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Math Help - Proving D is onto

  1. #1
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    Proving D is onto

    Let V be the collection of all functions whose domain is R<br />
and differentiable on the domain. Also, let U be the
    collection of all functions whose domain is Rand continuous on the domain.

    D: V\rightarrow U
     f(x) \mapsto Df(x)=f'(x)

    I know that I need to prove for any \vec{u}\epsilon U, D\vec{v}=\vec{v}\boldsymbol{'}=\vec{u} for some \vec{v}\epsilon V but I dont know how to go about writing the proof. Can anyone help me at least get a start on it? I'm sure I can figure it out with the proper hint.

    Also, every example I find online, has f(x) equaling some function (i.e. f(x)=x^2). In my case, I dont have a specific function so that is why I am stuck and confused.
    Last edited by Jiftim; April 18th 2010 at 09:26 AM. Reason: Adding information and Editing the domain thanks to HallsofIvy
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  2. #2
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    Quote Originally Posted by Jiftim View Post
    Let V be the collection of all functions whose domain is R^2<br />
and differentiable on the domain. Also, let U be the
    collection of all functions whose domain is R^2and continuous on the domain.

    D: V\rightarrow U
     f(x) \mapsto Df(x)=f'(x)

    I know that I need to prove for any \vec{u}\epsilon U, D\vec{v}=\vec{v}\boldsymbol{'}=\vec{u} for some \vec{v}\epsilon V but I dont know how to go about writing the proof. Can anyone help me at least get a start on it? I'm sure I can figure it out with the proper hint.

    Also, every example I find online, has f(x) equaling some function (i.e. f(x)=x^2). In my case, I dont have a specific function so that is why I am stuck and confused.
    Are you sure you have copied the problem correctly? If f is a function of two variables, then what do you mean by f'(x)? I can think of two equivalent conventions but neither of them is again a function on R^2.

    If these were on R, it would be easy. If u\in U then u is a continuous function- and so is integrable. Let v= \int u dx. what is Dv?

    There is a complication in that these are functions in R^2 and so it is really a function of two variables
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  3. #3
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    You are correct. My apologies. It should be in R and not in R^2. I also believe that the below is what I got using what you gave me. However, I guess I still dont see how exactly it proves it is "onto" and would like a worded explanation if you could. Perhaps it just seems too simple or perhaps my definition of what onto means is still blurry?


    Let \vec{v}=\int \vec{u}dx
    \Rightarrow D\vec{v}=D\int \vec{u}dx=\int \vec{u}'dx=\vec{u}
    Last edited by Jiftim; April 18th 2010 at 09:28 AM.
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  4. #4
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    A function f: V\to U is "onto" if and only if, for every U\in U, there exist v\in V such that f(v)= u.

    My point is that if u is a continuous function on R, then any of its anti-derivatives is a differentiable function, v, such that Dv= u.
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