The following proposition is from one of Gowers's articles:

Let $X$ be a vector space, and $x\in X$, $x\neq {\bf 0}$. Then there exists a linear map $g:X\to {\mathbb R}$ such that $g(x)\neq 0$.

The existence of this map can be proved as follows. Using the axiom of choice, one can show that the vector $x$ can be extended to a basis of $X$. Let $g(x)=1$, let $g(y)=0$ and extend linearly.

Here are my questions:

  • What does "extend linearly" mean in the proof?
  • I don't see any context about what $y$ is in his proof. So what is it here?

First of all, I'm assuming $X$ is an $\mathbb{R}$-vector space, then $\mathbb{R}$ is an $\mathbb{R}$-vector space and everything makes sense (remember that linear maps only make sense between a) vector spaces b) defined over the same field).

$x\in X$ is a non-zero vector, hence $\{x\}$ is an linearly independent set in $X$. Therefore (axiom of choice), we can extend it to a basis $B$ of $X$. This means that there exists $B\subset X$ a basis for $X$ such that $x\in B$.

Now, to define a linear map $g:X\to \mathbb{R}$ we can define it on a basis, and then extend by linearity. In this case, we do it like this:

let $g(x)=1$ for the $x$ we picked above. Now let $g(y)=0$ for every $y\in B\setminus \{x\}$.

This is a function $g: B\to \mathbb{R}$. It can be extended to a linear map $\tilde{g}: X\to \mathbb{R}$ like this:

Write $B=\{e_i\}_{i\in I}$. Every vector $v\in X$ is a linear combination of the vectors in $B$: there exist unique scalars $a_i$ such that $v=\sum_{i=1}^n a_i e_i$.

Then define the map $\tilde{g}:X\to \mathbb{R}$ as $$\tilde{g}(v)=\sum_{i=1}^n a_i g(e_i)$$

It is a linear map (check!)

(In fact, it's what one usually does in linear algebra: if $B$ is a basis for a vector space $V$, then given any other vector space $W$, for any function $g:B\to W$ there exists a unique linear map (called linear extension) $\tilde{g}: V\to W$ such that $\tilde{g}|_B=g$. In this case, $W$ is $\mathbb{R}$ as an $\mathbb{R}$-vector space).

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    $\begingroup$ What's a "l.i" set? $\endgroup$ – Jack Jul 13 '11 at 1:30
  • $\begingroup$ @Jack: sorry! It is a common shortcut for "linearly independent". I edited it out. $\endgroup$ – Bruno Stonek Jul 13 '11 at 1:31
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    $\begingroup$ I guess in your definition of $\tilde{g}$, there is no tilde in the right hand side. $\endgroup$ – Jack Jul 13 '11 at 1:38
  • $\begingroup$ @Jack: of course not! My bad. $\endgroup$ – Bruno Stonek Jul 13 '11 at 1:39
  • $\begingroup$ $X$ many not be finite-dimensional. Then what does $\sum_{i=1}^n a_ie_i$ mean? $\endgroup$ – Cantor Jan 28 '13 at 19:48

(1) If $x\neq 0$ is a vector in $X$, then there is a basis ${\cal B}$ of $X$ such that $x\in {\cal B}$ by the axiom of choice. Note that a linear functional $T:X\to \mathbb{R}$ is completely determined by its action on ${\cal B}$ (or, indeed, any basis of $X$).

(2) If $y\in {\cal B}$ but $y\neq x$, then we define $T(y)=0$; we also define $T(x)\neq 0$. Note that $T$ extends linearly to a linear functional on $X$. More precisely, if ${\cal B}=\{y_i\}_{i\in I}$ where $I$ is an index set and if $x=\Sigma_{i\in I} c_iy_i$ where all but finitely many coefficients in this sum are $0$, then we can define (and, in fact, are forced to define) $T(x)=\Sigma_{i\in I} c_iT(y_i)$.

Exercise 1: Let $V$ be a finite dimensional vector space and let $W$ be an arbitrary vector space. If $(v_1,\dots,v_n)$ is a basis of $V$ and if $(w_1,\dots,w_n)$ is an arbitrary tuple of vectors in $W$, prove that there exists a unique linear map $T:V\to W$ such that $T(v_i)=w_i$ for all $1\leq i\leq n$.

Exercise 2: Let $V$ be an arbitrary (real) vector space.

(a) Prove (using the axiom of choice) that $V$ has a basis ${\cal B}=\{y_i\}_{i\in I}$ (where $I$ is an index set).

(b) If $c_i\in \mathbb{R}$ for all $i\in I$, prove that there exists a unique linear functional $T:V\to \mathbb{R}$ such that $T(y_i)=c_i$ for all $i\in I$.

Challenge: Let $X$ be a (real) vector space. We say that $X$ is a normed linear space if there is a function $X\to [0,\infty)$ with the following properties (for notational convenience, we denote the image of $x\in X$ under this function by $\left\|x\right\|$):

  1. $\left\|x\right\|=0$ if and only if $x=0$.
  2. $\left\|\alpha\cdot x\right\|=\left|\alpha\right|\left\|x\right\|$ for all $x\in X$ and all scalars $\alpha\in \mathbb{R}$.
  3. $\left\|x+y\right\|\leq \left\|x\right\|+\left\|y\right\|$ for all $x,y\in X$.

A linear functional $g:X\to\mathbb{R}$ is said to be a bounded linear functional if

$\left\|g\right\|=\text{sup}\{\frac{\left|g(x)\right|}{\left\|x\right\|}:x\in X, x\neq 0\}<\infty$.

If $g:X\to\mathbb{R}$ is a bounded linear functional, then the norm of $g$ is defined to be the number $\left\|g\right\|$ (above).

Prove the Hahn-Banach theorem:

Let $X$ be a normed linear space (over $\mathbb{R}$, if you prefer) and let $Y\subseteq X$ be a subspace. If $f:Y\to \mathbb{R}$ is a bounded linear functional, prove that there exists a bounded linear functional $F:X\to\mathbb{R}$ such that $F(y)=f(y)$ for all $y\in Y$ and $\left\|F\right\|=\left\|f\right\|$.

I hope this helps!

  • $\begingroup$ What does your notation of a set being less than $\infty$ mean? I'm speaking in particular about your line $\left\|g\right\|=\{\frac{\left|g(x)\right|}{\left\|x\right\|}:x\in X, x\neq 0\}<\infty$ $\endgroup$ – Alex Ortiz Sep 7 '16 at 17:11
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    $\begingroup$ @AOrtiz Thanks for your message. In fact, there is a typo on that line and I should have written $\text{sup}$ before the set description. I have corrected it now. Thanks for the correction! $\endgroup$ – Amitesh Datta Sep 8 '16 at 8:20

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