Continuity of Linear Functionals

== Theorem ==
Let $V$ be a normed vector space, and let $L$ be a linear functional on $V$.
Then the following four statements are equivalent:
:$(1): \quad L$ is continuous
:$(2): \quad L$ is continuous at $\mathbf 0_V$
:$(3): \quad L$ is continuous at some point
:$(4): \quad L$ is bounded: $\exists c > 0: \forall v \in H: \size {L v} \le c \norm v$
== Proof ==
=== $(1)$ iff $(2)$ ===
==== $(1)$ implies $(2)$ ====
Clearly if $L$ is continuous, then in particular it is continuous at $\mathbf 0_V$. 
==== $(2)$ implies $(1)$ ====
If $L$ is continuous at $\mathbf 0_V$, then for all positive real numbers $\epsilon$ there exists some $\delta > 0$ such that: 
:$\forall x \in V: \norm {x - \mathbf 0_V} < \delta \implies \size {L x - \map L {\mathbf 0_V} } < \varepsilon$
That is, from Linear Functional fixes Zero Vector: 
:$\forall x \in V: \norm x < \delta \implies \size {L x} < \epsilon$
Fix $x \in V$ and let $y \in V$. 
We have from the definition of a linear functional: 
:$\size {L x - L y} = \size {\map L {x - y} }$
For any $y \in V$ with: 
:$\norm {x - y} < \delta$
we have: 
:$\size {\map L {x - y} } < \epsilon$
That is: 
:$\size {L x - L y} < \epsilon$
So:
:$L$ is continuous at $x \in V$. 
Since $x \in V$ was arbitrary, we have: 
:$L$ is continuous.
=== $(2)$ iff $(3)$ ===
==== $(2)$ implies $(3)$ ====
Clearly if $L$ is continuous at $\mathbf 0_V$, it is continuous at some point.
==== $(3)$ implies $(2)$ ====
Suppose that $L$ is continuous at $x_0 \in V$. 
Let $\epsilon$ be an arbitrary positive real number. 
Then, there exists $\delta > 0$ such that:
:$\forall x \in V: \norm {\paren {x + x_0} - x_0} < \delta \implies \size {\map L {x + x_0} - L x_0} < \epsilon$
That is: 
:$\forall x \in V: \norm x < \delta \implies \size {\map L {x + x_0} - L x_0} < \epsilon$
We have, from the definition of a linear functional:
:$\forall x \in V: \norm x < \delta \implies \size {L x} < \epsilon$
Hence: 
:$L$ is continuous at $\mathbf 0_V$. 
=== $(1)$ iff $(4)$ ===
==== $(1)$ implies $(4)$ ====
Since $L$ is continuous, it is in particular continuous at $\mathbf 0_V$.
So, there exists $\delta > 0$ such that:
:$\forall x \in V: \norm x < \delta \implies \size {L x} < 1$
Note that for any $x \in V$ with $x \ne \mathbf 0_V$, we have: 
:$\ds \norm {\frac x {\norm x} } = \frac {\norm x} {\norm x} = 1$
Thus: 
:$\ds \norm {\frac \delta 2 \times \frac x {\norm x} } = \frac \delta 2 < \delta$
So, we have: 
:$\ds \size {\map L {\frac \delta 2 \times \frac x {\norm x} } } < 1$ 
We can write: 
:$\ds x = \frac {\delta x} {2 \norm x} \times \frac {2 \norm x} \delta$ 
to obtain: 
So, for all $v \in V \setminus \set {\mathbf 0_V}$, we have: 
:$\size {L v} < c \norm v$
with: 
:$c = \dfrac {2} \delta$ 
For $v = \mathbf 0_V$, we have: 
:$\size {L v} =\size {L 0_V} = 0 = \dfrac 2 \delta \norm {0_V} = c \norm v$ 
Hence, $\exists c > 0$ such that: 
:$\size {L v} \le c \norm v$
for all $v \in V$. 
==== $(4)$ implies $(1)$ ====
Let $\epsilon$ be a positive real number.
Suppose that there exists $c > 0$ such that: 
:$\size {L v} \le c \norm v$
for $v \in V$.
Fix $x \in V$.
Then, for any $y \in V$, we have: 
:$\size {\map L {x - y} } \le c \norm {x - y}$ 
that is: 
:$\size {L x - L y} \le c \norm {x - y}$ 
So, whenever $y$ is such that: 
:$\norm {x - y} < \epsilon/c$
we have: 
:$\size {L x - Ly} < \epsilon$
That is:
:$L$ is continuous at $x \in V$. 
Since $x \in V$ was arbitrary, we have: 
:$L$ is continuous.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: Proposition $3.1$
Category:Linear Functionals
Category:Normed Vector Spaces