> [!definition] > > Let $E$ be a $\real$-[[Vector Space|vector space]], a mapping $\norm{\cdot}: E \to \real^+$ is a **quasi-norm** if: > 1. $\norm{x} = 0$ if and only if $x = 0$. > 2. $\norm{x + y} \le \norm{x + y}$ for all $x \in E$. > 3. $\norm{x} = \norm{-x}$ for all $x \in E$. > 4. $\lim_{\lambda \to 0}\norm{\lambda x} = 0$ for all $x \in E$. > 5. $\limv{n}\norm{\lambda x_n} = 0$ for any $\seq{x_n} \subset E$ with $\norm{x_n} \to 0$. > > The pair $(E, \norm{\cdot})$ is a **quasi-normed** space. In particular, the metric associated with each [[Fréchet Space|Fréchet space]] is a quasi-norm. > [!theorem] > > Let $K \subset \real$ be bounded and $\seq{x_n} \subset E$ with $\norm{x_n} \to 0$, then $\norm{\lambda x_n} \to 0$ uniformly across all $\lambda \in K$. In particular, this implies that > 6. For any $\seq{\lambda_n} \subset \real$ and $\seq{x_n} \subset E$, if $\lambda_n \to 0$ and $\norm{x_n} \to 0$ as $n \to \infty$, then $\norm{\lambda_n x_n} \to 0$ as $n \to \infty$. > 7. Using $d(x, y) = \norm{x - y}$, $E$ is a [[Topological Vector Space|topological vector space]]. > > *Proof*. For each $n \in \nat$, let $f_n(\lambda) = \norm{\lambda x}$, then $f_n \to 0$ almost everywhere. By [[Egoroff's Theorem]], there exists $A \subset \real$ of positive measure, on which $f_n \to 0$ uniformly. In which case, $A - A$ contains an open interval $(-\eps, \eps)$ centred at $0$. For any $\lambda \in (-\eps, \eps)$, there exists $\mu_1, \mu_2 \in A$ such that $\lambda = \mu_1 - \mu_2$ (where $(3)$ is used here), which yields > $ > f_n(\lambda) \le f_n(\mu_1) + f_n(\mu_2) \to 0 > $ > uniformly as $n \to \infty$. > > Now let $R > 0$, then there exists $N \in \nat$ such that $R < N\eps$. So for any $\lambda \in (-R, R)$, > $ > f_n(\lambda) \le Nf(\lambda/N) \to 0 > $ > uniformly as $\lambda/N \in (-\eps, \eps)$. > [!theorem] > > Let $B \subset E$, be bounded, then $\sup_{x \in B}\norm{x} < \infty$. > > Remark: the converse does not necessarily hold. > > *Proof*. Let $\eps > 0$, then there exists $R > 0$ such that $RB(0, \eps) \supset B$. Let $N \in \nat$ such that $N \ge R$, then for any $x \in NB(0, \eps)$, write $x = N \cdot x/N$ where $x/N \in B(0, \eps)$. This gives $\norm{x} \le N \norm{x/N} < N\eps$, which yields $B \subset NB(0, \eps) \subset B(0, N\eps)$. Therefore $\sup_{x \in B}\norm{x} < \infty$. > [!theorem] > > Let $\seq{x_n} \subset E$ such that $x_n \to 0$, then $\seq{x_n}$ is bounded. > > *Proof*. Let $\eps > 0$, then there exists $N \in \nat$ such that $x_n \subset B(0, \eps)$ for all $n \ge N$. From here, for each $n < N$, there exists $R_n \ge 0$ such that $x_n \in R_nB(0, \eps)$ by continuity of scalar multiplication. Choose $R = \max(1, \max_{1 \le n < N}R_n)$, then $\seq{x_n} \subset R_n B(0, \eps)$.