Product of two functions has Darboux Property (intermediate value property)
$begingroup$
Let $f, g:mathbb{R}rightarrowmathbb{R}$, such that
$fg$ has the Intermediate Value Property
$g$ has also this property
$f$ is injective
Is it true then that $f$ has also the Intermediate Value Property?
real-analysis calculus
edited Jan 18 at 16:38
bof
52.6k559121
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
$endgroup$
|
show 1 more comment
$begingroup$
Let $f, g:mathbb{R}rightarrowmathbb{R}$, such that
$fg$ has the Intermediate Value Property
$g$ has also this property
$f$ is injective
Is it true then that $f$ has also the Intermediate Value Property?
real-analysis calculus
edited Jan 18 at 16:38
bof
52.6k559121
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
$endgroup$
$begingroup$
If $g(x)=0$ the first two properties are trivial.
$endgroup$
– Yanko
Jan 18 at 14:03
$begingroup$
@Yanko ignoring that case
$endgroup$
– razvanelda
Jan 18 at 14:06
$begingroup$
@Yanko can you please give me a counter-example?
$endgroup$
– razvanelda
Jan 18 at 14:09
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I am quite sure that injectivity + IVT implies strict monotonicity, hence (with IVT again) implies continuity.
$endgroup$
– Mindlack
Jan 18 at 14:09
2
$begingroup$
Yes indeed, but since IVT does not imply continuity at all, this seems a rather strong statement. Indeed $f(x)=sgn(x)(|x|+1)$, and $g(x)=x$ is a counterexample.
$endgroup$
– Mindlack
Jan 18 at 14:15
|
show 1 more comment
$begingroup$
Let $f, g:mathbb{R}rightarrowmathbb{R}$, such that
$fg$ has the Intermediate Value Property
$g$ has also this property
$f$ is injective
Is it true then that $f$ has also the Intermediate Value Property?
real-analysis calculus
edited Jan 18 at 16:38
bof
52.6k559121
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
$endgroup$
Let $f, g:mathbb{R}rightarrowmathbb{R}$, such that
$fg$ has the Intermediate Value Property
$g$ has also this property
$f$ is injective
Is it true then that $f$ has also the Intermediate Value Property?
real-analysis calculus
real-analysis calculus
edited Jan 18 at 16:38
bof
52.6k559121
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
edited Jan 18 at 16:38
bof
52.6k559121
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
edited Jan 18 at 16:38
bof
52.6k559121
edited Jan 18 at 16:38
bof
52.6k559121
edited Jan 18 at 16:38
bof
52.6k559121
52.6k559121
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
asked Jan 18 at 14:00
razvaneldarazvanelda
915317
915317
$begingroup$
If $g(x)=0$ the first two properties are trivial.
$endgroup$
– Yanko
Jan 18 at 14:03
$begingroup$
@Yanko ignoring that case
$endgroup$
– razvanelda
Jan 18 at 14:06
$begingroup$
@Yanko can you please give me a counter-example?
$endgroup$
– razvanelda
Jan 18 at 14:09
$begingroup$
I am quite sure that injectivity + IVT implies strict monotonicity, hence (with IVT again) implies continuity.
$endgroup$
– Mindlack
Jan 18 at 14:09
2
$begingroup$
Yes indeed, but since IVT does not imply continuity at all, this seems a rather strong statement. Indeed $f(x)=sgn(x)(|x|+1)$, and $g(x)=x$ is a counterexample.
$endgroup$
– Mindlack
Jan 18 at 14:15
|
show 1 more comment
$begingroup$
If $g(x)=0$ the first two properties are trivial.
$endgroup$
– Yanko
Jan 18 at 14:03
$begingroup$
@Yanko ignoring that case
$endgroup$
– razvanelda
Jan 18 at 14:06
$begingroup$
@Yanko can you please give me a counter-example?
$endgroup$
– razvanelda
Jan 18 at 14:09
$begingroup$
I am quite sure that injectivity + IVT implies strict monotonicity, hence (with IVT again) implies continuity.
