avewright/proofwiki-math huggingface dataset:
Theorem: Open Convex Set in Hausdorff Locally Convex Space is Separated from Points outside Set by Continuous Linear Functional

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$ equipped with its standard topology.
Let $X^\ast$ be the topological dual of $X$.
Let $C$ be an open convex set such that ${\mathbf 0}_X \in C$. 
Let $x_0 \in X \setminus C$. 
Then there exists $f \in X^\ast$ such that: 
:$\map \Re {\map f x} < \map \Re {\map f {x_0} }$
for each $x \in X$.

Proof:
=== Case 1: $\GF = \R$ ===
Let $\mu_C$ be the Minkowski functional of $C$. 
From Minkowski Functional of Convex Absorbing Set is Sublinear Functional, $\mu_A$ is a sublinear functional.
Let $Y = \span \set {x_0}$. 
Define $g : Y \to \R$ by: 
:$\map g {\lambda x_0} = \lambda \map {\mu_C} {x_0}$
for each $\lambda \in \R$.
For $\lambda \in \R_{> 0}$, we have: 
:$\lambda \map {\mu_C} {x_0} = \map {\mu_C} {\lambda x_0}$
by {{SeminormAxiom|3}}.
For $\lambda \in \R_{\le 0}$, we have: 
:$\lambda \map {\mu_C} {x_0} \le 0 \le \map {\mu_C} {\lambda x_0}$
So we have: 
:$\map g y \le \map {\mu_C} y$
for all $y \in Y$. 
From Hahn-Banach Theorem: Real Vector Space, there exists a linear functional $f : X \to \R$  extending $g$ and satisfying: 
:$\map f x \le \map {\mu_C} x$
for each $x \in X$. 
We show that $f$ is continuous.
Since $C$ is an open neighborhood of ${\mathbf 0}_X$, $-C$ is is an open neighborhood of ${\mathbf 0}_X$ by Dilation of Open Set in Topological Vector Space is Open.
Then $U = C \cap \paren {-C}$ is an open neighborhood of ${\mathbf 0}_X$.
From Characterization of Continuous Linear Functionals on Topological Vector Space, it is enough to show that $f$ is bounded on $U$. 
Let $x \in U$.
Then $x \in C$.
From Minkowski Functional of Open Convex Set containing Zero Vector in Topological Vector Space recovers Set, we have: 
:$C = \set {y \in X : \map {\mu_C} x < 1}$
Hence we have that:
:$\map f x \le \map {\mu_C} x < 1$
Since $x \in U$, we also have $x \in -C$.
So:
:$\map f x = -\map f {-x} \ge -\map {\mu_C} x > -1$ 
Hence we deduce that: 
:$\cmod {\map f x} < 1$
for all $x \in U$.
So $f$ is bounded on $U$, and we deduce that $f$ is continuous by Characterization of Continuous Linear Functionals on Topological Vector Space.
{{qed|lemma}}
Theorem: Dual Operator is Well-Defined

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Then the dual operator $T : Y^\ast \to X^\ast$ is well-defined.

Proof:
We first show that $T^\ast$ is well-defined as a mapping $Y^\ast \to X^\ast$. 
That is, we want to show that $T^\ast f \in X^\ast$ for each $f \in Y^\ast$.
For $x, y \in X$ and $\lambda, \mu \in \GF$, we have: 
{{begin-eqn}}
{{eqn	| l = \map {\paren {T^\ast f} } {\alpha x + \beta y}
		| r = \map f {\map T {\alpha x + \beta y} }
}}
{{eqn	| r = \map f {\alpha T x + \beta T y}
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \alpha \map f {T x} + \beta \map f {T y}
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| r = \alpha \map {\paren {T^\ast f} } x + \beta \map {\paren {T^\ast f} } y
}}
{{end-eqn}}
So $T^\ast f$ is a linear functional.
We show that $T^\ast f$ is bounded.
For each $x \in X$, we have:
{{begin-eqn}}
{{eqn	| l = \cmod {\map {\paren {T^\ast f} } x}
		| r = \cmod {\map f {T x} }
}}
{{eqn	| o = \le
		| r = \norm f_{Y^\ast} \norm {T x}_Y
		| c = Fundamental Property of Norm on Bounded Linear Functional
}}
{{eqn	| o = \le
		| r = \norm f_{Y^\ast} \norm T_{\map B {X, Y} } \norm x_X
		| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{end-eqn}}
So $T^\ast f$ is bounded.
{{qed}}
Theorem: Dual Operator is Bounded Linear Transformation

