Suppose to have two real vector spaces $V$ and $W$ and an injective map $T:V\rightarrow W$ such that $T(\alpha v)=\alpha T(v)$ for all $v\in V$ and $\alpha \in\mathbb{R}$. Do there exist some conditions on the vector spaces or the map which guarantee that $T$ is also additive, that is $T(v+u)=T(v)+T(u)$?
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If $V$ is one dimensional, every homogeneous map $T:V\to W$ is necessarily linear and thus additive. If $W$ is one dimensional, it follows from the injectivity assumption on $T$ that $\dim(V)\leq1$ and so every homogeneous map is additive.
Let us then assume that both spaces have dimension two or higher but finite. If you are willing to accept the axiom of choice, there is always a homogeneous bijection $T:V\to W$ which is nonadditive. To see this, let $H_V$ and $H_W$ be unit hemispheres (with respect to any norm) on the two spaces. That is, for any point $x$ of unit norm on $V$ exactly one of $x$ and $-x$ is in $H_V$, and similarly in $W$. These sets have the same cardinality, so there is a bijection $f:H_V\to H_W$. Now if we define $T(rv)=rf(v)$ for all $v\in S_V$ and $r\in\mathbb R$, we get a homogeneous bijection. If it happens that $T$ is additive, we can always modify $f$ so that $T$ becomes nonadditive, for example by swapping the images of two points of $H_V$.
If $2\leq\dim(V)\leq\dim(W)$, the same construction can be done without choice so that $T$ is continuous. To do this, it suffices to produce such a map $T:\mathbb C\to\mathbb C$; we can write the spaces as $V=\mathbb R^2\times\mathbb R^n$ and $W=\mathbb R^2\times\mathbb R^n\times\mathbb R^m$ and then let $(x,y)\mapsto(Tx,y,0)$. For $0\leq\theta<\pi$ and $r\in\mathbb R$ let $T(re^{i\theta})=re^{i\sqrt{\pi\theta}}$. This mapping is homogeneous and bijective but not additive.
answered Dec 3, 2014 at 18:05
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$\begingroup$ Thank you very much for your answer Joonas, but actually I asked a little different question. You showed me that an homogeneous but non additive function does exist, and I get it, but I already have a map T and I would like to know if there is some theorem which guarantees its additivity under some suitable conditions. $\endgroup$– moppio89Commented Dec 4, 2014 at 10:12
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1$\begingroup$ @moppio89, I read your question like this: "Given $V,W,T$ of which $T$ homogeneous and injective, what assumptions on $V,W$ guarantee that $T$ is additive?" A necessary and sufficient assumption is that $\dim(V)\leq1$. If $\dim(V)\geq2$, there are always nonadditive homogeneous maps so you can never deduce additivity from homogeneity. If you want to ask what conditions on $T$ guarantee additivity, that is a different question. If you want conditions on the spaces, you will have to assume that $\dim(V)\leq1$ to be able to make that conclusion. $\endgroup$ Commented Dec 4, 2014 at 13:18
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1$\begingroup$ Yeah sorry, you are right, there doesn't exist any condition on the spaces which guarantee the additivity for $dim(V)>1$. I guess I woke up a little stupid this morning. . Anyhow in my question I was quite general, I wrote: "some conditions on the vector spaces or the map", what about some conditions on T? $\endgroup$– moppio89Commented Dec 4, 2014 at 13:39
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$\begingroup$ @moppio89, I seem to have skipped the words "or the map". I don't know what kind of a condition would be suitable if it should be simple but not immediately equivalent with additivity. Perhaps something like "the image of every convex set is convex" but I haven't checked this... $\endgroup$ Commented Dec 4, 2014 at 13:55
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$\begingroup$ Ok, I will try, thank you very much! $\endgroup$– moppio89Commented Dec 4, 2014 at 14:02