avewright/proofwiki-math huggingface dataset:
Theorem: Metric Projection of Point Lies on Boundary of Convex Set

Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\GF$.
Let $\norm {\, \cdot \,}_\HH$ be the inner product norm.
Let $\pi_C : \HH \to C$ be the metric projection of $\HH$ onto $C$. 
Let $\partial C$ be the topological boundary of $C$.
Let $x \in \HH \setminus C$. 
Then $\map {\pi_C} x \in \partial C$.

Proof:
Since $x \in \HH \setminus C$, we have $x \ne \map {\pi_C} x$.
Hence $\norm {x - \map {\pi_C} x}_\HH \ne 0$ from {{NormAxiomVector|1}}.
Suppose that $\map {\pi_C} x \in \Int C$, where $\Int C$ is the interior of $C$.
For $\lambda \in \closedint 0 1$, define:
:$x_\lambda = \lambda x + \paren {1 - \lambda} \map {\pi_C} x$
We have:
:$x_\lambda \to \map {\pi_C} x$ as $\lambda \to 0^+$.
Since $\map {\pi_C} x \in \Int C$, there exists $0 < \lambda < 1$ such that:
:$x_\lambda = \lambda x + \paren {1 - \lambda} \map {\pi_C} x \in \Int C$
In particular, $x_\lambda \in C$. 
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {x - x_\lambda}_\HH
		| r = \paren {1 - \lambda} \norm {x - \map {\pi_C} x}_\HH 
		| c = {{NormAxiomVector|2}}
}}
{{eqn	| o = <
		| r = \norm {x - \map {\pi_C} x}_\HH 
		| c = since $1 - \lambda < 0$, $\norm {x - \map {\pi_C} x}_\HH \ne 0$
}}
{{end-eqn}}
Since:
:$\ds \norm {x - \map {\pi_C} x}_\HH = \inf_{z \mathop \in C} \norm {x - z}_\HH$
this contradicts $x_\lambda \in C$.
Hence we have $\map {\pi_C} x \not \in \Int C$.
Since $\map {\pi_C} x \in C$, we must have $\map {\pi_C} x \in \partial C$. 
{{qed}}
Theorem: Bound on Inner Product of Differences with Metric Projection

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\R$.
Let $\norm {\, \cdot \,}_\HH$ be the inner product norm.
Let $C \subseteq \HH$ be a non-empty closed convex set.
Let $\pi_C : \HH \to C$ be the metric projection of $\HH$ onto $C$. 
Let $x \in \HH$ and $y \in C$.
Then we have:
:$\innerprod {x - \map {\pi_C} x} {y - \map {\pi_C} x} \le 0$

Proof:
Let $\alpha \in \openint 0 1$.
We have:
:$\alpha y + \paren {1 - \alpha} \map {\pi_C} x \in C$
since $C$ is convex and $y, \map {\pi_C} x \in C$.
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {x - \map {\pi_C} x}_\HH^2
		| r = \inf_{z \mathop \in C} \norm {x - z}_\HH^2 
		| c = {{Defof|Metric Projection}}
}}
{{eqn	| o = \le
		| r = \norm {x - \paren {\alpha y + \paren {1 - \alpha} \map {\pi_C} x} }_\HH^2 
}}
{{eqn	| r = \innerprod {\paren {x - \map {\pi_C} x} - \alpha \paren {y - \map {\pi_C} x} } {\paren {x - \map {\pi_C} x} - \alpha \paren {y - \map {\pi_C} x} }
}}
{{eqn	| r = \innerprod {x - \map {\pi_C} x} {x - \map {\pi_C} x} - \alpha \innerprod {x - \map {\pi_C} x} {y - \map {\pi_C} x} - \alpha \innerprod {y - \map {\pi_C} x} {x - \map {\pi_C} x} + \alpha^2 \innerprod {y - \map {\pi_C} x} {y - \map {\pi_C} x}
}}
{{eqn	| r = \norm {x - \map {\pi_C} x}^2 - 2 \alpha \innerprod {x - \map {\pi_C} x} {y - \map {\pi_C} x} + \alpha^2 \norm {y - \map {\pi_C} x}^2
		| c = {{Defof|Inner Product}}
}}
{{end-eqn}}
By rearranging, we have:
:$2 \alpha \innerprod {x - \map {\pi_C} x} {y - \map {\pi_C} x} \le \alpha^2 \norm {y - \map {\pi_C} x}^2$
Dividing through by $\alpha$, we have:
:$2 \innerprod {x - \map {\pi_C} x} {y - \map {\pi_C} x} \le \alpha \norm {y - \map {\pi_C} x}^2$
Taking $\alpha \to 0^+$, we have:
:$2 \innerprod {x - \map {\pi_C} x} {y - \map {\pi_C} x} \le 0$
Hence:
:$\innerprod {x - \map {\pi_C} x} {y - \map {\pi_C} x} \le 0$
{{qed}}
Theorem: Metric Projection onto Closed Convex Set is 1-Lipschitz

