Concave function (a estimate)
0
$begingroup$
Let $f$ be cancave and non negative, that is $f''(x)<0$ on $[0,1]$. For $0<mu<1/2$, show that $f(x)geq mu f(y)$, for any $xin [mu,1-mu]$, $yin [0,1]$.
Clearly, the minimum of $f$ over $[mu,1-mu]$ is $m=min {f(mu),f(1-mu)}$. Then how to control the maximum of $f$ over $[0,1]$ by $m/mu$?
calculus
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
$endgroup$
add a comment |
0
$begingroup$
Let $f$ be cancave and non negative, that is $f''(x)<0$ on $[0,1]$. For $0<mu<1/2$, show that $f(x)geq mu f(y)$, for any $xin [mu,1-mu]$, $yin [0,1]$.
Clearly, the minimum of $f$ over $[mu,1-mu]$ is $m=min {f(mu),f(1-mu)}$. Then how to control the maximum of $f$ over $[0,1]$ by $m/mu$?
calculus
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
$endgroup$
1
$begingroup$
As written, this makes no sense; you have a mixture of $a$ and $y$?
$endgroup$
– T. Bongers
Dec 7 '18 at 0:22
$begingroup$
@T.Bongers I have corrected it. Thanks.
$endgroup$
– xldd
Dec 7 '18 at 0:29
$begingroup$
This would imply that $f(1/2) ge frac 1 2 f(y)$ for all $y in [0, 1]$, or alternatively that $f(y) le 2 f(1/2)$. This might be an interesting starting point for your analysis, and it also shows that you're missing an assumption of positivity.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:31
$begingroup$
@T.Bongers How to prove then?
$endgroup$
– xldd
Dec 7 '18 at 3:58
add a comment |
0
0
0
$begingroup$
Let $f$ be cancave and non negative, that is $f''(x)<0$ on $[0,1]$. For $0<mu<1/2$, show that $f(x)geq mu f(y)$, for any $xin [mu,1-mu]$, $yin [0,1]$.
Clearly, the minimum of $f$ over $[mu,1-mu]$ is $m=min {f(mu),f(1-mu)}$. Then how to control the maximum of $f$ over $[0,1]$ by $m/mu$?
calculus
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
$endgroup$
Let $f$ be cancave and non negative, that is $f''(x)<0$ on $[0,1]$. For $0<mu<1/2$, show that $f(x)geq mu f(y)$, for any $xin [mu,1-mu]$, $yin [0,1]$.
Clearly, the minimum of $f$ over $[mu,1-mu]$ is $m=min {f(mu),f(1-mu)}$. Then how to control the maximum of $f$ over $[0,1]$ by $m/mu$?
calculus
calculus
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
edited Dec 7 '18 at 0:37
xldd
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
asked Dec 7 '18 at 0:18
xlddxldd
1,315510
1,315510
1
$begingroup$
As written, this makes no sense; you have a mixture of $a$ and $y$?
$endgroup$
– T. Bongers
Dec 7 '18 at 0:22
$begingroup$
@T.Bongers I have corrected it. Thanks.
$endgroup$
– xldd
Dec 7 '18 at 0:29
$begingroup$
This would imply that $f(1/2) ge frac 1 2 f(y)$ for all $y in [0, 1]$, or alternatively that $f(y) le 2 f(1/2)$. This might be an interesting starting point for your analysis, and it also shows that you're missing an assumption of positivity.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:31
$begingroup$
@T.Bongers How to prove then?
$endgroup$
– xldd
Dec 7 '18 at 3:58
add a comment |
1
$begingroup$
As written, this makes no sense; you have a mixture of $a$ and $y$?
$endgroup$
– T. Bongers
Dec 7 '18 at 0:22
$begingroup$
@T.Bongers I have corrected it. Thanks.
$endgroup$
– xldd
Dec 7 '18 at 0:29
$begingroup$
This would imply that $f(1/2) ge frac 1 2 f(y)$ for all $y in [0, 1]$, or alternatively that $f(y) le 2 f(1/2)$. This might be an interesting starting point for your analysis, and it also shows that you're missing an assumption of positivity.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:31
$begingroup$
@T.Bongers How to prove then?
$endgroup$
– xldd
Dec 7 '18 at 3:58
1
1
$begingroup$
As written, this makes no sense; you have a mixture of $a$ and $y$?
$endgroup$
– T. Bongers
Dec 7 '18 at 0:22
$begingroup$
As written, this makes no sense; you have a mixture of $a$ and $y$?
$endgroup$
– T. Bongers
Dec 7 '18 at 0:22
$begingroup$
@T.Bongers I have corrected it. Thanks.
$endgroup$
– xldd
Dec 7 '18 at 0:29
$begingroup$
@T.Bongers I have corrected it. Thanks.
$endgroup$
– xldd
Dec 7 '18 at 0:29
$begingroup$
This would imply that $f(1/2) ge frac 1 2 f(y)$ for all $y in [0, 1]$, or alternatively that $f(y) le 2 f(1/2)$. This might be an interesting starting point for your analysis, and it also shows that you're missing an assumption of positivity.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:31
$begingroup$
This would imply that $f(1/2) ge frac 1 2 f(y)$ for all $y in [0, 1]$, or alternatively that $f(y) le 2 f(1/2)$. This might be an interesting starting point for your analysis, and it also shows that you're missing an assumption of positivity.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:31
$begingroup$
@T.Bongers How to prove then?
$endgroup$
– xldd
Dec 7 '18 at 3:58
$begingroup$
@T.Bongers How to prove then?
$endgroup$
– xldd
Dec 7 '18 at 3:58
add a comment |
0
active
oldest
votes
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
});
}
});
draft saved
draft discarded
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%2f3029265%2fconcave-function-a-estimate%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3029265%2fconcave-function-a-estimate%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
1
$begingroup$
As written, this makes no sense; you have a mixture of $a$ and $y$?
$endgroup$
– T. Bongers
Dec 7 '18 at 0:22
$begingroup$
@T.Bongers I have corrected it. Thanks.
$endgroup$
– xldd
Dec 7 '18 at 0:29
$begingroup$
This would imply that $f(1/2) ge frac 1 2 f(y)$ for all $y in [0, 1]$, or alternatively that $f(y) le 2 f(1/2)$. This might be an interesting starting point for your analysis, and it also shows that you're missing an assumption of positivity.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:31
$begingroup$
@T.Bongers How to prove then?
$endgroup$
– xldd
Dec 7 '18 at 3:58