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I am trying to solve the following question: Let $E$ be a Banach space and $l \in E'$ be a nonzero linear functional. Prove that there exists a 1-dimensional subspace $L$ of $E$ such that every $x \in E$ admits the representation $x = x_1 + x_2$ with $x_1 \in L$ and $x_2 \in \ker(l)$.

I have a solution that goes as follows, however it makes no use of the fact that $E$ is Banach, and I am wondering if I have missed something.

Let $E$ be a Banach space and $l \in E'$ a nonzero linear functional.

Since $l$ is nonzero, there exists $z_0 \in E$ such that $l(z_0) \neq 0$. Define the 1-dimensional subspace $L := \mathrm{span}\{z_0\} = \{\lambda z_0 : \lambda \in \mathbb{K}\}$, where $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$.

Let $x \in E$ be arbitrary. We want to show that $x$ can be written as $x = x_1 + x_2$ with $x_1 \in L$ and $x_2 \in \ker l$.

Set $$ x_1 := \frac{l(x)}{l(z_0)} z_0 \in L. $$ Now consider $$ x_2 := x - x_1 = x - \frac{l(x)}{l(z_0)} z_0. $$ We claim that $x_2 \in \ker l$:

$$ l(x_2) = l\left(x - \frac{l(x)}{l(z_0)} z_0\right) = l(x) - \frac{l(x)}{l(z_0)} l(z_0) = l(x) - l(x) = 0. $$ Therefore, $x_2 \in \ker l$ as required.

user1590812
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2 Answers2

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It does not matter that $E$ is a Banach space. You do not need $E$ to be complete. In fact, you do not even need to have a norm on $E$. Your argument proves the following result.

Lemma: Let $E$ be a vector space over a field $\mathbb{K}$, and $l$ be a non-zero linear map from $E$ to $\mathbb{K}$. Then there exists a 1-dimensional subspace $L$ of $E$ such that $E = L \oplus (\ker l)$.

Gary
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cs89
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Slightly more abstractly, for any $\mathbb{K}$-vector space $E$ and any non-zero linear $l \colon E \to \mathbb{K}$, the sequence $0 \to \ker l \to E \to \mathbb{K} \to 0$ is exact, and short exact sequences split in the category of $\mathbb{K}$-vector spaces, finite dimensional or not.

Joshua P. Swanson
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