$fcirc fcirc f(x)=x^9$ then $f$ is increasing
up vote
5
down vote
favorite
The full statement is:
If $f:Bbb R to Bbb R$ is a continuous function and $fcirc fcirc f(x)=x^9$ then $f$ is increasing.
$(1)$ I was thinking about suppose that $f$ is decreasing (or constant) and then it is easy to get a contradiction.
After that I would like to state that
"If $f:Bbb R to Bbb R$ is continous and not increasing then there is an interval where $f$ is decreasing or constant."
and then I can use $(1)$ and get the result. But I still can't prove if the statement is true.
Any hint or any other solution?
real-analysis functions continuity
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
add a comment |
up vote
5
down vote
favorite
The full statement is:
If $f:Bbb R to Bbb R$ is a continuous function and $fcirc fcirc f(x)=x^9$ then $f$ is increasing.
$(1)$ I was thinking about suppose that $f$ is decreasing (or constant) and then it is easy to get a contradiction.
After that I would like to state that
"If $f:Bbb R to Bbb R$ is continous and not increasing then there is an interval where $f$ is decreasing or constant."
and then I can use $(1)$ and get the result. But I still can't prove if the statement is true.
Any hint or any other solution?
real-analysis functions continuity
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
4
$f$ is injective. A continuous injective function is either increasing or decreasing (I am not sure to have understood if this was what you were asking for though)
– Charles Madeline
Jul 20 '17 at 13:29
1
@charMD: That helps a lot! Thanks!
– Arnaldo
Jul 20 '17 at 13:34
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
The full statement is:
If $f:Bbb R to Bbb R$ is a continuous function and $fcirc fcirc f(x)=x^9$ then $f$ is increasing.
$(1)$ I was thinking about suppose that $f$ is decreasing (or constant) and then it is easy to get a contradiction.
After that I would like to state that
"If $f:Bbb R to Bbb R$ is continous and not increasing then there is an interval where $f$ is decreasing or constant."
and then I can use $(1)$ and get the result. But I still can't prove if the statement is true.
Any hint or any other solution?
real-analysis functions continuity
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
The full statement is:
If $f:Bbb R to Bbb R$ is a continuous function and $fcirc fcirc f(x)=x^9$ then $f$ is increasing.
$(1)$ I was thinking about suppose that $f$ is decreasing (or constant) and then it is easy to get a contradiction.
After that I would like to state that
"If $f:Bbb R to Bbb R$ is continous and not increasing then there is an interval where $f$ is decreasing or constant."
and then I can use $(1)$ and get the result. But I still can't prove if the statement is true.
Any hint or any other solution?
real-analysis functions continuity
real-analysis functions continuity
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
asked Jul 20 '17 at 13:18
Arnaldo
18.1k42246
18.1k42246
4
$f$ is injective. A continuous injective function is either increasing or decreasing (I am not sure to have understood if this was what you were asking for though)
– Charles Madeline
Jul 20 '17 at 13:29
1
@charMD: That helps a lot! Thanks!
– Arnaldo
Jul 20 '17 at 13:34
add a comment |
4
$f$ is injective. A continuous injective function is either increasing or decreasing (I am not sure to have understood if this was what you were asking for though)
– Charles Madeline
Jul 20 '17 at 13:29
1
@charMD: That helps a lot! Thanks!
– Arnaldo
Jul 20 '17 at 13:34
4
4
$f$ is injective. A continuous injective function is either increasing or decreasing (I am not sure to have understood if this was what you were asking for though)
– Charles Madeline
Jul 20 '17 at 13:29
$f$ is injective. A continuous injective function is either increasing or decreasing (I am not sure to have understood if this was what you were asking for though)
– Charles Madeline
Jul 20 '17 at 13:29
1
1
@charMD: That helps a lot! Thanks!
– Arnaldo
Jul 20 '17 at 13:34
@charMD: That helps a lot! Thanks!
– Arnaldo
Jul 20 '17 at 13:34
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
I think that your statement is true.
Let $a,b$ be points, $b>a$ so that $f(b) leq f(a)$. Let $c$ be the $x$ value where $f$ attains its maximum on the interval $[a,b]$. Clearly $c neq b$ (or at least we can take $c neq b$), and hence for some $epsilon> 0$, $f$ will be decreasing or constant on the interval: $(c, c + epsilon)$.
answered Jul 20 '17 at 13:39
Mohit
960513
Is this true? The Weierstrass function is not increasing or decreasing on any interval.
– Solomonoff's Secret
Jul 20 '17 at 13:56
add a comment |
up vote
0
down vote
Since $x mapsto x^9$ is one-to-one and onto, the function $f$ itself must also be one-to-one and onto. Together with continuity this means that $f$ is strictly monotone, so it must be strictly increasing.
As remarked in comments to some other answers, general continuous functions may fail to be monotone on any interval.
