If $f(y)=limsup_k|y-x_k|^2$ then $f$ is a strictly convex function
$begingroup$
Let $y in mathbb{R^n}$, and $(x_k)_{kin mathbb{N}}$ a sequence in $ mathbb{R^n}$ and $f(y)=limsup_k|y-x_k|^2$ ($|{cdot}|$ is the euclidean norm).
Then $f$ is a convex function.
But how can we prove that $f$ is a strictly convex function ?
convex-analysis convex-optimization
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
$endgroup$
add a comment |
$begingroup$
Let $y in mathbb{R^n}$, and $(x_k)_{kin mathbb{N}}$ a sequence in $ mathbb{R^n}$ and $f(y)=limsup_k|y-x_k|^2$ ($|{cdot}|$ is the euclidean norm).
Then $f$ is a convex function.
But how can we prove that $f$ is a strictly convex function ?
convex-analysis convex-optimization
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
$endgroup$
$begingroup$
$f(y)$ could be $infty$ everywhere
$endgroup$
– LinAlg
Dec 18 '18 at 19:49
$begingroup$
Why don't you show your proof for (weak) convexity? Then we can try to point out where you can replace $le$ by $<$.
$endgroup$
– Martin R
Dec 18 '18 at 19:49
add a comment |
$begingroup$
Let $y in mathbb{R^n}$, and $(x_k)_{kin mathbb{N}}$ a sequence in $ mathbb{R^n}$ and $f(y)=limsup_k|y-x_k|^2$ ($|{cdot}|$ is the euclidean norm).
Then $f$ is a convex function.
But how can we prove that $f$ is a strictly convex function ?
convex-analysis convex-optimization
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
$endgroup$
Let $y in mathbb{R^n}$, and $(x_k)_{kin mathbb{N}}$ a sequence in $ mathbb{R^n}$ and $f(y)=limsup_k|y-x_k|^2$ ($|{cdot}|$ is the euclidean norm).
Then $f$ is a convex function.
But how can we prove that $f$ is a strictly convex function ?
convex-analysis convex-optimization
convex-analysis convex-optimization
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
edited Dec 18 '18 at 19:41
Masacroso
13.1k41747
13.1k41747
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
asked Dec 18 '18 at 19:38
Anas BOUALIIAnas BOUALII
1388
1388
$begingroup$
$f(y)$ could be $infty$ everywhere
$endgroup$
– LinAlg
Dec 18 '18 at 19:49
$begingroup$
Why don't you show your proof for (weak) convexity? Then we can try to point out where you can replace $le$ by $<$.
$endgroup$
– Martin R
Dec 18 '18 at 19:49
add a comment |
$begingroup$
$f(y)$ could be $infty$ everywhere
$endgroup$
– LinAlg
Dec 18 '18 at 19:49
$begingroup$
Why don't you show your proof for (weak) convexity? Then we can try to point out where you can replace $le$ by $<$.
$endgroup$
– Martin R
Dec 18 '18 at 19:49
$begingroup$
$f(y)$ could be $infty$ everywhere
$endgroup$
– LinAlg
Dec 18 '18 at 19:49
$begingroup$
$f(y)$ could be $infty$ everywhere
$endgroup$
– LinAlg
Dec 18 '18 at 19:49
$begingroup$
Why don't you show your proof for (weak) convexity? Then we can try to point out where you can replace $le$ by $<$.
$endgroup$
– Martin R
Dec 18 '18 at 19:49
$begingroup$
Why don't you show your proof for (weak) convexity? Then we can try to point out where you can replace $le$ by $<$.
$endgroup$
– Martin R
Dec 18 '18 at 19:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm assuming that $(x_k)$ is a bounded sequence, so that $f(y)$ is finite for all $y in Bbb R$.
Generally, for $a, b in Bbb R^n$ and $lambda in Bbb R$ we have, with $cdot$ denoting the scalar product:
$$
Vert lambda a + (1-lambda) b Vert^2 = lambda^2 Vert a Vert^2 + 2 lambda(1-lambda) (a cdot b)+ lambda^2 Vert b Vert^2 \
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda)left( Vert a Vert^2 - 2 (a cdot b)+ Vert b Vert^2 right)\
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda) Vert a-b Vert^2 , .
$$
Now let $y, z in Bbb R^n$, $y ne z$, and $0<lambda< 1$. For all $k$ we have
$$
Vert lambda y + (1-lambda) z - x_k Vert^2 =
Vert lambda(y-x_k) + (1-lambda) (z-x_k) Vert^2 \
= lambda Vert y-x_k Vert^2 + (1-lambda) Vert z-x_k Vert^2
- lambda (1-lambda) Vert y-z Vert^2 , .
$$
It follows that
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
$endgroup$
$begingroup$
Why $f(y)=f(z)=0$ otherwise ?
