avewright/proofwiki-math huggingface dataset:
Theorem: Phillips-Sobczyk Theorem

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {\ell_\infty} {\N, \GF}, \norm {\, \cdot \,}_\infty}$ be the space of bounded sequences.
Let $\struct {\map {c_0} {\N, \GF}, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Then there does not exist a bounded projection $P : \map {\ell_\infty} {\N, \GF} \to \map {c_0} {\N, \GF}$.
That is, $\map {c_0} {\N, \GF}$ is not topologically complemented in $\map {\ell_\infty} {\N, \GF}$.

Proof:
{{AimForCont}} there exists a bounded projection $P : \map {\ell_\infty} {\N, \GF} \to \map {c_0} {\N, \GF}$.
Let $I : \map {\ell_\infty} {\N, \GF} \to \map {\ell_\infty} {\N, \GF}$ be the identity mapping.
Then $T = I - P$ is a bounded linear transformation.
Since $P$ is a projection onto $\map {c_0} {\N, \GF}$, we have:
:$P \xi = \xi$ for each $\xi \in \map {c_0} {\N, \GF}$
and hence:
:$T \xi = {\mathbf 0}_{\ell_\infty}$ for each $\xi \in \map {c_0} {\N, \GF}$.
From Bounded Linear Operator on Space of Bounded Sequences Vanishing on Zero-Limit Sequences also Vanishes on Bounded Sequences with certain Support, there exists a countably infinite set $A \subseteq \N$ such that:
:$T \xi = {\mathbf 0}_{\ell_\infty}$ for each $\xi \in \map {\ell_\infty} {A, \GF}$
where $\map {\ell_\infty} {A, \GF}$ denotes the space of bounded sequences supported on $A$.
Define $z : \N \to \GF$: 
:$\ds \map z n = \begin{cases}1 & n \in A \\ 0 & n \not \in A\end{cases}$
We have $\map z n = 0$ for all $n \in \N \setminus A$, hence $z \in \map {\ell_\infty} {A, \GF}$.
We therefore have that $T z = {\mathbf 0}_{\ell_\infty}$. 
Hence $P z = z$. 
Hence, from the definition of a projection, $z \in \map {c_0} {\N, \GF}$.
Since $A$ is countably infinite, for each $n \in \N$ there exists $k \in A$ with $k > n$.
Hence $z \not \in \map {c_0} {\N, \GF}$.
This is a contradiction.
Hence there does not exist a bounded projection $P : \map {\ell_\infty} {\N, \GF} \to \map {c_0} {\N, \GF}$.
Hence $\map {c_0} {\N, \GF}$ is not topologically complemented in $\map {\ell_\infty} {\N, \GF}$.
{{qed}}
{{Namedfor|Ralph Saul Phillips|name2 = Andrew Sobczyk|cat = Phillips|cat2 = Sobczyk}}
Theorem: Weak-* Compact Set in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable

Let $X$ be a separable normed vector space.
Let $\sequence {x_n}_{n \in \N}$ be a countable everywhere dense subset of $X$.
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X$.
Let $\struct {K, w^\ast}$ be a compact subspace of $X^\ast$.
Define $d : X^\ast \times X^\ast \to \R$ by:
:$\ds \map d {f, g} = \sum_{n \mathop = 1}^\infty \frac 1 {2^n} \min \set {\norm {\map f {x_n} - \map g {x_n} }_X, 1}$ for each $f, g \in X^\ast$. 
From Initial Topology Generated by Countable Family of Functions Separating Points is Metrizable, this is a metric. 
Let $d_K$ be the restriction of $d$ to $K$. 
Then $\struct {K, w^\ast}$ is metrizable with metric $d_K$.
Further
:$\map d {f - g, {\mathbf 0}_{X^\ast} } = \map d {f, g}$ whenever $f, g \in X^\ast$.