$endgroup$
– Mindlack
Jan 18 at 14:09
2
$begingroup$
Yes indeed, but since IVT does not imply continuity at all, this seems a rather strong statement. Indeed $f(x)=sgn(x)(|x|+1)$, and $g(x)=x$ is a counterexample.
$endgroup$
– Mindlack
Jan 18 at 14:15
$begingroup$
If $g(x)=0$ the first two properties are trivial.
$endgroup$
– Yanko
Jan 18 at 14:03
$begingroup$
If $g(x)=0$ the first two properties are trivial.
$endgroup$
– Yanko
Jan 18 at 14:03
$begingroup$
@Yanko ignoring that case
$endgroup$
– razvanelda
Jan 18 at 14:06
$begingroup$
@Yanko ignoring that case
$endgroup$
– razvanelda
Jan 18 at 14:06
$begingroup$
@Yanko can you please give me a counter-example?
$endgroup$
– razvanelda
Jan 18 at 14:09
$begingroup$
@Yanko can you please give me a counter-example?
$endgroup$
– razvanelda
Jan 18 at 14:09
$begingroup$
I am quite sure that injectivity + IVT implies strict monotonicity, hence (with IVT again) implies continuity.
$endgroup$
– Mindlack
Jan 18 at 14:09
$begingroup$
I am quite sure that injectivity + IVT implies strict monotonicity, hence (with IVT again) implies continuity.
$endgroup$
– Mindlack
Jan 18 at 14:09
2
2
$begingroup$
Yes indeed, but since IVT does not imply continuity at all, this seems a rather strong statement. Indeed $f(x)=sgn(x)(|x|+1)$, and $g(x)=x$ is a counterexample.
$endgroup$
– Mindlack
Jan 18 at 14:15
$begingroup$
Yes indeed, but since IVT does not imply continuity at all, this seems a rather strong statement. Indeed $f(x)=sgn(x)(|x|+1)$, and $g(x)=x$ is a counterexample.
$endgroup$
– Mindlack
Jan 18 at 14:15
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Here is a counterexample where $g$ and $fg$ have a strong form of the intermediate value property, namely, each of them takes on all real values in every interval.
Lemma. There are sets $A,Bsubseteqmathbb R$ and a function $varphi:mathbb Rtomathbb R$ such that $Acup B=mathbb R$, $Acap B=emptyset$, and $varphi(Acap I)=varphi(Bcap I)=mathbb R$ for every interval $I$.
Proof. Let $I_1,I_2,I_3,dots$ enumerate the open intervals with rational endpoints. Construct pairwise disjoint Cantor sets $A_1,B_1,A_2,B_2,A_3,B_3,dots$ with $A_n,B_nsubseteq I_n$. Let $A=bigcup_{n=1}^infty A_n$ and $B=mathbb Rsetminus Asupseteqbigcup_{n=1}^infty B_n$. Define $varphi:mathbb Rtomathbb R$ so that $varphi(A_n)=varphi(B_n)=mathbb R$ for every $n$.
Let $A,Bsubseteqmathbb R$ and $varphi:mathbb Rtomathbb R$ be as in the Lemma.
Let $h:mathbb Rto(0,1)$ be injective. Define $f:mathbb Rto(1,2)cup(3,4)$ by setting
$$f(x)=begin{cases}
h(x)+1quadtext{ if }quad xin A,\
h(x)+3quadtext{ if }quad xin B.
end{cases}$$
Then $f$ is injective and does not have the intermediate value property.
Finally, define $g:mathbb Rtomathbb R$ by setting
$$g(x)=begin{cases}
varphi(x)quadtext{ if }quad xin A,\
frac{varphi(x)}{f(x)}quad text{ if }quad xin B.
end{cases}$$
Then for every interval $I$ we have
$$f(I)supseteq f(Acap I)=varphi(Acap I)=mathbb R$$
and
$$fg(I)supset fg(Bcap I)=varphi(Bcap I)=mathbb R.$$
answered Jan 19 at 7:33
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
Fix a set $A$ of real numbers (for simplicity say $A=(0,1)$).
