Mixing
How to prove this result about convexity?
$begingroup$
I am reading this result which I am not able to prove:
$$ frac{f'(x)}{f'(f^{-1}(f(x)+a))}-1 $$ is negative for all x and, and for all $ageq1$ if and only if $f$ is convex. $f'$ is the derivative of $f$ wrt $x$ and $f^{-1}$ is the inverse function. I am actually blanking and do not have an idea about how to prove this result. Can anyone help?
convex-analysis
asked Jan 30 at 18:36
ApiApi
12
$endgroup$
add a comment |
$begingroup$
I am reading this result which I am not able to prove:
$$ frac{f'(x)}{f'(f^{-1}(f(x)+a))}-1 $$ is negative for all x and, and for all $ageq1$ if and only if $f$ is convex. $f'$ is the derivative of $f$ wrt $x$ and $f^{-1}$ is the inverse function. I am actually blanking and do not have an idea about how to prove this result. Can anyone help?
convex-analysis
asked Jan 30 at 18:36
ApiApi
12
$endgroup$
$begingroup$
Convex functions are not necessarily invertible, e.g. for $f(x) = x^2$ we do not have an inverse function.
$endgroup$
– gerw
Jan 31 at 14:34
add a comment |
$begingroup$
I am reading this result which I am not able to prove:
$$ frac{f'(x)}{f'(f^{-1}(f(x)+a))}-1 $$ is negative for all x and, and for all $ageq1$ if and only if $f$ is convex. $f'$ is the derivative of $f$ wrt $x$ and $f^{-1}$ is the inverse function. I am actually blanking and do not have an idea about how to prove this result. Can anyone help?
convex-analysis
asked Jan 30 at 18:36
ApiApi
12
$endgroup$
I am reading this result which I am not able to prove:
$$ frac{f'(x)}{f'(f^{-1}(f(x)+a))}-1 $$ is negative for all x and, and for all $ageq1$ if and only if $f$ is convex. $f'$ is the derivative of $f$ wrt $x$ and $f^{-1}$ is the inverse function. I am actually blanking and do not have an idea about how to prove this result. Can anyone help?
convex-analysis
convex-analysis
asked Jan 30 at 18:36
ApiApi
12
asked Jan 30 at 18:36
ApiApi
12
edited Jan 30 at 18:42
Api
asked Jan 30 at 18:36
ApiApi
12
asked Jan 30 at 18:36
ApiApi
12
asked Jan 30 at 18:36
ApiApi
12
12
$begingroup$
Convex functions are not necessarily invertible, e.g. for $f(x) = x^2$ we do not have an inverse function.
$endgroup$
– gerw
Jan 31 at 14:34
add a comment |
$begingroup$
Convex functions are not necessarily invertible, e.g. for $f(x) = x^2$ we do not have an inverse function.
$endgroup$
– gerw
Jan 31 at 14:34
$begingroup$
Convex functions are not necessarily invertible, e.g. for $f(x) = x^2$ we do not have an inverse function.
$endgroup$
– gerw
Jan 31 at 14:34
$begingroup$
Convex functions are not necessarily invertible, e.g. for $f(x) = x^2$ we do not have an inverse function.
$endgroup$
– gerw
Jan 31 at 14:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try drawing a picture, say of $f(x) = x^2$. To make it invertible, you'll need to restrict your attention to either the right half ($x ge 0$) or the left half $(x le 0)$. The top and bottom of that fraction can be seen as slopes of two different tangent lines. Which is steeper?
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help!
$endgroup$
– Api
Jan 30 at 19:38
$begingroup$
the 'only if' and the technicalities around $a geq 1$ are still nice challenges
$endgroup$
– LinAlg
Jan 30 at 19:40
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%2f3093932%2fhow-to-prove-this-result-about-convexity%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
$begingroup$
Try drawing a picture, say of $f(x) = x^2$. To make it invertible, you'll need to restrict your attention to either the right half ($x ge 0$) or the left half $(x le 0)$. The top and bottom of that fraction can be seen as slopes of two different tangent lines. Which is steeper?
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help!
$endgroup$
– Api
Jan 30 at 19:38
$begingroup$
the 'only if' and the technicalities around $a geq 1$ are still nice challenges
$endgroup$
– LinAlg
Jan 30 at 19:40
add a comment |
$begingroup$
Try drawing a picture, say of $f(x) = x^2$. To make it invertible, you'll need to restrict your attention to either the right half ($x ge 0$) or the left half $(x le 0)$. The top and bottom of that fraction can be seen as slopes of two different tangent lines. Which is steeper?
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help!
$endgroup$
– Api
Jan 30 at 19:38
$begingroup$
the 'only if' and the technicalities around $a geq 1$ are still nice challenges
$endgroup$
– LinAlg
Jan 30 at 19:40
add a comment |
$begingroup$
Try drawing a picture, say of $f(x) = x^2$. To make it invertible, you'll need to restrict your attention to either the right half ($x ge 0$) or the left half $(x le 0)$. The top and bottom of that fraction can be seen as slopes of two different tangent lines. Which is steeper?
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
Try drawing a picture, say of $f(x) = x^2$. To make it invertible, you'll need to restrict your attention to either the right half ($x ge 0$) or the left half $(x le 0)$. The top and bottom of that fraction can be seen as slopes of two different tangent lines. Which is steeper?
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 18:52
Nathaniel MayerNathaniel Mayer
1,863516
1,863516
$begingroup$
I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help!
$endgroup$
– Api
Jan 30 at 19:38
$begingroup$
the 'only if' and the technicalities around $a geq 1$ are still nice challenges
$endgroup$
– LinAlg
Jan 30 at 19:40
add a comment |
$begingroup$
I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help!
$endgroup$
– Api
Jan 30 at 19:38
$begingroup$
the 'only if' and the technicalities around $a geq 1$ are still nice challenges
$endgroup$
– LinAlg
Jan 30 at 19:40
$begingroup$
I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help!
$endgroup$
– Api
Jan 30 at 19:38
$begingroup$
I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help!
$endgroup$
– Api
Jan 30 at 19:38
$begingroup$
the 'only if' and the technicalities around $a geq 1$ are still nice challenges
$endgroup$
– LinAlg
Jan 30 at 19:40
$begingroup$
the 'only if' and the technicalities around $a geq 1$ are still nice challenges
$endgroup$
– LinAlg
Jan 30 at 19:40
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%2f3093932%2fhow-to-prove-this-result-about-convexity%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
$begingroup$
Convex functions are not necessarily invertible, e.g. for $f(x) = x^2$ we do not have an inverse function.
$endgroup$
– gerw
Jan 31 at 14:34