Continuity of Linear Functionals

Theorem
Let $H$ be a Hilbert space, and let $L$ be a linear functional on $H$.

Then the following four statements are equivalent:


 * $(1): \quad L$ is continuous
 * $(2): \quad L$ is continuous at $\mathbf 0_H$
 * $(3): \quad L$ is continuous at some point
 * $(4): \quad \exists c > 0: \forall h \in H: \size {L h} \le c \norm h$

$(1)$ implies $(2)$
Clearly if $L$ is continuous, then in particular it is continuous at $\mathbf 0_H$.

$(2)$ implies $(1)$
If $L$ is continuous at $\mathbf 0_H$, then for all positive real numbers $\epsilon$ there exists some $\delta > 0$ such that:


 * for all $x \in H$ with $\norm {x - \mathbf 0_H} < \delta$ we have $\size {L x - \map L {\mathbf 0_H} } < \varepsilon$.

That is, from Linear Functional fixes Zero Vector:


 * for all $x \in H$ with $\norm x < \delta$ we have $\size {L x} < \epsilon$.

Fix $x \in H$ and let $y \in H$.

We have from the definition of a linear functional:


 * $\size {L x - L y} = \size {\map L {x - y} }$

For any $y \in H$ 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 $x \in H$.

Since $x \in H$ was arbitrary, we have:


 * $L$ is continuous.

$(2)$ implies $(3)$
Clearly if $L$ is continuous at $\mathbf 0_H$, it is continuous at some point.

$(3)$ implies $(2)$
Suppose that $L$ is continuous at $x_0 \in H$.

Let $\epsilon$ be a positive real number.

Then, there exists $\delta > 0$ such that:


 * for all $x \in H$ 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 $x \in H$ 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 $x \in H$ with $\norm x < \delta$ we have $\size {L x} < \epsilon$.

That is:


 * $L$ is continuous at $\mathbf 0_H$.

$(1)$ implies $(4)$
Since $L$ is continuous, it is in particular continuous at $\mathbf 0_H$.

So, there exists $\delta > 0$ such that:


 * for all $x \in H$ with $\norm x < \delta$, we have $\size {L x} < 1$

Note that for any $x \in H$ with $x \ne \mathbf 0_H$, 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:

So, for all $h \in H \setminus \set {\mathbf 0_H}$, we have:


 * $\size {L h} < c \norm h$

with:


 * $c = \dfrac {2 \norm h} \delta$

For $h = \mathbf 0_H$, we have:


 * $\size {L h} = c \norm h$

So, we have:


 * $\size {L h} \le c \norm h$

for all $h \in H$.

$(4)$ implies $(1)$
Let $\epsilon$ be a positive real number.

Suppose that there exists $c > 0$ such that:


 * $\size {L h} \le c \norm h$

for $h \in H$.

Fix $x \in H$.

Then, for any $y \in H$, 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 $x \in H$.

Since $x \in H$ was arbitrary, we have:


 * $L$ is continuous.