Different domains-different story

Algebra

Do there exist functions f,g such that f(x)+g(y)=x^2+xy+y^2 for all x,y ?

The problem with this problem is that the domains of f,g are not specified.

For instance, if f,g :\mathbb{R}\rightarrow\mathbb{R} then, putting x=0 we get g(y)=y^2 for all y and putting y=0 we get f(x)=x^2 for all x. Thus in order to have the given equation true, we must have xy=0 to hold for all x,y which is certainly impossible. Hence there does not exist such f,g:\mathbb{R}\rightarrow\mathbb{R}.

On the other hand, let \mathbb{A}=\Big\{\left( \begin{array} {cc} a & 0 \\ 0 & 0 \end{array} \right) : a\in \mathbb{R}\Big\}
And let \mathbb{B}=\Big\{\left( \begin{array} {cc} 0 & 0 \\ a & 0 \end{array} \right) : a\in \mathbb{R}\Big\}
Now define f :\mathbb{A}\to \mathbb{A} such that f(A)=A^2 \hspace{3mm}\forall A\in \mathbb{A} And define g :\mathbb{B}\to \mathbb{B} such that g(B)=B^2 \hspace{3mm}\forall B\in \mathbb{B}.
(Its easy to verify that f :\mathbb{A}\to \mathbb{A} and g :\mathbb{B}\to \mathbb{B})
Next note that, AB=O\hspace{3mm} \forall A\in \mathbb{A}, B\in \mathbb{B}.
Hence, \forall A\in \mathbb{A}, B\in \mathbb{B}, we do have
f(A)+g(B)=A^2+AB+B^2.

Published

5 thoughts on “Different domains-different story”

  3. But in that case domains of $f$ and $g$ are different. This result fails to hold if the domains of both $f$ and $g$ are $\Bbb A$ or $\Bbb B.$ Can you give an example where this result holds with the domain of both the functions being same?

    Liked by 1 person

    Reply

    1. Thank you for asking this question. Yes, I can give one such example: let C be the set of all 2 by 2 matrices with c_11 = c_21 = c_22 = 0 (and c_12 can be any real number). Check that XY = O holds for every X, Y in C.

      Like

      Reply

  4. Well if I say that xy=Ax+By this means A/x+B/y=1….(i) That means if we restrict the domain to S= R\{0} and ensure for all x,y € S, There exist A,B € R such that (i) holds , then we may probably have a solution in R.

    Like

    Reply

Leave a comment