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Math Help - A dicontinuous additive but not homogenuous function

  1. #1
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    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
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: A dicontinuous additive but not homogenuous function

    We are given that the function must satisfy the following conditions
    \\(i) f(x+y) = f(x) + f(y)\\(ii) f(ax) \neq af(x)

    We can rewrite f(ax) = f(ax) + f(0) by property (i) and we know that 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 f(0)=c for some non-zero constant, then f(0+0) = f(0) = c and f(0) + f(0) = 2c, a contradiction. so 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)
    Last edited by MacstersUndead; November 19th 2012 at 10:01 PM.
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  3. #3
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    Re: A dicontinuous additive but not homogenuous function

    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.
    Thanks from MacstersUndead
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  4. #4
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    Re: 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).

    Quote Originally Posted by MacstersUndead View Post
    We are given that the function must satisfy the following conditions
    \\(i) f(x+y) = f(x) + f(y)\\(ii) f(ax) \neq af(x)

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