Choose a function $f:Arightarrow mathbb{R}$ that is injective and does not have IVT. Extend $f$ to $mathbb{R}$ arbitrarily such that $f$ is still injective and continuous on $A^c$ (make sure it is also continuous on $0,1$). Now let $g:mathbb{R}rightarrowmathbb{R}$ be zero on $A$ and continuous.
Then $fg$ is continuous (as it is continuous on $A^c$ and zero on $A$) and so has an IVT.
$g$ is contiuous so it has the IVT.
But $f$ is constructed not to have the IVT on $A$.
answered Jan 18 at 14:14
YankoYanko
8,5742830
$endgroup$
$begingroup$
Why is $fg$ continuous?
$endgroup$
– razvanelda
Jan 18 at 14:15
$begingroup$
@razvanelda because $fg$ is continuous on $A^c$ as product of continuous functions. Also it's zero on $A$. Showing continuity on $0,1$ is easy because as $x$ goes to zero $g(x)$ goes to zero and $f(x)$ goes to $f(0)$ which is constant. The same with $x$ goes to $1$.
$endgroup$
– Yanko
Jan 18 at 14:16
add a comment |
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2 Answers
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$begingroup$
Here is a counterexample where $g$ and $fg$ have a strong form of the intermediate value property, namely, each of them takes on all real values in every interval.
Lemma. There are sets $A,Bsubseteqmathbb R$ and a function $varphi:mathbb Rtomathbb R$ such that $Acup B=mathbb R$, $Acap B=emptyset$, and $varphi(Acap I)=varphi(Bcap I)=mathbb R$ for every interval $I$.
Proof. Let $I_1,I_2,I_3,dots$ enumerate the open intervals with rational endpoints. Construct pairwise disjoint Cantor sets $A_1,B_1,A_2,B_2,A_3,B_3,dots$ with $A_n,B_nsubseteq I_n$. Let $A=bigcup_{n=1}^infty A_n$ and $B=mathbb Rsetminus Asupseteqbigcup_{n=1}^infty B_n$. Define $varphi:mathbb Rtomathbb R$ so that $varphi(A_n)=varphi(B_n)=mathbb R$ for every $n$.
Let $A,Bsubseteqmathbb R$ and $varphi:mathbb Rtomathbb R$ be as in the Lemma.
Let $h:mathbb Rto(0,1)$ be injective. Define $f:mathbb Rto(1,2)cup(3,4)$ by setting
$$f(x)=begin{cases}
h(x)+1quadtext{ if }quad xin A,\
h(x)+3quadtext{ if }quad xin B.
end{cases}$$
Then $f$ is injective and does not have the intermediate value property.
Finally, define $g:mathbb Rtomathbb R$ by setting
$$g(x)=begin{cases}
varphi(x)quadtext{ if }quad xin A,\
frac{varphi(x)}{f(x)}quad text{ if }quad xin B.
end{cases}$$
Then for every interval $I$ we have
$$f(I)supseteq f(Acap I)=varphi(Acap I)=mathbb R$$
and
$$fg(I)supset fg(Bcap I)=varphi(Bcap I)=mathbb R.$$
answered Jan 19 at 7:33
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
Here is a counterexample where $g$ and $fg$ have a strong form of the intermediate value property, namely, each of them takes on all real values in every interval.
Lemma. There are sets $A,Bsubseteqmathbb R$ and a function $varphi:mathbb Rtomathbb R$ such that $Acup B=mathbb R$, $Acap B=emptyset$, and $varphi(Acap I)=varphi(Bcap I)=mathbb R$ for every interval $I$.
Proof. Let $I_1,I_2,I_3,dots$ enumerate the open intervals with rational endpoints. Construct pairwise disjoint Cantor sets $A_1,B_1,A_2,B_2,A_3,B_3,dots$ with $A_n,B_nsubseteq I_n$. Let $A=bigcup_{n=1}^infty A_n$ and $B=mathbb Rsetminus Asupseteqbigcup_{n=1}^infty B_n$. Define $varphi:mathbb Rtomathbb R$ so that $varphi(A_n)=varphi(B_n)=mathbb R$ for every $n$.
