Prove sup((fog)(R)) <= sup(f(R)) for f bounded on R

[orir]

\(\displaystyle f(x)\) is a bounded function in \(\displaystyle R\).
i need to proof that:
sup(f

g)(\(\displaystyle R\))

sup f(\(\displaystyle R\))


the only thing that comes to mind for me is the sentence of the lim of composition of functions, but i don't think it has something to do with this question..

thanks!

 

[pka]

\(\displaystyle f(x)\) is a bounded function in \(\displaystyle R\).
i need to proof that:
sup(f

g)(\(\displaystyle R\))

sup f(\(\displaystyle R\))

We must make some other assumptions: \(\displaystyle \text{Dom}(f)=\text{Dom}(g)=\mathbb{R}\).

Note that \(\displaystyle (\exists B>0)(\forall x\in\mathbb{R})[|f(x)|\le B].\)

Therefore, \(\displaystyle (\forall x\in\mathbb{R})[|f\circ g(x)|\le B].\) thus \(\displaystyle \sup(f\circ g)\) exists.

Because \(\displaystyle \text{Rng}(f\circ g)\subset\text{Rng}(f)\) you can finish?

 

[orir]

We must make some other assumptions: \(\displaystyle \text{Dom}(f)=\text{Dom}(g)=\mathbb{R}\).

Note that \(\displaystyle (\exists B>0)(\forall x\in\mathbb{R})[|f(x)|\le B].\)

Therefore, \(\displaystyle (\forall x\in\mathbb{R})[|f\circ g(x)|\le B].\) thus \(\displaystyle \sup(f\circ g)\) exists.

Because \(\displaystyle \text{Rng}(f\circ g)\subset\text{Rng}(f)\) you can finish?

i didn't realize how you concluded that , \(\displaystyle (\forall x\in\mathbb{R})[|f\circ g(x)|\le B].\)."?
and, what is it Rng?

 

[pka]

i didn't realize how you concluded that , \(\displaystyle (\forall x\in\mathbb{R})[|f\circ g(x)|\le B].\)."?
and, what is it Rng?

Some use Rng for range as others use codomain( i.e. the set of images)

Now \(\displaystyle (\forall x\in\mathbb{R})[f\circ g(x)\in \text{Rng}(f)]\) which is bounded by B.

 

[orir]

Some use Rng for range as others use codomain( i.e. the set of images)

Now \(\displaystyle (\forall x\in\mathbb{R})[f\circ g(x)\in \text{Rng}(f)]\) which is bounded by B.

so if \(\displaystyle f\circ g(x)\in \text{Rng}(f)\) i can now conclude that \(\displaystyle sup (f\circ g(x)) <= sup f({R})\)?? is that it?

 

[pka]

so if \(\displaystyle f\circ g(x)\in \text{Rng}(f)\) i can now conclude that \(\displaystyle sup (f\circ g(x)) <= sup f({R})\)?? is that it?

Surely you know that if \(\displaystyle A\) is a bounded set in \(\displaystyle \mathbb{R}\) and \(\displaystyle B\subset A\) then \(\displaystyle \sup(B)\le\sup(A)~?\)

If this has not been done in your course, then why was this question given?

 

[orir]

Surely you know that if \(\displaystyle A\) is a bounded set in \(\displaystyle \mathbb{R}\) and \(\displaystyle B\subset A\) then \(\displaystyle \sup(B)\le\sup(A)~?\)

If this has not been done in your course, then why was this question given?

yes, it has done in my course. thank you.
but again for more clarification, how exactly did we prove that \(\displaystyle f \circ g(R) \subset f(R) \)?

 

[daon2]

yes, it has done in my course. thank you.
but again for more clarification, how exactly did we prove that \(\displaystyle f \circ g(R) \subset f(R) \)?

The range of g is a subset of R and so is a subset of the domain of f. It is elementary that A is a subset of B implies f(A) is a subset of f(B).