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Then the dual operator $T^\ast : Y^\ast \to X^\ast$ is a bounded linear transformation.
Further: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$

Proof:
Let $f, g \in Y^\ast$ and $\alpha, \beta \in \GF$.
Then we have, for each $x \in X$: 
{{begin-eqn}}
{{eqn	| l = \map {\paren {\map {T^\ast} {\alpha f + \beta g} } } x
		| r = \map {\paren {\alpha f + \beta g} } {T x}
}}
{{eqn	| r = \alpha \map f {T x} + \beta \map g {T x}
}}
{{eqn	| r = \alpha \map {\paren {T^\ast f} } x + \beta \map {\paren {T^\ast g} } x
}}
{{end-eqn}}
So we have: 
:$\map {T^\ast} {\alpha f + \beta g} = \alpha T^\ast f + \beta T^\ast g$
In Dual Operator is Well-Defined, it was shown that for each $f \in Y^\ast$ and $x \in X$ we have: 
:$\cmod {\map {\paren {T^\ast f} } x} \le \norm f_{Y^\ast} \norm T_{\map B {X, Y} } \norm x_X$
So we have: 
:$\norm {T^\ast f}_{X^\ast} \le \norm T_{\map B {X, Y} } \norm f_{Y^\ast}$
for $f \in Y^\ast$. 
Hence: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \le \norm T_{\map B {X, Y} }$
We finally need to show that $\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$.
Pick a sequence $\sequence {x_n}_{n \in \N}$ such that $\norm {x_n}_X = 1$ and: 
:$\ds \norm T_{\map B {X, Y} } - \frac 1 n \le \norm {T x_n}_Y \le \norm T_{\map B {X, Y} }$
so that: 
:$\norm {T x_n}_Y \to \norm T_{\map B {X, Y} }$
For each $n \in \N$, let $f_n \in Y^\ast$ be a support functional for $T x_n$.
Then we have: 
:$\map {\paren {T^\ast f_n} } {x_n} = \map {f_n} {T x_n} = \norm {T x_n}_Y$
so that: 
:$\ds \cmod {\map {\paren {T^\ast f_n} } {x_n} } \ge \paren {\norm T_{\map B {X, Y} } - \frac 1 n} \norm {x_n}_X$
We therefore have: 
:$\ds \paren {\norm T_{\map B {X, Y} } - \frac 1 n} \norm {f_n}_{Y^\ast} \le \norm {T^\ast f_n}_{X^\ast}$
so that: 
:$\ds \norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \ge \norm T_{\map B {X, Y} } - \frac 1 n$
for each $n \in \N$.
Then: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \ge \norm T_{\map B {X, Y} }$
With that, we obtain: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$
{{qed}}
Theorem: Normed Vector Space is Separable iff Weakly Separable

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.
Let $w$ be the weak topology on $X$.
Then $\struct {X, \norm {\, \cdot \,} }$ is separable {{iff}} $\struct {X, w}$ is separable.

Proof:
=== Necessary Condition ===
This follows from Separable Topological Space remains Separable in Coarser Topology.
{{qed}}
Theorem: Separable Topological Space remains Separable in Coarser Topology

Let $\struct {X, \tau_2}$ be a separable topological space. 
Let $\tau_1$ be a topology on $X$ such that:
:$\tau_1 \subseteq \tau_2$
That is, such that $\tau_1$ is coarser than $\tau_2$. 
Then $\struct {X, \tau_1}$ is separable.

