Examples of nonlinear transformations with only multiplicative or additive property but not both.
Solution 1
Any map (on a vector space) that is continuous and has the additive property will also have the multiplicative property.
Indeed, suppose $f:V \longrightarrow E$ has the additive property, that is, $f(x+y)=f(x)+f(y)$ for all $x,y \in V$. Then $f(0)=f(0+0)=f(0)+f(0)$, so $f(0)=0$.
Now, take any $v \neq 0$ in $V$. We have that $0=f(0)=f(v+(v))=f(v)+f(v)$, so $f(v)=f(v)$. Moreover, for any $n \in \mathbb{N}$ it holds that
$$f(n\cdot v)=\underbrace{f(v)+f(v)+\dots+f(v)}_{n \text{ times}}=n\cdot f(v)$$
Well, this is not enough, but we can do better. Take $\frac{p}{q} \in \mathbb{Q}$ with $p,q \in \mathbb{Z}$. Then
$$p\cdot f(v)= f(p\cdot v) = f \left( \underbrace{\frac{p}{q}\cdot v+\frac{p}{q}\cdot v+\dots+\frac{p}{q}\cdot v}_{q \text{ times}} \right) =q\cdot f\left(\frac{p}{q}\cdot v\right)\\ \Rightarrow \frac{p}{q}\cdot f(v)=f\left(\frac{p}{q}\cdot v\right)$$
Now that we've established the multiplicative property in a dense subset of $\mathbb{R}$, continuity takes care of the rest, so that if $f$ is continuous it holds that for any $\alpha \in \mathbb{R}$ and $v\in V$
$$f(\alpha \cdot v) = \alpha \cdot f(v)$$
We can use a similar idea for a map that has the multiplicative property, but that will only work along a onedimensional subspace.
EDIT: For an example of a (continuous) map that is multiplicative but not additive, consider $f:\mathbb{R}^3\longrightarrow \mathbb{R}$ given by $f(x,y,z)=\sqrt[3]{xyz}$. We have that:
$$f(\alpha x, \alpha y, \alpha z)=\sqrt[3]{(\alpha x)(\alpha y)(\alpha z)}=\sqrt[3]{\alpha^3\cdot xyz}=\alpha\cdot\sqrt[3]{xyz}=\alpha\cdot f(x,y,z)$$
so $f$ is indeed multiplicative. However, $f(1,1,0)=f(0,0,1)=0$, so $f(1,1,0)+f(0,0,1)=0$, but
$$f\Big((1,1,0)+(0,0,1)\Big)=f(1,1,1)= 1$$
It follows that $f$ is not additive.
Solution 2
Regarding additive, nonmultiplicative maps there are no simple examples. You need the Axiom of choice to construct such. You may look e.g. in Cauchy functional equation.
For example if $f:{\Bbb R}\rightarrow {\Bbb R}$ is measurable and additive then it is also multiplicative. Except for the 1 dimensional case, it is easy to construct multiplicative nonadditive maps.
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Y.X.
Updated on August 01, 2022Comments

Y.X. over 1 year
I know that a linear map must have both of these two properties. But I find it not easy to build a counterexample that have only one of these properties. Could someone give me some such examples? Thanks so much.

gmajal about 7 yearsThere is an example given here link