Proof:
Let $\sequence {x_n^\wedge }_{n \in \N}$ be the evaluation maps associated with $\sequence {x_n}_{n \in \N}$.
From Image of Everywhere Dense Subset of Normed Vector Space Separates Points of Normed Dual Space, $\sequence {x_n^\wedge }_{n \in \N}$ separates the points of $X^\ast$.
Let $\sigma$ be the initial topology generated by $\sequence {x_n^\wedge \restriction_K}_{n \in \N}$.
From Initial Topology Generated by Countable Family of Functions Separating Points is Metrizable, $\sigma$ is metrizable with metric $e$ satisfying:
:$\ds \map e {f, g} = \sum_{n \mathop = 1}^\infty \frac {\min \set {\norm {\map f {x_n} - \map g {x_n} }_X, 1} } {2^n}$ for each $f, g \in K$. 
In particular $e = d_K$. 
We can write:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\min \set {\norm {\map f {x_n} - \map g {x_n} }_X, 1} } {2^n} = \sum_{n \mathop = 1}^\infty \frac {\min \set {\norm {\map {\paren {f - g} } {x_n} + \map { {\mathbf 0}_{X^\ast} } {x_n} }_X, 1} } {2^n}$
Hence we have:
:$\map d {f - g, {\mathbf 0}_{X^\ast} } = \map d {f, g}$ whenever $f, g \in X^\ast$. 
Note that the weak-$\ast$ topology on $K$ is generated by $\set {y^\wedge \restriction_K : y \in X}$ from Subspace Topology on Initial Topology is Initial Topology on Restrictions.
So $\sigma \subseteq w^\ast$.
Hence the identity mapping $\iota : \struct {K, w^\ast} \to \struct {K, \sigma}$ is continuous.
By assumption, $\struct {K, w^\ast}$ is compact.
Since $\sigma$ is metrizable, it is Hausdorff from Metric Space is Hausdorff.
Hence from Continuous Bijection from Compact to Hausdorff is Homeomorphism, $\iota$ is a homeomorphism.
Hence $\sigma = w^\ast$, and $\struct {K, w^\ast}$ is metrizable with metric $d_K$.
{{qed}}
Category:Weak-* Topologies
Category:Metrizable Spaces
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Theorem: Distance from Point to Compact Set in Metric Space is Attained

Let $\struct {X, d}$ be a metric space.
Let $A \subseteq X$ be compact.
Let $\map d {\cdot, A}$ denote the $d$-distance to $A$.
Let $x \in X$.
Then there exists $\alpha \in A$ such that:
:$\map d {x, A} = \map d {x, \alpha}$

Proof:
From Distance Function of Metric Space is Continuous, the mapping:
:$\tuple {x, a} \mapsto \map d {x, a}$ is continuous.
From Vertical Section of Continuous Function is Continuous, the mapping $f : A \to \hointr 0 \infty$ defined by:
:$\map f a = \map d {x, a}$ for each $a \in A$
is continuous.
From {{Corollary|Continuous Image of Compact Space is Compact|3}}:
:$f$ is bounded and there exists $\alpha \in A$ such that:
::$\ds \map f \alpha = \inf_{a \mathop \in A} \map f a$
That is:
:$\ds \map d {x, \alpha} = \inf_{a \mathop \in A} \map d {x, a} = \map d {x, A}$
{{qed}}
Category:Metric Spaces
p30u1bojkprvv8rf3f1774mni8mxnx2
Theorem: Join of Meets Precedes Meet with Join

Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $x, y, z \in S$.
Then:
:$\paren {x \wedge y} \vee \paren {x \wedge z} \preceq x \wedge \paren {y \vee z}$

Proof:
By Meet Precedes Operands and Join Succeeds Operands:
:$x \wedge y \preceq x$
:$x \wedge z \preceq x$
:$x \wedge y \preceq y \preceq y \vee z$
:$x \wedge z \preceq z \preceq y \vee z$
so by definition of join:
:$\paren {x \wedge y} \vee \paren {x \wedge z} \preceq x$
:$\paren {x \wedge y} \vee \paren {x \wedge z} \preceq y \vee z$
hence by definition of meet:
:$\paren {x \wedge y} \vee \paren {x \wedge z} \preceq x \wedge \paren {y \vee z}$
{{qed}}
Category:Lattices (Order Theory)
nrmz0vfymflxr2dip89eoyqrk4bvluj
Theorem: Series whose Terms Converge to Zero is not necessarily Convergent

Let $\sequence {a_n}$ be a sequence in any of the standard number fields $\Q$, $\R$, or $\C$.
Suppose that $\sequence {a_n}$ converges to zero.
Then it is not necessarily the case that the series $\ds \sum_{n \mathop = 1}^\infty a_n$ is itself convergent.