Proof:
Let $\cl_1$ and $\cl_2$ denote the topological closure in $\struct {X, \tau_1}$ and $\struct {X, \tau_2}$ respectively.
Let $D$ be a countable everywhere dense subset of $\struct {X, \tau_2}$.
From Topological Closure in Coarser Topology is Larger, we have: 
:$\map {\cl_2} D \subseteq \map {\cl_1} D$
Since $D$ is everywhere dense in $\struct {X, \tau_2}$, we have $\map {\cl_2} D = X$.
So:
:$\map {\cl_1} D = X$
So $D$ is everywhere dense in $\struct {X, \tau_1}$.
{{qed}}
Category:Separable Spaces
3afj20cbu21rhq9vlgy0bz2fbimxrhr
Theorem: Point in Set Closure iff Limit of Net

Let $\struct {X, \tau}$ be a topological space.
Let $E \subseteq X$ be a subset. 
Let $\map \cl E$ be the closure of $E$ in $\struct {X, \tau}$.
Then $x \in \map \cl E$ {{iff}} there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $E$ converging to $x$.

Proof:
=== Necessary Condition ===
Let $x \in \map \cl E$.
Let $\Lambda$ be the set of open neighborhoods of $x$ in $\struct {X, \tau}$. 
From Open Neighborhoods of Point form Directed Ordering, $\struct {\Lambda, \supseteq}$ is a directed set.
For each $U \in \Lambda$, we have $U \cap E \ne \O$ from the definition of closure.
So for each $U \in \Lambda$ we can pick $x_U \in U \cap E$, forming a net $\family {x_U}_{U \in \Lambda}$ valued in $E$.
We show that $\family {x_U}_{U \in \Lambda}$ converges to $x$.
Let $V$ be an open neighborhood of $x$ in $\struct {X, \tau}$.
Then if $U \supseteq V$, we have $x_V \in V \cap E \subseteq U \cap E \subseteq U$.
Hence $\family {x_U}_{U \in \Lambda}$ converges to $x$.
{{qed|lemma}}
Theorem: Characterization of Openness in terms of Nets

Let $\struct {X, \tau}$ be a topological space.
Let $U \subseteq X$. 
Then $U \in \tau$ {{iff}} for each:
:$x \in U$
:directed set $\tuple {\Lambda, \preceq}$
:net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$
there exists $\lambda \in \Lambda$ with $x_\lambda \in U$.

Proof:
=== Necessary Condition ===
Suppose that $U \in \tau$. 
Take $x \in U$, a directed set $\tuple {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$.
By the definition of convergence for net, we have: 
:there exists $\lambda_0 \in \Lambda$ such that for all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$ we have $x_\lambda \in U$.
In particular, $x_{\lambda_0} \in U$.
{{qed|lemma}}
Theorem: Characterization of Continuity in terms of Nets

Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_Y}$ be topological spaces.
Let $f : \struct {X, \tau_X} \to \struct {Y, \tau_Y}$ be a function.
Let $x \in X$. 
Then $f$ is continuous at $x$ {{iff}}:
:for every directed set $\struct {\Lambda, \preceq}$ and net $\family {x_\lambda}_{\lambda \in \Lambda}$ such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, \tau_X}$, we have that: 
::the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$ in $\struct {Y, \tau_Y}$.

Proof:
=== Necessary Condition ===
Suppose that $f$ is continuous at $x$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$.
Let $U$ be an open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$.
Then there exists an open neighborhood $V$ of $x$ in $\struct {X, \tau_X}$ such that: 
:$f \sqbrk V \subseteq U$
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda_0 \in \Lambda$ such that: 
:$x_n \in V$ for all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$. 
Then: 
:$\map f {x_n} \in U$
for $\lambda_0 \preceq \lambda$.
Since $U$ was an arbitrary open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$, we have that $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$. 
{{qed|lemma}}
Theorem: Existence of Upper Bound of Finite Subset of Directed Set

Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\FF \subseteq \Lambda$ be a finite subset.
Then $\FF$ has an upper bound in $\struct {\Lambda, \preceq}$.