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\R$.
Let $\norm {\, \cdot \,}_\HH$ be the inner product norm.
Let $C \subseteq \HH$ be a non-empty closed convex set.
Let $\pi_C : \HH \to C$ be the metric projection of $\HH$ onto $C$. 
Then $\pi_C$ is $1$-Lipschitz.

Proof:
Let $x, y \in \HH$.
From Bound on Inner Product of Differences with Metric Projection, we have:
:$\innerprod {x - \map {\pi_C} x} {\map {\pi_C} y - \map {\pi_C} x} \le 0$
since $\map {\pi_C} y \in C$.
Similarly:
:$\innerprod {y - \map {\pi_C} y} {\map {\pi_C} x - \map {\pi_C} y} \le 0$
since $\map {\pi_C} x \in C$.
That is, by moving a factor of $-1$:
{{begin-eqn}}
{{eqn	| l = \innerprod {y - \map {\pi_C} y} {\map {\pi_C} x - \map {\pi_C} y}
		| r = \innerprod {-\paren {\map {\pi_C} y - y} } {\map {\pi_C} x - \map {\pi_C} y}
}}
{{eqn	| r = -\innerprod {\map {\pi_C} y - y} {\map {\pi_C} x - \map {\pi_C} y}
}}
{{eqn	| r = \innerprod {\map {\pi_C} y - y} {\map {\pi_C} y - \map {\pi_C} x}
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn	| l = 0
		| o = \ge
		| r = \innerprod {x - \map {\pi_C} x} {\map {\pi_C} y - \map {\pi_C} x} + \innerprod {\map {\pi_C} y - y} {\map {\pi_C} y - \map {\pi_C} x}
}}
{{eqn	| r = \innerprod {x - y} {\map {\pi_C} y - \map {\pi_C} x} + \innerprod {\map {\pi_C} y - \map {\pi_C} x} {\map {\pi_C} y - \map {\pi_C} x}
}}
{{eqn	| r = \innerprod {x - y} {\map {\pi_C} y - \map {\pi_C} x} + \norm {\map {\pi_C} y - \map {\pi_C} x}^2_\HH 
}}
{{end-eqn}}
Hence we have:
:$\norm {\map {\pi_C} y - \map {\pi_C} x}^2_\HH \le \innerprod {y - x} {\map {\pi_C} y - \map {\pi_C} x}$
By the Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces, we then have:
:$\norm {\map {\pi_C} y - \map {\pi_C} x}^2_\HH \le \norm {y - x}_\HH \norm {\map {\pi_C} y - \map {\pi_C} x}_\HH$
If $\map {\pi_C} y \ne \map {\pi_C} x$, we can divide through $\norm {\map {\pi_C} y - \map {\pi_C} x}_\HH$ to obtain:
:$\norm {\map {\pi_C} y - \map {\pi_C} x}_\HH \le \norm {y - x}_\HH$
This holds trivially if $\map {\pi_C} y = \map {\pi_C} x$.
Hence we have:
:$\norm {\map {\pi_C} y - \map {\pi_C} x}_\HH \le \norm {y - x}_\HH$ for all $x, y \in \HH$.
{{qed}}
Theorem: Topological Dual Space of Product of Topological Vector Spaces