Let $A,Bsubseteqmathbb R$ and $varphi:mathbb Rtomathbb R$ be as in the Lemma.
Let $h:mathbb Rto(0,1)$ be injective. Define $f:mathbb Rto(1,2)cup(3,4)$ by setting
$$f(x)=begin{cases}
h(x)+1quadtext{ if }quad xin A,\
h(x)+3quadtext{ if }quad xin B.
end{cases}$$
Then $f$ is injective and does not have the intermediate value property.
Finally, define $g:mathbb Rtomathbb R$ by setting
$$g(x)=begin{cases}
varphi(x)quadtext{ if }quad xin A,\
frac{varphi(x)}{f(x)}quad text{ if }quad xin B.
end{cases}$$
Then for every interval $I$ we have
$$f(I)supseteq f(Acap I)=varphi(Acap I)=mathbb R$$
and
$$fg(I)supset fg(Bcap I)=varphi(Bcap I)=mathbb R.$$
answered Jan 19 at 7:33
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
Here is a counterexample where $g$ and $fg$ have a strong form of the intermediate value property, namely, each of them takes on all real values in every interval.
Lemma. There are sets $A,Bsubseteqmathbb R$ and a function $varphi:mathbb Rtomathbb R$ such that $Acup B=mathbb R$, $Acap B=emptyset$, and $varphi(Acap I)=varphi(Bcap I)=mathbb R$ for every interval $I$.
Proof. Let $I_1,I_2,I_3,dots$ enumerate the open intervals with rational endpoints. Construct pairwise disjoint Cantor sets $A_1,B_1,A_2,B_2,A_3,B_3,dots$ with $A_n,B_nsubseteq I_n$. Let $A=bigcup_{n=1}^infty A_n$ and $B=mathbb Rsetminus Asupseteqbigcup_{n=1}^infty B_n$. Define $varphi:mathbb Rtomathbb R$ so that $varphi(A_n)=varphi(B_n)=mathbb R$ for every $n$.
Let $A,Bsubseteqmathbb R$ and $varphi:mathbb Rtomathbb R$ be as in the Lemma.
Let $h:mathbb Rto(0,1)$ be injective. Define $f:mathbb Rto(1,2)cup(3,4)$ by setting
$$f(x)=begin{cases}
h(x)+1quadtext{ if }quad xin A,\
h(x)+3quadtext{ if }quad xin B.
end{cases}$$
Then $f$ is injective and does not have the intermediate value property.
Finally, define $g:mathbb Rtomathbb R$ by setting
$$g(x)=begin{cases}
varphi(x)quadtext{ if }quad xin A,\
frac{varphi(x)}{f(x)}quad text{ if }quad xin B.
end{cases}$$
Then for every interval $I$ we have
$$f(I)supseteq f(Acap I)=varphi(Acap I)=mathbb R$$
and
$$fg(I)supset fg(Bcap I)=varphi(Bcap I)=mathbb R.$$
answered Jan 19 at 7:33
bofbof
52.6k559121
$endgroup$
Here is a counterexample where $g$ and $fg$ have a strong form of the intermediate value property, namely, each of them takes on all real values in every interval.
Lemma. There are sets $A,Bsubseteqmathbb R$ and a function $varphi:mathbb Rtomathbb R$ such that $Acup B=mathbb R$, $Acap B=emptyset$, and $varphi(Acap I)=varphi(Bcap I)=mathbb R$ for every interval $I$.
Proof. Let $I_1,I_2,I_3,dots$ enumerate the open intervals with rational endpoints. Construct pairwise disjoint Cantor sets $A_1,B_1,A_2,B_2,A_3,B_3,dots$ with $A_n,B_nsubseteq I_n$. Let $A=bigcup_{n=1}^infty A_n$ and $B=mathbb Rsetminus Asupseteqbigcup_{n=1}^infty B_n$. Define $varphi:mathbb Rtomathbb R$ so that $varphi(A_n)=varphi(B_n)=mathbb R$ for every $n$.
Let $A,Bsubseteqmathbb R$ and $varphi:mathbb Rtomathbb R$ be as in the Lemma.
