n \mathop \ge 1}$ be a sequence in $U$ which converges to $\lambda_0$.
Define the sequence of functions $f_n: X \to \R$, for $n = 0$ and $n \ge 1$ by $\map {f_n} x = \map f {\lambda_n, x}$.
By hypothesis $(1)$, for each $n \ge 1$, the function $f_n$ is $\mu$-integrable.
By hypothesis $(2)$, and by Sequential Continuity is Equivalent to Continuity in Metric Space, we have:
:$\ds \map {f_0} x = \lim_{n \mathop \to \infty} \map {f_n} x$
for $\mu$-almost all $x \in X$.
By hypothesis $(3)$, we have:
:$\size {\map {f_n} x} \le \map g x$
for all $x \in X$ and all $n \ge 1$.
Therefore by Lebesgue's Dominated Convergence Theorem:
:$\ds \lim_{n \mathop \to \infty} \int f_n \rd \mu = \int f_0 \rd \mu$
That is:
:$\ds \lim_{n \mathop \to \infty} \int \map f {\lambda_n, x} \rd \mu = \int \map f {\lambda_0, x} \rd \mu$
or in terms of $h$:
:$\ds \lim_{n \mathop \to \infty} \map h {\lambda_n} = \map h {\lambda_0}$
Since the sequence $\sequence {\lambda_n}_{n \mathop \ge 1}$ was arbitrary, by Sequential Continuity is Equivalent to Continuity in Metric Space this shows that $h$ is continuous at $\lambda_0 \in U$.
But $\lambda_0 \in U$ was arbitrary, so $h$ is continuous in $U$.
== Sources ==
* : $11.4$
Category:Measure Theory
Continuous Bijection from Compact to Hausdorff is Homeomorphism

== Theorem ==
Let $T_1$ be a compact space.
Let $T_2$ be a Hausdorff space.
Let $f: T_1 \to T_2$ be a continuous bijection.
Then $f$ is a homeomorphism.
=== Corollary ===
== Proof ==
Let $g = f^{-1}$.
We need to show that $g: T_2 \to T_1$ is continuous.
For any $V \subseteq T_1$, we have $g^{-1} \sqbrk V = \map f V$.
We are to show that if $V$ is closed in $T_1$, then $g^{-1} \sqbrk V$ is closed in $T_2$.
Suppose $V$ is closed in $T_1$.
Since $T_1$ is compact, $V$ is compact by Closed Subspace of Compact Space is Compact.
So $f \sqbrk V$ is compact from Continuous Image of Compact Space is Compact.
Since $T_2$ is Hausdorff, $f \sqbrk V$ is closed by Compact Subspace of Hausdorff Space is Closed.
But $f \sqbrk V = g^{-1} \sqbrk V$, so $g^{-1} \sqbrk V$ is closed.
From Continuity Defined from Closed Sets, it follows that $g$ is continuous.
Thus by definition, $f$ is a homeomorphism.
== Sources ==
* : $4.17$
* : $5$: Compact spaces: $5.9$: An inverse function theorem: Theorem $5.9.1$
Category:Homeomorphisms (Topological Spaces)
Category:Compact Spaces
Category:Hausdorff Spaces
Category:Continuous Mappings
Continuous Bijection from Compact to Hausdorff is Homeomorphism/Corollary

== Corollary to Continuous Bijection from Compact to Hausdorff is Homeomorphism ==
Let $T_1$ be a compact space.
Let $T_2$ be a Hausdorff space.
Let $f: T_1 \to T_2$ be a continuous injection.
Then $f$ determines a homeomorphism from $T_1$ to $f \sqbrk {T_1}$.
That is, $f$ is an embedding of $T_1$ into $T_2$.
== Proof ==
Follows from Continuous Bijection from Compact to Hausdorff is Homeomorphism and Continuity of Composite with Inclusion.
== Sources ==
* : $5$: Compact spaces: $5.9$: An inverse function theorem: Theorem $5.9.2$
Category:Homeomorphisms (Topological Spaces)
Category:Compact Spaces
Category:Hausdorff Spaces
Category:Continuous Mappings
Continuous Closed Surjective Mapping is Quotient Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a continuous closed surjective mapping.
Then $f$ is a quotient mapping.
== Proof ==
Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.
For $X \subseteq S$, let $\relcomp S X$ denote the relative complement of $X$ is $S$.
By definition of closed set, $\relcomp {S_1} {f^{-1} \sqbrk U}$ is closed in $T_1$.
By definition of closed mapping, $f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} }$ is closed in $T_2$.
By definition of closed set, $\relcomp {S_2} {f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} } }$ is open in $T_2$.
Then:
By definition of quotient mapping, it follows that $f$ is a quotient mapping.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Closed Mappings
Category:Quotient Topology
Continuous Complex-Valued Function Vanishing at Infinity is Bounded and Attains Supremum

