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 $3.1$