# Extending a “linear” map to $\mathrm{span}(S)$

This question stems from the comments and answers of Alex G here. As can be seen in the question or the comments to his answer, we take $V,W$ to be real vector spaces, and say that for $S\subset V$, a function $f:S\rightarrow W$ is "linear", we'll say on $S$, if the following two conditions hold:

1. for each $x,y \in S$, such that $x+y \in S$, we have $f(x+y)=f(x)+f(y)$.
2. for each $x \in S$ and each $\lambda \in \Bbb R$ such that $\lambda \cdot x \in S$, we have $f(\lambda \cdot x)=\lambda f(x)$.

Now it turns out this doesn't seem a useful concept of linearity as it admits functions such as the one to follow (which alex mentions) with no linear extension, or it isn't the restriction of a linear map.

Let $S:=\{(1,y)\mid y \in \mathbb{R}\}\subset \Bbb R^2$, and define $f:S\rightarrow \Bbb R$ by $f((x,y))=y^2$. Now this function satisfies the first condition above vacuously, and for the second condition the only admissible $\lambda$ is $\lambda=1$. So $f$ is "linear", but clearly can't be the restriction of a linear map from $\Bbb R^2$ to $\Bbb R$ or any of it's subspaces.

What I'm curious about is how to add condition on the set $S$ so that if possible it will imply a possibly unique extension of $f$ which is linear on $\mathrm{span}(S)$. I'm curious if this is possible or if the only way will be for $S$ to be a subspace of $V$.

So to start, we may want $1.$ to actually hold for at least one pair of vectors in $S$, and maybe for $2.$, that there exists at least one $x\in S$ and $\lambda_x\neq 1 \in \Bbb R$ so that $\lambda_x \cdot x \in S$ and $f(\lambda_x\cdot x)=\lambda f(x)$. They're just some ideas $\ldots$ Or maybe something like convexity?

But to start, I believe we should assume $f$ "linear" on $S$, and let $M\subset S$ be a maximal linearly independent subset set of $S$. Then $\mathrm{span}(M)=\mathrm{span}(S)$. So for each $s\in S$ we can write $s=\sum_i \alpha_i m_i$, uniquely with $\alpha_i \in \Bbb R$ and $m_i \in M$. Now we can't really do much from here though, since we don't know that for each $s$ that the vectors $\alpha_i m_i$ are also in $S$.

So back to the question is there some way to add conditions on the sets $S$ so that there is an extension of $f$ to $\mathrm{span}(S)$?

Update

So if $S$ is a basis you get a unique linear map. If $S$ and $f$ are defined as above, there's no linear map which restricts to $S$ to give $f$. Is there some way to distinguish these cases?

• You don't need $S$ to be a subspace for this to work. For instance, set $S$ equal to a basis of $V$ and let $f: S \to W$ be any function at all. Then $f$ extends to a unique linear map $\tilde{f}: V \to W$. – Alex G. Aug 18 '15 at 12:50
• @AlexG. yeah but really again there's no pairs of vectors from the basis which actually satisfy the first condition... So it's vacuous, and similarly the second condition again requires $\lambda=1$ as in your counterexample. So I'm curious is there a way to separate the counterexamples from the examples with extra conditions, or are would we need to require $S$ to be a subspace. – snulty Aug 18 '15 at 12:54
• @AlexG. or maybe it's a modification of "linearity" on $S$ that's needed$\ldots$ But yes thank you for pointing out that $S$ a basis clearly works well! :) – snulty Aug 18 '15 at 12:58
• When you say " the only $\lambda$ admissible is $1$, you're saying it's not linear, as every $\lambda$ must be admissible for the function to be linear (point 2 say " for each $\lambda$ ") – Tryss Aug 18 '15 at 13:23
• @Tryss But it only has to hold if $\lamda\cdot x \in S$, which won't be true for every $\lambda$. I don't think my quantifiers are wrong but I'm open to correction – snulty Aug 18 '15 at 13:32