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

Printable View

- Nov 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 - Nov 19th 2012, 09:49 PMMacstersUndeadRe: A dicontinuous additive but not homogenuous function
We are given that the function must satisfy the following conditions

$\displaystyle \\(i) f(x+y) = f(x) + f(y)\\(ii) f(ax) \neq af(x)$

We can rewrite $\displaystyle f(ax) = f(ax) + f(0)$ by property (i) and we know that $\displaystyle f(ax) + f(0) \neq af(x)$ 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 $\displaystyle f(0)=c$ for some non-zero constant, then $\displaystyle f(0+0) = f(0) = c$ and $\displaystyle f(0) + f(0) = 2c$, a contradiction. so $\displaystyle f(0)=0$

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) - Nov 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.

- Nov 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).