Characterization of Continuous Linear Functionals on Topological Vector Space

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a topological vector space over $\Bbb F$.

Let $f$ be a linear functional on $X$ such that $\map f x \ne 0$ for some $x \in X$.

The following statements are equivalent:

$(1) \quad$ $f$ is continuous

$(2) \quad$ $\ker f$ is closed, where $\ker f$ is the kernel of $f$

$(3) \quad$ $\ker f$ is not everywhere dense in $X$

$(4) \quad$ $f$ is bounded in a open neighborhood of ${\mathbf 0}_X$.

Proof

$(1)$ implies $(2)$

Suppose that $f$ is continuous.

From the definition of the kernel, we have:

$\ker f = f^{-1} \sqbrk {\set 0}$

Since $\set 0$ is closed in $\GF$, we have that:

$f^{-1} \sqbrk {\set 0}$ is closed

from Continuity Defined from Closed Sets.

$\Box$

$(2)$ implies $(3)$

Suppose that $\ker f$ is closed.

Since $\map f x \ne 0$ for some $x \in X$, we have $\ker f \ne X$.

From Set is Closed iff Equals Topological Closure, we have:

$\paren {\ker f}^- = \ker f \ne X$

where $\paren {\ker f}^-$ denotes the closure of $\ker f$.

So $\ker f$ is not everywhere dense in $X$.

$\Box$

$(3)$ implies $(4)$

Suppose that $\ker f$ is not everywhere dense in $X$.

From Everywhere Dense iff Interior of Complement is Empty, it follows that the interior of $X \setminus \ker f$ is non-empty.

That is, there exists a non-empty open set $W$ disjoint from $\ker f$.

Let $x \in W$.

From Translation of Open Set in Topological Vector Space is Open, $U = W - x$ is an open neighborhood of ${\mathbf 0}_X$.

From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists a balanced open neighborhood $V$ of ${\mathbf 0}_X$ with $V \subseteq U$ with:

$\paren {x + V} \cap \ker f = \O$

From Image of Balanced Set under Linear Transformation is Balanced, $f \sqbrk V$ is a balanced subset of $\Bbb F$.

Applying:

Balanced Subset of Real Numbers is Bounded or Entire Space if $\Bbb F = \R$

Balanced Subset of Complex Plane is Bounded or Entire Space if $\Bbb F = \C$

we have that $f \sqbrk V$ is:

a bounded subset of $\R$ if $\Bbb F = \R$

a bounded subset of $\C$ if $\Bbb F = \C$

or $f \sqbrk V = \Bbb F$.

If $f \sqbrk V$ is a bounded subset of $\R$ or a bounded subset of $\C$, $(4)$ immediately follows since $V$ is an open neighborhood of ${\mathbf 0}_X$.

Now suppose that $f \sqbrk V = \Bbb F$.

Then there exists $y \in V$ such that:

$\map f y = -\map f x$

Then, we have $\map f {x + y} = 0$ from linearity, and so $x + y \in \ker f$.

But since $y \in V$, we have $x + y \in x + V$.

So:

$x + y \in \paren {x + V} \cap \ker f$.

This contradicts that $\paren {x + V} \cap \ker f = \O$.

So we must have that $f \sqbrk V$ is a bounded subset of $\R$ or a bounded subset of $\C$, so $(4)$ holds.

$\Box$

$(4)$ implies $(1)$

Suppose that $f$ is bounded in a open neighborhood of ${\mathbf 0}_X$.

From Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin, it suffices to establish that $f$ is continuous at ${\mathbf 0}_X$.

Let $V$ be an open neighborhood of $0 \in \Bbb F$.

Using:

the definition of an open set in $\R$ if $\Bbb F = \R$

the definition of an open set in $\C$ if $\Bbb F = \C$

there exists $r > 0$ such that if $\alpha \in \Bbb F$ has $\cmod \alpha < r$ we have $\alpha \in V$.

Since $f$ is bounded in a open neighborhood of ${\mathbf 0}_X$, there exists an open neighborhood $U$ of ${\mathbf 0}_X$ and $M > 0$ such that:

$\cmod {\map f x} < M$ for each $x \in U$.

Set:

$\ds W = \frac r M V$

From Dilation of Open Set in Topological Vector Space is Open, $W$ is an open neighborhood of ${\mathbf 0}_X$.

For $x \in W$, we then have:

$\cmod {\map f x} < r$

So:

$\map f x \in V$ for $x \in W$.

That is:

$f \sqbrk W \subseteq V$

Since $V$ was arbitrary, we have that $f$ is continuous at ${\mathbf 0}_X$ as desired.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.18$: Theorem