Let $h:mathbb Rto(0,1)$ be injective. Define $f:mathbb Rto(1,2)cup(3,4)$ by setting
$$f(x)=begin{cases}
h(x)+1quadtext{ if }quad xin A,\
h(x)+3quadtext{ if }quad xin B.
end{cases}$$
Then $f$ is injective and does not have the intermediate value property.
Finally, define $g:mathbb Rtomathbb R$ by setting
$$g(x)=begin{cases}
varphi(x)quadtext{ if }quad xin A,\
frac{varphi(x)}{f(x)}quad text{ if }quad xin B.
end{cases}$$
Then for every interval $I$ we have
$$f(I)supseteq f(Acap I)=varphi(Acap I)=mathbb R$$
and
$$fg(I)supset fg(Bcap I)=varphi(Bcap I)=mathbb R.$$
answered Jan 19 at 7:33
bofbof
52.6k559121
answered Jan 19 at 7:33
bofbof
52.6k559121
answered Jan 19 at 7:33
bofbof
52.6k559121
answered Jan 19 at 7:33
bofbof
52.6k559121
52.6k559121
add a comment |
add a comment |
$begingroup$
Fix a set $A$ of real numbers (for simplicity say $A=(0,1)$).
Choose a function $f:Arightarrow mathbb{R}$ that is injective and does not have IVT. Extend $f$ to $mathbb{R}$ arbitrarily such that $f$ is still injective and continuous on $A^c$ (make sure it is also continuous on $0,1$). Now let $g:mathbb{R}rightarrowmathbb{R}$ be zero on $A$ and continuous.
Then $fg$ is continuous (as it is continuous on $A^c$ and zero on $A$) and so has an IVT.
$g$ is contiuous so it has the IVT.
But $f$ is constructed not to have the IVT on $A$.
answered Jan 18 at 14:14
YankoYanko
8,5742830
$endgroup$
$begingroup$
Why is $fg$ continuous?
$endgroup$
– razvanelda
Jan 18 at 14:15
$begingroup$
@razvanelda because $fg$ is continuous on $A^c$ as product of continuous functions. Also it's zero on $A$. Showing continuity on $0,1$ is easy because as $x$ goes to zero $g(x)$ goes to zero and $f(x)$ goes to $f(0)$ which is constant. The same with $x$ goes to $1$.
$endgroup$
– Yanko
Jan 18 at 14:16
add a comment |
$begingroup$
Fix a set $A$ of real numbers (for simplicity say $A=(0,1)$).
Choose a function $f:Arightarrow mathbb{R}$ that is injective and does not have IVT. Extend $f$ to $mathbb{R}$ arbitrarily such that $f$ is still injective and continuous on $A^c$ (make sure it is also continuous on $0,1$). Now let $g:mathbb{R}rightarrowmathbb{R}$ be zero on $A$ and continuous.
Then $fg$ is continuous (as it is continuous on $A^c$ and zero on $A$) and so has an IVT.
$g$ is contiuous so it has the IVT.
But $f$ is constructed not to have the IVT on $A$.
answered Jan 18 at 14:14
YankoYanko
8,5742830
$endgroup$
$begingroup$
Why is $fg$ continuous?
$endgroup$
– razvanelda
Jan 18 at 14:15
$begingroup$
@razvanelda because $fg$ is continuous on $A^c$ as product of continuous functions. Also it's zero on $A$. Showing continuity on $0,1$ is easy because as $x$ goes to zero $g(x)$ goes to zero and $f(x)$ goes to $f(0)$ which is constant. The same with $x$ goes to $1$.
$endgroup$
– Yanko
Jan 18 at 14:16
add a comment |
$begingroup$
Fix a set $A$ of real numbers (for simplicity say $A=(0,1)$).
Choose a function $f:Arightarrow mathbb{R}$ that is injective and does not have IVT. Extend $f$ to $mathbb{R}$ arbitrarily such that $f$ is still injective and continuous on $A^c$ (make sure it is also continuous on $0,1$). Now let $g:mathbb{R}rightarrowmathbb{R}$ be zero on $A$ and continuous.
