Mixing
Is there a continuous $f$ satisfying $f(f(x))=-x^3+x$?
Is there a continuous function $f$ defined on real number satisfying:
$$f(f(x))=-x^3+x.$$
I'm shame to say it's my homework and I've spend several hours on it. Also, I've tried to construct a function, but all failed.
Any hint will be appreciated, at least I want to konw whether $f$ exist.
continuity functional-equations
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
|
show 1 more comment
Is there a continuous function $f$ defined on real number satisfying:
$$f(f(x))=-x^3+x.$$
I'm shame to say it's my homework and I've spend several hours on it. Also, I've tried to construct a function, but all failed.
Any hint will be appreciated, at least I want to konw whether $f$ exist.
continuity functional-equations
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
2
You have spent several hours on it, so please include in your post all results you derived during these several hours, so that others attempting the question do not take several hours to derive things you already know from your work.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:41
If we restrict to an interval, $frac{-1}{sqrt 3} < x <frac{1}{sqrt 3}, $ we can construct a $C^infty$ function $f$ that does this. My guess is that we cannot extend the function to the whole line, critical points and decreasing functions cause trouble
– Will Jagy
Nov 22 '18 at 2:47
$f(-1)$, $f(0)$ and $f(1)$ will all be zeroes of $f(x)$
– Seth
Nov 22 '18 at 3:35
I'm sorry for being away.I've thought f must be a bijection,so unbounded,moreover f(f) is monotonic when x is big enough.If f is monotonic when x is big enough,then f(f) must be monotonically increasing.But it's just a assumption.My homework always has an elegant solution,so I think it won't take too long.Sorry for my poor English and ugly typing style.
– Oolong milk tea
Nov 22 '18 at 4:45
@Fume There is no way $f$ can be a bijection. If it is one-to-one then, since $f(f(0))=f(f(1))$ we get $0=1$.
– Kavi Rama Murthy
Nov 22 '18 at 5:48
|
show 1 more comment
Is there a continuous function $f$ defined on real number satisfying:
$$f(f(x))=-x^3+x.$$
I'm shame to say it's my homework and I've spend several hours on it. Also, I've tried to construct a function, but all failed.
Any hint will be appreciated, at least I want to konw whether $f$ exist.
continuity functional-equations
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
Is there a continuous function $f$ defined on real number satisfying:
$$f(f(x))=-x^3+x.$$
I'm shame to say it's my homework and I've spend several hours on it. Also, I've tried to construct a function, but all failed.
Any hint will be appreciated, at least I want to konw whether $f$ exist.
continuity functional-equations
continuity functional-equations
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
edited Nov 22 '18 at 2:32
Tianlalu
3,09621038
3,09621038
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
asked Nov 22 '18 at 2:22
Oolong milk teaOolong milk tea
787
787
2
You have spent several hours on it, so please include in your post all results you derived during these several hours, so that others attempting the question do not take several hours to derive things you already know from your work.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:41
If we restrict to an interval, $frac{-1}{sqrt 3} < x <frac{1}{sqrt 3}, $ we can construct a $C^infty$ function $f$ that does this. My guess is that we cannot extend the function to the whole line, critical points and decreasing functions cause trouble
– Will Jagy
Nov 22 '18 at 2:47
$f(-1)$, $f(0)$ and $f(1)$ will all be zeroes of $f(x)$
– Seth
Nov 22 '18 at 3:35
I'm sorry for being away.I've thought f must be a bijection,so unbounded,moreover f(f) is monotonic when x is big enough.If f is monotonic when x is big enough,then f(f) must be monotonically increasing.But it's just a assumption.My homework always has an elegant solution,so I think it won't take too long.Sorry for my poor English and ugly typing style.
– Oolong milk tea
Nov 22 '18 at 4:45
@Fume There is no way $f$ can be a bijection. If it is one-to-one then, since $f(f(0))=f(f(1))$ we get $0=1$.