Let $\struct {K, \tau_K}$ be a topological field.
Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_Y}$ be topological vector spaces over $K$.
Let $\struct {X \times Y, \tau}$ be the direct product of $X$ and $Y$ equipped with the product topology.
Let $\paren {X \times Y}^\ast$, $X^\ast$ and $Y^\ast$ denote the topological dual spaces.
Then $\phi \in \paren {X \times Y}^\ast$ {{iff}} there exists $\phi_X \in X^\ast$ and $\phi_Y \in Y^\ast$ such that:
:$\map \phi {x, y} = \map {\phi_X} x + \map {\phi_Y} y$ for each $\tuple {x, y} \in X \times Y$. 
That is, $\paren {X \times Y}^\ast$ is isomorphic to the direct product $X^\ast \times Y^\ast$.

Proof:
=== Necessary Condition ===
Let $\phi \in \paren {X \times Y}^\ast$.
For each $\tuple {x, y} \in X \times Y$, we have:
:$\map \phi {x, y} = \map \phi {x, {\mathbf 0}_Y} + \map \phi { {\mathbf 0}_X, y}$
since $\phi$ is linear.
Define $\phi_X : X \to K$ by:
:$\map {\phi_X} x = \map \phi {x, {\mathbf 0}_Y}$ for each $x \in X$.
Define $\phi_Y : Y \to K$ by:
:$\map {\phi_Y} y = \map \phi { {\mathbf 0}_X, y}$ for each $y \in Y$.
We show that $\phi_X \in X^\ast$ and $\phi_Y \in Y^\ast$. 
From Horizontal Section of Linear Transformation is Linear Transformation, we have that $\phi_X$ is linear.
From Vertical Section of Linear Transformation is Linear Transformation, we have that $\phi_Y$ is linear.
From Horizontal Section of Continuous Function is Continuous, we have that $\phi_X$ is continuous.
From Vertical Section of Continuous Function is Continuous, we have that $\phi_Y$ is continuous.
Hence $\phi_X \in X^\ast$ and $\phi_Y \in Y^\ast$ and:
:$\map \phi {x, y} = \map {\phi_X} x + \map {\phi_Y} y$ for each $\tuple {x, y} \in X \times Y$. 
{{qed|lemma}}
Theorem: Intersection of Epigraphs is Epigraph of Supremum

Let $X$ be a set.
Let $\family {f_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of functions $f_\alpha : X \to \overline \R$.
Let $\map {\operatorname {epi} } {f_\alpha}$ be the epigraph of $f_\alpha$ for each $\alpha \in I$.
Let $\ds \map {\operatorname {epi} } {\sup_{\alpha \mathop \in I} f_\alpha}$ be the epigraph of the pointwise supremum $\ds \sup_{\alpha \mathop \in I} f_\alpha$.
Then:
:$\ds \map {\operatorname {epi} } {\sup_{\alpha \mathop \in I} f_\alpha} = \bigcap_{\alpha \mathop \in I} \map {\operatorname {epi} } {f_\alpha}$

Proof:
We have:
{{begin-eqn}}
{{eqn	| l = \bigcap_{\alpha \mathop \in I} \map {\operatorname {epi} } {f_\alpha}
		| r = \bigcap_{\alpha \mathop \in I} \set {\tuple {x, c} \in X \times \R : \map {f_\alpha} x \le c}
		| c = {{Defof|Epigraph}}
}}
{{eqn	| r = \set {\tuple {x, c} \in X \times \R : \map {f_\alpha} x \le c \text { for each } \alpha \in I}
}}
{{eqn	| r = \set {\tuple {x, c} \in X \times \R : \sup_{\alpha \mathop \in I} \map {f_\alpha} x \le c}
}}
{{eqn	| r = \map {\operatorname {epi} } {\sup_{\alpha \mathop \in I} f_\alpha}
		| c = {{Defof|Epigraph}}
}}
{{end-eqn}}
{{qed}}
Category:Epigraphs
eqki5fy27bn59ao9scqxukm5886cfr5
Theorem: Pointwise Infimum of Family of Concave Real Functions is Concave

