# Thread: Getting a continuous function P(x,y)=f(x,y)g(x,y)

1. ## Getting a continuous function P(x,y)=f(x,y)g(x,y)

Let f :[a,b] ---->R be a continuous function and likewise let g:[c,d] -----> R be another continuous function. Where R represents the set of real numbers, [a,b] and [c,d] are subsets of R

Next we define a function P : [a,b] x [c,d] -----> R by P(x,y)=f(x)g(y). Is it possible to "construct" the continuous functions f and g such that (x1 , y1) # (x2 , y2) implies P(x1 , y1) # P(x2 , y2)

Note that f and g have to be continuous functions defined on their respective domains, namely [a,b] and [c,d]

I have "struggled" with this issue for months and cannot get a concert answer

Thanks

2. ## Re: Getting a continuous function P(x,y)=f(x,y)g(x,y)

what does the symbol # mean in this context?

3. ## Re: Getting a continuous function P(x,y)=f(x,y)g(x,y)

Dear Romsek

Thanks for your reply, the symbol # just means " is not equal to".

Best regards

KLHON

4. ## Re: Getting a continuous function P(x,y)=f(x,y)g(x,y)

It is not possible. Continuous functions possess the Intermediate Value property. If $f (x)$ is the constant function, then $P (x_1,y)=P (x_2,y)$ for all $x_1,x_2\in [a,b]$. So, WLOG, assume $f (b)>f (a)$. Also WLOG, assume $g (d)>g (c)$. Then pick $(x,y)$ such that $f(a)<f (x)<f (b)$ and $g (c)<g (y)<g (d)$. You need each function to be strictly increasing or decreasing to get an injection. Anyway, it should be evident that as you vary x and y slightly, you can get the same value through multiplication. This is just the beginning of the solution, obviously. I'll post more tomorrow if I have time.

5. ## Re: Getting a continuous function P(x,y)=f(x,y)g(x,y)

Suppose $P(x,y)=0$ for some $(x,y) \in [a,b]\times [c,d]$. Then either $f(x)=0$ or $g(y)=0$. WLOG, assume it is $f(x)$. Then $P(x,y_1) = P(x,y_2) = 0$ for all $y_1,y_2 \in [c,d]$. This means that neither $f(x)$ nor $g(y)$ ever equal zero. So, they are always positive or always negative. WLOG, assume they are always positive (and strictly increasing). This gives $P(a,c)$ is the minimum value of $P$ and $P(b,d)$ is the maximum value.

Define the function $P_a(y): [c,d] \to \mathbb{R}$ by $P_a(y) = f(a)g(y) = P(a,y)$. This is a strictly increasing function. Similarly, the function $P_c(x):[a,b] \to \mathbb{R}$ defined by $P_c(x) = f(x)g(c)$ is a strictly increasing function. Note that $P_a(c) = P_c(a)$. There is some interval over which they match, meaning you cannot have $(x_1,y_1) \neq (x_2,y_2) \Longrightarrow P(x_1,y_1) \neq P(x_2,y_2)$.