Existence of Totally Non-monotonic Continuous Real-valued Function in $[0, 1]$
Definition:
Let $f: [0, 1] rightarrow mathbb{R}$ be a function. Say $f$ is Totally-Nonmonotonic iff for any $a, b$ in [$0, 1$], $f$ is NOT monotonic on [$a, b$].
Question:
Let $f: [0, 1] rightarrow mathbb{R}$ (both of which are equipped with the topology generated by the usual distance) be a continuous function and let $epsilon$ be a real positive real number. Prove that there exists a continuous totally-nonmonotonic function $g: [0, 1] rightarrow mathbb{R}$ such that |$f(t) - g(t)$| $leq epsilon$ for every $t in [0, 1]$.
I am tempted to apply Stone-Weirstrass Theorem here but then realize that although the set of all totally-nonmonotonic continuous functions does separate points in $C[X, mathbb{R}]$, however, it is not a subalgebra because, assuming $h$ is a totally-nonmonotonic continuous function, both $h$ and $-h$ are in the set but $h + (-h)$ is not in the set. I think the purpose here is to prove that the set of all totally-nonmonotonic continuous function is dense in $C[X, mathbb{R}]$. Could you anybody provide me some hints if you have any ideas?
Any responses will be appreciated.
real-analysis monotone-functions
edited Dec 26 at 2:46
twnly
32818
asked Dec 26 at 1:09
Sanae Kochiya
386
add a comment |
Definition:
Let $f: [0, 1] rightarrow mathbb{R}$ be a function. Say $f$ is Totally-Nonmonotonic iff for any $a, b$ in [$0, 1$], $f$ is NOT monotonic on [$a, b$].
Question:
Let $f: [0, 1] rightarrow mathbb{R}$ (both of which are equipped with the topology generated by the usual distance) be a continuous function and let $epsilon$ be a real positive real number. Prove that there exists a continuous totally-nonmonotonic function $g: [0, 1] rightarrow mathbb{R}$ such that |$f(t) - g(t)$| $leq epsilon$ for every $t in [0, 1]$.
I am tempted to apply Stone-Weirstrass Theorem here but then realize that although the set of all totally-nonmonotonic continuous functions does separate points in $C[X, mathbb{R}]$, however, it is not a subalgebra because, assuming $h$ is a totally-nonmonotonic continuous function, both $h$ and $-h$ are in the set but $h + (-h)$ is not in the set. I think the purpose here is to prove that the set of all totally-nonmonotonic continuous function is dense in $C[X, mathbb{R}]$. Could you anybody provide me some hints if you have any ideas?
Any responses will be appreciated.
real-analysis monotone-functions
edited Dec 26 at 2:46
twnly
32818
asked Dec 26 at 1:09
Sanae Kochiya
386
I am very sorry for my mistakes, Ross. My daughter just accidentally quick the "post" bottom before I finished .....
– Sanae Kochiya
Dec 26 at 1:21
One suggestion I have is playing around with the Weierstrass function (or some other continuous nowhere-monotonic function).
– MathematicsStudent1122
Dec 26 at 2:18
For b <= a, [a,b] is empty or singleton and f is monotonic over [a,b].
– William Elliot
Dec 26 at 2:18
add a comment |
Definition:
Let $f: [0, 1] rightarrow mathbb{R}$ be a function. Say $f$ is Totally-Nonmonotonic iff for any $a, b$ in [$0, 1$], $f$ is NOT monotonic on [$a, b$].
Question:
Let $f: [0, 1] rightarrow mathbb{R}$ (both of which are equipped with the topology generated by the usual distance) be a continuous function and let $epsilon$ be a real positive real number. Prove that there exists a continuous totally-nonmonotonic function $g: [0, 1] rightarrow mathbb{R}$ such that |$f(t) - g(t)$| $leq epsilon$ for every $t in [0, 1]$.
I am tempted to apply Stone-Weirstrass Theorem here but then realize that although the set of all totally-nonmonotonic continuous functions does separate points in $C[X, mathbb{R}]$, however, it is not a subalgebra because, assuming $h$ is a totally-nonmonotonic continuous function, both $h$ and $-h$ are in the set but $h + (-h)$ is not in the set. I think the purpose here is to prove that the set of all totally-nonmonotonic continuous function is dense in $C[X, mathbb{R}]$. Could you anybody provide me some hints if you have any ideas?
