> [!definition] > > Let $\cx$ be a [[Normed Vector Space|normed space]] and $\cm$ be a closed subspace. The [[Quotient Space|quotient space]] $\cx/\cm$ is a normed space under the **quotient norm** > $ > \norm{\ol{x}}_q = \inf_{y \in \ol{x}}\norm{y} > $ > *Proof*. > > ### Triangle Inequality > > The quotient norm satisfies the triangle inequality. > > *Proof*. Let $\ol{x}, \ol{y} \in \cx/\cm$. then for any $x' \in \ol{x}$ and $y' \in \ol{y}$, > $ > \norm{\ol{x + y}}_q = \norm{\ol{x' + y'}}_q \le \norm{x' + y'} \le \norm{x'} + \norm{y'} > $ > Since this applies to every estimate used for $\norm{\ol{x}}_q$ and $\norm{\ol{y}}_q$, $\norm{\ol{x + y}}_q \le \norm{\ol{x}}_q + \norm{\ol{y}}_q$. > > ### Scaling > > Let $\ol{x} \in \cx/\cm$ and $\lambda \in \complex$, then $\norm{\ol{\lambda x}}_q = \abs{\lambda}\norm{\ol{x}}_q$. > > *Proof*. Let $x' \in \ol{x}$, then > $ > \begin{align*} > \norm{\ol{\lambda x}}_q = \inf_{y \in \ol{\lambda x}}{\norm{y}} \le \norm{\lambda x'} &= \abs{\lambda}\norm{x'} > \end{align*} > $ > Since this applies to every estimate used for $\norm{\ol{x}}_q$, we have $\norm{\ol{\lambda x}}_q \le \abs{\lambda} \norm{\ol{x}}_q$. If $\lambda = 0$, then $\norm{\ol{\lambda x}}_q = 0 = \abs{\lambda}\norm{\ol{x}}_q$. If not, applying the result again yields > $ > \begin{align*} > \norm{\ol{\lambda^{-1}\lambda x}}_q&\le \abs{\lambda^{-1}}\norm{\ol{\lambda x}}_q \\ > \norm{\ol{x}}_q &\le \abs{\lambda^{-1}}\norm{\ol{\lambda x}}_q \\ > \abs{\lambda} \norm{\ol{x}}_q &\le \abs{\lambda} \abs{\lambda^{-1}}\norm{\ol{\lambda x}}_q \\ > \abs{\lambda} \norm{\ol{x}}_q &\le \norm{\ol{\lambda x}}_q > \end{align*} > $ > > > ### Norm > > Let $\ol{x} \in \cx/\cm: \ol{x} \ne 0$. If $\cm$ is closed, then $\norm{\ol{x}}_q > 0$. > > *Proof*. If $X$ is a metric space and $M$ is a subset, then $x \not\in \ol{M}$ if and only if $d(x, M) = \inf_{y \in M}d(x, m) > 0$. To see this, $x \not\in \ol{M}$ if and only if it is not an [[Topological Closure|adherent point]] of $\ol{M}$, which is only true when there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset M^c$. This is equivalent to when $d(x, m) > \varepsilon > 0$ for all $m \in M$, which only happens when $d(x, M) \ge \varepsilon > 0$. > > Now, if $\cm$ is closed, then there exists $\varepsilon > 0$ such that $\norm{x - m} > \varepsilon$ for all $m \in \cm$. Therefore $\norm{\ol{x}}_q \ge \varepsilon > 0$. # Continuity of Projection > [!theorem] > > Let $\cx$ be a normed space and $\cm$ be a properly closed subspace. Then for any $\varepsilon > 0$ there exists $x \in \cx$ such that $\norm{x} = 1$ and $\norm{x + \cm}_q \ge 1 - \varepsilon$. > > *Proof*. Let $x + \cm \in \cx/\cm$ such that $\norm{x + \cm}_q = 1$, which exists since $\cm$ is a proper subspace. If $\cm = \bracs{0}$, then $\norm{x} = 1$ as well and we are done. > > If $\cm \supsetneq \bracs{0}$, let $x' = (1 - \varepsilon)x$, then $\norm{x' + \cm}_q = 1 - \varepsilon$. Let $y \in x' + \cm$ such that $\norm{y} < 1$, and $m \in \cm: m \ne 0$. Consider the function $\phi: \real \to \real$ by $\phi(t) = \norm{y + tm}$, then $\phi$ is continuous with $\phi(0) = \norm{y} < 1$. Moreover, > $ > \phi(t) = \norm{y + tm} \ge \norm{tm} - \norm{y} = t\norm{m} - \norm{y} > $ > there exists $t \in \real$ such that $t\norm{m} > \norm{y} + 1$ and $\phi(t) > 1$. By the intermediate value theorem, there exists $t' \in [0, t]$ such that $\phi(t) = 1$. Taking $y' = y + t'm$, then $\norm{y'} = 1$ and $\norm{y' + \cm} = 1 - \varepsilon$. > [!theorem]- > > Let $\cx$ be a normed space and $\cm$ be a properly closed subspace. Then for any $\varepsilon > 0$ there exists $x \in \cx$ such that $\norm{x} = 1$ and $\norm{\ol{x}}_q > 1 - \varepsilon$. > > *Proof*. > > ![[norm_shifting.png]] > > Let $\ol{x} \in \cx/\cm: \norm{\ol{x}}_q = 1$ and $\varepsilon > 0$. Choose $\delta > 0: \frac{1}{1 + \delta} > 1- \varepsilon$. Then there exists $x' \in \ol{x}$ such that $1 \le \norm{x'} < 1 + \delta$. Let > $ > \hat{x'} = \frac{x - m}{\norm{x - m}} \quad \norm{\hat{x'}} = 1 > $ > then > $ > \begin{align*} > \norm{\ol{\hat{x'}}}_q &= \norm{\frac{\ol{x - m}}{\norm{x - m}}}_q \\ > &= \frac{\norm{\ol{x - m}}_q}{\norm{x - m}} \\ > &= \frac{1}{\norm{x - m}} \\ > &> \frac{1}{1 + \delta} \\ > &> 1 - \varepsilon > \end{align*} > $ > [!theorem] > > Let $\cx$ be a normed space and $\cm$ be a closed subspace, then the map $\pi: \cx \to \cx/\cm$ with $x \mapsto \ol{x}$ has [[Bounded Linear Map|operator norm]] $1$. > > *Proof*. Let $x \in \cx$ such that $\norm{x} = 1$ and $\norm{\ol{x}}_q > 1 - \varepsilon$, then > $ > 1 - \varepsilon< \norm{\ol{x}}_q \le \norm{\pi} \le \norm{x - 0} = \norm{x} = 1 > $ > Since this applies to all $\varepsilon > 0$, $\norm{\pi} = 1$. # Preserving Completeness > [!theorem] > > Let $\cx$ be a [[Banach Space|Banach space]] and $\cm$ be a closed subspace, then $\cx/\cm$ is complete. > > *Proof*. Let $\seq{\ol{x_n}} \subset \cx/\cm$ be a sequence such that $\sum_{n \in \nat}\norm{\ol{x_n}}_q < \infty$. Let $\varepsilon > 0$ and choose for each $n \in \nat$, $m_n \in \cm$ such that $\norm{x_n + m_n} < \norm{\ol{x_n}}_q + \varepsilon/2^n$. Then since > $ > \sum_{n \in \nat}\norm{x_n + m_n} \le 2\varepsilon + \sum_{n \in \nat}\norm{\ol{x_n}}_q > $ > and $\cx$ is complete, $\sum_{n = 1}^{\infty}(x_n + m_n) = x$ for some $x \in \cx$. In which case, > $ > \begin{align*} > \norm{\sum_{n = 1}^{N}\ol{x_n} - \ol{x}}_q &= > \norm{\sum_{n = 1}^{N}\ol{x_n} - \sum_{n = 1}^{N}\ol{x_n - m_n} - \sum_{n = N + 1}^{\infty}\ol{x_n - m_n}}_q \\ > &= \norm{\sum_{n = N + 1}^{\infty}\ol{x_n - m_n}}_q \\ > &\le \sum_{n = N + 1}^{\infty}\norm{\ol{x_n - m_n}}_q \to 0 > \end{align*} > $ > and we have $\sum_{n = 1}^{N}\ol{x_n} \to \ol{x}$. # Relation to the Quotient Topology > [!theorem] > > Let $\cx$ be a normed space and $\cm$ be a closed subspace. The norm topology on $\cx/\cm$ is the same as the [[Quotient Topology|quotient]] of the norm topology on $\cx$. > > *Proof*. For notation clarity, denote $\topo$ as the norm topology on $\cx$, $\ol{\topo}$ as the quotient norm topology on $\cx/\cm$, and $\tilde \topo$ as the quotient topology on $\cx/\cm$. > > ### Norm Topology in Quotient Topology > > Let $U \in \ol{\topo}$, then $U \in \tilde \topo$ ($\ol\topo \subset \tilde \topo$). > > *Proof*. Since the projection map $\pi: \cx \to \cx/\cm$ is continuous