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:18
$begingroup$
@AnasBOUALII: My original answer was indeed wrong. – Have a look at the new version, it should be correct now.
$endgroup$
– Martin R
Dec 19 '18 at 2:40
$begingroup$
yeah, well i didnt get the second identity used, where u took $z$ diffrent from $y$.
$endgroup$
– Anas BOUALII
Dec 19 '18 at 15:35
$begingroup$
@AnasBOUALII: I have applied the first identity with $a = y-x_k$ and $b=z-x_k$.
$endgroup$
– Martin R
Dec 19 '18 at 15:58
$begingroup$
yeah that s obvious, but I asked how we prove it wasnt clair at first(the identity).
$endgroup$
– Anas BOUALII
Dec 19 '18 at 18:34
|
show 4 more comments
$begingroup$
we first prove that $g_k(x)=||x-x_k||^2$ is a convex function.
Then we will have $g_k(tx+(1-t)y)leq tg_k(x)+(1-t)g_k(y)$ for all $tin [0,1] $ and $x,yin mathbb{R^n}.$
which means $$||tx+(1-t)y-x_k|| leq ||x-x_k||+ ||y-x_k||$$
then $$limsup_k||tx+(1-t)y-x_k|| leq limsup_k||x-x_k||+ limsup_k||y-x_k||$$
Hence, $f$ is convex.
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
$endgroup$
$begingroup$
But after we pass to the limit all strict inequality become $leq$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:03
1
$begingroup$
@Masacroso: Anas is the OP. This might be a misunderstanding from my above comment. What I meant is that Anas adds his/her proof of weak convexity to the question.
$endgroup$
– Martin R
Dec 18 '18 at 20:05
$begingroup$
@Martin lol, thanks, I didnt noticed. Yes, this must be written in the text body of the original question
$endgroup$
– Masacroso
Dec 18 '18 at 20:07
$begingroup$
This is not a complete ''answer", just the convexity of $f$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:09
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%2f3045601%2fif-fy-limsup-k-y-x-k-2-then-f-is-a-strictly-convex-function%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
$begingroup$
I'm assuming that $(x_k)$ is a bounded sequence, so that $f(y)$ is finite for all $y in Bbb R$.
Generally, for $a, b in Bbb R^n$ and $lambda in Bbb R$ we have, with $cdot$ denoting the scalar product:
$$
Vert lambda a + (1-lambda) b Vert^2 = lambda^2 Vert a Vert^2 + 2 lambda(1-lambda) (a cdot b)+ lambda^2 Vert b Vert^2 \
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda)left( Vert a Vert^2 - 2 (a cdot b)+ Vert b Vert^2 right)\
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda) Vert a-b Vert^2 , .
$$
Now let $y, z in Bbb R^n$, $y ne z$, and $0<lambda< 1$. For all $k$ we have
$$
Vert lambda y + (1-lambda) z - x_k Vert^2 =
Vert lambda(y-x_k) + (1-lambda) (z-x_k) Vert^2 \
= lambda Vert y-x_k Vert^2 + (1-lambda) Vert z-x_k Vert^2
- lambda (1-lambda) Vert y-z Vert^2 , .
$$
It follows that
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
$endgroup$
$begingroup$
Why $f(y)=f(z)=0$ otherwise ?
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:18
$begingroup$
@AnasBOUALII: My original answer was indeed wrong. – Have a look at the new version, it should be correct now.
$endgroup$
– Martin R
Dec 19 '18 at 2:40
$begingroup$
yeah, well i didnt get the second identity used, where u took $z$ diffrent from $y$.
$endgroup$
– Anas BOUALII
Dec 19 '18 at 15:35
$begingroup$
@AnasBOUALII: I have applied the first identity with $a = y-x_k$ and $b=z-x_k$.
$endgroup$
– Martin R
Dec 19 '18 at 15:58
$begingroup$
yeah that s obvious, but I asked how we prove it wasnt clair at first(the identity).
$endgroup$
– Anas BOUALII
Dec 19 '18 at 18:34
|
show 4 more comments
$begingroup$
I'm assuming that $(x_k)$ is a bounded sequence, so that $f(y)$ is finite for all $y in Bbb R$.
