Example of a function which is nearly a linear transformation

Question

Let $V$ and $W$ be vector spaces over the field $\mathbb{F}.$ Let $f$ be a function from $V$ to $W.$ Now $f$ will be called a linear transformation if

\begin{align} &\tag1 f(\alpha + \beta) = f(\alpha) + f(\beta)\,\, \forall \alpha,\beta \in V \\ &f(c\alpha) = cf(\alpha)\,\, \forall c\in \mathbb{F}\tag2 \end{align}

I am interested in finding examples of functions where :

(a) the first condition $(1)$ fails and second condition $(2)$ holds

(b) the second condition $(2)$ fails and first condition $(1)$ holds

I have two examples for the (b) part:

Consider $f : M_{n\times n}(\mathbb{C}) \rightarrow M_{n\times n}(\mathbb{C}) $ with the mapping $A \to A^*$ where $A^*$ is the conjugate transpose of $A.$

Consider $f : \mathbb{C} \rightarrow \mathbb{C}$ wih the mapping $z \to \overline{z}.$

So far I haven't been able to find an example for (a).Please help me wind this up. Also if you find more examples for (b), please list them too.

Also see the bottom of: https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation — rubikscube09, Aug 29 '18 at 18:49
Notice that your second example is just the special case $n = 1$ of your first example. — Travis Willse, Aug 29 '18 at 19:00

Mohammad Riazi-Kermani · Accepted Answer · 2018-08-29T20:59:01.683

2

Let the field be the the field of complex numbers and the vector space be the vector space of complex numbers over the field of complex numbers. The function $f(z)=Re(z)$ satisfies $$f(z+w)=f(z)+f(w)$$ but fails to satisfy, f(Cz)=Cf(z) for a complex number C.

Let consider the vector space of complex numbers over the real field.

Define $f(z)=(\text {Sgn} (Re(z))|z|$,where Sgn stands for the signum function. Then $$f(z+w)=f(z)+f(w)$$ fails but $f(cz)=cf(z)$ holds.

edited Aug 29 '18 at 20:59

answered Aug 29 '18 at 18:54

Mohammad Riazi-Kermani

69,891

Thanks, but have you got any example for part (a) of the problem? – Bijesh K.S Aug 29 '18 at 18:56
1

Check my new edited version. I added an example of part (a) – Mohammad Riazi-Kermani Aug 29 '18 at 19:04
Brilliant. Been looking for this for a while. Thanks – Bijesh K.S Aug 29 '18 at 19:06
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@BijeshK.S Thanks for your attention and understanding – Mohammad Riazi-Kermani Aug 29 '18 at 19:07
Wait a sec. .looks like $f(cz) = c^2f(z) $. ? – Bijesh K.S Aug 29 '18 at 19:49
Thanks, check my newly edited version please. – Mohammad Riazi-Kermani Aug 29 '18 at 21:00
This works. Thanks – Bijesh K.S Aug 30 '18 at 07:35

score 0 · Answer 2 · answered Aug 29 '18 at 20:02

Hint Condition (b) is equivalent to property that the restriction of $f$ to any line $\{c \alpha : c \in \Bbb F\}$, $\alpha \in V - \{0\}$, is linear (check this!). So, any such function $f$ is determined by choosing a value at one nonzero point on each line, and generically most of these choices will give nonlinear $f$, i.e., functions that satisfy (b) but not (a).

Additional hint Take $f : \Bbb F \to \Bbb F^2$ to be the map $f(x, y) = x$ for $y = 0$ and $f(x, y) = 0$ otherwise.

Example of a function which is nearly a linear transformation

2 Answers2