Any responses will be appreciated.
real-analysis monotone-functions
edited Dec 26 at 2:46
twnly
32818
asked Dec 26 at 1:09
Sanae Kochiya
386
Definition:
Let $f: [0, 1] rightarrow mathbb{R}$ be a function. Say $f$ is Totally-Nonmonotonic iff for any $a, b$ in [$0, 1$], $f$ is NOT monotonic on [$a, b$].
Question:
Let $f: [0, 1] rightarrow mathbb{R}$ (both of which are equipped with the topology generated by the usual distance) be a continuous function and let $epsilon$ be a real positive real number. Prove that there exists a continuous totally-nonmonotonic function $g: [0, 1] rightarrow mathbb{R}$ such that |$f(t) - g(t)$| $leq epsilon$ for every $t in [0, 1]$.
I am tempted to apply Stone-Weirstrass Theorem here but then realize that although the set of all totally-nonmonotonic continuous functions does separate points in $C[X, mathbb{R}]$, however, it is not a subalgebra because, assuming $h$ is a totally-nonmonotonic continuous function, both $h$ and $-h$ are in the set but $h + (-h)$ is not in the set. I think the purpose here is to prove that the set of all totally-nonmonotonic continuous function is dense in $C[X, mathbb{R}]$. Could you anybody provide me some hints if you have any ideas?
Any responses will be appreciated.
real-analysis monotone-functions
real-analysis monotone-functions
edited Dec 26 at 2:46
twnly
32818
asked Dec 26 at 1:09
Sanae Kochiya
386
edited Dec 26 at 2:46
twnly
32818
asked Dec 26 at 1:09
Sanae Kochiya
386
edited Dec 26 at 2:46
twnly
32818
edited Dec 26 at 2:46
twnly
32818
edited Dec 26 at 2:46
twnly
32818
32818
asked Dec 26 at 1:09
Sanae Kochiya
386
asked Dec 26 at 1:09
Sanae Kochiya
386
asked Dec 26 at 1:09
Sanae Kochiya
386
386
I am very sorry for my mistakes, Ross. My daughter just accidentally quick the "post" bottom before I finished .....
– Sanae Kochiya
Dec 26 at 1:21
One suggestion I have is playing around with the Weierstrass function (or some other continuous nowhere-monotonic function).
– MathematicsStudent1122
Dec 26 at 2:18
For b <= a, [a,b] is empty or singleton and f is monotonic over [a,b].
– William Elliot
Dec 26 at 2:18
add a comment |
I am very sorry for my mistakes, Ross. My daughter just accidentally quick the "post" bottom before I finished .....
– Sanae Kochiya
Dec 26 at 1:21
One suggestion I have is playing around with the Weierstrass function (or some other continuous nowhere-monotonic function).
– MathematicsStudent1122
Dec 26 at 2:18
For b <= a, [a,b] is empty or singleton and f is monotonic over [a,b].
– William Elliot
Dec 26 at 2:18
I am very sorry for my mistakes, Ross. My daughter just accidentally quick the "post" bottom before I finished .....
– Sanae Kochiya
Dec 26 at 1:21
I am very sorry for my mistakes, Ross. My daughter just accidentally quick the "post" bottom before I finished .....
– Sanae Kochiya
Dec 26 at 1:21
One suggestion I have is playing around with the Weierstrass function (or some other continuous nowhere-monotonic function).
– MathematicsStudent1122
Dec 26 at 2:18
One suggestion I have is playing around with the Weierstrass function (or some other continuous nowhere-monotonic function).
– MathematicsStudent1122
Dec 26 at 2:18
For b <= a, [a,b] is empty or singleton and f is monotonic over [a,b].
– William Elliot
Dec 26 at 2:18
For b <= a, [a,b] is empty or singleton and f is monotonic over [a,b].
– William Elliot
Dec 26 at 2:18
add a comment |
2 Answers
2
active
oldest
votes
Build a continuous and nowhere monotonic $h$ on $[0,1]$ with $h(0)=0$,$h(1)=1$ and $|h|le1$ as done here. For fixed $epsilon>0$ there is some $n$ s.t. $|x-y|<1/n$ implies $|f(x)-f(y)|<epsilon$. Let $I_k=[frac kn,frac{k+1}n]$, $k=0,...,n-1$ partition $[0,1]$ and set $h_k(x)=(f(frac{k+1}n)-f(frac kn))h(nx-k)+f(frac kn)$. Observe $g(x)=sum_k chi_{I_k}h_k(x)$ is continuous and nowhere monotonic. Now any $xin [0,1]$ belongs to some $I_k$ so $|f(x)-g(x)|le|f(x)-f(k/n)|+|f(k/n)-g(x)|le 2epsilon$.
answered Dec 26 at 2:32
Guacho Perez
3,88911131
I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help!