– Kavi Rama Murthy
Nov 22 '18 at 5:48
|
show 1 more comment
2
You have spent several hours on it, so please include in your post all results you derived during these several hours, so that others attempting the question do not take several hours to derive things you already know from your work.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:41
If we restrict to an interval, $frac{-1}{sqrt 3} < x <frac{1}{sqrt 3}, $ we can construct a $C^infty$ function $f$ that does this. My guess is that we cannot extend the function to the whole line, critical points and decreasing functions cause trouble
– Will Jagy
Nov 22 '18 at 2:47
$f(-1)$, $f(0)$ and $f(1)$ will all be zeroes of $f(x)$
– Seth
Nov 22 '18 at 3:35
I'm sorry for being away.I've thought f must be a bijection,so unbounded,moreover f(f) is monotonic when x is big enough.If f is monotonic when x is big enough,then f(f) must be monotonically increasing.But it's just a assumption.My homework always has an elegant solution,so I think it won't take too long.Sorry for my poor English and ugly typing style.
– Oolong milk tea
Nov 22 '18 at 4:45
@Fume There is no way $f$ can be a bijection. If it is one-to-one then, since $f(f(0))=f(f(1))$ we get $0=1$.
– Kavi Rama Murthy
Nov 22 '18 at 5:48
2
2
You have spent several hours on it, so please include in your post all results you derived during these several hours, so that others attempting the question do not take several hours to derive things you already know from your work.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:41
You have spent several hours on it, so please include in your post all results you derived during these several hours, so that others attempting the question do not take several hours to derive things you already know from your work.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:41
If we restrict to an interval, $frac{-1}{sqrt 3} < x <frac{1}{sqrt 3}, $ we can construct a $C^infty$ function $f$ that does this. My guess is that we cannot extend the function to the whole line, critical points and decreasing functions cause trouble
– Will Jagy
Nov 22 '18 at 2:47
If we restrict to an interval, $frac{-1}{sqrt 3} < x <frac{1}{sqrt 3}, $ we can construct a $C^infty$ function $f$ that does this. My guess is that we cannot extend the function to the whole line, critical points and decreasing functions cause trouble
– Will Jagy
Nov 22 '18 at 2:47
$f(-1)$, $f(0)$ and $f(1)$ will all be zeroes of $f(x)$
– Seth
Nov 22 '18 at 3:35
$f(-1)$, $f(0)$ and $f(1)$ will all be zeroes of $f(x)$
– Seth
Nov 22 '18 at 3:35
I'm sorry for being away.I've thought f must be a bijection,so unbounded,moreover f(f) is monotonic when x is big enough.If f is monotonic when x is big enough,then f(f) must be monotonically increasing.But it's just a assumption.My homework always has an elegant solution,so I think it won't take too long.Sorry for my poor English and ugly typing style.
– Oolong milk tea
Nov 22 '18 at 4:45
I'm sorry for being away.I've thought f must be a bijection,so unbounded,moreover f(f) is monotonic when x is big enough.If f is monotonic when x is big enough,then f(f) must be monotonically increasing.But it's just a assumption.My homework always has an elegant solution,so I think it won't take too long.Sorry for my poor English and ugly typing style.
– Oolong milk tea
Nov 22 '18 at 4:45
@Fume There is no way $f$ can be a bijection. If it is one-to-one then, since $f(f(0))=f(f(1))$ we get $0=1$.
– Kavi Rama Murthy
Nov 22 '18 at 5:48
@Fume There is no way $f$ can be a bijection. If it is one-to-one then, since $f(f(0))=f(f(1))$ we get $0=1$.
– Kavi Rama Murthy
Nov 22 '18 at 5:48
|
show 1 more comment
2 Answers
2
active
oldest
votes
The Range of $f$
If the range of $f$ is a proper subset of $mathbb{R}$, then the range of $f circ f$ is a proper subset of a proper subset of $mathbb{R}$, which is itself a proper subset of $mathbb{R}$.
The range of $-x^3 + x$ is $mathbb{R}$, so it follows that the range of $f$ is $mathbb{R}$.
Suppose $f$ is not monotonic.
It has a range of $mathbb{R}$. and is continuous. Therefore there have to be (at least) three disjoint intervals values $A,B,C$ such that $forall ain A|exists bin B,cin C|f(a)=f(b)=f(c)$. Assume we're talking about a "maximal" $A,B,C$, in the sense that there's not a similar triplet of which these three intervals are subsets.
${x|exists ain A | f(a)=x}$ is an interval; call it $O$.