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $f : X \to \C$ be a continuous function that vanishes at infinity.
Then $f$ is bounded and there exists $x_\ast \in X$ such that:
:$\ds \map f {x_\ast} = \sup_{x \mathop \in X} \cmod {\map f x}$
== Proof ==
From the definition of a function vanishing at infinity:
:there exists a compact set $F \subseteq X$ such that:
::$\cmod {\map f x} < 1$ for all $x \in X \setminus F$.
From Continuous Function on Compact Space is Bounded:
:$f$ is bounded on $F$. 
That is, there exists a real number $M > 0$ such that:
:$\cmod {\map f x} \le M$ for all $x \in F$. 
Hence, we have:
:$\cmod {\map f x} \le \max \set {1, M}$ for each $x \in X$.
So $f$ is bounded.
Now suppose that $\map f x = 0$ for each $x \in X$. 
Then $f$ clearly attains its bounds.
Now take $f$ such that:
:$\map f x \ne 0$ for some $x \in X$. 
Let:
:$\ds M = \sup_{x \mathop \in X} \cmod {\map f x} > 0$ 
which is finite from the above. 
Since $f$ vanishes at infinity, there exists a compact set $K \subseteq X$ such that:
:$\ds \cmod {\map f x} < \frac M 2$ for each $x \in X \setminus K$. 
By the definition of supremum:
:for each $M/2 > \epsilon > 0$ there exists $x_0 \in X$ such that $M - \epsilon < \cmod {\map f {x_0} } \le M$ 
Since:
:$\ds \cmod {\map f x} < \frac M 2$ for each $x \in X \setminus K$
the points $x_0$ described above must fall in $K$. 
Hence we have:
:$\ds \sup_{x \mathop \in X} \cmod {\map f x} = \sup_{x \mathop \in K} \cmod {\map f x}$ 
From Complex Modulus Function is Continuous and Composite of Continuous Mappings is Continuous, $\cmod f$ is continuous.
Consider the restriction $\cmod f \restriction K : K \to \R$. 
From Restriction of Continuous Mapping is Continuous: Topological Spaces, $\cmod f \restriction K$ is continuous.
Hence from Continuous Image of Compact Space is Compact: Corollary 3, $\cmod f \restriction K$ attains its bounds. 
Hence there exists $x_0 \in K$ such that $\cmod {\map f {x_0} } = M$, hence the result.
Category:Complex-Valued Functions Vanishing at Infinity
Continuous Complex Function is Complex Riemann Integrable

