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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

Thomas Andrews
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Wooster
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  • Is there anything wrong with using a basis? – Matt Samuel Nov 16 '14 at 17:35
  • Well the Banach space isn't assumed finite dimensional and I haven't actually had any notion of a basis in the course yet so I'm really trying to find a solution without using one! – Wooster Nov 16 '14 at 17:37
  • The second reason is a good one, but the first reason is not. Every vector space has a basis whether it is finite dimensional or not. – Matt Samuel Nov 16 '14 at 17:38
  • Oh okay, I haven't seen a definition of a basis for the infinite dimensional case before! Thanks – Wooster Nov 16 '14 at 17:39
  • @MattS Sure, if you insist on the axiom of choice. That's a big hammer, if it is not necessary. – Thomas Andrews Nov 16 '14 at 17:39

2 Answers2

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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 $y\in\ker f$.

Martin Argerami
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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 $$ y=x-f(x)f(x_0)^{-1}x_0 $$

egreg
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