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $C \subseteq X$ be a non-empty convex set.
Let $\family {f_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of concave functions $f_\alpha : C \to \overline \R$.
Let:
:$\ds f = \inf_{\alpha \mathop \in I} f_\alpha$
be the pointwise infimum of $\family {f_\alpha}_{\alpha \mathop \in I}$.
Then $f$ is concave.

Proof:
Let $\operatorname {hypo}$ denote hypograph.
From Intersection of Hypographs is Hypograph of Infimum, we have:
:$\ds \map {\operatorname {hypo} } f = \bigcap_{\alpha \mathop \in I} \map {\operatorname {hypo} } {f_\alpha}$ 
From Function is Concave iff Hypograph is Convex, we have that:
:$\map {\operatorname {hypo} } {f_\alpha}$ is convex.
From Intersection of Convex Sets is Convex Set (Vector Spaces):
:$\ds \bigcap_{\alpha \mathop \in I} \map {\operatorname {hypo} } {f_\alpha}$ is convex.
From Function is Concave iff Hypograph is Convex, $f$ is concave.
{{qed}}
Category:Hypographs
Category:Concave Real Functions
q2pykcua6kda3a687t17snf11m9grh7
Theorem: Affine Functional is Convex and Concave

Let $X$ be a vector space over $\R$.
Let $f : X \to \R$ be an affine functional on $X$.
Then $f$ is convex and concave.

Proof:
Let $g : X \to \R$ be a linear functional be such that:
:$\map f x = \map g x + c$ for all $x \in X$
for some $c \in \R$. 
Let $x, y \in X$ and $t \in \closedint 0 1$.
Then we have:
{{begin-eqn}}
{{eqn	| l = t \map f x + \paren {1 - t} \map f y
		| r = t \paren {\map g x + c} + \paren {1 - t} \paren {\map g y + c}
}}
{{eqn	| r = t \map g x + \paren {1 - t} \map g y + c 
}}
{{eqn	| r = \map g {t x + \paren {1 - t} y} + c
}}
{{eqn	| r = \map f {t x + \paren {1 - t} y}
}}
{{end-eqn}}
Hence we have both:
:$\map f {t x + \paren {1 - t} y} \le t \map f x + \paren {1 - t} \map f y$ for each $x, y \in X$ and $t \in \closedint 0 1$
and:
:$t \map f x + \paren {1 - t} \map f y \le \map f {t x + \paren {1 - t} y}$ for each $x, y \in X$ and $t \in \closedint 0 1$.
Hence $f$ is both convex and concave.
{{qed}}
Category:Affine Functionals
Category:Convex Real Functions
Category:Concave Real Functions
sh13qnjpkjtjw7izp7dx3mr8jeghz37
Theorem: Concave Envelope is Concave Majorant of Function

Let $\struct {X, \tau}$ be a topological vector space over $\R$.
Let $C \subseteq X$ be a non-empty closed convex set.
Let $f : C \to \R$ be a function.
Let $\hat f : C \to \overline \R$ be the concave envelope of $f$.
Then $\hat f$ is concave and $f \le \hat f$.