with respect to $\ol\topo$, $\pi^{-1}(U) \in \topo$ for any $U \in \ol\topo$. Therefore $\pi^{-1}(U) \in \tilde \topo$. > > > ### Translation by Subspace > > Let $E \subset \cx/\cm$ and $F \subset \pi^{-1}(E)$, then $F + m \subset \pi^{-1}(E)$ for all $m \in \cm$. > > *Proof*. Let $x \in \pi^{-1}(E)$, then $\pi(x + m) = \ol{x + m} = \ol{x} \in E$ for any $m \in \cm$, and $x + m \in \pi^{-1}(E)$. Therefore if $\pi(F) \subset E$, then $\pi(F + m) \subset E$ as well and $F + m \subset \pi^{-1}(E)$. > > > ### Preimage of Neighbourhood > > > ![[quotient_preimage-light.png|400]] > > Let $\ol{x} \in \cx/\cm$ and $\varepsilon > 0$, then > $ > \pi^{-1}(B(\ol{x}, \varepsilon)) = \bigcup_{y \in \ol{x}}B(y, \varepsilon) = \bigcup_{m \in \cm}B(x + m, \varepsilon) > $ > *Proof*. Let $z \in \cx$, if there exists $y \in \ol{x}$ such that $\norm{y - z} < \varepsilon$, then > $ > \norm{\ol{x} - \ol{z}}_q = \norm{\ol{y - z}}_q \le \norm{y - z} < \varepsilon > $ > and $\ol{z} \in B(\ol{x}, \varepsilon)$. Let $\ol{z} \in \cx/\cm$ such that $\norm{\ol{x} - \ol{z}}_q < \varepsilon$, then there exists $m \in \cm$ such that > $ > \norm{\ol{x} - \ol z}_q = \norm{\ol{x - z}}_q \le \norm{x - z + m} < \varepsilon > $ > and $z \in B(x + m, \varepsilon)$. > > > ### Quotient Topology in Norm Topology > > Let $U \in \tilde \topo$, then for any $\ol{x} \in U$, there exists $\varepsilon > 0$ such that $B(\ol{x}, \varepsilon) \subset U$. Therefore $U \in \ol\topo$. > > *Proof*. Since $U \in \tilde \topo$, $\pi^{-1}(U) \in \topo$ is open. > > Let $\ol{x} \in U$, then $x \in \pi^{-1}(U)$ and there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset \pi^{-1}(U)$. From earlier, we know that $B(x + m, \varepsilon) \subset \pi^{-1}(U)$ for all $m \in \cm$, which means that > $ > \pi^{-1}(B(\ol{x}, \varepsilon)) = \bigcup_{m \in \cm}B(x + m, \varepsilon) \subset \pi^{-1}(U) > $ > and $B(\ol{x}, \varepsilon) \subset \pi^{-1}(U)$. # "Quotient Map" > [!theorem] > > Let $\cx$ and $\cy$ be normed spaces and $T \in L(\cx, \cy)$ be a continuous linear map, then > 1. $\ker (T)$ a closed subspace of $\cx$. > 2. There is a unique $S \in L(\cx/\ker(T), \cy)$ such that $T = S \circ \pi$. > 3. $\norm{S} = \norm{T}$. > > *Proof*. Let $\seq{x_n} \subset \ker(T)$ be a convergent sequence with $x_n \to x$, then $Tx_n \to Tx = 0$ and $x \in \ker(T)$. Therefore $\ker(T)$ is closed. > > > ### First Isomorphism Theorem Flashbacks > > Let $S: \cx/\ker(T) \to \cy$ by $(x + \ker(T)) \mapsto Tx$. First verify that this is well-defined: let $x + m_1, x + m_2 \in x + \ker(T)$, then > $ > T(x + m_1) = Tx + Tm_1 = Tx = Tx + Tm_2 = T(x + m_2) > $ > and with > $ > (S \circ \pi)x = S(x + \ker(T)) = Tx > $ > we have $T = S \circ \pi$. > > > ### Continuity > > Let $x \in \cx$, then > $ > \norm{S(x + \ker(T))} = \norm{Tx} > $ > and > $ > \norm{S} = \sup_{x \ne 0}\frac{\norm{S(x + \ker(T))}}{\norm{x}} = \sup_{x \ne 0}\frac{\norm{Tx}}{\norm{x}} = \norm{T} > $ # Seminorm and Non-Closed Subspaces > [!theorem] > > Let $\cx$ be a vector space and $p: \cx \to \real$ be a [[Seminorm|seminorm]], then $\ker(p)$ is a linear subspace, and $\norm{\cdot}: \cx/\ker(p) \to \real$ with $x + \ker(p) \mapsto p(x)$ is a norm on $\cx/\ker(p)$. > > *Proof*. > > ### Linear Subspace > > Let $\lambda \in F$, $x \in \ker(p)$, and $y \in \ker(p)$, then > $ > p(\lambda x + y) \le \lambda p(x) + p(y) = 0 > $ > therefore $\ker(p)$ is a linear subspace. > > > ### Well-Defined > > Now define the norm as above. First verify that it is well-defined. Let $x + m \in x + \ker(p)$, then > $ > p(x)= p(x) - p(m)\le p(x + m) \le p(x) + p(m) = p(x) > $ > and the value of the function does not depend on the choice of the representative. > > > ### Norm > > Firstly, let $x \not\in \ker(p)$, then $p(x + \ker(p)) = p(x) \ne 0$. Now, let $\lambda \in F$, then > $ > \begin{align*} > p(\lambda [x + \ker(p)]) &= p(\lambda x + \ker(p)) \\ > &= p(\lambda x) = \abs{\lambda}p(x) \\ > &= \abs{\lambda}p(x + \ker(p)) > \end{align*} > $ > Lastly, let $x, y \in \cx$, then > $ > \begin{align*} > p([x + \ker(p)] + [y + \ker(p)]) &= p(x + y + \ker(p))\\ > &= p(x + y) \le p(x) + p(y) \\ > &= p(x + \ker(p)) + p(y + \ker(p)) > \end{align*} > $ > [!theorem] > > Let $\cx$ be a normed space and $\cm$ be a subspace, then the map $p:\cx/\cm \to \real$ with $x + \cm \mapsto \inf_{m \in \cm}\norm{x + m}$ is a seminorm on $\cx/\cm$, and > $ > (\cx/\cm)/\ker(p) \iso \cx/\ol{\cm} > $ > isometrically. > > *Proof*. Since $\cm$ is not guaranteed to be closed, $p$ is a seminorm with $\ker(p) = \bracs{m + \cm :m \in \ol\cm}$. > > ### Mapping > > Let $x \in \cx$, then denote $\ol{x} = x + \cm$, $\tilde x = (x + \cm) + \ker(p)$, and $\ol{\ol{x}} = x + \ol{\cm}$. Consider the map > $ > (\cx/\cm)/\ker(p) \to \cx/\cm \quad \tilde x \mapsto \ol{\ol{x}} > $ > > > ### Well-Defined > > Let $x, y \in \cx$. If $\tilde x = \tilde y$, then $\ol{\ol{x}} = \ol{\ol{y}}$. > > *Proof*. Denote the norm on $(\cx/\cm)/\ker(p)$ as $\norm{\cdot}_p$ and the norm on $\cx/\ol{\cm}$ as $\norm{\cdot}_{quotient}$. Since $p(x) = 0$ if and only if $x \in \ol{\cm}$, > $ > \ker(\norm{\cdot }_{p}) = \bracs{\ol{x}: x \in \ol{\cm}} > $ > If $\tilde x = \tilde y$, then $\ol{x - y} \in \ker(\norm{\cdot}_p)$ with $\ol{x - y} = \ol{m}$ for some $m \in \ol{m}$. Sharing the coset means that there exists $n \in \cm$ such that $x - y = m + n$. As $\ol{\cm}$ is a linear subspace with $\cm \subset \ol{\cm}$, $x - y \in \ol{\cm}$, and $\ol{\ol{x}} = \ol{\ol{y}}$. > > > ### Linear Isomorphism > > Let $\lambda \in \complex$ and $x, y \in \cx$, then > $ > \begin{align*} > \phi(\widetilde{\lambda x + y}) = \ol{\ol{\lambda x + y}} = \lambda\ol{\ol{x}} + \ol{\ol{y}} > = \lambda\phi(\tilde x) + \phi(\tilde y) > \end{align*} > $ > it is linear. For any $\ol{\ol{x}} \in \cx/\ol{\cm}$, there exists $\tilde{x}$ such that $\phi(\tilde x) = \ol{\ol{x}}$, and $\phi$ is surjective. If $\tilde x \in \ker(\phi)$, then $\ol{\ol{x}} = 0$ and $\norm{\ol{\ol{x}}}_{quotient} = 0$, meaning that $x \in \ol{\cm}$, $\norm{\tilde x}_p = 0$ and $\tilde x = 0$, therefore $\phi$ is injective. > > > ### Isometry > > Now, let $\tilde x \in (\cx/\cm)/\ker(p)$, then > $ > \norm{\tilde x}_{p} = p(x) = \inf_{m \in \cm}\norm{x + m} = \inf_{m \in \ol{\cm}}\norm{x + m} = \norm{\ol{\ol{x}}}_{quotient} > $ > and $\phi$ is an isometry.