Proof:
We induct on the cardinality of $\FF$.
Let $n \in \Z_{\ge 0}$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:if $\FF \subseteq \Lambda$ has $\card \FF = n$, then $\FF$ has an upper bound.
Theorem: Characterization of Convergent Net in Weak Topology

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space over $\GF$ admitting a weak topology $w$. 
Let $X^\ast$ be the topological dual of $X$. 
Let $x \in X$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net in $X$. 
Then: 
:$\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, w}$
{{iff}}:
:for each $f \in X^\ast$, the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$.

Proof:
=== Necessary Condition ===
Suppose that: 
:$\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, w}$.
Let $f \in X^\ast$. 
From Characterization of Continuity of Linear Functional in Weak Topology, $f : \struct {X, w} \to \GF$ is continuous.
From Characterization of Continuity in terms of Nets, we have that: 
:for each $f \in X^\ast$, the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$.
{{qed|lemma}}
Theorem: Characterization of Convergent Net in Weak-* Topology

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space.
Let $X^\ast$ be the topological dual of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X$.
Let $f \in X^\ast$. 
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {f_\lambda}_{\lambda \in \Lambda}$ be a net. 
Then:
:$\family {f_\lambda}_{\lambda \in \Lambda}$ converges to $f$ in $\struct {X^\ast, w^\ast}$
{{iff}}:
:for each $x \in X$, the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ in $\GF$.

Proof:
=== Necessary Condition ===
Suppose that: 
:$\family {f_\lambda}_{\lambda \in \Lambda}$ converges to $f$ in $\struct {X^\ast, w^\ast}$
Let $x \in X$. 
From Characterization of Continuity of Linear Functional in Weak-* Topology, $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Characterization of Continuity in terms of Nets, we have that: 
:for each $f \in X^\ast$, the net $\family {\map {x^\wedge} {f_\lambda} }_{\lambda \in \Lambda} = \family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$.
{{qed|lemma}}
Theorem: Characterization of Hausdorff Property in terms of Nets

Let $\struct {X, \tau}$ be a topological space.
We have that $\struct {X, \tau}$ is Hausdorff {{iff}} convergent nets have a unique limit. 
That is, {{iff}} whenever $\struct {\Lambda, \preceq}$ is a directed set, $x, y \in X$ and $\family {x_\lambda}_{\lambda \in \Lambda}$ is a net converging to $x$ and $y$ in $\struct {X, \tau}$, we have $x = y$.

Proof:
=== Necessary Condition ===
Suppose that $\struct {X, \tau}$ is Hausdorff.
Let $\struct {\Lambda, \preceq}$ is a directed set, $x, y \in X$ and $\family {x_\lambda}_{\lambda \in \Lambda}$ is a net converging to $x$ and $y$ in $\struct {X, \tau}$.
{{AimForCont}} suppose that $x \ne y$. 
Since $\struct {X, \tau}$ is Hausdorff there exists open neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that $U \cap V = \O$.
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda_x \in \Lambda$ such that: 
:$x_\lambda \in U$ for $\lambda_x \preceq \lambda$.
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $y$, there exists $\lambda_y \in \Lambda$ such that: 
:$x_\lambda \in V$ for $\lambda_y \preceq \lambda$.
Since $\struct {\Lambda, \preceq}$ is a directed set, there exists $\lambda_\ast \in \Lambda$ such that: 
:$\lambda_x \preceq \lambda_\ast$ and $\lambda_y \preceq \lambda_\ast$. 
Then: 
:$x_{\lambda_\ast} \in U$
and:
:$x_{\lambda_\ast} \in V$
But then $x_{\lambda_\ast} \in U \cap V = \O$, contradiction. 
So we have $x = y$.
{{qed|lemma}}
Theorem: Characterization of Closedness in terms of Nets

Let $\struct {X, \tau}$ be a topological space.
Let $E \subseteq X$ be a subset. 
Then $E$ is closed {{iff}} for each $x \in X$ we have $x \in E$ {{iff}}:
:there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $E$ converging to $x$.