Proof:
;Proof by Counterexample
Consider the harmonic sequence:
:$\sequence {a_n} = \sequence {\dfrac 1 n}$
from Sequence of Reciprocals is Null Sequence:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$
However, from Harmonic Series is Divergent, $\ds s = \sum_{n \mathop = 1}^\infty \dfrac 1 n$ is divergent.
Hence the result.
{{qed}}
Theorem: Closed Subspace of Separable Banach Space Linearly Isomorphic to Space of Zero-Limit Sequences is Topologically Complemented

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a separable Banach space over $\GF$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Let $E$ be a closed vector subspace of $X$ that is linearly isomorphic to $\map {C_0} \GF$.
Then $E$ is topologically complemented in $X$.

Proof:
Let $T : E \to \map {c_0} \GF$ be a linear isomorphism.
Then $T^{-1} : \map {c_0} \GF \to E$ is a linear isomorphism.
From Sobczyk's Theorem, there exists a bounded linear transformation $\widetilde T : X \to \map {c_0} \GF$ extending $T$.
Let $P = T^{-1} \widetilde T : X \to E$.
Then $P$ is a bounded linear transformation.
We show that $P^2 x = P x$ for each $x \in X$.
That is:
:$\map {\paren {T^{-1} \widetilde T T^{-1} \widetilde T} } x =  \map {\paren {T^{-1} \widetilde T} } x$ for each $x \in X$.
Note that since $T^{-1} \widetilde T x \in E$.
Hence, since $\widetilde T$ extends $T$ from $E$:
:$\widetilde T T^{-1} \widetilde T x = T T^{-1} \widetilde T x = T x$
Hence:
:$\map {\paren {T^{-1} \widetilde T T^{-1} \widetilde T} } x =  \map {\paren {T^{-1} \widetilde T} } x$ for each $x \in X$.
That is:
:$P^2 x = P x$ for each $x \in X$.
Hence $P$ is a bounded projection.
Since $T^{-1} \sqbrk {\map {c_0} \GF} \subseteq E$, we have:
:$P \sqbrk X \subseteq E$
Further, for each $x \in E$, we have:
:$P x = T^{-1} \widetilde T x = T^{-1} T x = x$
Hence $P \sqbrk X = E$. 
Hence $P$ is a bounded projection $X \to E$ with $P \sqbrk X = E$.
{{qed}}
Theorem: Injective Closed Subspace of Banach Space is Topologically Complemented

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $E$ be a closed vector subspace of $X$ that is injective.
Then $E$ is topologically complemented in $X$.

Proof:
Define $I : E \to E$ by:
:$I x = x$ for each $x \in E$.
Since $E$ is an injective Banach space, there exists a bounded linear transformation $P : X \to E$ that extends $I$.
Then, for each $x \in E$, we have $P x = I x = x$, and hence $P^2 x = P x$.
So $P$ is also a projection.
Hence $E$ is topologically complemented in $X$.
{{qed}}
Category:Topologically Complemented Subspaces of Banach Spaces
Category:Injective Banach Spaces
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Theorem: Space of Zero-Limit Sequences is not Injective

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Then $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ is not injective.

Proof:
Let $\struct {\map {\ell_\infty} \GF, \norm {\, \cdot \,}_\infty}$ be the space of bounded sequences.
From Space of Zero-Limit Sequences with Supremum Norm forms Banach Space, $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ is indeed a Banach space.
From Subspace of Complete Metric Space is Closed iff Complete, it follows that $\map {c_0} \GF$ is closed in $\struct {\map {\ell_\infty} \GF, \norm {\, \cdot \,}_\infty}$.
{{AimForCont}} $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ is injective.
From Injective Closed Subspace of Banach Space is Topologically Complemented, $\map {c_0} \GF$ is then topologically complemented in $\map {\ell_\infty} \GF$.
However, from the Phillips-Sobczyk Theorem, $\map {c_0} \GF$ is not topologically complemented in $\map {\ell_\infty} \GF$, a contradiction.
Hence $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ is not injective. 
{{qed}}
Category:Injective Banach Spaces
Category:Space of Zero-Limit Sequences
a0p82v7c9rsl3fighh1yly2jy5oxvos
Theorem: Characterization of Isomorphism in Loc*

