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

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- November 19th 2012, 04:11 AMraedA 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 - November 19th 2012, 09:49 PMMacstersUndeadRe: A dicontinuous additive but not homogenuous function
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) - November 20th 2012, 12:28 AMemakarovRe: A dicontinuous additive but not homogenuous function
See Cauchy's functional equation:

Quote:

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.

- November 20th 2012, 03:55 AMraedRe: A dicontinuous additive but not homogenuous function
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).