Seminorms and Norms on Vector Spaces

# Seminorms and Norms on Vector Spaces

 Definition: Let $E$ be a vector space. A Seminorm on $E$ is a function $p$ on $E$ that has the following properties: (1) $p(x) \geq 0$ for all $x \in E$. (2) $p(\lambda x) = |\lambda| p(x)$ for all $x \in E$ and for all $\lambda \in \mathbf{F}$. (3) $p(x + y) \leq p(x) + p(y)$ for all $x, y \in E$.
 Definition: Let $E$ be a vector space. A Norm on $E$ is a function $p$ on $E$ that has the following properties: (1) $p(x) \geq 0$ for all $x \in E$ AND $p(x) = 0$ if and only if $x = o$. (2) $p(\lambda x) = |\lambda| p(x)$ for all $x \in E$ and for all $\lambda \in \mathbf{F}$. (3) $p(x + y) \leq p(x) + p(y)$ for all $x, y \in E$.

Thus, every norm is a seminorm with the additional property that $p(x) = 0$ if and only if $x = o$.

For an example of a seminorm that is not a norm, consider the vector space $\mathbb{R}^2$ over $\mathbb{R}$. Let $p$ be the function defined for all $(x, y) \in \mathbb{R}^2$ by:

(1)
\begin{align} \quad p(x, y) := |x| \end{align}

Then clearly $p(x, y) \geq 0$ for all $(x, y) \in \mathbb{R}^2$. Furthermore:

(2)
\begin{align} \quad p(\lambda(x, y)) = p(\lambda x, \lambda y) = |\lambda x| = |\lambda||x| = |\lambda| p(x, y) \end{align}

for all $(x, y) \in \mathbb{R}^2$ and for all $\lambda \in \mathbf{F}$. Lastly:

(3)
\begin{align} \quad p((x_1, y_1) + (x_2, y_2)) = p(x_1 + x_2, y_1 + y_2) = |x_1 + x_2| \leq |x_1| + |x_2| = p(x_1, y_1) + p(x_2, y_2) \end{align}

for all $(x_1, y_1), (x_2, y_2) \in \mathbb{R}^2$. So indeed, $p$ is a seminorm on $\mathbb{R}^2$. However, it is NOT a norm on $\mathbb{R}^2$, for $(0, 1) \in \mathbb{R}^2$ is such that $o = (0, 0) \neq (0, 1)$, but $p(0, 1) = 0$.

 Proposition 1: Let $E$ be a vector space and let $p$ be a seminorm on $E$. Then $p^{-1}(0)$ is a subspace of $E$.
• Proof: Clearly $p^{-1}(0) \subseteq E$.
• Let $x, y \in p^{-1}(0)$. Then $p(x) = 0$ and $p(y) = 0$. Since $p$ is a seminorm on $E$ we have that:
(4)
\begin{align} 0 \leq \quad p(x + y) \leq p(x) + p(y) = 0 + 0 = 0 \end{align}
• So $p(x + y) = 0$ and consequently, $(x + y) \in p^{-1}(0)$. So $p^{-1}(0)$ is closed under vector addition.
• Now let $x \in p^{-1}(0)$ and let $\lambda \in \mathbf{F}$. Then $p(x) = 0$. Again, since $p$ is a seminorm on $E$ we have that:
(5)
\begin{align} \quad p(\lambda x) = |\lambda| p(x) = |\lambda| 0 = 0 \end{align}
• So $\lambda x \in p^{-1}(0)$. So $p^{-1}(0)$ is closed under scalar multiplication too.
• Hence $p^{-1}(0)$ is a vector subspace of $E$. $\blacksquare$
 Proposition 2: Let $E$ be a vector space and let $p$ be a seminorm on $E$. Then $|p(x) - p(y)| \leq p(x - y)$ for all $x, y \in E$.
• Proof: Let $x, y \in E$. Since $p$ is a seminorm, by (3) we have that:
(6)
\begin{align} p(x) = p(y + x - y) \leq p(y) + p(x - y) \quad \mathrm{and} \quad p(y) = p(x + y - x) \leq p(x) + p(y - x) = \end{align}
• for all $x, y \in E$. Combining these inequalities yields:
(7)
\begin{align} p(y) - p(y - x) \leq p(x) \leq p(y) + p(x - y) \end{align}
• for all $x, y \in E$, or equivalently:
(8)
\begin{align} \quad -p(y - x) \leq p(x) - p(y) \leq p(x - y) \end{align}
• for all $x , y \in E$. But $p(y - x) = p(-1(x - y)) = |-1| p(x - y) = p(x - y)$ by (2). Thus:
(9)
\begin{align} \quad |p(x) - p(y)| \leq p(x - y) \end{align}
• for all $x, y \in E$. $\blacksquare$
 Proposition 3: Let $E$ be a vector space. If $p$ and $q$ are seminorms on $E$ with the property that $p(x) < 1$ implies that $q(x) \leq 1$ then $q(x) \leq p(x)$ for all $x \in E$.
• Proof: Suppose instead that $q(x) > p(x)$ for some $x \in E$. Let $e \in E$ and a number $\alpha > 0$ such that:
(10)
\begin{align} 0 \leq p(e) < \alpha < q(e) \end{align}
• Dividing each side of the inequality by $\alpha$ and using (2), we get that:
(11)
\begin{align} \quad 0 \leq p \left ( \frac{e}{\alpha} \right ) < 1 < q \left ( \frac{e}{\alpha} \right ) \end{align}
• But then $p \left ( \frac{e}{\alpha} \right ) < 1$ does not imply that $q \left ( \frac{e}{\alpha} \right ) \leq 1$ - a contradiction. So the assumtpion that $q(x) > p(x)$ for some $x \in E$ is false. So $q(x) \leq p(x)$ for all $x \in E$. $\blacksquare$