# when is f o g = g o f?

• Jan 19th 2008, 12:06 AM
tombrownington
when is f o g = g o f?
If f and g are one-to-one functions on a set A, and for any elements x and y belonging to A if:

f(x)+f(y)=f(x+y) &

g(x)+g(y)=g(x+y)

is it true that f o g = g o f ? If so, please show why.

Otherwise what are sufficient conditions for any functions m and p to commute, i.e. m o p = p o m.
• Jan 19th 2008, 09:42 AM
topsquark
Quote:

Originally Posted by tombrownington
If f and g are one-to-one functions on a set A, and for any elements x and y belonging to A if:

f(x)+f(y)=f(x+y) &

g(x)+g(y)=g(x+y)

is it true that f o g = g o f ? If so, please show why.

Otherwise what are sufficient conditions for any functions m and p to commute, i.e. m o p = p o m.

We call any function $p(x + y) = p(x) + p(y)$ a linear function in its arguments. That is to say, we may write the function as
$p(x) = ax$
where a is some (presumably) non-zero constant.

So
$f(x) = ax$

$g(x) = bx$

Thus
$(f \circ g)(x) = f(bx) = a(bx) = abx$

$(g \circ f)(x) = g(ax) = b(ax) = bax$

In order for these to be equal we require that ba = ab. Which may/may not depend on the commutivity of A under multiplication, depending on the specific values of a and b.

This condition (ba = ab) is certainly sufficient to show that your linear f and g functions are commutative over composition, but I do not know what the necessary conditions are for this to be true for general operators.

-Dan
• Jan 20th 2008, 01:27 AM
tombrownington
[QUOTE=topsquark;100021]We call any function $p(x + y) = p(x) + p(y)$ a linear function in its arguments. That is to say, we may write the function as
$p(x) = ax$
where a is some (presumably) non-zero constant.

the functions you have just described are particular cases. say we didn't know anything about the functions - or any other functions - except the details provided, can you still say that g o f = f o g?
and if it is so - which i suspect it is - how does one prove it.
• Jan 20th 2008, 04:09 AM
Opalg
Quote:

Originally Posted by tombrownington
If f and g are one-to-one functions on a set A, and for any elements x and y belonging to A if:

f(x)+f(y)=f(x+y) &

g(x)+g(y)=g(x+y)

is it true that f o g = g o f ? If so, please show why.

If the functions f and g are continuous, then what topsquark says is correct. The functions must be of the form f(x)=ax and g(x)=bx, and the composite functions fog and gof are equal (provided that multiplication on A is commutative).

If you do not assume that f and g are continuous then the situation is very different. For example, suppose that A is the set of all real numbers of the form r+s√2, where r and s are rational. Define functions f and g on A by f(r+s√2)=s+r√2 and g(r+s√2)=2r+s√2. Then f and g are both one-to-one and additive, and you can check that fog(r+s√2)=s+2r√2 but gof(r+s√2)=2s+r√2. So in this case fog is not equal to gof.

If you want functions defined on the whole of $\mathbb{R}$, the situation is the same as in the previous paragraph. In fact, there is a theorem which says that any one-to-one additive functions defined on the set A can be extended to be defined on the whole real line. (It's not an elementary theorem, though. Its proof requires the axiom of choice.)

Quote:

Originally Posted by tombrownington
Otherwise what are sufficient conditions for any functions m and p to commute, i.e. m o p = p o m.

As far as I know, there are no such conditions except in very special cases.
• Jan 20th 2008, 10:36 AM
ThePerfectHacker
Quote:

Originally Posted by Opalg
If you want functions defined on the whole of $\mathbb{R}$, the situation is the same as in the previous paragraph. In fact, there is a theorem which says that any one-to-one additive functions defined on the set A can be extended to be defined on the whole real line. (It's not an elementary theorem, though. Its proof requires the axiom of choice.)

One of my very first posts.This is maybe what you mean?