If $O$ isn't disjoint with $(Acup Bcup C)$,
Define $O' = Acap O neqemptyset$. (or $B$ or $C$; this part's symmetric)
Consider $a_0in O'$. Because $a_0in O$, we know there are $a_{-1}^a,b_{-1}^a,c_{-1}^a|f(a_{-1}^a)=f(b_{-1}^a)=f(c_{-1}^a)=a_0$.
Because $a_0in A$, we know there exist $b_0,c_0|f(a_0)=f(b_0)=f(c_0)$. There must be $b_{-1}^b|f(b_{-1}^b)=b_0$ and $c_{-1}^c|f(c_{-1}^c)=c_0$.
Because $b_0neq c_0neq a_0$, and $a_{-1}^aneq b_{-1}^aneq c_{-1}^a$, we now have five distinct numbers $xin{a_{-1}^a,b_{-1}^a,c_{-1}^a,b_{-1}^b,c_{-1}^c}$ such that $f(f(x)) = f(a_0)$
But if $f(f(x))=-x^3+x$, then there could be at most three such points. A contradiction.
If $O$ is disjoint with $(Acup Bcup C)$,
then $notexists xin(Acup Bcup C)|f^{-1}(x)in(Acup Bcup C)$.
Then we have
$forall ain A|exists bin B,cin C|f(f(a))=f(f((b))=ff(((c))$
and we have intervals $A_{-1},B_{-1},C_{-1}$ where
$forall a_{-1}in A_{-1}|exists b_{-1}in B_{-1},c_{-1}in C_{-1}|f(f(a_{-1}))=f(f((b))=f(f((c))in O$
But $(Acup Bcup C)$ is distinct from $(A_{-1}cup B_{-1}cup C_{-1})$, and $-x^3+x$ has only one such triplet of overlapping intervals, another contradiction. (It's important here that $A,B,C$ were "maximal", as mentioned.)
If $f$ is monotonic?
If $f$ is monotonic, then $fcirc f$ is also monotonic, but $-x^3+x$ isn't, so that's not an option either.
We conclude that no such function exists.
Your teacher probably won't give you many points for the above proof; it needs a lot of cleaning up. But I haven't found any holes in it yet.
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
add a comment |
One of my classmate show me the answer below.I think it's probably true.
First it's easy to see while $xto+infty$ ,$f(f(x))to-infty$,while $xto-infty$ ,$f(f(x))to+infty$.
If $limsup_{xto+infty}f(x)=+infty$,then $limsup_{xto+infty}f(f(x))=+infty$,a contradiction.
(Notice $limsup_{xto+infty}f(x)$ always exists.)
The same for $limsup_{xto-infty}f(x)=-infty$
And for $limsup_{xto+infty}f(x)=-infty$,$limsup_{xto+infty}f(f(x))=-infty$ can't be true.
So $A=limsup_{xto+infty}f(x)$ is bounded.
The same for $B=liminf_{xto+infty}f(x)$,then $liminf_{xto+infty}f(x)$ is bounded.
Choose any $epsilon>0$, there is a $M$,while $x>M$,$B-epsilon<f(x)<A+epsilon$.
So $f(f(x))to+infty$ can't be true since $f$ is continuous on $mathbb{R}$,bounded on the interval.
Sorry for my poor English and ugly typing.
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008665%2fis-there-a-continuous-f-satisfying-ffx-x3x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The Range of $f$
If the range of $f$ is a proper subset of $mathbb{R}$, then the range of $f circ f$ is a proper subset of a proper subset of $mathbb{R}$, which is itself a proper subset of $mathbb{R}$.
The range of $-x^3 + x$ is $mathbb{R}$, so it follows that the range of $f$ is $mathbb{R}$.
Suppose $f$ is not monotonic.
It has a range of $mathbb{R}$. and is continuous. Therefore there have to be (at least) three disjoint intervals values $A,B,C$ such that $forall ain A|exists bin B,cin C|f(a)=f(b)=f(c)$. Assume we're talking about a "maximal" $A,B,C$, in the sense that there's not a similar triplet of which these three intervals are subsets.
${x|exists ain A | f(a)=x}$ is an interval; call it $O$.
