# Let C be the set of all real functions that are continuous on the closed interval [0,

• Dec 9th 2011, 10:28 PM
Brjakewa
Let C be the set of all real functions that are continuous on the closed interval [0,
I was studying for my final when I stumbled across this question, I am completely stuck and would appreciate any and all help for how you'd go about proving this or finding a counterexample if its false.

Let C be the set of all real functions that are continuous on the closed interval [0,1]. Define the function: A:C-> as follows: For each f in C, A(f) = the integral from 1 down to 0 f(x)dx.

Is the function A an injection? Is it a surjection? Justify your conclusions.

If it is true, or if it is an injection or surjection the justifications should be proofs.
• Dec 10th 2011, 01:07 AM
emakarov
Re: Let C be the set of all real functions that are continuous on the closed interval
Quote:

Define the function: A:C-> as follows: For each f in C, A(f) = the integral from 1 down to 0 f(x)dx.
What is the codomain of A?

So, $A(f)=\int_1^0 f(x)\,dx=\int_0^1(-f(x))\,dx$. If $-f(x)\ge0$, this is the area under the graph of $-f(x)$. Since $C$ consists of all continuous function, and not only positive ones, it does not matter much that there is a minus in front of $f$; in the first approximation you can think about the area under a continuous line above the x-axis.

So, do you think that any two lines produce different areas? Think about the area of a triangle and rectangle. Can you make a rectangle with any given area?