Proof:
We have:
:$\map {\hat f} x = \inf \set {\map g x : g \ge f \text { and } g \text { is affine and continuous} }$ for each $x \in C$.
If there exists no continuous affine functional $g \ge f$, then $\map {\hat f} x = \infty$ for each $x \in C$.
Then, from Positive Infinity Function is Convex and Concave, $\hat f$ is concave.
We then immediately have $\map f x \le \map {\hat f} x$ for all $x \in C$.
Now suppose that there exists a continuous affine functional $g \ge f$.
From Pointwise Infimum of Family of Concave Real Functions is Concave, we have that $\hat f$ is concave.
Further, for all $\map g x \in \set {\map g x : g \ge f \text { and } g \text { is affine and continuous} }$, we have:
:$\map g x \ge \map f x$
Hence taking the infimum over $\map g x$ we have:
:$\map {\hat f} x \ge \map f x$
{{qed}}
Category:Concave Envelopes
7m45khvcrt5ok655yshcrwudxnzeu4r
Theorem: Composite of Isomorphisms of Categories is Isomorphism

Let $\mathbf C, \mathbf D$ and $\mathbf E$ be metacategories.
Let $F: \mathbf C \to \mathbf D$ and $G: \mathbf D \to \mathbf E$ be isomorphisms of categories.
Let $GF: \mathbf C \to \mathbf E$ be the composition of $G$ with $F$.
Let $F^{-1}: \mathbf D \to \mathbf C$ and $G^{-1}: \mathbf E \to \mathbf D$ be the inverse isomorphisms of $F$ and $G$ respectively.
Then:
:$G F$ is also an isomorphism of categories.
with inverse:
:$\paren{GF}^{-1} = F^{-1} G^{-1}$
where $F^{-1}: \mathbf D \to \mathbf C$ and $G^{-1}: \mathbf E \to \mathbf D$ denote the inverse isomorphisms of $F$ and $G$ respectively.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren{F^{-1} G^{-1} } \paren{G F}
      | r = F^{-1} \paren{G^{-1} \paren{G F} }
      | c = Composition of Functors is Associative
}}
{{eqn | r = F^{-1} \paren{\paren{G^{-1} G} F}
      | c = Composition of Functors is Associative
}}
{{eqn | r = F^{-1} \paren{\operatorname{id}_{\mathbf D} F}
      | c = {{Defof|Inverse Isomorphism of Categories}}
}}
{{eqn | r = F^{-1} F
      | c = {{Defof|Identity Functor}}
}}
{{eqn | r = \operatorname{id}_{\mathbf C}
      | c = {{Defof|Inverse Isomorphism of Categories}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \paren{G F} \paren{F^{-1} G^{-1} }
      | r = G \paren{F \paren{F^{-1} G^{-1} } }
      | c = Composition of Functors is Associative
}}
{{eqn | r = G \paren{\paren{F F^{-1} } G^{-1} }
      | c = Composition of Functors is Associative
}}
{{eqn | r = G \paren{\operatorname{id}_{\mathbf D} G^{-1} }
      | c = {{Defof|Inverse Isomorphism of Categories}}
}}
{{eqn | r = G G^{-1}
      | c = {{Defof|Identity Functor}}
}}
{{eqn | r = \operatorname{id}_{\mathbf E}
      | c = {{Defof|Inverse Isomorphism of Categories}}
}}
{{end-eqn}}
It follows that $GF$ is an isomorphism of categories with inverse $F^{-1} G^{-1}$ by definition.
{{qed}}
Category:Isomorphisms of Categories
gfm3kinotydxfrdyndsc5f1ivxl7q9s
Theorem: Convex Envelope is Convex Minorant of Function

Let $\struct {X, \tau}$ be a topological vector space over $\R$.
Let $C \subseteq X$ be a non-empty closed convex set.
Let $f : C \to \R$ be a function.
Let $\check f : C \to \overline \R$ be the convex envelope of $f$.
Then $\check f$ is convex and $\check f \le f$.