Let $\mathbf{Loc_*}$ denote the category of locales with localic mappings.
Let $\mathbf{Frm}$ denote the category of frames.
Let $L_1 = \struct{S_1, \preceq_1}$ and $L_2 = \struct{S_2, \preceq_2}$ be locales.
Let $f:L_1 \to L_2$ be a localic mapping of $\mathbf{Loc_*}$.
Let $\loweradjoint f : L_2 \to L_1$ be the lower adjoint of $f$ in $\mathbf{Frm}$.
{{TFAE}}:
:$(1) \quad f$ is an isomorphism of $\mathbf{Loc_*}$
:$(2) \quad \loweradjoint f$ is a frame isomorphism
:$(3) \quad \loweradjoint f$ is an order isomorphism
:$(4) \quad f$ is an order isomorphism
:$(5) \quad f$ is a frame isomorphism

Proof:
=== Statement $(1)$ Iff Statement (2) ===
{{:Characterization of Isomorphism in Loc*/Isomorphism iff Lower Adjoint is Frame Isomorphism}}{{qed|lemma}}
Theorem: Injective Separable Banach Space is Isometrically Isomorphic to Topologically Complemented Subspace of Space of Bounded Sequences

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a injective separable Banach space over $\GF$.
Let $\struct {\map {\ell_\infty} \GF, \norm {\, \cdot \,}_\infty}$ be the space of bounded sequences.
Then there exists a topologically complemented subspace $E$ of $\map {\ell_\infty} \GF$ that is isometrically isomorphic to $X$.

Proof:
From Separable Normed Vector Space Isometrically Isomorphic to Linear Subspace of Space of Bounded Sequences, there exists a linear isometry $T : X \to \map {\ell_\infty} \GF$ such that $T : X \to T \sqbrk X$ is an isometric isomorphism.
Hence $T \sqbrk X$ is a closed vector subspace of $\map {\ell_\infty} \GF$.
We show that $T \sqbrk X$ is a topologically complemented subspace of $\map {\ell_\infty} \GF$.
We will then have the demand with $E = T \sqbrk X$.
Then $S = T^{-1} : T \sqbrk X \to X$ is an isometric isomorphism.
Since $X$ is injective, there exists a bounded linear transformation $\widetilde S : \map {\ell_\infty} \GF \to X$ extending $S$.
Let $P = T \widetilde S : \map {\ell_\infty} \GF \to T \sqbrk X$.
Then $P$ is a bounded linear transformation.
Since $S$ is a surjection, so is $\widetilde S$. 
Since $T$ is a surjection, we have that $P \sqbrk {\map {\ell_\infty} \GF} = T \sqbrk X$ from Composite of Surjections is Surjection.
We finish by showing that $P^2 = P$. 
Since $P x \in T \sqbrk X$ for each $x \in \map {\ell_\infty} \GF$, it is enough to show that $P x = x$ for $x \in T \sqbrk X$.
For $x \in T \sqbrk X$, we have $\widetilde S x = S x = T^{-1} x$.
Hence $T \widetilde S x = T T^{-1} x = x$. 
Hence we have $P^2 = P$. 
So $P$ is a bounded projection with $P \sqbrk {\map {\ell_\infty} \GF} = T \sqbrk X$.
Hence $T \sqbrk X$ is topologically complemented in $\map {\ell_\infty} \GF$.
We therefore have the demand with $E = T \sqbrk X$.
{{qed}}
Theorem: Characterization of Isomorphism in Loc*/Order Isomorphism iff Lower Adjoint is Order Isomorphism

Let $\mathbf{Loc_*}$ denote the category of locales with localic mappings.
Let $\mathbf{Frm}$ denote the category of frames.
Let $L_1 = \struct{S_1, \preceq_1}$ and $L_2 = \struct{S_2, \preceq_2}$ be locales.
Let $f:L_1 \to L_2$ be a localic mapping of $\mathbf{Loc_*}$.
Let $\loweradjoint f : L_2 \to L_1$ be the lower adjoint of $f$ in $\mathbf{Frm}$.
Then:
:$\loweradjoint f$ is an order isomorphism
{{iff}}:
:$f$ is an order isomorphism

