Proving if $f$ is strictly increasing and onto, there exists a strictly increasing and onto function $g$ such...
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Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.
I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.
real-analysis
edited Nov 23 at 3:16
user302797
19.7k92252
asked Nov 22 at 19:10
Nykis
436
|
show 6 more comments
up vote
1
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Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.
I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.
real-analysis
edited Nov 23 at 3:16
user302797
19.7k92252
asked Nov 22 at 19:10
Nykis
436
I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13
@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14
@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16
@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17
Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17
|
show 6 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.
I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.
real-analysis
edited Nov 23 at 3:16
user302797
19.7k92252
asked Nov 22 at 19:10
Nykis
436
Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.
I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.
real-analysis
real-analysis
edited Nov 23 at 3:16
user302797
19.7k92252
asked Nov 22 at 19:10
Nykis
436
edited Nov 23 at 3:16
user302797
19.7k92252
asked Nov 22 at 19:10
Nykis
436
edited Nov 23 at 3:16
user302797
19.7k92252
edited Nov 23 at 3:16
user302797
19.7k92252
edited Nov 23 at 3:16
user302797
19.7k92252
19.7k92252
asked Nov 22 at 19:10
Nykis
436
asked Nov 22 at 19:10
Nykis
436
asked Nov 22 at 19:10
Nykis
436
436
I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13
@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14
@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16
@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17
Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17
|
show 6 more comments
I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13
@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14
@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16
@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17
Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17
I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13
I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13
@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14
@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14
@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16
@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16
@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17
@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17
Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17
Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17
|
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).
But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
$$h : (c,d) to mathbb R
$$
given by
$$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
$$
This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.
Now define
$$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
$$
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
add a comment |
up vote
1
down vote
Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).
But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
$$h : (c,d) to mathbb R
$$
given by
$$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
$$
This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.
Now define
$$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
$$
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
add a comment |
up vote
5
down vote
accepted
You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).
But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
$$h : (c,d) to mathbb R
$$
given by
$$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
$$
This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.
Now define
$$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
$$
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).
But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
$$h : (c,d) to mathbb R
$$
given by
$$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
$$
This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.
Now define
$$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
$$
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).
But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
$$h : (c,d) to mathbb R
$$
given by
$$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
$$
This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.
Now define
$$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
$$
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
answered Nov 22 at 19:28
Lee Mosher
47.7k33681
47.7k33681
add a comment |
add a comment |
up vote
1
down vote
Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
add a comment |
up vote
1
down vote
Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
answered Nov 22 at 19:13
Rob Arthan
28.7k42865
28.7k42865
add a comment |
add a comment |
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I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13
@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14
@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16
@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17
Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17