Proof:
We have:
:$\map {\check f} x = \sup \set {\map g x : g \le f \text { and } g \text { is affine and continuous} }$ for each $x \in C$.
If there exists no continuous affine functional $g \le f$, then $\map {\check f} x = -\infty$ for each $x \in C$.
Then, from Negative Infinity Function is Convex and Concave, $\hat f$ is convex.
We then immediately have $\map {\check f} x \le \map f x$ for each $x \in C$. 
Now suppose that there exists a continuous affine functional $g \le f$.
From Pointwise Infimum of Family of Concave Real Functions is Concave, we have that $\check f$ is concave.
Further, for all $\map g x \in \set {\map g x : g \le f \text { and } g \text { is affine and continuous} }$, we have:
:$\map g x \le \map f x$
Hence taking the supremum over $\map g x$ we have:
:$\map {\check f} x \le \map f x$
{{qed}}
Category:Convex Envelopes
lhn2udl4dtutmfzu8m6bxqfuvljbwn6
Theorem: Characterization of Locally Convex Spaces

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space.
Then $\tau$ is the standard topology of a locally convex space $\struct {X, \PP}$ {{iff}}:
:$\tau$ has a local basis $\BB_{ {\mathbf 0}_X}$ of ${\mathbf 0}_X$ consisting of absorbing, balanced, convex sets.

Proof:
=== Necessary Condition ===
Suppose that $\tau$ is standard topology of a locally convex space $\struct {X, \PP}$.
Let $U \in \tau$ be an open neighborhood of ${\mathbf 0}_X$. 
From Open Sets in Standard Topology of Locally Convex Space, for each $y \in X$ there exists a finite set $F \subseteq \PP$ and $\epsilon > 0$ such that:
:$\set {x \in X : \map p x < \epsilon \text { for each } p \in F} \subseteq U$
where:
:$\set {x \in X : \map p x < \epsilon \text { for each } p \in F}$ is open.
From Open Ball with respect to Seminorm is Convex, Balanced and Absorbing:
:$\set {x \in X : \map p x < \epsilon}$ is absorbing, balanced, convex for each $p \in F$. 
From Finite Intersection of Absorbing Sets is Absorbing:
:$\set {x \in X : \map p x < \epsilon \text { for each } p \in F}$ is absorbing.
From Intersection of Balanced Sets in Vector Space is Balanced:
:$\set {x \in X : \map p x < \epsilon \text { for each } p \in F}$ is balanced.
From Intersection of Convex Sets is Convex Set (Vector Spaces):
:$\set {x \in X : \map p x < \epsilon \text { for each } p \in F}$ is convex.
Let $\map \FF \PP$ be the set of finite subsets of $\PP$.
We have shown that:
:$\BB_{ {\mathbf 0}_X} = \set {\set {x \in X : \map p x < \epsilon \text { for each } p \in F} : F \in \map \FF \PP}$
is a local basis for ${\mathbf 0}_X$ consisting of absorbing, balanced, convex sets.
{{qed|lemma}}
Theorem: Cartesian Product of Convex Sets is Convex in Direct Product

Let $\GF \in \set {\R, \C}$.
Let $I$ be a set. 
Let $\family {X_\alpha}_{\alpha \in I}$ be an $I$-indexed family of vector spaces over $\GF$.
Let:
:$\ds X = \prod_{\alpha \mathop \in I} X_\alpha$ be the direct product of $\family {X_\alpha}_{\alpha \in I}$.
For each $\alpha \in I$, let $C_\alpha \subseteq X_\alpha$ be a non-empty convex set.
Let:
:$\ds C = \prod_{\alpha \mathop \in I} C_\alpha$
Then:
:$C$ is a convex set.

Proof:
Let $x, y \in C$ and $t \in \closedint 0 1$.
Let $\alpha \in I$.
Then $\map x \alpha, \map y \alpha \in C_\alpha$.
We have:
:$\map {\paren {t x + \paren {1 - t} y} } \alpha = t \map x \alpha + \paren {1 - t} \map y \alpha \in C_\alpha$
since $C_\alpha$ is convex.
Hence:
:$\map {\paren {t x + \paren {1 - t} y} } \alpha \in C_\alpha$ for each $\alpha \in I$
we have:
:$\ds t x + \paren {1 - t} y \in \prod_{\alpha \mathop \in I} C_\alpha = C$
Hence $C$ is convex.
{{qed}}
Category:Direct Product of Vector Spaces
Category:Convex Sets (Vector Spaces)
ryrksi2e9nv1634do6atur2nz7p0zcj
Theorem: Direct Product of Locally Convex Spaces is Locally Convex

