Bounds on the second derivative of a convex function.
$begingroup$
Edit: This question was answered in https://mathoverflow.net/questions/265256/how-bad-can-the-second-derivative-of-a-convex-function-be
Let $a<b$ and suppose that $f:(a,b)to mathbb{R}$ is a increasing convex function. From the Alexandrov’s theorem, $f$ is twice differentiable almost everywhere. My question is the following
Does $f''in L_{loc}^1(a,b)$?
If the answer to the previous question is negative, consider the set
$$
A={xin (a,b): mbox{there is no} delta>0, mbox{such that} fin L^1(x-delta,x+delta)}.
$$
What is the measure of $A$?
No ideas up to now. Any reference is appreciated.
For an accessible reference, take a look here.
real-analysis measure-theory
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
$endgroup$
add a comment |
$begingroup$
Edit: This question was answered in https://mathoverflow.net/questions/265256/how-bad-can-the-second-derivative-of-a-convex-function-be
Let $a<b$ and suppose that $f:(a,b)to mathbb{R}$ is a increasing convex function. From the Alexandrov’s theorem, $f$ is twice differentiable almost everywhere. My question is the following
Does $f''in L_{loc}^1(a,b)$?
If the answer to the previous question is negative, consider the set
$$
A={xin (a,b): mbox{there is no} delta>0, mbox{such that} fin L^1(x-delta,x+delta)}.
$$
What is the measure of $A$?
No ideas up to now. Any reference is appreciated.
For an accessible reference, take a look here.
real-analysis measure-theory
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
$endgroup$
$begingroup$
I know that this is an old question, but I would like to drop a comment. I guess that we have something like $f'(d) - f'(c) ge int_c^d f''(x) , mathrm{d}x ge 0$ for $[c,d] subset (a,b)$. This should give your assertion.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
Or speaking differently: The second derivative of a convex function can be identified with a positive distribution, i.e., a measure. Then, your $f''$ should be the absolutely continuous part of this measure. Again, it belongs to $L^1_{text{loc}}(a,b)$.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
@gerw take a look in mathoverflow.net/questions/265256/…
$endgroup$
– Tomás
Dec 3 '18 at 20:01
$begingroup$
I didn't know that this question was crossposted. You should link both questions.
$endgroup$
– gerw
Dec 3 '18 at 20:21
add a comment |
$begingroup$
Edit: This question was answered in https://mathoverflow.net/questions/265256/how-bad-can-the-second-derivative-of-a-convex-function-be
Let $a<b$ and suppose that $f:(a,b)to mathbb{R}$ is a increasing convex function. From the Alexandrov’s theorem, $f$ is twice differentiable almost everywhere. My question is the following
Does $f''in L_{loc}^1(a,b)$?
If the answer to the previous question is negative, consider the set
$$
A={xin (a,b): mbox{there is no} delta>0, mbox{such that} fin L^1(x-delta,x+delta)}.
$$
What is the measure of $A$?
No ideas up to now. Any reference is appreciated.
For an accessible reference, take a look here.
real-analysis measure-theory
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
$endgroup$
Edit: This question was answered in https://mathoverflow.net/questions/265256/how-bad-can-the-second-derivative-of-a-convex-function-be
Let $a<b$ and suppose that $f:(a,b)to mathbb{R}$ is a increasing convex function. From the Alexandrov’s theorem, $f$ is twice differentiable almost everywhere. My question is the following
Does $f''in L_{loc}^1(a,b)$?
If the answer to the previous question is negative, consider the set
$$
A={xin (a,b): mbox{there is no} delta>0, mbox{such that} fin L^1(x-delta,x+delta)}.
$$
What is the measure of $A$?
No ideas up to now. Any reference is appreciated.