Proof:
From Set is Closed iff Equals Topological Closure, we have that $E$ is closed {{iff}}:
:$x \in E$ {{iff}} $x \in \map \cl E$
From Point in Set Closure iff Limit of Net, for $x \in X$ we have $x \in \map \cl E$ {{iff}}:
:there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $E$ converging to $x$.
{{qed}}
Category:Nets (Set Theory)
96ap4fppu75ez0sc725d8lptgwlyqev
Theorem: Dual Operator is Weak-* to Weak-* Continuous

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ and $\struct {Y^\ast, \norm {\, \cdot \,}_{Y^\ast} }$ be the normed duals of $X$ and $Y$ respectively.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$.
Then $T^\ast$ is $\struct {w^\ast, w^\ast}$-continuous.

Proof:
From Continuity in Initial Topology, it is enough to show that for each $\Phi \in \struct {X^\ast, w^\ast}^\ast$, we have: 
:$\Phi \circ T^\ast : \struct {Y^\ast, w^\ast} \to \GF$ is Continuous.
From Characterization of Continuity of Linear Functional in Weak-* Topology, we have $\Phi = x^\wedge$ for some $x \in X$, where $x^\wedge$ is the evaluation linear transformation evaluated at $x$.
For $f \in Y^\ast$, we have: 
{{begin-eqn}}
{{eqn	| l = \map {\paren {x^\wedge \circ T^\ast} } f
		| r = \map {x^\wedge} {T^\ast f}
}}
{{eqn	| r = \map {\paren {T^\ast f} } x
		| c = {{Defof|Evaluation Linear Transformation on Normed Vector Space}}
}}
{{eqn	| r = \map {\paren {f \circ T} } x
}}
{{eqn	| r = \map f {T x}
}}
{{eqn	| r = \map {\paren {T x}^\wedge} f
		| c = {{Defof|Evaluation Linear Transformation on Normed Vector Space}}
}}
{{end-eqn}}
So we have: 
:$x^\wedge \circ T^\ast = \paren {T x}^\wedge$
with $T x \in Y$. 
From Characterization of Continuity of Linear Functional in Weak-* Topology , $\paren {T x}^\ast : \struct {Y^\ast, w^\ast} \to \GF$ is continuous.
So $x^\wedge \circ T^\ast : \struct {Y^\ast, w^\ast} \to \GF$ is continuous for each $x \in X$.
So from Continuity in Initial Topology, $T^\ast$ is $\struct {w^\ast, w^\ast}$-continuous.
{{qed}}
Category:Dual Operators
Category:Weak-* Topologies
3vo709v0iojuxi98m2hj1lbc5nehm4k
Theorem: Second Dual Operator Extends Operator

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^{\ast \ast}$ and $Y^{\ast \ast}$ be the second normed duals of $X$ and $Y$ respectively. 
Let $\iota_X : X \to X^{\ast \ast}$ be the evaluation linear transformation on $X$.
Let $\iota_Y : Y \to Y^{\ast \ast}$ be the evaluation linear transformation on $Y$.
Let $T^{\ast \ast} : X^{\ast \ast} \to Y^{\ast \ast}$ be the second dual operator of $T$.
Then $T^{\ast \ast}$ extends $T$ in the sense that: 
:$\map {T^{\ast \ast} } {\iota_X x} = \map {\iota_Y} {T x}$

Proof:
It is shown in Dual Operator is Weak-* to Weak-* Continuous that for each $x \in X$ we have: 
:$x^\wedge \circ T^\ast = \paren {T x}^\wedge$
where $x^\wedge = \iota_X x$ and $\paren {T x}^\wedge = \map {\iota_Y} {T x}$. 
From the definition of the second dual operator and the evaluation linear transformations, this is precisely the claim: 
:$\map {T^{\ast \ast} } {\iota_X x} = \map {\iota_Y} {T x}$
{{qed}}
Category:Second Dual Operators
eqynlalcjkag0u19i4r35dcr7oogpun
Theorem: Characterization of Dual Operator

