Trying to show, in $\mathbb{R}^2$, that a linear extension $g$ of a linear functional $f$ is in fact linear

banach-spacesfunctional-analysislinear algebra

How do we obtain a linear extension of a linear functional $f$ where $f$ is defined on some subspace $Z$ of $X$? I always see something like the following phrase, 'let $E$ be the set of all linear extensions $g$ of $f$…but how do we construct such a $g$?

Here is what I tried – To make things simple, consider $X = \mathbb{R}^2 = \{(x_1,x_2)\ | x_1,x_2 \in \mathbb{R}\}$ and $Z = \{(x_1,0)\}$. Suppose $f$ is a linear functional on $Z$. Define $g$ as a extension of $f$ such that $g:X\to \mathbb{R}$ and $g|_Z = f$.

Now I want to show $g$ is in fact linear. Let $a = (a_1,0) + (0,a_2)$ and $b = (b_1,0) + (0,b_2)$ for $a,b\in \mathbb{R}^2$. Then we have
\begin{align}
a + b & = (a_1,0) + (0,a_2) + (b_1,0) + (0,b_2) \\
& = (a_1+b_1,0) + (0,a_2+b_2).
\end{align}

Now I want to show linearity of the extension $g$ so I apply it to $a+b$:
\begin{align}
g(a+b) & = g((a_1+b_1,0) + (0,a_2+b_2)) \\
& = \dots
\end{align}
Here is where I am stuck. If I could say that $f = 0$ on $X\setminus Z$, then I could replace $g$ with $f$ and using the linearity of $f$ I could show $g$ is linear. But all I know is that $f$ is defined on $Z$ so it seems I can't do this.

So how do I should $g$ is linear?

Best Answer

g is not zero on the complement! that is a big mistake a lot of people make! In fact, your definition actually involves PICKING a basis of your subspace and then extending to all of $V$, if $V$ is your vector space! After that you define $g(b_i):= f(b_i)$ if $b_i$ basivector of $Z$ and zero else. Hence $g$ is not $0$ on $V\setminus Z$ but only on the span of the additional basis vectors.

This is one of the many cases where people forget where and how important choices can be!

if you in fact want to set $g$ zero on the set theoretic complement for a non zero $f$, it will not be linear! let $v \in V\setminus Z $ and $w \in V\setminus Z$ s.t. $0 \neq v+w \in Z$ and $f(v+w)\neq 0$ (those exist by taking an element $v\in V \setminus Z$ and setting $w=v-z$ for $0\neq z \in Z\setminus \ker(f)$) then $$g(v-w)=g(z)=f(z) \neq 0=0+0 =g(v) + g(w) $$ (assuming that $f$ is not the zero map)

Related Question