If $O$ isn't disjoint with $(Acup Bcup C)$,
Define $O' = Acap O neqemptyset$. (or $B$ or $C$; this part's symmetric)
Consider $a_0in O'$. Because $a_0in O$, we know there are $a_{-1}^a,b_{-1}^a,c_{-1}^a|f(a_{-1}^a)=f(b_{-1}^a)=f(c_{-1}^a)=a_0$.
Because $a_0in A$, we know there exist $b_0,c_0|f(a_0)=f(b_0)=f(c_0)$. There must be $b_{-1}^b|f(b_{-1}^b)=b_0$ and $c_{-1}^c|f(c_{-1}^c)=c_0$.
Because $b_0neq c_0neq a_0$, and $a_{-1}^aneq b_{-1}^aneq c_{-1}^a$, we now have five distinct numbers $xin{a_{-1}^a,b_{-1}^a,c_{-1}^a,b_{-1}^b,c_{-1}^c}$ such that $f(f(x)) = f(a_0)$
But if $f(f(x))=-x^3+x$, then there could be at most three such points. A contradiction.
If $O$ is disjoint with $(Acup Bcup C)$,
then $notexists xin(Acup Bcup C)|f^{-1}(x)in(Acup Bcup C)$.
Then we have
$forall ain A|exists bin B,cin C|f(f(a))=f(f((b))=ff(((c))$
and we have intervals $A_{-1},B_{-1},C_{-1}$ where
$forall a_{-1}in A_{-1}|exists b_{-1}in B_{-1},c_{-1}in C_{-1}|f(f(a_{-1}))=f(f((b))=f(f((c))in O$
But $(Acup Bcup C)$ is distinct from $(A_{-1}cup B_{-1}cup C_{-1})$, and $-x^3+x$ has only one such triplet of overlapping intervals, another contradiction. (It's important here that $A,B,C$ were "maximal", as mentioned.)
If $f$ is monotonic?
If $f$ is monotonic, then $fcirc f$ is also monotonic, but $-x^3+x$ isn't, so that's not an option either.
We conclude that no such function exists.
Your teacher probably won't give you many points for the above proof; it needs a lot of cleaning up. But I haven't found any holes in it yet.
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
add a comment |
The Range of $f$
If the range of $f$ is a proper subset of $mathbb{R}$, then the range of $f circ f$ is a proper subset of a proper subset of $mathbb{R}$, which is itself a proper subset of $mathbb{R}$.
The range of $-x^3 + x$ is $mathbb{R}$, so it follows that the range of $f$ is $mathbb{R}$.
Suppose $f$ is not monotonic.
It has a range of $mathbb{R}$. and is continuous. Therefore there have to be (at least) three disjoint intervals values $A,B,C$ such that $forall ain A|exists bin B,cin C|f(a)=f(b)=f(c)$. Assume we're talking about a "maximal" $A,B,C$, in the sense that there's not a similar triplet of which these three intervals are subsets.
${x|exists ain A | f(a)=x}$ is an interval; call it $O$.
If $O$ isn't disjoint with $(Acup Bcup C)$,
Define $O' = Acap O neqemptyset$. (or $B$ or $C$; this part's symmetric)
Consider $a_0in O'$. Because $a_0in O$, we know there are $a_{-1}^a,b_{-1}^a,c_{-1}^a|f(a_{-1}^a)=f(b_{-1}^a)=f(c_{-1}^a)=a_0$.
Because $a_0in A$, we know there exist $b_0,c_0|f(a_0)=f(b_0)=f(c_0)$. There must be $b_{-1}^b|f(b_{-1}^b)=b_0$ and $c_{-1}^c|f(c_{-1}^c)=c_0$.
Because $b_0neq c_0neq a_0$, and $a_{-1}^aneq b_{-1}^aneq c_{-1}^a$, we now have five distinct numbers $xin{a_{-1}^a,b_{-1}^a,c_{-1}^a,b_{-1}^b,c_{-1}^c}$ such that $f(f(x)) = f(a_0)$
But if $f(f(x))=-x^3+x$, then there could be at most three such points. A contradiction.
If $O$ is disjoint with $(Acup Bcup C)$,
then $notexists xin(Acup Bcup C)|f^{-1}(x)in(Acup Bcup C)$.
