Thread: A dicontinuous additive but not homogenuous function

Dear Friends,

Can any one help me in finding a discontinuous additive (i.e. f(x+y)=f(x)+f(y) ) function on the real numbers R such which is not homogeneous (i.e. f(ax) is not equal to af(x) for any a in R)


Regards
Raed

We are given that the function must satisfy the following conditions


We can rewrite by property (i) and we know that by property (ii).
Property (ii) can also be interpreted as "if the function undergoes a horizontal stretch by a factor of 1/a, then the function is not the same as if it were vertically stretched by a factor of a"

It should be noted also that additivity implies homogeneity for an rational constant a and for continuous functions. Also note that if for some non-zero constant, then and , a contradiction. so

Let
f(x) = 0 if x=0
-1 if x<0
1 if x>0

f(x+0) = f(x) = f(x) + 0 = f(x) + f(0)
For x and y with the same sign, then f(x+y) = f(x) + f(y)
Without loss of generality if x>y and x is positive and y is negative and x not equal to y, x+y > 0 and f(x+y) = 1 f(x)+f(y) = 1-1 = 0

So we can't use a simple piecewise function...

I'm stumped for now but tell me what you have so far. I'll sleep on it.

EDIT: The irony is that I almost forgot about a certain concept called "Freshman's Dream". I just don't know how to formally write it down. Maybe this doesn't make any sense.

Let f(x) = x^2 but f(x+y) = x^2 + y^2 (in other words, it's foiling x+y but it has no middle terms) = f(x) + f(y)
f(ax) = a^2*x^2 = a^2f(x)

See Cauchy's functional equation:

Over the rational numbers, it can be shown using elementary algebra that there is a single family of solutions, namely f(x) = cx for any arbitrary rational number c. Over the real numbers, this is still a family of solutions; however there can exist other solutions that are extremely complicated.

Thank you very much for your reply

In fact using just the additive property we can prove that f(r)=cr where c=f(1), where r is a rational number. Now I think the issue is how to define such a function over the irrationals so that the function over R is additive but not homogeneous over R (precisely over the irrationals).

Originally Posted by MacstersUndead

We are given that the function must satisfy the following conditions


We can rewrite by property (i) and we know that by property (ii).
Property (ii) can also be interpreted as "if the function undergoes a horizontal stretch by a factor of 1/a, then the function is not the same as if it were vertically stretched by a factor of a"

It should be noted also that additivity implies homogeneity for an rational constant a and for continuous functions. Also note that if for some non-zero constant, then and , a contradiction. so

Let
f(x) = 0 if x=0
-1 if x<0
1 if x>0

f(x+0) = f(x) = f(x) + 0 = f(x) + f(0)
For x and y with the same sign, then f(x+y) = f(x) + f(y)
Without loss of generality if x>y and x is positive and y is negative and x not equal to y, x+y > 0 and f(x+y) = 1 f(x)+f(y) = 1-1 = 0

So we can't use a simple piecewise function...

I'm stumped for now but tell me what you have so far. I'll sleep on it.

EDIT: The irony is that I almost forgot about a certain concept called "Freshman's Dream". I just don't know how to formally write it down. Maybe this doesn't make any sense.

Let f(x) = x^2 but f(x+y) = x^2 + y^2 (in other words, it's foiling x+y but it has no middle terms) = f(x) + f(y)
f(ax) = a^2*x^2 = a^2f(x)