How to prove the non-existence of continuous solution for a functional equation?
$begingroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
$endgroup$
add a comment |
$begingroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
$endgroup$
add a comment |
$begingroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
$endgroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
edited Dec 1 '18 at 16:37
Yuhang
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
835118
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
$endgroup$
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%2f3021515%2fhow-to-prove-the-non-existence-of-continuous-solution-for-a-functional-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
$endgroup$
add a comment |
$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
$endgroup$
add a comment |
$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
$endgroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
answered Dec 1 '18 at 17:16
jmerryjmerry
4,454514
4,454514
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.
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%2f3021515%2fhow-to-prove-the-non-existence-of-continuous-solution-for-a-functional-equation%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