Proof:
==== Necessary Condition ====
Let $\loweradjoint f$ be an order isomorphism.
From Inverse of Order Isomorphism is Order Isomorphism:
:$\paren{\loweradjoint f}^{-1}$ is an order isomorphism
From Order Isomorphism forms Galois Connection:
:$\tuple{\paren{\loweradjoint f}^{-1}, \loweradjoint f}$ is a Galois connection
By definition of localic mapping:
:$\tuple{f, \loweradjoint f}$ is a Galois connection
From Galois Connection is Unique for Given Lower Adjoint:
:$f = \paren{\loweradjoint f}^{-1}$
It follows that $f$ is an order isomorphism.
{{qed|lemma}}
==== Sufficient Condition ====
Let $f$ be an order isomorphism.
From Order Isomorphism forms Galois Connection:
:$\tuple{f, f^{-1}}$ is a Galois connection
By definition of localic mapping:
:$\tuple{f, \loweradjoint f}$ is a Galois connection
From Galois Connection is Unique for Given Upper Adjoint:
:$\loweradjoint f = f^{-1}$
From Inverse of Order Isomorphism is Order Isomorphism:
:$\loweradjoint f$ is an order isomorphism
Theorem: Sequence in Metric Space with Summable Difference between Consecutive Terms is Cauchy Sequence

Let $\struct {X, d}$ be a metric space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\struct {X, d}$ such that:
:$\ds \sum_{k \mathop = 1}^\infty \map d {x_k, x_{k + 1} } < \infty$
Then $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence.

Proof:
Since:
:$\ds \sum_{k \mathop = 1}^\infty \map d {x_k, x_{k + 1} } < \infty$
we have that:
:$\ds \sequence {\sum_{k \mathop = 1}^n \map d {x_k, x_{k + 1} } }_{n \mathop \in \N}$ is a Cauchy sequence.
Let $\epsilon > 0$.
Then there exists $N \in \N$ such that:
:$\ds \size {\sum_{k \mathop = 1}^{n - 1} \map d {x_k, x_{k + 1} } - \sum_{k \mathop = 1}^{m - 1} \map d {x_k, x_{k + 1} } } < \epsilon$ for $n > m \ge N + 1$.
Hence, we have:
:$\ds \sum_{k \mathop = m}^n \map d {x_k, x_{k + 1} } < \epsilon$ for $n > m \ge N$.
By a repeated application of {{Metric-space-axiom|3}}, we have:
:$\ds \map d {x_n, x_m} \le \sum_{k \mathop = m}^{n - 1} \map d {x_k, x_{k + 1} } < \epsilon$
for $n > m \ge N + 1$.
Hence:
:$\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence.
{{qed}}
Category:Sequence in Metric Space with Summable Difference between Consecutive Terms is Cauchy Sequence
Category:Metric Spaces
bkj6p7v5nmredss3e6pmzk55aiiqqp7
Theorem: Convolution Integral is Commutative

Let $f$ and $g$ be real functions which are integrable.
Let $\map f t * \map g t$ denote the convolution integral of $f$ and $g$.
Then:
:$\map f t * \map g t = \map g t * \map f t$
That is, the convolution integral is commutative.

Proof:
{{begin-eqn}}
{{eqn | l = \map f t * \map g t
      | r = \int_{-\infty}^\infty \map f u \map g {t - u} \rd u
      | c = {{Defof|Convolution Integral}}
}}
{{eqn | r = -\int_\infty^{-\infty} \map f {t - v} \map g v \rd v
      | c = Integration by Substitution with $\begin {cases} v & = & t - u \\ \d v & = & -\d u \end {cases}$
}}
{{eqn | r = \int_{-\infty}^\infty \map g v \map f {t - v} \rd v
      | c = Reversal of Limits of Definite Integral
}}
{{eqn | r = \map g t * \map f t
      | c = {{Defof|Convolution Integral}}
}}
{{end-eqn}}
{{qed}}
Theorem: Conway Circle is Circle

Let $\triangle ABC$ be a triangle.
Let $\KK$ be the Conway circle of $ABC$.
Then $\KK$ is indeed a circle.

