How to decompose a normed vector space into direct sums with a kernel of functions.

Solution 1:

Suppose $G \in N(F)^{0}$. Then $G(x)=0$ whenever $F(x)=0$. Fix $x$ such that $F(x) \neq 0$ and pick any $y \in X$. Then $F(y-cx)=0$ if $c= \frac {F(Y)} {F(x)}$. Hence, $G(y-cx)=0$. Thus, $G(y)=cG(x)=\frac {F(y)} {F(x)}G(x)$. This is true for all $y$ which means $G=aF$ where $a=\frac {G(x)} {F(x)}$. This proves that $N(F)^{0}$ is one-dimensional.

Proof without using the Lemma:

Just pick any $x$ with $F(x) \neq 0$. Let $M$ be the span of $x$. Then $X$ is the direct sum of $N(F)$ and $M$: $y \in X$ implies $y-cx \in N(F)$ where $c =\frac {F(y)} {F(x)}$. Now $y=(y-cx)+cx \in N(F)+M$. Thus, $X=N(F)+M$. I will let you verify that $N(F) \cap M=\{0\}$.