# Proving D is onto

• Apr 18th 2010, 06:46 AM
Jiftim
Proving D is onto
Let $\displaystyle V$ be the collection of all functions whose domain is $\displaystyle R$ and differentiable on the domain. Also, let $\displaystyle U$ be the
collection of all functions whose domain is $\displaystyle R$and continuous on the domain.

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

I know that I need to prove for any $\displaystyle \vec{u}\epsilon U$, $\displaystyle D\vec{v}=\vec{v}\boldsymbol{'}=\vec{u}$ for some $\displaystyle \vec{v}\epsilon V$ 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 $\displaystyle f(x)$ equaling some function (i.e. $\displaystyle f(x)=x^2$). In my case, I dont have a specific function so that is why I am stuck and confused.
• Apr 18th 2010, 09:03 AM
HallsofIvy
Quote:

Originally Posted by Jiftim
Let $\displaystyle V$ be the collection of all functions whose domain is $\displaystyle R^2$ and differentiable on the domain. Also, let $\displaystyle U$ be the
collection of all functions whose domain is $\displaystyle R^2$and continuous on the domain.

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

I know that I need to prove for any $\displaystyle \vec{u}\epsilon U$, $\displaystyle D\vec{v}=\vec{v}\boldsymbol{'}=\vec{u}$ for some $\displaystyle \vec{v}\epsilon V$ 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 $\displaystyle f(x)$ equaling some function (i.e. $\displaystyle f(x)=x^2$). 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 $\displaystyle R^2$.

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

There is a complication in that these are functions in $\displaystyle R^2$ and so it is really a function of two variables
• Apr 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 $\displaystyle \vec{v}=\int \vec{u}dx$
$\displaystyle \Rightarrow D\vec{v}=D\int \vec{u}dx=\int \vec{u}'dx=\vec{u}$
• Apr 19th 2010, 04:08 AM
HallsofIvy
A function $\displaystyle f: V\to U$ is "onto" if and only if, for every $\displaystyle U\in U$, there exist $\displaystyle v\in V$ 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.