Continuity of Seminorms on Vector Spaces

# Continuity of Seminorms on Vector Spaces

Definition: Let $E$ be a vector space. If $p : E \to [0, \infty)$ is a seminorm on $E$ then $p$ is continuous on $E$ if and only if $p$ is continuous at the origin. |

**Proof:**$\Rightarrow$ If $p$ is continuous on $E$ then trivially $p$ is continuous at the origin.

- $\Leftarrow$ Suppose that $p$ is continuous at the origin $o \in E$. Then for each $\epsilon > 0$, $(-\epsilon, \epsilon)$ is a neighbourhood of $p(o) = 0$, and so by the continuity of $p$ at $o$, there exists a neighbourhood $U$ of the origin such that:

\begin{align} \quad p(U) \subseteq (-\epsilon, \epsilon) \end{align}

- Or equivalently, for all $u \in U$, $p(u) < \epsilon$. Let $a \in E$. Then given any $\epsilon > 0$, choose a neighbourhood $U$ as above. Then $U + a$ is a neighbourhood of $a$, and moreover, for all $x \in a + U$ write $x = a + u$ with $u \in U$. Then:

\begin{align} \quad |p(x) - p(a)| \leq p(x - a) = p(a + u - a) = p(u) < \epsilon \end{align}

- So $p$ is continuous at $a$. Thus $p$ is continuous on $E$. $\blacksquare$

Proposition 2: Let $E$ be a vector space. If $A \subseteq E$ is an absolutely convex and absorbent subset of $E$, then the gauge of $A$, $p_A$, is continuous if and only if $A$ is a neighbourhood of the origin, in which case $\mathrm{int}(A) = \{ x : p_A(x) < 1 \}$ and $\overline{A} = \{ x : p(x) \leq 1 \}$. |

**Proof:**$\Rightarrow$ Suppose that $p_A$ is continuous. Since $(-1, 1)$ is an open set in $\mathbb{R}$, we see that $\{ x : p_A(x) < 1 \} = p^{-1}((-1, 1))$ is an open set such that:

\begin{align} \quad \{ x : p_A(x) < 1 \} \subseteq A \end{align}

- So $A$ is a neighbourhood of the origin.

**Claim 1: $\mathrm{int} ( \{ x : p_A(x) \leq 1 \}) = \{ x : p_A(x) < 1 \}$:**Again, since $p_A$ is continuous, $\{ x : p_A(x) < 1 \} = p_A^{-1}((-1, 1))$ is open and is a subset of $\{ x : p_A(x) \leq 1 \}$, so that:

\begin{align} \{ x : p_A(x) < 1 \} \subseteq \mathrm{int} ( \{ x : p_A(x) \leq 1 \}) \end{align}

- For the reverse inclusion, let $y \in \mathrm{int} ( \{ x : p_A(x) \leq 1 \})$. Then there exists a neighbourhood $W$ of the origin such that $y + W \subseteq \{ x : p_A(x) \leq 1 \}$. Since $W$ is absorbent, by one of the propositions on the Properties of Absorbent Sets of Vectors page, there exists a $\lambda > 0$ with $0 < \lambda < 1$ and $\lambda y \in W$. Then $(1 + \lambda)y = y + \lambda x \in y + W$ and thus $(1 + \lambda)y \in \{ x : p_A(x) \leq 1 \}$. Therefore $p_A((1 + \lambda)y) = (1 + \lambda) p_A(y) \leq 1$, implying that $p_A(y) < 1$. So $y \in \{ x : p_A(x) < 1 \}$, which shows that:

\begin{align} \{ x : p_A(x) < 1 \} \supseteq \mathrm{int} (\{ x : p_A(x) \leq 1 \}) \end{align}

- So indeed claim 1 is true. $\mathrm{int} (\{ x : p_A(x) \leq 1 \}) = \{ x : p_A(x) < 1 \}$.

**Claim 2: $\mathrm{cl}( \{ x : p_A(x) < 1 \}) = \{ x : p_A(x) \leq 1 \}$:**Yet again, since $p_A$ is continuous and since $[-1, 1]$ is a closed set in $\mathbb{R}$ we see that $\{ x : p_A(x) \leq 1 \} = p_A^{-1}([-1, 1])$ is closed and is a superset of $\{ x : p_A(x) < 1$. Thus:

\begin{align} \mathrm{cl} ( \{ x : p_A(x) < 1 \} ) \subseteq \{ x : p_A(x) \leq 1 \} \end{align}

- Now for the reverse inclusion, let $y \in \{ x : p_A(x) \leq 1 \}$. If $W$ is a neighbourhood of the origin, then $W$ is necessarily an absorbent set. By one of the propositions on the page mentioned above, for the given $y$, there exists a $\lambda$ with $0 < \lambda < 1$ such that $y \in -\lambda W$.

- We now show that $(y + W) \cap \{ x : p(x) \leq 1 \} \neq \emptyset$. Indeed, the point $(1 - \lambda)x = x - \lambda x \in x + W$, and furthermore $p_A((1 - \lambda)x) = (1 - \lambda) p(x) \leq 1 - \lambda < 1$, so that $(1 - \lambda)x \in \{ x : p_A(x) < 1 \}$. Hence, for each point $y \in \{ x : p(x) \leq 1 \}$, the intersection of every neighbourhood of $y$ with $\{ x : p_A(x) \leq 1 \}$ is nonempty, and thus:

\begin{align} \quad \mathrm{cl} ( \{ x : p_A(x) < 1 \} ) \supseteq \{ x : p_A(x) \leq 1 \} \end{align}

- So indeed claim 2 is true. $\mathrm{cl} ( \{ x : p_A(x) < 1 \}) = \{ x : p_A(x) \leq 1 \}$.

- Lastly, since $A$ is absolutely convex and absorbent, we have that $\{ x : p_A(x) < 1 \} \subseteq A \subseteq \{ x : p_A(x) \leq 1 \}$. Taking the interior of each set in this chain shows us that $\mathrm{int}(A) = \{ x : p_A(x) < 1 \}$, and taking the closure of each set in this chain shows us that $\overline{A} = \{ x : p_A(x) \leq 1 \}$.

- $\Leftarrow$ Since $A$ is absolutely convex and absorbent we have that:

\begin{align} \quad A \subseteq \{ x : p_A(x) < 1 \} \end{align}

- So for each $\epsilon > 0$, we have that:

\begin{align} \quad \epsilon A \subseteq \{ x : p_A(x) < \epsilon \} \end{align}

- So given $\epsilon > 0$, $(-\epsilon, \epsilon)$ is a neighbourhood of $p_A(o) = 0$ and $\epsilon A$ is a neighbourhood of the origin such that $p_A(x) < \epsilon$ for all $x \in \epsilon A$. So $p_A$ is continuous at the origin and by the previous proposition, is continuous on $E$.