Proof:
{{ProofWanted|Easy enough to get from the internet if anyone wants to post this up, it's a fun little thing}}
Theorem: Center of Conway Circle is Incenter

Let $\triangle ABC$ be a triangle.
Let $\KK$ be the Conway circle of $ABC$.
Then the center of $\KK$ coincides with the incenter of $\triangle ABC$.

Proof:
{{ProofWanted|As with Conway Circle is Circle the proof of one leads to the other}}
Theorem: Function is Convex iff Strict Epigraph is Convex

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Consider $X \times \GF$ as the direct product of $X$ and $\GF$.
Let $f : X \to \R$ be a function.
Let $\map {\operatorname {epi}_S} f$ be the strict epigraph of $f$.
Then $f$ is convex {{iff}} $\map {\operatorname {epi}_S} f$ is convex in $X \times \GF$.

Proof:
=== Necessary Condition ===
Suppose that $f$ is convex.
Let $t \in \openint 0 1$. 
Let $\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$.
Then:
:$\map f {x_1} < \alpha_1$
and:
:$\map f {x_2} < \alpha_2$ 
We therefore have, since $t > 0$:
:$t \map f {x_1} + \paren {1 - t} \map f {x_2} < t \alpha_1 + \paren {1 - t} \alpha_2$
Since $f$ is convex, we have:
:$\map f {t x_1 + \paren {1 - t} x_2} \le t \map f {x_1} + \paren {1 - t} \map f {x_2}$
Hence:
:$\map f {t x_1 + \paren {1 - t} x_2} < t \alpha_1 + \paren {1 - t} \alpha_2$ 
Hence:
:$\tuple {t x_1 + \paren {1 - t} x_2, t \alpha_1 + \paren {1 - t} \alpha_2} \in \map {\operatorname {epi}_S} f$
so that:
:$t \tuple {x_1, \alpha_1} + \paren {1 - t} \tuple {x_2, \alpha_2} \in \map {\operatorname {epi}_S} f$
whenever:
:$\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi}_S} f$
Hence $\map {\operatorname {epi}_S} f$ is convex in $X \times \GF$.
{{qed|lemma}}
Theorem: Function is Concave iff Hypograph is Convex

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Consider $X \times \GF$ as the direct product of $X$ and $\GF$.
Let $f : X \to \R$ be a function.
Let $\map {\operatorname {hypo} } f$ be the hypograph of $f$.
Then $f$ is concave {{iff}} $\map {\operatorname {hypo} } f$ is convex in $X \times \GF$.

Proof:
=== Necessary Condition ===
Suppose that $f$ is concave.
Let $t \in \openint 0 1$. 
Let $\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$.
Then:
:$\map f {x_1} \ge \alpha_1$
and:
:$\map f {x_2} \ge \alpha_2$ 
We therefore have, since $t > 0$:
:$t \map f {x_1} + \paren {1 - t} \map f {x_2} \ge t \alpha_1 + \paren {1 - t} \alpha_2$
Since $f$ is concave, we have:
:$\map f {t x_1 + \paren {1 - t} x_2} \ge t \map f {x_1} + \paren {1 - t} \map f {x_2}$
Hence:
:$\map f {t x_1 + \paren {1 - t} x_2} \ge t \alpha_1 + \paren {1 - t} \alpha_2$ 
Hence:
:$\tuple {t x_1 + \paren {1 - t} x_2, t \alpha_1 + \paren {1 - t} \alpha_2} \in \map {\operatorname {hypo} } f$
so that:
:$t \tuple {x_1, \alpha_1} + \paren {1 - t} \tuple {x_2, \alpha_2} \in \map {\operatorname {hypo} } f$
whenever:
:$\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {hypo} } f$
Hence $\map {\operatorname {hypo} } f$ is convex in $X \times \GF$.
{{qed|lemma}}
Theorem: Function is Concave iff Strict Hypograph is Convex

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Consider $X \times \GF$ as the direct product of $X$ and $\GF$.
Let $f : X \to \R$ be a function.
Let $\map {\operatorname {hypo} } f$ be the hypograph of $f$.
Then $f$ is convex {{iff}} $\map {\operatorname {hypo} } f$ is convex in $X \times \GF$.

