The following is exercise III.1.13 from Analysis I by Amann and Escher.

Suppose that $V$ and $W$ are normed vector spaces and $f : V → W$ is a continuous group homomorphism from $(V, +)$ to $(W, +)$. Prove that $f$ is linear. (Hint: If $\mathbb{K = R}$, $x \in V$ and $q ∈ \mathbb{Q}$, then $f(qx) = qf(x)$. See also Exercise $6$.)

Here $\mathbb{K}$ means the base field for $V$ and $W$ and $\mathbb{K=C}$ or $\mathbb{R}$ (which is a convention stated in the text).

Now if $\mathbb{K=C}$ then by the hint we can only show that $f(a+bi)=af(1)+bf(i)$ for all $a,b\in\mathbb{R}$, but I think $f(i)$ need not equal $if(1)$, and so the result is false. For instance, take $V=W=\mathbb{C}$, regarded as vector spaces over $\mathbb{C}$ and equipped with the usual Euclidean norm. Let $f$ be the conjugation, i.e., $f:a+bi\mapsto a-bi$. I do think this is a continuous group homomorphism, but it isn't linear. Am I right?

  • $\begingroup$ Why is your conjugation $f$ not linear? $\endgroup$ – Mark Neuhaus Jan 21 '18 at 12:43
  • $\begingroup$ @MarkNeuhaus It's not linear over $\mathbb{C}$. $\endgroup$ – Colescu Jan 21 '18 at 12:43
  • $\begingroup$ right... Its not. $\endgroup$ – Mark Neuhaus Jan 21 '18 at 12:44

Yes, you are right. The proof in the real case works because $\mathbb Q$ is dense in $\mathbb R$. But that argument doesn't work in $\mathbb C$.


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