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \C$ be a continuous complex function.
Then $f$ is complex Riemann integrable over $\closedint a b$.
== Proof ==
Define the real function $x: \closedint a b \to \R$ by:
:$\forall t \in \closedint a b : \map x t = \map \Re {\map f t}$
Define the real function $y: \closedint a b \to \R$ by:
:$\forall t \in \closedint a b : \map y t = \map \Im {\map f t}$
where:
:$\map \Re {\map f t}$ denotes the real part of the complex number $\map f t$
:$\map \Im {\map f t}$ denotes the imaginary part of $\map f t$.
From Real and Imaginary Part Projections are Continuous, it follows that $\Re: \C \to \R$ and $\Im: \C \to \R$ are continuous functions.
From Composite of Continuous Mappings is Continuous, it follows that $x$ and $y$ are continuous.
From Continuous Real Function is Darboux Integrable, it follows that $x$ and $y$ are Darboux integrable over $\closedint a b$.
By definition, it follows that $f$ is complex Riemann integrable.
== Sources ==
* $\S 2.1$
Category:Complex Analysis
Continuous Composition of Measurable Functions into Second Countable Space is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\struct {X_i, \tau_i}$ for $i = 1, \ldots, n$ and $\struct {Y, \tau_Y}$ be topological spaces such that $X_1, \ldots, X_n$ are second countable. 
Let $f_i: \struct {X, \Sigma} \to \struct {X_i, \map \BB {X_i, \tau_i} }$, $i = 1, \ldots, n$ be measurable functions where $\map \BB {X_i, \tau_i}$ denotes the Borel-sigma algebra.
Let $F: \struct {\ds \prod_{i \mathop = 1}^n X_i, \tau} \to \struct {Y, \tau_Y}$ be continuous where $\tau$ denotes the product topology on the finite Cartesian product $\ds \prod_{i \mathop = 1}^n X_i$. 
Then the composition $h: \struct {X, \Sigma} \to \struct {Y, \map \BB {Y, \tau_Y} }$, $x \mapsto \map F {\map {f_1} x, \ldots, \map {f_n} x}$ is measurable.
== Proof ==
By Mapping Measurable iff Measurable on Generator, it suffices to check that $h$ is measurable on open sets.
Thus let $U \in \tau_Y$ be given.
As $F$ is continuous by assumption, the pre-image $F^{-1} \sqbrk U$ is in $\tau$. 
By:
:definition of basis
:Countable Product of Second-Countable Spaces is Second-Countable
there exists an at most countable indexing set $J \subseteq \N$ and indexed family of open sets $\family {U^i_j}_{j \mathop \in J} \subseteq \tau_i$, $i = 1, \ldots, n$ such that:
:$F^{-1} \sqbrk U = \ds \bigcup_{j \mathop \in J} \prod_{i \mathop = 1}^n U^i_j$
Putting things together:
By assumption each $f_i$ is measurable.
By definition of the Borel-sigma algebra:
:$\tau_i \subseteq \map \BB {X_i, \tau_i}$
Hence it follows that every $f_i^{-1} \sqbrk {U^i_j}$ is in $\Sigma$.
Therefore by Sigma-Algebra Closed under Countable Intersection:
:$\ds \bigcap_{i \mathop = 1}^n f_i^{-1} \sqbrk {U^i_j}$
is also in $\Sigma$.
Finally, we note that by definition of $\sigma$-algebra, $\Sigma$ contains at most countable unions of measurable sets.
Hence it follows that:
:$h^{-1} \sqbrk U = \ds \bigcup_{j \mathop \in J} \bigcap_{i \mathop = 1}^n f_i^{-1} \sqbrk {U^i_j}$ is in $\Sigma$
This concludes the proof. 
== Sources ==
* : $10.18$
::This statement found in Rudin is concerned with the special case $X = Y = \R$ and $n = 2$; however the idea of his proof generalizes to the above setting.
Category:Measure Theory
Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $I$ be a closed ideal.
Let $x \in I$ be normal.
Let $\map {\sigma_A} x \subseteq \R$ be the spectrum of $x$ in $A$.
Let $K$ be a compact set such that $\map {\sigma_A} x \cup \set 0 \subseteq K$. 
Let $f : K \to \R$ be a continuous function with $\map f 0 = 0$.
Let $\map f x$ be obtained from the continuous functional calculus, restricting $f$ to $\map {\sigma_A} x$.
Then $\map f x \in I$. 
=== Corollary ===
== Proof ==
From Closed Ideal of C*-Algebra is Self-Adjoint, we have $x^\ast \in I$. 
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map \CC K$.
From Subalgebra Generated by Inclusion and Conjugate is Everywhere Dense in Space of Continuous Functions on Compact Subset of Complex Numbers, the set:
:$\ds \AA = \set {z \mapsto \sum_{j \mathop = 1}^N \alpha_j z^{\lambda_j} \overline z^{\mu_j} : n \in \N, \, \alpha_j \in \C, \, \lambda_j, \mu_j \ge 0} \subseteq \map \CC K$
is everywhere dense in $\map \CC K$.
Take $\sequence {p_n}_{n \mathop \in \N}$ in $\AA$ so that:
:$\norm {p_n - f}_\infty \to 0$ as $n \to \infty$.
We therefore have $\map {p_n} 0 \to \map f 0 = 0$. 
Set $q_n = p_n - \map {p_n} 0$. 
Hence $\sequence {q_n}_{n \mathop \in \N}$ is such that:
:$\norm {q_n - f}_\infty \to 0$
and there exists $N_n$, $\alpha_{j, n}$, $\lambda_{j, n}$ and $\mu_{j, n}$ such that:
:$\ds \map {q_n} z = \sum_{j \mathop = 1}^{N_n} \alpha_{j, n} z^{\lambda_{j, n} } \overline z^{\mu_{j, n} }$
where for each $j$ and $n$ we have $\lambda_{j, n} \ge 1$ or $\mu_{j, n} \ge 1$. 
We have:
:$\ds \map {q_n} x = \sum_{j \mathop = 1}^{N_n} \alpha_{j, n} x^{\lambda_{j, n} } \paren {x^\ast}^{\mu_{j, n} }$
Hence $\map {q_n} x \in I$ for each $n \in \N$ since $I$ is an ideal. 
Since the continuous functional calculus is an linear isometry, we have:
:$\norm {\map {q_n} x - \map f x}_\infty \to 0$ as $n \to \infty$.
Hence since $I$ is closed, we have $\map f x \in I$.
Category:Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal
Category:Continuous Functional Calculus
Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal/Corollary