Then we have
$forall ain A|exists bin B,cin C|f(f(a))=f(f((b))=ff(((c))$
and we have intervals $A_{-1},B_{-1},C_{-1}$ where
$forall a_{-1}in A_{-1}|exists b_{-1}in B_{-1},c_{-1}in C_{-1}|f(f(a_{-1}))=f(f((b))=f(f((c))in O$
But $(Acup Bcup C)$ is distinct from $(A_{-1}cup B_{-1}cup C_{-1})$, and $-x^3+x$ has only one such triplet of overlapping intervals, another contradiction. (It's important here that $A,B,C$ were "maximal", as mentioned.)
If $f$ is monotonic?
If $f$ is monotonic, then $fcirc f$ is also monotonic, but $-x^3+x$ isn't, so that's not an option either.
We conclude that no such function exists.
Your teacher probably won't give you many points for the above proof; it needs a lot of cleaning up. But I haven't found any holes in it yet.
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
add a comment |
The Range of $f$
If the range of $f$ is a proper subset of $mathbb{R}$, then the range of $f circ f$ is a proper subset of a proper subset of $mathbb{R}$, which is itself a proper subset of $mathbb{R}$.
The range of $-x^3 + x$ is $mathbb{R}$, so it follows that the range of $f$ is $mathbb{R}$.
Suppose $f$ is not monotonic.
It has a range of $mathbb{R}$. and is continuous. Therefore there have to be (at least) three disjoint intervals values $A,B,C$ such that $forall ain A|exists bin B,cin C|f(a)=f(b)=f(c)$. Assume we're talking about a "maximal" $A,B,C$, in the sense that there's not a similar triplet of which these three intervals are subsets.
${x|exists ain A | f(a)=x}$ is an interval; call it $O$.
If $O$ isn't disjoint with $(Acup Bcup C)$,
Define $O' = Acap O neqemptyset$. (or $B$ or $C$; this part's symmetric)
Consider $a_0in O'$. Because $a_0in O$, we know there are $a_{-1}^a,b_{-1}^a,c_{-1}^a|f(a_{-1}^a)=f(b_{-1}^a)=f(c_{-1}^a)=a_0$.
Because $a_0in A$, we know there exist $b_0,c_0|f(a_0)=f(b_0)=f(c_0)$. There must be $b_{-1}^b|f(b_{-1}^b)=b_0$ and $c_{-1}^c|f(c_{-1}^c)=c_0$.
Because $b_0neq c_0neq a_0$, and $a_{-1}^aneq b_{-1}^aneq c_{-1}^a$, we now have five distinct numbers $xin{a_{-1}^a,b_{-1}^a,c_{-1}^a,b_{-1}^b,c_{-1}^c}$ such that $f(f(x)) = f(a_0)$
But if $f(f(x))=-x^3+x$, then there could be at most three such points. A contradiction.
If $O$ is disjoint with $(Acup Bcup C)$,
then $notexists xin(Acup Bcup C)|f^{-1}(x)in(Acup Bcup C)$.
Then we have
$forall ain A|exists bin B,cin C|f(f(a))=f(f((b))=ff(((c))$
and we have intervals $A_{-1},B_{-1},C_{-1}$ where
$forall a_{-1}in A_{-1}|exists b_{-1}in B_{-1},c_{-1}in C_{-1}|f(f(a_{-1}))=f(f((b))=f(f((c))in O$
But $(Acup Bcup C)$ is distinct from $(A_{-1}cup B_{-1}cup C_{-1})$, and $-x^3+x$ has only one such triplet of overlapping intervals, another contradiction. (It's important here that $A,B,C$ were "maximal", as mentioned.)
If $f$ is monotonic?
If $f$ is monotonic, then $fcirc f$ is also monotonic, but $-x^3+x$ isn't, so that's not an option either.
We conclude that no such function exists.
Your teacher probably won't give you many points for the above proof; it needs a lot of cleaning up. But I haven't found any holes in it yet.
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
The Range of $f$
If the range of $f$ is a proper subset of $mathbb{R}$, then the range of $f circ f$ is a proper subset of a proper subset of $mathbb{R}$, which is itself a proper subset of $mathbb{R}$.
The range of $-x^3 + x$ is $mathbb{R}$, so it follows that the range of $f$ is $mathbb{R}$.
Suppose $f$ is not monotonic.
