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How can we prove that a continuous additive homomorphism $ \Phi \colon \mathbb{R}^{n}\to \mathbb{R}^{m} $ is $\mathbb{R}$-linear. i.e. satisfies $ \Phi (rv)=r \Phi (v)$ for $ r\in \mathbb{R} $ and $ v \in \mathbb{R}^{n} $?

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  • $\begingroup$ they are not homework problems $\endgroup$ – the code Mar 4 '12 at 4:40
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    $\begingroup$ Then, when you post them, please say (i) in what context you are encountering these problems; and (ii) what your thoughts on the problems so far are. When one sees the same person posting multiple homework-level problems with little or no work showing (in three of the four you provided no work), it sure looks like homework. $\endgroup$ – Arturo Magidin Mar 4 '12 at 4:40
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    $\begingroup$ I think it is pretty absurd to offer a bounty for «credible and/or official sources» when there are two proofs offered! If what is wanted is more details, then more details should be asked for, but asking for «credible and/or official sources» is very weird :) $\endgroup$ – Mariano Suárez-Álvarez Mar 19 '12 at 6:11
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    $\begingroup$ To make the question more fun, drop the continuity hypothesis and replace it by Lebesgue measurability. $\endgroup$ – Ryan Budney Mar 22 '12 at 5:50
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    $\begingroup$ @Mariano: I don't know about yours, but my degree says I have "all rights and privileges pertaining thereto"; I'd say we are both "official sources", to say nothing of our credibility! $\endgroup$ – Arturo Magidin Mar 23 '12 at 5:42
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HINT. Use induction and additivity to show that it is $\mathbb{N}$-linear; then use that to show that it is $\mathbb{Z}$-linear, and from there to show that it is $\mathbb{Q}$-linear.

Once you know it is $\mathbb{Q}$-linear, use continuity to deduce it is $\mathbb{R}$-linear.

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Let $V=\mathbb R^n$ and let $f:V\to V$ be a continuous homomorphism. Let $\{e_1,\dots,e_n\}$ be a basis of $V$ over $\mathbb R$ and let $g:V\to V$ be the unique $\mathbb R$-linear map such that $g(e_i)=f(e_i)$ for all $i\in\{1,\dots,n\}$, which is of course continuous.

The difference $h=f-g$ is also continuous, so its kernel $K=\ker h$ is therefore a closed subgroup of $V$ which contains all the $e_i$. If $\frac pq\in\mathbb Q$ with $p$, $q$ integral, then $q\cdot h(\frac pqe_i)=h(pe_i)=ph(e_i)=0$ for all $i$, so in fact $h(\frac pqe_i)=0$ because $V$ has no torsion: we thus see that $H$ contains the subgroup $\mathbb Q e_1+\mathbb Q e_2+\cdots+\mathbb Q e_n$. Since $H$ is closed, this implies that $H=V$.

It follows from this that in fact $f=g$.

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