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $\struct {A, \ast, \norm {\, \cdot \,} }$.
Let $A_0 = \set {\tuple {a, 0} : a \in A}$. 
Let $x \in I$ be normal.
Let $\map {\sigma_{A_+} } x \subseteq \R$ be the spectrum of $\tuple {x, 0}$ in $A_+$.
Let $f : \map {\sigma_{A_+} } x \to \R$ be a continuous function with $\map f 0 = 0$.
Let $\map f {\tuple {x, 0} }$ be obtained from the continuous functional calculus.
Then $\map f {\tuple {x, 0} } \in A_0$.
Hence there exists $u \in A$ such that $\map f {\tuple {x, 0} } = \tuple {u, 0}$. 
== Proof ==
From Normed Algebra Embeds into Unitization as Closed Ideal, $A_0$ is a closed ideal of $A$.
We have that $\tuple {x, 0}$ is normal in $A_+$.
The result follows from Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal.
Category:Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal
Continuous Function applied to Normal Element of Unital C*-Algebra is Hermitian iff Function is Real-Valued

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $f : \map {\sigma_A} x \to \C$ be continuous.
Let $\map {\Theta_x} f = \map f x$ be the continuous functional calculus of $x$ applied to $f$. 
Then $\map f x$ is Hermitian $\map f z \in \R$ for each $z \in \map {\sigma_A} x$. 
== Proof ==
From Normal Element of C*-Algebra is Hermitian iff Spectrum is Real we have:
:$\map f x$ is Hermitian $\map {\sigma_A} {\map f x} \subseteq \R$.
From the Spectral Mapping Theorem, we have:
:$\map {\sigma_A} {\map f x} = f \sqbrk {\map {\sigma_A} x}$
Hence $\map f x$ is Hermitian $f \sqbrk {\map {\sigma_A} x} \subseteq \R$.
Hence $\map f x$ is Hermitian $\map f z \in \R$ for each $z \in \map {\sigma_A} x$.
Category:Continuous Functional Calculus
Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $f : \map {\sigma_A} x \to \C$ be continuous.
Let $\map {\Theta_x} f = \map f x$ be the continuous functional calculus of $x$ applied to $f$. 
Then $\map f x$ is positive $\map f z \in \hointr 0 \infty$ for each $z \in \map {\sigma_A} x$. 
=== Corollary ===
== Proof ==
From the definition of a positive element, we have that $\map f x$ is positive Hermitian and $\map {\sigma_A} {\map f x} \subseteq \hointr 0 \infty$. 
Note that $\map {\sigma_A} {\map f x} \subseteq \hointr 0 \infty$ implies that $\map {\sigma_A} {\map f x} \subseteq \R$. 
From Normal Element of C*-Algebra is Hermitian iff Spectrum is Real we have that $\map f x$ is Hermitian if $\map {\sigma_A} {\map f x} \subseteq \hointr 0 \infty$.
So $\map f x$ is positive $\map {\sigma_A} {\map f x} \subseteq \hointr 0 \infty$.
From the Spectral Mapping Theorem, we have:
:$\map {\sigma_A} {\map f x} = f \sqbrk {\map {\sigma_A} x}$
Hence we have $\map {\sigma_A} {\map f x} \subseteq \hointr 0 \infty$ $\map f z \in \hointr 0 \infty$ for each $z \in \map {\sigma_A} x$.
Category:Positive Elements of C*-Algebras
Category:Continuous Functional Calculus
Category:Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative
Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative/Corollary