It has a range of $mathbb{R}$. and is continuous. Therefore there have to be (at least) three disjoint intervals values $A,B,C$ such that $forall ain A|exists bin B,cin C|f(a)=f(b)=f(c)$. Assume we're talking about a "maximal" $A,B,C$, in the sense that there's not a similar triplet of which these three intervals are subsets.
${x|exists ain A | f(a)=x}$ is an interval; call it $O$.
If $O$ isn't disjoint with $(Acup Bcup C)$,
Define $O' = Acap O neqemptyset$. (or $B$ or $C$; this part's symmetric)
Consider $a_0in O'$. Because $a_0in O$, we know there are $a_{-1}^a,b_{-1}^a,c_{-1}^a|f(a_{-1}^a)=f(b_{-1}^a)=f(c_{-1}^a)=a_0$.
Because $a_0in A$, we know there exist $b_0,c_0|f(a_0)=f(b_0)=f(c_0)$. There must be $b_{-1}^b|f(b_{-1}^b)=b_0$ and $c_{-1}^c|f(c_{-1}^c)=c_0$.
Because $b_0neq c_0neq a_0$, and $a_{-1}^aneq b_{-1}^aneq c_{-1}^a$, we now have five distinct numbers $xin{a_{-1}^a,b_{-1}^a,c_{-1}^a,b_{-1}^b,c_{-1}^c}$ such that $f(f(x)) = f(a_0)$
But if $f(f(x))=-x^3+x$, then there could be at most three such points. A contradiction.
If $O$ is disjoint with $(Acup Bcup C)$,
then $notexists xin(Acup Bcup C)|f^{-1}(x)in(Acup Bcup C)$.
Then we have
$forall ain A|exists bin B,cin C|f(f(a))=f(f((b))=ff(((c))$
and we have intervals $A_{-1},B_{-1},C_{-1}$ where
$forall a_{-1}in A_{-1}|exists b_{-1}in B_{-1},c_{-1}in C_{-1}|f(f(a_{-1}))=f(f((b))=f(f((c))in O$
But $(Acup Bcup C)$ is distinct from $(A_{-1}cup B_{-1}cup C_{-1})$, and $-x^3+x$ has only one such triplet of overlapping intervals, another contradiction. (It's important here that $A,B,C$ were "maximal", as mentioned.)
If $f$ is monotonic?
If $f$ is monotonic, then $fcirc f$ is also monotonic, but $-x^3+x$ isn't, so that's not an option either.
We conclude that no such function exists.
Your teacher probably won't give you many points for the above proof; it needs a lot of cleaning up. But I haven't found any holes in it yet.
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
answered Nov 22 '18 at 5:53
ShapeOfMatterShapeOfMatter
1513
1513
add a comment |
add a comment |
One of my classmate show me the answer below.I think it's probably true.
First it's easy to see while $xto+infty$ ,$f(f(x))to-infty$,while $xto-infty$ ,$f(f(x))to+infty$.
If $limsup_{xto+infty}f(x)=+infty$,then $limsup_{xto+infty}f(f(x))=+infty$,a contradiction.
(Notice $limsup_{xto+infty}f(x)$ always exists.)
The same for $limsup_{xto-infty}f(x)=-infty$
And for $limsup_{xto+infty}f(x)=-infty$,$limsup_{xto+infty}f(f(x))=-infty$ can't be true.
So $A=limsup_{xto+infty}f(x)$ is bounded.
The same for $B=liminf_{xto+infty}f(x)$,then $liminf_{xto+infty}f(x)$ is bounded.
Choose any $epsilon>0$, there is a $M$,while $x>M$,$B-epsilon<f(x)<A+epsilon$.
So $f(f(x))to+infty$ can't be true since $f$ is continuous on $mathbb{R}$,bounded on the interval.
Sorry for my poor English and ugly typing.
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
add a comment |
One of my classmate show me the answer below.I think it's probably true.
First it's easy to see while $xto+infty$ ,$f(f(x))to-infty$,while $xto-infty$ ,$f(f(x))to+infty$.
If $limsup_{xto+infty}f(x)=+infty$,then $limsup_{xto+infty}f(f(x))=+infty$,a contradiction.
(Notice $limsup_{xto+infty}f(x)$ always exists.)
