Problem Let $X$ be a normed vector space and let $F\neq 0$ be any element of $X'$. ($X'$ is the dual space of $X$.) Show that there is a one-dimensional subspace $M$ of $X$ such that $X=N(F)\oplus M$.
My view I think I will use the following lemma. We have already proved this.
Lemma : Let $X$ be a normed vector space and let $R$ be a closed subspace such that $R^0$ is of finite dimension $n$. ($R^0$ is the set of annihilators of $R$.) Then there is an $n$-dimensional subspace $M$ of $X$ such that $X=R\oplus M$.
Since $N(F)$ is a closed subspace of $X$, we can conclude from the Lemma that the dimension of $N(F)^0$ is one. I think $N(F)^0$ will be a one-dimensional subspace spaned by $F$, but I can't prove this. Is there a mistake in the policy? Please help me.
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 $N(F) \cap M=\{0\}$.