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $f, g, h : \map {\sigma_A} x \to \R$ be continuous such that:
:$\map g z \le \map f z$ for all $z \in \map {\sigma_A} x$. 
Let $\map {\Theta_x} f = \map f x$, $\map {\Theta_x} g = \map g x$ be the continuous functional calculus of $x$ applied to $f$ and $g$ respectively.
Let $\le_A$ be the canonical ordering of $A$. 
Then:
:$\map g x \le_A \map f x$
== Proof ==
Since $g \le f$ we have:
:$\map {\paren {f - g} } z \in \R_{\ge 0}$ for all $z \in \map {\sigma_A} x$.
Hence from Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative, we have:
:$\map {\Theta_x} {f - g} \ge_A {\mathbf 0}_A$
Since $\Theta_x$ is linear, we have:
:$\map {\Theta_x} f - \map {\Theta_x} g \ge_A {\mathbf 0}_A$
and hence:
:$\map f x \ge_A \map g x$
Category:Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative
Continuous Function applied to Normal Element of Unital C*-Algebra is Unitary iff Function is Valued in Unit Circle

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $f : \map {\sigma_A} x \to \C$ be continuous.
Let $\map {\Theta_x} f = \map f x$ be the continuous functional calculus of $x$ applied to $f$. 
Then $\map f x$ is unitary $\cmod {\map f z} = 1$ for each $z \in \map {\sigma_A} x$. 
== Proof ==
From Normal Element of Unital C*-Algebra is Unitary iff Spectrum is Subset of Unit Circle we have:
:$\map f x$ is unitary $\map {\sigma_A} {\map f x} \subseteq \set {z \in \C : \cmod z = 1}$.
From the Spectral Mapping Theorem, we have:
:$\map {\sigma_A} {\map f x} = f \sqbrk {\map {\sigma_A} x}$
Hence $\map f x$ is unitary $f \sqbrk {\map {\sigma_A} x} \subseteq \set {z \in \C : \cmod z = 1}$.
Hence $\map f x$ is unitary $\cmod {\map f z} = 1$ for each $z \in \map {\sigma_A} x$.
Category:Continuous Functional Calculus
Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set

== Theorem ==
Let $\struct {X, \tau}$ be a compact Hausdorff space. 
Let $f : X \to X$ be a continuous function.
Then there exists a non-empty subset $A \subseteq X$ such that: 
:$f \sqbrk A = A$
== Proof ==
Define a sequence of sets $\sequence {X_i}_{i \mathop \in \N}$ by:
:$X_i = \begin{cases} X & : i = 1 \\ f \sqbrk {X_{i - 1} } & : i \ge 2 \end{cases}$
Since:
:$f \sqbrk X \subseteq X$
we have: 
:$X_i \subseteq X$
for each $i \in \N$. 
Define:
:$\ds A = \bigcap_{n \mathop = 1}^\infty X_n$
We have: 
We aim to show that:
:$f \sqbrk A = A$
=== Lemma 1 ===
 
Since $X_{i + 1} \subseteq X_i$ for all $i \in \N$, we can apply Intersection of Nested Closed Subsets of Compact Space is Non-Empty to obtain that: 
:$A$ is non-empty.
We then have:
so:
:$f \sqbrk A \subseteq A$
It remains to show that:
:$A \subseteq f \sqbrk A$
=== Lemma 2 ===
 
We then have:
:$A = f \sqbrk A$ 
as required.
Category:Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set
Category:Compact Spaces
Category:Hausdorff Spaces