The same for $limsup_{xto-infty}f(x)=-infty$
And for $limsup_{xto+infty}f(x)=-infty$,$limsup_{xto+infty}f(f(x))=-infty$ can't be true.
So $A=limsup_{xto+infty}f(x)$ is bounded.
The same for $B=liminf_{xto+infty}f(x)$,then $liminf_{xto+infty}f(x)$ is bounded.
Choose any $epsilon>0$, there is a $M$,while $x>M$,$B-epsilon<f(x)<A+epsilon$.
So $f(f(x))to+infty$ can't be true since $f$ is continuous on $mathbb{R}$,bounded on the interval.
Sorry for my poor English and ugly typing.
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
add a comment |
One of my classmate show me the answer below.I think it's probably true.
First it's easy to see while $xto+infty$ ,$f(f(x))to-infty$,while $xto-infty$ ,$f(f(x))to+infty$.
If $limsup_{xto+infty}f(x)=+infty$,then $limsup_{xto+infty}f(f(x))=+infty$,a contradiction.
(Notice $limsup_{xto+infty}f(x)$ always exists.)
The same for $limsup_{xto-infty}f(x)=-infty$
And for $limsup_{xto+infty}f(x)=-infty$,$limsup_{xto+infty}f(f(x))=-infty$ can't be true.
So $A=limsup_{xto+infty}f(x)$ is bounded.
The same for $B=liminf_{xto+infty}f(x)$,then $liminf_{xto+infty}f(x)$ is bounded.
Choose any $epsilon>0$, there is a $M$,while $x>M$,$B-epsilon<f(x)<A+epsilon$.
So $f(f(x))to+infty$ can't be true since $f$ is continuous on $mathbb{R}$,bounded on the interval.
Sorry for my poor English and ugly typing.
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
One of my classmate show me the answer below.I think it's probably true.
First it's easy to see while $xto+infty$ ,$f(f(x))to-infty$,while $xto-infty$ ,$f(f(x))to+infty$.
If $limsup_{xto+infty}f(x)=+infty$,then $limsup_{xto+infty}f(f(x))=+infty$,a contradiction.
(Notice $limsup_{xto+infty}f(x)$ always exists.)
The same for $limsup_{xto-infty}f(x)=-infty$
And for $limsup_{xto+infty}f(x)=-infty$,$limsup_{xto+infty}f(f(x))=-infty$ can't be true.
So $A=limsup_{xto+infty}f(x)$ is bounded.
The same for $B=liminf_{xto+infty}f(x)$,then $liminf_{xto+infty}f(x)$ is bounded.
Choose any $epsilon>0$, there is a $M$,while $x>M$,$B-epsilon<f(x)<A+epsilon$.
So $f(f(x))to+infty$ can't be true since $f$ is continuous on $mathbb{R}$,bounded on the interval.
Sorry for my poor English and ugly typing.
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
edited Nov 27 '18 at 8:08
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
answered Nov 27 '18 at 5:27
Oolong milk teaOolong milk tea
787
787
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008665%2fis-there-a-continuous-f-satisfying-ffx-x3x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
You have spent several hours on it, so please include in your post all results you derived during these several hours, so that others attempting the question do not take several hours to derive things you already know from your work.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:41
If we restrict to an interval, $frac{-1}{sqrt 3} < x <frac{1}{sqrt 3}, $ we can construct a $C^infty$ function $f$ that does this. My guess is that we cannot extend the function to the whole line, critical points and decreasing functions cause trouble
– Will Jagy
Nov 22 '18 at 2:47
$f(-1)$, $f(0)$ and $f(1)$ will all be zeroes of $f(x)$
– Seth
Nov 22 '18 at 3:35
I'm sorry for being away.I've thought f must be a bijection,so unbounded,moreover f(f) is monotonic when x is big enough.If f is monotonic when x is big enough,then f(f) must be monotonically increasing.But it's just a assumption.My homework always has an elegant solution,so I think it won't take too long.Sorry for my poor English and ugly typing style.
– Oolong milk tea
Nov 22 '18 at 4:45
@Fume There is no way $f$ can be a bijection. If it is one-to-one then, since $f(f(0))=f(f(1))$ we get $0=1$.
– Kavi Rama Murthy
Nov 22 '18 at 5:48