– Sanae Kochiya
yesterday
add a comment |
A standard application of Baire Category Theorem says that the set of functions in $C[0,1]$ that are differntiable at at least one point is of first category. Hence its complement is dense. It follows that any $f in C[0,1]$ can be approximated uniformly by nowhere differentiable functions. Any nowhere differentiable function is nowhere monotonic.
answered 2 days ago
Kavi Rama Murthy
49.6k31854
First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim?
– Sanae Kochiya
yesterday
Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya
– Kavi Rama Murthy
yesterday
Very appreciate your help. I got the pdf and will check it by myself.
– Sanae Kochiya
yesterday
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
Build a continuous and nowhere monotonic $h$ on $[0,1]$ with $h(0)=0$,$h(1)=1$ and $|h|le1$ as done here. For fixed $epsilon>0$ there is some $n$ s.t. $|x-y|<1/n$ implies $|f(x)-f(y)|<epsilon$. Let $I_k=[frac kn,frac{k+1}n]$, $k=0,...,n-1$ partition $[0,1]$ and set $h_k(x)=(f(frac{k+1}n)-f(frac kn))h(nx-k)+f(frac kn)$. Observe $g(x)=sum_k chi_{I_k}h_k(x)$ is continuous and nowhere monotonic. Now any $xin [0,1]$ belongs to some $I_k$ so $|f(x)-g(x)|le|f(x)-f(k/n)|+|f(k/n)-g(x)|le 2epsilon$.
answered Dec 26 at 2:32
Guacho Perez
3,88911131
I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help!
– Sanae Kochiya
yesterday
add a comment |
Build a continuous and nowhere monotonic $h$ on $[0,1]$ with $h(0)=0$,$h(1)=1$ and $|h|le1$ as done here. For fixed $epsilon>0$ there is some $n$ s.t. $|x-y|<1/n$ implies $|f(x)-f(y)|<epsilon$. Let $I_k=[frac kn,frac{k+1}n]$, $k=0,...,n-1$ partition $[0,1]$ and set $h_k(x)=(f(frac{k+1}n)-f(frac kn))h(nx-k)+f(frac kn)$. Observe $g(x)=sum_k chi_{I_k}h_k(x)$ is continuous and nowhere monotonic. Now any $xin [0,1]$ belongs to some $I_k$ so $|f(x)-g(x)|le|f(x)-f(k/n)|+|f(k/n)-g(x)|le 2epsilon$.
answered Dec 26 at 2:32
Guacho Perez
3,88911131
I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help!
– Sanae Kochiya
yesterday
add a comment |
Build a continuous and nowhere monotonic $h$ on $[0,1]$ with $h(0)=0$,$h(1)=1$ and $|h|le1$ as done here. For fixed $epsilon>0$ there is some $n$ s.t. $|x-y|<1/n$ implies $|f(x)-f(y)|<epsilon$. Let $I_k=[frac kn,frac{k+1}n]$, $k=0,...,n-1$ partition $[0,1]$ and set $h_k(x)=(f(frac{k+1}n)-f(frac kn))h(nx-k)+f(frac kn)$. Observe $g(x)=sum_k chi_{I_k}h_k(x)$ is continuous and nowhere monotonic. Now any $xin [0,1]$ belongs to some $I_k$ so $|f(x)-g(x)|le|f(x)-f(k/n)|+|f(k/n)-g(x)|le 2epsilon$.
answered Dec 26 at 2:32
Guacho Perez
3,88911131
Build a continuous and nowhere monotonic $h$ on $[0,1]$ with $h(0)=0$,$h(1)=1$ and $|h|le1$ as done here. For fixed $epsilon>0$ there is some $n$ s.t. $|x-y|<1/n$ implies $|f(x)-f(y)|<epsilon$. Let $I_k=[frac kn,frac{k+1}n]$, $k=0,...,n-1$ partition $[0,1]$ and set $h_k(x)=(f(frac{k+1}n)-f(frac kn))h(nx-k)+f(frac kn)$. Observe $g(x)=sum_k chi_{I_k}h_k(x)$ is continuous and nowhere monotonic. Now any $xin [0,1]$ belongs to some $I_k$ so $|f(x)-g(x)|le|f(x)-f(k/n)|+|f(k/n)-g(x)|le 2epsilon$.