Let $\GF \in \set {\R, \C}$.
Let $I$ be a non-empty set.
For each $\alpha \in I$, let $\struct {X_\alpha, \PP_\alpha}$ be a locally convex space over $\GF$ with standard topology $\tau_\alpha$.
That is, $\struct {X_\alpha, \PP_\alpha}$ is a locally convex topological vector space.
Let:
:$\ds X = \prod_{\alpha \mathop \in I} X_\alpha$ be the direct product of $\family {X_\alpha}_{\alpha \mathop \in I}$. 
Let $\tau$ be the product topology on $X$.
Then $\tau$ is the standard topology of a locally convex space $\struct {X, \PP}$.
That is, $\struct {X, \tau}$ is a locally convex topological vector space.

Proof:
From Characterization of Locally Convex Spaces: Corollary, it is enough to establish that $\struct {X, \tau}$ has a local basis of ${\mathbf 0}_X$ consisting of convex sets.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$ in $\struct {X, \tau}$.
By Natural Basis of Product Topology, there exists a finite set $F \subseteq I$ such that:
:$\ds \prod_{\alpha \mathop \in I} U_\alpha \subseteq U$
where $U_\alpha = X$ for $\alpha \not \in F$.
Since each $\struct {X_\alpha, \tau_\alpha}$ is locally convex, for each $\alpha \in F$ there exists a convex open neighborhood $V_\alpha$ of ${\mathbf 0}_{X_\alpha}$ with $V_\alpha \subseteq U_\alpha$.
Defining $V_\alpha = X$ for $\alpha \not \in F$, we have:
:$\ds \prod_{\alpha \mathop \in I} V_\alpha \subseteq U$
From Cartesian Product of Convex Sets is Convex in Direct Product:
:$\ds \prod_{\alpha \mathop \in I} V_\alpha$ is convex.
Let $\BB_{ {\mathbf 0}_X}$ be the set of Cartesian products:
:$\ds \prod_{\alpha \mathop \in I} V_\alpha$
where $V_\alpha$ is a convex open neighborhood of ${\mathbf 0}_{X_\alpha}$ for each $\alpha \in I$ and $V_\alpha \ne X$ for only finitely many $\alpha$.
We have shown that $\BB_{ {\mathbf 0}_X}$ is a local basis of $X_\alpha$ consisting of convex sets.
Hence by Characterization of Locally Convex Spaces: Corollary, $\tau$ is the standard topology of a locally convex space $\struct {X, \PP}$.
{{qed}}
Category:Locally Convex Spaces
qiz7uxbmrm131tc8ug35dworivpb42r
Theorem: Characterization of Subgradient in terms of Maximizer of Linear Functional

Let $\innerprod X {X'}_X$ be a dual system over $\R$.
Let $f : X \to \overline \R$ be a proper convex function.
Let $\Dom f$ be the effective domain of $f$. 
Define the canonical duality $\innerprod \cdot \cdot_{X \times \R} : \paren {X \times \R} \times \paren {X' \times \R} \to \R$ by:
:$\innerprod {\tuple {x, \alpha} } {\tuple {f, \lambda} } = \innerprod x f_X + \lambda \alpha$ for each $\tuple {\tuple {x, \alpha}, \tuple {f, \lambda} } \in \paren {X \times \R} \times \paren {X' \times \R}$. 
Let $x \in \Dom f$.
Let $x' \in X'$.
As an abuse of notation, understand $\tuple {x', -1}$ as the linear functional:
:$\tuple {x, \alpha} \mapsto \innerprod {\tuple {x, \alpha} } {\tuple {x', -1} }_{X \times \R}  = \innerprod x {x'}_X - \alpha$
Let $\map {\operatorname {epi} } f$ be the epigraph of $f$. 
Then $x'$ is a subgradient for $f$ {{iff}} $\tuple {x, \map f x}$ is a maximum of $\tuple {x', -1}$ over $\map {\operatorname {epi} } f$.