For an accessible reference, take a look here.
real-analysis measure-theory
real-analysis measure-theory
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
edited Dec 4 '18 at 15:12
Tomás
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
asked Mar 20 '17 at 18:03
TomásTomás
15.8k32177
15.8k32177
$begingroup$
I know that this is an old question, but I would like to drop a comment. I guess that we have something like $f'(d) - f'(c) ge int_c^d f''(x) , mathrm{d}x ge 0$ for $[c,d] subset (a,b)$. This should give your assertion.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
Or speaking differently: The second derivative of a convex function can be identified with a positive distribution, i.e., a measure. Then, your $f''$ should be the absolutely continuous part of this measure. Again, it belongs to $L^1_{text{loc}}(a,b)$.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
@gerw take a look in mathoverflow.net/questions/265256/…
$endgroup$
– Tomás
Dec 3 '18 at 20:01
$begingroup$
I didn't know that this question was crossposted. You should link both questions.
$endgroup$
– gerw
Dec 3 '18 at 20:21
add a comment |
$begingroup$
I know that this is an old question, but I would like to drop a comment. I guess that we have something like $f'(d) - f'(c) ge int_c^d f''(x) , mathrm{d}x ge 0$ for $[c,d] subset (a,b)$. This should give your assertion.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
Or speaking differently: The second derivative of a convex function can be identified with a positive distribution, i.e., a measure. Then, your $f''$ should be the absolutely continuous part of this measure. Again, it belongs to $L^1_{text{loc}}(a,b)$.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
@gerw take a look in mathoverflow.net/questions/265256/…
$endgroup$
– Tomás
Dec 3 '18 at 20:01
$begingroup$
I didn't know that this question was crossposted. You should link both questions.
$endgroup$
– gerw
Dec 3 '18 at 20:21
$begingroup$
I know that this is an old question, but I would like to drop a comment. I guess that we have something like $f'(d) - f'(c) ge int_c^d f''(x) , mathrm{d}x ge 0$ for $[c,d] subset (a,b)$. This should give your assertion.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
I know that this is an old question, but I would like to drop a comment. I guess that we have something like $f'(d) - f'(c) ge int_c^d f''(x) , mathrm{d}x ge 0$ for $[c,d] subset (a,b)$. This should give your assertion.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
Or speaking differently: The second derivative of a convex function can be identified with a positive distribution, i.e., a measure. Then, your $f''$ should be the absolutely continuous part of this measure. Again, it belongs to $L^1_{text{loc}}(a,b)$.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
Or speaking differently: The second derivative of a convex function can be identified with a positive distribution, i.e., a measure. Then, your $f''$ should be the absolutely continuous part of this measure. Again, it belongs to $L^1_{text{loc}}(a,b)$.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
@gerw take a look in mathoverflow.net/questions/265256/…
$endgroup$
– Tomás
Dec 3 '18 at 20:01
$begingroup$
@gerw take a look in mathoverflow.net/questions/265256/…
$endgroup$
– Tomás
Dec 3 '18 at 20:01
$begingroup$
I didn't know that this question was crossposted. You should link both questions.
$endgroup$
– gerw
Dec 3 '18 at 20:21
$begingroup$
I didn't know that this question was crossposted. You should link both questions.
$endgroup$
– gerw
Dec 3 '18 at 20:21
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
});
}
});
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%2f2195394%2fbounds-on-the-second-derivative-of-a-convex-function%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
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%2f2195394%2fbounds-on-the-second-derivative-of-a-convex-function%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
$begingroup$
I know that this is an old question, but I would like to drop a comment. I guess that we have something like $f'(d) - f'(c) ge int_c^d f''(x) , mathrm{d}x ge 0$ for $[c,d] subset (a,b)$. This should give your assertion.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
Or speaking differently: The second derivative of a convex function can be identified with a positive distribution, i.e., a measure. Then, your $f''$ should be the absolutely continuous part of this measure. Again, it belongs to $L^1_{text{loc}}(a,b)$.
$endgroup$
– gerw
Nov 22 '18 at 12:36
$begingroup$
@gerw take a look in mathoverflow.net/questions/265256/…
$endgroup$
– Tomás
Dec 3 '18 at 20:01
$begingroup$
I didn't know that this question was crossposted. You should link both questions.
$endgroup$
– gerw
Dec 3 '18 at 20:21