Find Functions: f o g = Iℝ (ℝ→ℝ)

[ver_mathstats]

Homework Statement

Find the functions:
f: (0, ∞) → ℝ and g: ℝ → ( 0, ∞) such that f o g = I_ℝ (I_ℝ denotes identity function on ℝ).

Homework Equations

The Attempt at a Solution

I am having trouble working backwards. I know that (f o g)(x) is f(g(x)). I am unsure if this is correct but would f o g be ℝ →ℝ be correct? And then that is the identity function? Making the identity function x?. I am confused on how exactly to approach this problem however. Any help is appreciated. Thank you.

 

[fresh_42]

I am confused, too, as there is no single solution, so "Find the functions" without further requirements doesn't make sense.
Can you at least name a few examples for $f(g(x))=x\,?$

 

[ver_mathstats]

I am confused, too, as there is no single solution, so "Find the functions" without further requirements doesn't make sense.
Can you at least name a few examples for $f(g(x))=x\,?$

My apologies it's a homework question and I meant to put "find functions" rather than "find the functions", a few examples of f(g(x)) = x would be f(x) = x+4 and g(x) = x-4, and another example would be f(x) = x² and g(x) = √x.

 

[PeroK]

My apologies it's a homework question and I meant to put "find functions" rather than "find the functions", a few examples of f(g(x)) = x would be f(x) = x+4 and g(x) = x-4,

Can you see what's wrong with this example?

Hint: check all the criteria in the question.

Sorry, I misread the question.

 

[Delta2]

I think the general solution is to take f(x) a random 1-1 function, and to take as $g=f^{-1}$ the inverse of f which is well defined since f is 1-1. So it will hold $$f(g(x))=f(f^{-1}(x))=x$$.

 

[PeroK]

I think the general solution is to take f(x) a random 1-1 function, and to take as $g=f^{-1}$ the inverse of f which is well defined since f is 1-1. So it will hold $$f(g(x))=f(f^{-1}(x))=x$$.

That might not quite be correct either! See the example above.

 

[Delta2]

Ok I think I see the problem, it has to do with the domain/image of the functions.

How about the functions $g(x)=x^{2n}$ $f(x)={x}^{\frac{1}{2n}}$ I think those two qualify all the criteria correct?

 

[PeroK]

Ok I think I see the problem, it has to do with the domain/image of the functions.

How about the functions $g(x)=x^2$ $f(x)=\sqrt{x}$ I think those two qualify all the criteria correct?

Yes, of course.

The problem as stated is quite awkward, because you have to ensure that both functions are always positive. I'm not really sure what the point of the question is, unless it is to give a few examples.

 

[Delta2]

Actually I am wrong cause in my example it is $f(g(x))=|x|$... would work only if we consider $g:R^{+}\to R^{+}$ but the statement requires g to has domain R.

I think one point of the question is to understand that f,g must be some sort of inverse pair $f,g=f^{-1}$ but with the proper domains.

The only valid example I can think with well known functions is to take $f(x)=\ln x$. You got in mind any other examples @PeroK?

 

[PeroK]

Actually I am wrong cause in my example it is $f(g(x))=|x|$... would work only if we consider $g:R^{+}\to R^{+}$ but the statement requires g to has domain R.

I think one point of the question is to understand that f,g must be some sort of inverse pair $f,g=f^{-1}$ but with the proper domains.

The only valid example I can think with well known functions is to take $f(x)=\ln x$. You got in mind any other examples @PeroK?

$\ln x$ isn't allowed either, as it's not always positive. Also, $f$ doesn't have to be 1-1, only $g$. E.g. if $g(x) = e^x$, then $f(x) = \ln x$ doesn't work, but you can still find a suitable $f$.

 

[Delta2]

$\ln x$ isn't allowed either, as it's not always positive. .

The statement doesn't require that f is positive, only g must be positive.

 

[PeroK]

The statement doesn't require that f is positive, only g must be positive.

Yes, okay, I missed that!

Ignore all my posts!

 

[fresh_42]

The easiest solution, i.e. which requires no thoughts about the domains, is probably a linear function. These are already $\mathfrak{c}$ many.

 

[haruspex]

The easiest solution, i.e. which requires no thoughts about the domains, is probably a linear function. These are already $\mathfrak{c}$ many.

Wouldn't linear run into a problem with g>0?
How about just requiring g to be monotonic with an asymptotic value of zero in one direction and unbounded in the other?

 

[fresh_42]

Wouldn't linear run into a problem with g>0?

Why that? $c > 0 \Longleftrightarrow 1/c > 0$

 

[Delta2]

Why that? $c > 0 \Longleftrightarrow 1/c > 0$

Because g has as domain the whole real line, so for x negative we can't have g(x) and g(-x) both positive.

 

[fresh_42]

Because g has as domain the whole real line, so for x negative we can't have g(x) and g(-x) both positive.

And who said we had to choose the function on the entire line?

 

[Delta2]

Well to be honest I am not sure . The statement of the problem shows as domain for g the whole $\mathbb{R}$. I am not sure if that kind of statement implies that the domain can be a subset of $\mathbb{R}$.

 

[fresh_42]

Well to be honest I am not sure . The statement of the problem shows as domain for g the whole $\mathbb{R}$. I am not sure if that kind of statement implies that the domain can be a subset of $\mathbb{R}$.

Nothing what an absolute value can't fix.

 

[Delta2]

Nothing what an absolute value can't fix.

It seems to fix one problem but creates another, then we will have $f(g(x))=|x|$ but the statement requires that $f(g(x))=x$ I made a similar mistake too in post #7,#9.

 

[fresh_42]

I obviously hadn't considered all aspects. Say $M:=(0,\infty)$. Then we have $f\, : \,M \longrightarrow \mathbb{R}$ and $g\, : \,\mathbb{R}\longrightarrow M$ and $f(g(x))=x$ for all $x\in \mathbb{R}$.

This means that $g$ is injective: $g(a)=g(b) \Longrightarrow a=f(g(a))=f(g(b))=b$. So we need an embedding of the real numbers in $M$. I think this has to be solved first.

 

[fresh_42]

I think I have a solution and the embedding was the key: The negative numbers go to $(0,1/2)$ and the positive to $(1/2,\infty)$ under $g$.