Proof:
We have that $x'$ is a subgradient for $f$ {{iff}}:
:$\map f y \ge \map f x + \innerprod {y - x} {x'}$ for each $y \in \Dom f$. 
We can write this as:
:$\innerprod x {x'}_X - \map f x \ge \innerprod y {x'}_X - \map f y$ for each $y \in \Dom f$.
Theorem: Distance Travelled during Time Interval

Let $P$ be a particle in space.
Let $\map {\mathbf r} t$ denote the displacement from $\map {\mathbf r} 0$ of $P$ at time $t$ where $0 \le t \le T$ for some instant $T$.
Then the total distance travelled by $P$ between $t = 0$ and $t = T$ is given by:
:$\map s T = \ds \int_0^T \size {\dfrac {\d \mathbf r} {\d t} } \rd t$
where $\size {\dfrac {\d \mathbf r} {\d t} }$ is the magnitude of $\dfrac {\d \mathbf r} {\d t}$.

Proof:
At a given time $t$, the particle $P$ has an instantaneous velocity
:$\map {\mathbf v} t = \dfrac {\d \mathbf r} {\d t}$
During the infinitesimal time interval $\d t$, $P$ moves a distance of:
{{begin-eqn}}
{{eqn | l = \d s
      | r = \size {\map {\mathbf v} t \d t}
      | c = magnitude of displacement is distance
}}
{{eqn | r = \size {\map {\mathbf v} t} \d t
      | c = Absolute Value Function is Completely Multiplicative and $\d t$ is positive
}}
{{end-eqn}}
Therefore, the total distance travelled from $t = 0$ to $t = T$ is given by:
{{begin-eqn}}
{{eqn | l = \map s T
      | r = \int_0^T \rd s
}}
{{eqn | r = \int_0^T \size {\map {\mathbf v} t} \rd t
}}
{{eqn | r = \int_0^T \size {\frac {\d \mathbf r} {\d t} } \rd t
}}
{{end-eqn}}
{{qed}}
Theorem: Perpendicular is Shortest Straight Line from Point to Plane

Let $\PP$ be a plane.
Let $C$ be a point which is not on $\PP$.
Let $D$ be a point on $\PP$ such that $CD$ is perpendicular to $\PP$.
Then the length of $CD$ is less than the length of all other line segments that can be drawn from $C$ to $\PP$.

Proof:
Let $E$ on $\PP$ such that $E$ is different from $D$.
Then $CDE$ forms a right triangle where $CE$ is the hypotenuse.
By Pythagoras's Theorem:
:$CD^2 + DE^2 = CE^2$
and so $CD < CE$.
{{qed}}
Theorem: Vector Equation of Straight Line in Space/Formulation 1

Let $\mathbf a$ and $\mathbf b$ denote the position vectors of two points in space.
Let $L$ be a straight line in space passing through $\mathbf a$ which is parallel to $\mathbf b$.
Let $\mathbf r$ be the position vector of an arbitrary point on $L$.
Then:
:$\mathbf r = \mathbf a + t \mathbf b$
for some real number $t$, which may be positive or negative, or even $0$ if $\mathbf r = \mathbf a$.

Proof:
:300px
Let $a$ and $b$ be points as given, with their position vectors $\mathbf a$ and $\mathbf b$ respectively.
Let $P$ be an arbitrary point on the straight line $L$ passing through $\mathbf a$ which is parallel to $\mathbf b$.
By the parallel postulate, $L$ exists and is unique.
Let $\mathbf r$ be the position vector of $P$.
Let $\mathbf r = \mathbf a + \mathbf x$ for some $\mathbf x$.
Then we have:
:$\mathbf x = \mathbf r - \mathbf a$
As $\mathbf a$ and $\mathbf r$ are both on $L$, it follows that $\mathbf x$ is parallel to $\mathbf b$.
That is:
:$\mathbf x = t \mathbf b$
for some real number $t$.
Hence the result.
{{qed}}