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 $\mathbf 0_V$.
$\Box$
$(2)$ implies $(1)$
If $L$ is continuous at $\mathbf 0_V$, then for all positive real numbers $\epsilon$ there exists some $\delta > 0$ such that:
- for all $x \in V$ with $\norm {x - \mathbf 0_V} < \delta$ we have $\size {L x - \map L {\mathbf 0_V} } < \varepsilon$.
That is, from Linear Functional fixes Zero Vector:
- for all $x \in V$ with $\norm x < \delta$ we have $\size {L x} < \epsilon$.
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.
$\Box$
$(2)$ iff $(3)$
$(2)$ implies $(3)$
Clearly if $L$ is continuous at $\mathbf 0_V$, it is continuous at some point.
$\Box$
$(3)$ implies $(2)$
Suppose that $L$ is continuous at $x_0 \in V$.
Let $\epsilon$ be a positive real number.
Then, there exists $\delta > 0$ such that:
- for all $x \in V$ such that $\norm {\paren {x + x_0} - x_0} < \delta$, we have $\size {\map L {x + x_0} - L x_0} < \epsilon$.
That is:
- for all $x \in V$ with $\norm x < \delta$, we have $\size {\map L {x + x_0} - L x_0} < \epsilon$
By the definition of a linear functional, we therefore have:
- for all $x \in V$ with $\norm x < \delta$ we have $\size {L x} < \epsilon$.
That is:
- $L$ is continuous at $\mathbf 0_V$.
$\Box$
$(1)$ 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:
- for all $x \in V$ with $\norm x < \delta$, we have $\size {L x} < 1$
Note that for any $x \in V$ with $x \ne \mathbf 0_V$, we have:
- $\ds \norm {\frac x {\norm x} } = \frac {\norm x} {\norm x} = 1$
so that:
- $\ds \norm {\frac \delta 2 \times \frac x {\norm x} } = \frac \delta 2 < \delta$
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:
\(\ds \size {L x}\) | \(=\) | \(\ds \size {\map L {\frac {\delta x} {2 \norm x} \times \frac {2 \norm x} \delta} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \norm x} \delta \size {\map L {\frac {\delta x} {2 \norm x} } }\) | Definition of Linear Functional | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac {2 \norm x} \delta\) |
So, for all $v \in V \setminus \set {\mathbf 0_V}$, we have:
- $\size {L v} < c \norm v$
with:
- $c = \dfrac {2 \norm v} \delta$
For $v = \mathbf 0_V$, we have:
- $\size {L v} = c \norm v$
So, we have:
- $\size {L v} \le c \norm v$
for all $v \in V$.
$\Box$
$(4)$ 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.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: Proposition $3.1$