> [!definition] > > Let $\cx$ be a [[Banach Space|Banach space]]. A [[Closed Set|closed]] subspace $\cm$ of $\cx$ is said to **split** whenever there exists another closed subspace $\cn$ such that $\cx = \cm \oplus \cn$ is their [[Direct Sum|inner direct sum]]. > [!definition] > > Let $\cx$ and $\cy$ be Banach spaces. An injective [[Bounded Linear Map|bounded linear map]] $\lambda \in L(\cx, \cy)$ **splits** if $\lambda(\cx)$ splits $\cy$. > [!theorem] > > Let $\cm$ be a closed subspace and $x \in \cx \setminus \cm$, then $\cm + \complex x$ is also closed. > > *Proof*. By the [[Hahn-Banach Theorem|Hahn-Banach theorem]], there exists a continuous [[Linear Functional|linear functional]] $f$ such that $f(\cm) = \bracs{0}$ and $f(x) = \delta > 0$. Let $\seq{z_n} \subset \cm + \complex x$ that converges to $z \in \cx$, then $\frac{f(z_n)}{\delta}x \to \frac{f(z)}{\delta}$. Take > $ > y_n = z_n - \frac{f(z_n)}{\delta}x \quad y = z - \frac{f(z)}{\delta}x > $ > then > $ > \begin{align*} > y_n &= z_n - \frac{f(z_n)}{\delta} x \\ > &= (m_n + \lambda_n x) - \frac{f(m_n + \lambda_n x)}{\delta}x \\ > &= m_n + \lambda_n x - \lambda_n x \\ > &= m_n \in \cm > \end{align*} > $ > As $y_n$ is a linear combination of two convergent sequences, it also converges. Since $\cm$ is closed, $y_n \to y$ is convergent, and $z = \frac{f(z)}{\delta}x + y \in \cm + \complex x$. > [!theorem] > > Every finite-dimensional subspace is closed. > > *Proof, by induction on dimension of subspace*. Let $\seqf{e_i}$ be a [[Basis|basis]] for the subspace $\cm$, then $\span(\emptyset) = \bracs{0}$ is [[Closed Set|closed]]. Suppose that $\span(\bracs{e_i}_1^k)$ is closed, then > $ > \span(\bracs{e_i}_1^{k + 1}) = \span(\bracs{e_i}_1^k) + \complex e_{k + 1} > $ > is also closed. Therefore $\span(\bracs{e_i}_1^{k})$ is closed for all $k \le n$, and $\cm$ is closed. > [!theorem] > > Every finite-dimensional subspace splits. > > *Proof*. Let $\cm$ be a finite-dimensional subspace and $\seqf{e_i}$ be a basis. For each $e_i$, let $\phi_i: \cm \to \complex$ be such that $\phi_i(e_i) = 1$ and $\phi_i(e_j) = 0$ for all $j \ne i$. Use the Hahn-Banach theorem to extend each $\phi_i$ to a continuous linear functional on $\cx$. > > Let $z \in \cx$, decompose > $ > z = \sum_{i = 1}^{n}\phi_i(z) \cdot e_i + \braks{z - \sum_{i = 1}^{n}\phi_i(z) \cdot e_i} > $ > then > $ > \phi_j\braks{z - \sum_{i = 1}^{n}\phi_i(z) \cdot e_i} = \phi_j(z) - \phi_j(z) = 0 > $ > therefore the latter component is in the kernel of all $\phi_j$. For any sequence $\seq{z_n}$ in $\ker (\phi_j)$, if $z_n \to z$, then $\phi_j(z_n) \to \phi(z) = 0$. Therefore each $\ker(\phi_j)$ is closed. Let > $ > \cn = \bigcap_{i = 1}^{n}\ker (\phi_i) > $ > then $\cn$ is a closed subspace of $\cx$. If $z \in \cm$ and $\cn$, then $\phi_j(z) = 0$ for all $j$, and $z = \sum_{i = 1}^{n}\phi_i(z) \cdot e_i = 0$. So $\cx$ is the inner direct sum of $\cm$ and $\cn$. > [!theorem] > > Let $\cx$ and $\cy$ be [[Banach Space|Banach spaces]], $T \in L(\cx, \cy)$ be a [[Bounded Linear Map|continuous linear map]]. Then the [[Quotient Norm|quotient space]] $\cx/\ker (T) \iso T(\cx)$ if and only if $T(\cx)$ is closed. > > *Proof*. Since $\ker(T)$ is closed, there exists $S \in L(\cx/\ker(T), T(\cx))$ with $T = S \circ \pi$ where $\pi: \cx \to \cx/\ker(T)$ > > ### Bijection > > $S$ is a bijective linear map. > > *Proof*. Let $x \in \cx$ such that $x + \ker(T) \in \ker (S)$. Since $T = S \circ \pi$, $T(x) = S(x + \ker(T)) = 0$ and $x \in \ker(T)$, meaning that $x + \ker(T) = \ker(T)$. Therefore $S$ is injective. > > Let $T(x) \in T(\cx)$, then $T(x) = S(x + \ker(T))$, $T(x) \in S(\cx/\ker(T))$. Therefore $S$ is surjective. > > > ### Continuity > > $S^{-1} \in L(T(\cx), \cx/\ker(T))$ if and only if $T(\cx)$ is closed. > > *Proof*. If $T(\cx)$ is closed, then it is also a Banach space. Since $\cx$ is complete, so is $\cx/\ker(T)$. By the [[Open Mapping Theorem]], $S$ is an isomorphism. If $S$ is an isomorphism, then $T(\cx)$ is complete (any Cauchy sequence $S(x_n + \cm)$ converges to $Sx$ if $x_n \to x$), and therefore closed. > [!theorem] > > Let $E$ be a Banach space such that $E = E_1 \oplus E_2$. Let $U$ be [[Open Set|open]] in $E_1$, then for any $y \in E_2$, the set $U \times y \subset E$ is [[Locally Closed|locally closed]]. > > *Proof*. Since $U \times E_2$ is open and $U \times \bracs{y}^c$ is open, $U \times E_2 \setminus U \times \bracs{y}^c$ is closed in the [[Relative Topology|relative topology]] on $U \times E_2$. Moreover, for any open sets of $U \times E_2$ intersecting $U$, $U \times \bracs{y}^c$ is also open, and the set would be closed in the relative topology.