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
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',
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%2f2365006%2ff-circ-f-circ-fx-x9-then-f-is-increasing%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
up vote
1
down vote
I think that your statement is true.
Let $a,b$ be points, $b>a$ so that $f(b) leq f(a)$. Let $c$ be the $x$ value where $f$ attains its maximum on the interval $[a,b]$. Clearly $c neq b$ (or at least we can take $c neq b$), and hence for some $epsilon> 0$, $f$ will be decreasing or constant on the interval: $(c, c + epsilon)$.
answered Jul 20 '17 at 13:39
Mohit
960513
Is this true? The Weierstrass function is not increasing or decreasing on any interval.
– Solomonoff's Secret
Jul 20 '17 at 13:56
add a comment |
up vote
1
down vote
I think that your statement is true.
Let $a,b$ be points, $b>a$ so that $f(b) leq f(a)$. Let $c$ be the $x$ value where $f$ attains its maximum on the interval $[a,b]$. Clearly $c neq b$ (or at least we can take $c neq b$), and hence for some $epsilon> 0$, $f$ will be decreasing or constant on the interval: $(c, c + epsilon)$.
answered Jul 20 '17 at 13:39
Mohit
960513
Is this true? The Weierstrass function is not increasing or decreasing on any interval.
– Solomonoff's Secret
Jul 20 '17 at 13:56
add a comment |
up vote
1
down vote
up vote
1
down vote
I think that your statement is true.
Let $a,b$ be points, $b>a$ so that $f(b) leq f(a)$. Let $c$ be the $x$ value where $f$ attains its maximum on the interval $[a,b]$. Clearly $c neq b$ (or at least we can take $c neq b$), and hence for some $epsilon> 0$, $f$ will be decreasing or constant on the interval: $(c, c + epsilon)$.
answered Jul 20 '17 at 13:39
Mohit
960513
I think that your statement is true.
Let $a,b$ be points, $b>a$ so that $f(b) leq f(a)$. Let $c$ be the $x$ value where $f$ attains its maximum on the interval $[a,b]$. Clearly $c neq b$ (or at least we can take $c neq b$), and hence for some $epsilon> 0$, $f$ will be decreasing or constant on the interval: $(c, c + epsilon)$.
answered Jul 20 '17 at 13:39
Mohit
960513
answered Jul 20 '17 at 13:39
Mohit
960513
answered Jul 20 '17 at 13:39
Mohit
960513
answered Jul 20 '17 at 13:39
Mohit
960513
960513
Is this true? The Weierstrass function is not increasing or decreasing on any interval.
– Solomonoff's Secret
Jul 20 '17 at 13:56
add a comment |
Is this true? The Weierstrass function is not increasing or decreasing on any interval.
– Solomonoff's Secret
Jul 20 '17 at 13:56
Is this true? The Weierstrass function is not increasing or decreasing on any interval.
– Solomonoff's Secret
Jul 20 '17 at 13:56
Is this true? The Weierstrass function is not increasing or decreasing on any interval.
– Solomonoff's Secret
Jul 20 '17 at 13:56
add a comment |
up vote
0
down vote
Since $x mapsto x^9$ is one-to-one and onto, the function $f$ itself must also be one-to-one and onto. Together with continuity this means that $f$ is strictly monotone, so it must be strictly increasing.
As remarked in comments to some other answers, general continuous functions may fail to be monotone on any interval.
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
add a comment |
up vote
0
down vote
Since $x mapsto x^9$ is one-to-one and onto, the function $f$ itself must also be one-to-one and onto. Together with continuity this means that $f$ is strictly monotone, so it must be strictly increasing.
As remarked in comments to some other answers, general continuous functions may fail to be monotone on any interval.
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
add a comment |
up vote
0
down vote
up vote
0
down vote
Since $x mapsto x^9$ is one-to-one and onto, the function $f$ itself must also be one-to-one and onto. Together with continuity this means that $f$ is strictly monotone, so it must be strictly increasing.
As remarked in comments to some other answers, general continuous functions may fail to be monotone on any interval.
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
Since $x mapsto x^9$ is one-to-one and onto, the function $f$ itself must also be one-to-one and onto. Together with continuity this means that $f$ is strictly monotone, so it must be strictly increasing.
As remarked in comments to some other answers, general continuous functions may fail to be monotone on any interval.
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
answered Nov 20 at 21:01
Lukas Geyer
13.2k1454
13.2k1454
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%2f2365006%2ff-circ-f-circ-fx-x9-then-f-is-increasing%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
4
$f$ is injective. A continuous injective function is either increasing or decreasing (I am not sure to have understood if this was what you were asking for though)
– Charles Madeline
Jul 20 '17 at 13:29
1
@charMD: That helps a lot! Thanks!
– Arnaldo
Jul 20 '17 at 13:34