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Let $f$ be a linear functional on the vector space $X$ over $F$, $f \neq 0$, and $N$ is the nullspace of $f$. Show that there exists a vector $y$ such that every $x \in X$ can be written uniquely in the form $x=\lambda y+z$, where $z \in N$ and $\lambda \in F$.

Since $f \neq 0$, there exists a point $y \in X \backslash N$ such that $d(y,N)=d>0$. Then by one of the consequences of Hahn-Banach theorem, $f$ is a bounded linear functional with the property such that $||f||=1$, $f(y)=d$ and $f(x)=0$ for all $x \in N$.

Let $x \in X$. Then either $x \in N$ or $x \in X \backslash N$. If $x \in N$, then $x=z$. If $x \in X \backslash N$, then $x=\lambda y \Rightarrow f(x-\lambda y)=f(0)=0 \Rightarrow f(x)=f(\lambda y)+f(z)=f(\lambda y+z)$

Since bounded linear functional is one-to-one, we have $x=\lambda y+z$ for $z \in N$ and $\lambda \in F$

Uniqueness: Suppose $x=\lambda_1 y+z_1=\lambda_2 y+z_2 \Rightarrow (\lambda_1 -\lambda_2) f(y)=0 \Rightarrow \lambda_1 =\lambda_2 \Rightarrow z_1=z_2$

Can anyone check my proof?

Remark: $d(y,N)=\inf_{n \in N}{||y-n||}$. $F$ here denotes either real or complex number.

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And what is $d(y, N)$? Problem statement doesn't mention anything about a metric on $X$. Also, to the best of my knowledge, there's no Hahn-Banach theorem for the general field $F$. And even if there is, the problem can be solved without it using only very basic tools of linear algebra. – Dan Shved Sep 30 '13 at 15:14
Nope, the remark doesn't help. My point is, there is no such thing as distance. It is simply not defined in $X$. The only structure $X$ has, according to the problem statement, is that of a linear space over field $F$. – Dan Shved Sep 30 '13 at 15:22
You can add vectors from $X$ to each other. You can multiply them by scalars from $F$. You cannot measure distances between points in $X$. – Dan Shved Sep 30 '13 at 15:23
  • Let $y \in X$ such that $f(y) \neq 0$. It exists because of the condition $f \neq 0$.

  • For every $x \in X$, let $z = x - \frac{f(x)}{f(y)}y $. By linearity of $f$ one has $f(z) = f(x) -f(x) = 0$, that is $z \in N$. Furthermore $x = \lambda y + z$ with $\lambda = \frac{f(x)}{f(y)}$.

  • Uniqueness of the decomposition has been proved in the OP.

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For any $x\in X$, choose $\lambda = f(x)$, then

  1. If $\lambda = 0$, you are done.
  2. If $\lambda \neq 0$, then note that any non-zero linear functional is automatically surjective (why?), so choose $y\in X$ such that $f(y) = 1$ and consider $$ z = x-\lambda y $$ Now $f(z) = 0$ and you're done.
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This is not what was asked. Your $y$ depends $x$. What is needed is one fixed $y$, such that for each $x$ the appropriate $\lambda$ and $z$ can be found. – Dan Shved Sep 30 '13 at 17:03
@DanShved : $y$ does not depend on $x$. $y$ is chosen so that $f(y) = 1$, that is all. – Prahlad Vaidyanathan Sep 30 '13 at 17:05
OK, it actually doesn't, my bad. Still, your phrasing is unfortunate. You say "for any $x \in X$", and only then you pick an $y$. It creates the impression that you choose an $y$ for each $x$ separately. – Dan Shved Sep 30 '13 at 17:08
Also, it is not clear that in fact you use this very same $y$ even for the case $\lambda = 0$, which you do. – Dan Shved Sep 30 '13 at 17:09

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