# A dicontinuous additive but not homogenuous function

• Nov 19th 2012, 04:11 AM
raed
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
• Nov 19th 2012, 09:49 PM
Re: 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 AM
emakarov
Re: 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 AM
raed
Re: A dicontinuous additive but not homogenuous function

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).

Quote:

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)