answered Dec 26 at 2:32
Guacho Perez
3,88911131
edited Dec 26 at 4:29
answered Dec 26 at 2:32
Guacho Perez
3,88911131
answered Dec 26 at 2:32
Guacho Perez
3,88911131
answered Dec 26 at 2:32
Guacho Perez
3,88911131
3,88911131
I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help!
– Sanae Kochiya
yesterday
add a comment |
I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help!
– Sanae Kochiya
yesterday
I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help!
– Sanae Kochiya
yesterday
I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help!
– Sanae Kochiya
yesterday
add a comment |
A standard application of Baire Category Theorem says that the set of functions in $C[0,1]$ that are differntiable at at least one point is of first category. Hence its complement is dense. It follows that any $f in C[0,1]$ can be approximated uniformly by nowhere differentiable functions. Any nowhere differentiable function is nowhere monotonic.
answered 2 days ago
Kavi Rama Murthy
49.6k31854
First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim?
– Sanae Kochiya
yesterday
Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya
– Kavi Rama Murthy
yesterday
Very appreciate your help. I got the pdf and will check it by myself.
– Sanae Kochiya
yesterday
add a comment |
A standard application of Baire Category Theorem says that the set of functions in $C[0,1]$ that are differntiable at at least one point is of first category. Hence its complement is dense. It follows that any $f in C[0,1]$ can be approximated uniformly by nowhere differentiable functions. Any nowhere differentiable function is nowhere monotonic.
answered 2 days ago
Kavi Rama Murthy
49.6k31854
First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim?
– Sanae Kochiya
yesterday
Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya
– Kavi Rama Murthy
yesterday
Very appreciate your help. I got the pdf and will check it by myself.
– Sanae Kochiya
yesterday
add a comment |
A standard application of Baire Category Theorem says that the set of functions in $C[0,1]$ that are differntiable at at least one point is of first category. Hence its complement is dense. It follows that any $f in C[0,1]$ can be approximated uniformly by nowhere differentiable functions. Any nowhere differentiable function is nowhere monotonic.
answered 2 days ago
Kavi Rama Murthy
49.6k31854
A standard application of Baire Category Theorem says that the set of functions in $C[0,1]$ that are differntiable at at least one point is of first category. Hence its complement is dense. It follows that any $f in C[0,1]$ can be approximated uniformly by nowhere differentiable functions. Any nowhere differentiable function is nowhere monotonic.
answered 2 days ago
Kavi Rama Murthy
49.6k31854
answered 2 days ago
Kavi Rama Murthy
49.6k31854
answered 2 days ago
Kavi Rama Murthy
49.6k31854
answered 2 days ago
Kavi Rama Murthy
49.6k31854
49.6k31854
First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim?
– Sanae Kochiya
yesterday
Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya
– Kavi Rama Murthy
yesterday
Very appreciate your help. I got the pdf and will check it by myself.
– Sanae Kochiya
yesterday
add a comment |
First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim?
– Sanae Kochiya
yesterday
Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya
– Kavi Rama Murthy
yesterday
Very appreciate your help. I got the pdf and will check it by myself.
– Sanae Kochiya
yesterday
First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim?
– Sanae Kochiya
yesterday
First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim?
– Sanae Kochiya
yesterday
Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya
– Kavi Rama Murthy
yesterday
Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya
– Kavi Rama Murthy
yesterday
Very appreciate your help. I got the pdf and will check it by myself.
– Sanae Kochiya
yesterday
Very appreciate your help. I got the pdf and will check it by myself.
– Sanae Kochiya
yesterday
add a comment |
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I am very sorry for my mistakes, Ross. My daughter just accidentally quick the "post" bottom before I finished .....
– Sanae Kochiya
Dec 26 at 1:21
One suggestion I have is playing around with the Weierstrass function (or some other continuous nowhere-monotonic function).
– MathematicsStudent1122
Dec 26 at 2:18
For b <= a, [a,b] is empty or singleton and f is monotonic over [a,b].
– William Elliot
Dec 26 at 2:18