Banach space is sum of $ker f$ and $X \ ker(f)$.

Question

I'm trying to show that if $f$ is an element of the dual space $X^*$ of a Banach space, $X$, and $x_0 \in X-ker(f)$, then every element in $X$ can be expressed as $x = \lambda x_0 + y$ with $y \in ker(f)$. I feel like this should be trivial to prove, and I can certainly do it in the finite dimensional case, however I'm having trouble showing it without resorting to a basis.

Thanks for any help

Answer 1

Let $f\in X^*$. If $f=0$ there's nothing to prove, just take $x_0=0$ and $y=x$.

So suppose $f\ne0$ and let $f(x_0)\ne0$; then, for $x\in X$, set $$ y=x-f(x)f(x_0)^{-1}x_0 $$

Answer 2

Assume $f\ne0$.

Choose $x_0\in X$ with $f(x_0)=1$ (it has to exist, otherwise $f=0$). Now take any other $x\in X$. Let $y=x-f(x)x_0$. Then $$ f(y)=f(x)=f(x)=0, $$ so $y\in\ker f$. That is, $$ x=\lambda x_0+y, $$ with $\lambda=f(x)$ and $y\in\ker f$.