Then $fg$ is continuous (as it is continuous on $A^c$ and zero on $A$) and so has an IVT.
$g$ is contiuous so it has the IVT.
But $f$ is constructed not to have the IVT on $A$.
answered Jan 18 at 14:14
YankoYanko
8,5742830
$endgroup$
Fix a set $A$ of real numbers (for simplicity say $A=(0,1)$).
Choose a function $f:Arightarrow mathbb{R}$ that is injective and does not have IVT. Extend $f$ to $mathbb{R}$ arbitrarily such that $f$ is still injective and continuous on $A^c$ (make sure it is also continuous on $0,1$). Now let $g:mathbb{R}rightarrowmathbb{R}$ be zero on $A$ and continuous.
Then $fg$ is continuous (as it is continuous on $A^c$ and zero on $A$) and so has an IVT.
$g$ is contiuous so it has the IVT.
But $f$ is constructed not to have the IVT on $A$.
answered Jan 18 at 14:14
YankoYanko
8,5742830
answered Jan 18 at 14:14
YankoYanko
8,5742830
answered Jan 18 at 14:14
YankoYanko
8,5742830
answered Jan 18 at 14:14
YankoYanko
8,5742830
8,5742830
$begingroup$
Why is $fg$ continuous?
$endgroup$
– razvanelda
Jan 18 at 14:15
$begingroup$
@razvanelda because $fg$ is continuous on $A^c$ as product of continuous functions. Also it's zero on $A$. Showing continuity on $0,1$ is easy because as $x$ goes to zero $g(x)$ goes to zero and $f(x)$ goes to $f(0)$ which is constant. The same with $x$ goes to $1$.
$endgroup$
– Yanko
Jan 18 at 14:16
add a comment |
$begingroup$
Why is $fg$ continuous?
$endgroup$
– razvanelda
Jan 18 at 14:15
$begingroup$
@razvanelda because $fg$ is continuous on $A^c$ as product of continuous functions. Also it's zero on $A$. Showing continuity on $0,1$ is easy because as $x$ goes to zero $g(x)$ goes to zero and $f(x)$ goes to $f(0)$ which is constant. The same with $x$ goes to $1$.
$endgroup$
– Yanko
Jan 18 at 14:16
$begingroup$
Why is $fg$ continuous?
$endgroup$
– razvanelda
Jan 18 at 14:15
$begingroup$
Why is $fg$ continuous?
$endgroup$
– razvanelda
Jan 18 at 14:15
$begingroup$
@razvanelda because $fg$ is continuous on $A^c$ as product of continuous functions. Also it's zero on $A$. Showing continuity on $0,1$ is easy because as $x$ goes to zero $g(x)$ goes to zero and $f(x)$ goes to $f(0)$ which is constant. The same with $x$ goes to $1$.
$endgroup$
– Yanko
Jan 18 at 14:16
$begingroup$
@razvanelda because $fg$ is continuous on $A^c$ as product of continuous functions. Also it's zero on $A$. Showing continuity on $0,1$ is easy because as $x$ goes to zero $g(x)$ goes to zero and $f(x)$ goes to $f(0)$ which is constant. The same with $x$ goes to $1$.
$endgroup$
– Yanko
Jan 18 at 14:16
add a comment |
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$begingroup$
If $g(x)=0$ the first two properties are trivial.
$endgroup$
– Yanko
Jan 18 at 14:03
$begingroup$
@Yanko ignoring that case
$endgroup$
– razvanelda
Jan 18 at 14:06
$begingroup$
@Yanko can you please give me a counter-example?
$endgroup$
– razvanelda
Jan 18 at 14:09
$begingroup$
I am quite sure that injectivity + IVT implies strict monotonicity, hence (with IVT again) implies continuity.
$endgroup$
– Mindlack
Jan 18 at 14:09
2
$begingroup$
Yes indeed, but since IVT does not imply continuity at all, this seems a rather strong statement. Indeed $f(x)=sgn(x)(|x|+1)$, and $g(x)=x$ is a counterexample.
$endgroup$
– Mindlack
Jan 18 at 14:15