Problem 6

Exercise 7.6 (One-stop Banach space decomposition)   Let $X_0$ be a one-dimensional subspace of a Banach space $X$. Summon a closed subspace $X_1$ such that $X=X_0\dot{+}X_1$.

Proof. Select a non-zero $z\in X_0$ and $\phi : X \to \mathbb{R}$ such that $\phi(z)=1$ by extending the linear functional

$$\operatorname{span}\{z\} = X_0 \to \mathbb{R}$$ (7.22)

$$\lambda z \mapsto \lambda\vert z\vert$$ (7.23)

to a continuous linear functional via the Hahn-Banach theorem. Let $X_1=\ker\phi$ which is a closed subspace because $\phi$ is a continuous linear functional. We will prove that

$$X = X_0\dot{+}X_1$$ (7.24)

A sufficient condition is that $X_0$ and the kernel are complemented: $X_0\cap X_1=\{0\}$ and $X_0+X_1=X$.

To show $X_0\cap X_1=\{0\}$, suppose $x\in X_0$ and $\phi(x)=0$. Then $\phi(x)=\phi(\lambda z)=\lambda = 0$, implying $x=0$. To show $X_0+X_1=X$, let $x\in X$. We will break $x$ into an $X_0$ summand and a kernel summand:

$$x = \phi(x)z + (x - \phi(x)z)$$ (7.25)

Because $z\in X_0$ and $\phi(x)$ is a scalar, it is certain that $\phi(x)z\in X_0$. To verify that the second term lies in the kernel, simply evaluate

$$\phi(x - \phi(x)z) = \phi(x) - \phi(x)\phi(z) = \phi(x)-\phi(x)=0$$ (7.26)

The decomposition is defined for any $x\in X$, so we are done. $\qedsymbol$