Proof:
=== Necessary Condition ===
Suppose that $f$ is convex.
Let $t \in \openint 0 1$. 
Let $\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$.
Then:
:$\map f {x_1} > \alpha_1$
and:
:$\map f {x_2} > \alpha_2$ 
We therefore have, since $t > 0$:
:$t \map f {x_1} + \paren {1 - t} \map f {x_2} > t \alpha_1 + \paren {1 - t} \alpha_2$
Since $f$ is concave, we have:
:$\map f {t x_1 + \paren {1 - t} x_2} \ge t \map f {x_1} + \paren {1 - t} \map f {x_2}$
Hence:
:$\map f {t x_1 + \paren {1 - t} x_2} > t \alpha_1 + \paren {1 - t} \alpha_2$ 
Hence:
:$\tuple {t x_1 + \paren {1 - t} x_2, t \alpha_1 + \paren {1 - t} \alpha_2} \in \map {\operatorname {hypo}_S} f$
so that:
:$t \tuple {x_1, \alpha_1} + \paren {1 - t} \tuple {x_2, \alpha_2} \in \map {\operatorname {hypo}_S} f$
whenever:
:$\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {hypo}_S} f$
Hence $\map {\operatorname {hypo}_S} f$ is convex in $X \times \GF$.
{{qed|lemma}}
Theorem: Real Constant Function is Convex and Concave

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $c \in \R$.
Let $f : X \to \R$ be a real function by:
:$\map f x = c$
Then $f$ is convex and concave.

Proof:
Let $x, y \in X$ and $t \in \openint 0 1$.
We have:
:$\map f {t x + \paren {1 - t} y} = c$
We also have:
{{begin-eqn}}
{{eqn	| l = t \map f x + \paren {1 - t} \map f y
		| r = t c + \paren {1 - t} c
}}
{{eqn	| r = c
}}
{{eqn	| r = \map f {t x + \paren {1 - t} y}
}}
{{end-eqn}}
Hence we have:
:$\map f {t x + \paren {1 - t} y} \le t \map f x + \paren {1 - t} \map f y$
giving that $f$ is convex and:
:$\map f {t x + \paren {1 - t} y} \ge t \map f x + \paren {1 - t} \map f y$
giving that $f$ is concave.
{{qed}}
Category:Convex Real Functions
Category:Concave Real Functions
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Theorem: Non-Negative Multiple of Convex Real Function is Convex

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $f : X \to \R$ be a convex real function.
Let $\lambda \ge 0$.
Then $\lambda f$ is convex.

Proof:
Let $x, y \in X$ and $t \in \openint 0 1$.
Then:
:$\map f {t x + \paren {1 - t} y} \le t \map f x + \paren {1 - t} \map f y$
Multiplying through by $\lambda$ we have:
:$\lambda \map f {t x + \paren {1 - t} y} \le t \lambda \map f x + \paren {1 - t} \lambda \map f y$
Hence $\lambda f$ is convex.
{{qed}}
Category:Convex Real Functions
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Theorem: Sum of Convex Real Functions is Convex

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $f, g : X \to \R$ be a convex real function.
Then $f + g$ is a convex real function.

Proof:
Let $x, y \in X$ and $t \in \openint 0 1$. 
Since $f$ is convex, we have:
:$\map f {t x + \paren {1 - t} y} \le t \map f x + \paren {1 - t} \map f y$
Since $g$ is convex, we have:
:$\map g {t x + \paren {1 - t} y} \le t \map g x + \paren {1 - t} \map g y$
Hence:
{{begin-eqn}}
{{eqn	| l = \map {\paren {f + g} } {t x + \paren {1 - t} y}
		| r = \map f {t x + \paren {1 - t} y} + \map g {t x + \paren {1 - t} y}
}}
{{eqn	| o = \le
		| r = t \map f x + \paren {1 - t} \map f y + t \map g x + \paren {1 - t} \map g y
}}
{{eqn	| r = t \map {\paren {f + g} } x + \paren {1 - t} \map {\paren {f + g} } y
}}
{{end-eqn}}
Hence $f + g$ is convex.
{{qed}}
Category:Convex Real Functions
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