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Thread: well defined

  1. #1
    Dec 2008

    well defined

    Let $\displaystyle f,g:R \rightarrow R$ be Lebesgue integrable functions. Let $\displaystyle \phi(x,y)=f(x-y)g(y)$. Show that $\displaystyle \phi$ is well defined.

    i am not sure what it means by a function is well defined. would someone tell me what i have to show to say that it is well defined? any help would be appreciated.
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  2. #2
    Junior Member Dark Sun's Avatar
    Apr 2009
    San Francisco, California
    Well-defined means that for each input of a function, there is one output. For example, a circle is not a well-defined function.

    First, note that since $\displaystyle f,g$ are Lebesgue integrable, they are well defined.

    Also, $\displaystyle x=x',y=y'$ if and only if $\displaystyle (x,y)=(x',y')$.

    Then, we have that since $\displaystyle f,g$ are well-defined, then $\displaystyle f(x-y)=f(x'-y')$, on account of $\displaystyle x-y=x'-y'$. Similarly, $\displaystyle g(y)=g(y')$.

    Hence $\displaystyle \phi (x,y)=f(x-y)g(y)=f(x'-y')g(y')=\phi (x',y')$, and $\displaystyle \phi $ is well-defined.
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