Proving D is onto

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April 18th 2010, 06:46 AM
Jiftim

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 $R$ and 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.
April 18th 2010, 09:03 AM
HallsofIvy

Quote:

Originally Posted by Jiftim

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 $R^2$ and 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
April 18th 2010, 09:17 AM
Jiftim

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 $\vec{v}=\int \vec{u}dx$
$\Rightarrow D\vec{v}=D\int \vec{u}dx=\int \vec{u}'dx=\vec{u}$
April 19th 2010, 04:08 AM
HallsofIvy

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.