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.
Let $X^{\ast \ast}$ and $Y^{\ast \ast}$ be the second normed duals of $X$ and $Y$ respectively. 
Let $T : Y^\ast \to X^\ast$ be a bounded linear transformation.
Let $\iota_X : X \to X^{\ast \ast}$ be the evaluation linear transformation for $X$. 
Let $\iota_Y : Y \to Y^{\ast \ast}$ be the evaluation linear transformation for $Y$.
{{TFAE}}
:$(1) \quad$ $T$ is $\struct {w^\ast, w^\ast}$-continuous
:$(2) \quad$ there exists a bounded linear transformation $S : X \to Y$ such that $S^\ast = T$, where $S^\ast$ denotes the dual operator of $S$
:$(3) \quad$ $T^\ast \sqbrk {\iota_X X} \subseteq \iota_Y Y$

Proof:
=== $(1)$ implies $(2)$ ===
Suppose that $T$ is $\struct {w^\ast, w^\ast}$-continuous.
From Characterization of Continuity of Linear Functional in Weak-* Topology , we have that: 
:$x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Composite of Continuous Mappings is Continuous:
:$x^\wedge \circ T : \struct {Y^\ast, w^\ast} \to \GF$ is continuous.
Applying Characterization of Continuity of Linear Functional in Weak-* Topology , there exists $S x \in Y$ such that: 
:$x^\wedge \circ T = \paren {S x}^\wedge$
We first show that $S : X \to Y$ is linear. 
We have:
{{begin-eqn}}
{{eqn | l = \paren {\map S {\alpha x + \beta y} }^\wedge
      | r = \paren {\alpha x + \beta y}^\wedge \circ T
}}
{{eqn | r = \alpha \paren {x^\wedge \circ T} + \beta \paren {y^\wedge \circ T}
      | c = Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual
}}
{{eqn | r = \alpha S x + \beta S y
}}
{{end-eqn}}
We now show that $S$ is bounded. 
We have, for each $x \in X$: 
{{begin-eqn}}
{{eqn | l = \norm {S x}_Y
      | r = \norm {\paren {S x}^\wedge}_{Y^{\ast \ast} }
}}
{{eqn | r = \norm {x^\wedge \circ T}_{Y^{\ast \ast} }
}}
{{eqn | o = \le
      | r = \norm {x^\wedge}_{X^{\ast \ast} } \norm T_{\map B {Y^\ast, X^\ast} }
      | c = Norm on Bounded Linear Transformation is Submultiplicative
}}
{{eqn | r = \norm T_{\map B {Y^\ast, X^\ast} } \norm x_X
      | c = Evaluation Linear Transformation on Normed Vector Space is Linear Isometry
}}
{{end-eqn}}
So $S$ is bounded. 
Finally, for $f \in Y^\ast$ and $x \in X$ we have: 
{{begin-eqn}}
{{eqn | l = \map f {S x}
      | r = \map {\paren {S x}^\wedge} f
}}
{{eqn | r = \map {\paren {x^\wedge \circ T} } f
}}
{{eqn | r = \map {x^\wedge} {T f}
}}
{{eqn | r = \map {\paren {T f} } x
      | c = {{Defof|Evaluation Linear Transformation on Normed Vector Space}}
}}
{{end-eqn}}
That is: 
:$f \circ S = T f$ 
for each $f \in Y^\ast$. 
From the definition of the dual operator we have: 
:$S^\ast f = T f$
for each $f \in Y^\ast$.
So $S^\ast = T$. 
{{qed|lemma}}
Theorem: Characterization of Norm-to-Weak and Weak-to-Weak Continuity of Linear Transformations

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $w_X$ be the weak topology on $X$.
Let $w_Y$ be the weak topology on $Y$.
Let $T : X \to Y$ be a linear transformation. 
{{TFAE}}
:$(1): \quad$ $T$ is $\struct {\norm {\, \cdot \,}_X, \norm {\, \cdot \,}_Y}$-continuous
:$(2): \quad$ $T$ is $\struct {\norm {\, \cdot \,}_X, w_Y}$-continuous
:$(3): \quad$ $T$ is $\struct {w_X, w_Y}$-continuous.

Proof:
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.