Generally, for $a, b in Bbb R^n$ and $lambda in Bbb R$ we have, with $cdot$ denoting the scalar product:
$$
Vert lambda a + (1-lambda) b Vert^2 = lambda^2 Vert a Vert^2 + 2 lambda(1-lambda) (a cdot b)+ lambda^2 Vert b Vert^2 \
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda)left( Vert a Vert^2 - 2 (a cdot b)+ Vert b Vert^2 right)\
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda) Vert a-b Vert^2 , .
$$
Now let $y, z in Bbb R^n$, $y ne z$, and $0<lambda< 1$. For all $k$ we have
$$
Vert lambda y + (1-lambda) z - x_k Vert^2 =
Vert lambda(y-x_k) + (1-lambda) (z-x_k) Vert^2 \
= lambda Vert y-x_k Vert^2 + (1-lambda) Vert z-x_k Vert^2
- lambda (1-lambda) Vert y-z Vert^2 , .
$$
It follows that
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
$endgroup$
$begingroup$
Why $f(y)=f(z)=0$ otherwise ?
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:18
$begingroup$
@AnasBOUALII: My original answer was indeed wrong. – Have a look at the new version, it should be correct now.
$endgroup$
– Martin R
Dec 19 '18 at 2:40
$begingroup$
yeah, well i didnt get the second identity used, where u took $z$ diffrent from $y$.
$endgroup$
– Anas BOUALII
Dec 19 '18 at 15:35
$begingroup$
@AnasBOUALII: I have applied the first identity with $a = y-x_k$ and $b=z-x_k$.
$endgroup$
– Martin R
Dec 19 '18 at 15:58
$begingroup$
yeah that s obvious, but I asked how we prove it wasnt clair at first(the identity).
$endgroup$
– Anas BOUALII
Dec 19 '18 at 18:34
|
show 4 more comments
$begingroup$
I'm assuming that $(x_k)$ is a bounded sequence, so that $f(y)$ is finite for all $y in Bbb R$.
Generally, for $a, b in Bbb R^n$ and $lambda in Bbb R$ we have, with $cdot$ denoting the scalar product:
$$
Vert lambda a + (1-lambda) b Vert^2 = lambda^2 Vert a Vert^2 + 2 lambda(1-lambda) (a cdot b)+ lambda^2 Vert b Vert^2 \
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda)left( Vert a Vert^2 - 2 (a cdot b)+ Vert b Vert^2 right)\
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda) Vert a-b Vert^2 , .
$$
Now let $y, z in Bbb R^n$, $y ne z$, and $0<lambda< 1$. For all $k$ we have
$$
Vert lambda y + (1-lambda) z - x_k Vert^2 =
Vert lambda(y-x_k) + (1-lambda) (z-x_k) Vert^2 \
= lambda Vert y-x_k Vert^2 + (1-lambda) Vert z-x_k Vert^2
- lambda (1-lambda) Vert y-z Vert^2 , .
$$
It follows that
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
$endgroup$
I'm assuming that $(x_k)$ is a bounded sequence, so that $f(y)$ is finite for all $y in Bbb R$.
Generally, for $a, b in Bbb R^n$ and $lambda in Bbb R$ we have, with $cdot$ denoting the scalar product:
$$
Vert lambda a + (1-lambda) b Vert^2 = lambda^2 Vert a Vert^2 + 2 lambda(1-lambda) (a cdot b)+ lambda^2 Vert b Vert^2 \
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda)left( Vert a Vert^2 - 2 (a cdot b)+ Vert b Vert^2 right)\
= lambda Vert a Vert^2 + lambda Vert b Vert^2 - lambda (1-lambda) Vert a-b Vert^2 , .
$$
Now let $y, z in Bbb R^n$, $y ne z$, and $0<lambda< 1$. For all $k$ we have
$$
Vert lambda y + (1-lambda) z - x_k Vert^2 =
Vert lambda(y-x_k) + (1-lambda) (z-x_k) Vert^2 \
= lambda Vert y-x_k Vert^2 + (1-lambda) Vert z-x_k Vert^2
- lambda (1-lambda) Vert y-z Vert^2 , .
$$
It follows that
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
edited Dec 19 '18 at 20:49
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
answered Dec 18 '18 at 20:10
Martin RMartin R
30.5k33558
30.5k33558
$begingroup$
Why $f(y)=f(z)=0$ otherwise ?
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:18
$begingroup$
@AnasBOUALII: My original answer was indeed wrong. – Have a look at the new version, it should be correct now.
$endgroup$
– Martin R
Dec 19 '18 at 2:40
$begingroup$
yeah, well i didnt get the second identity used, where u took $z$ diffrent from $y$.
$endgroup$
– Anas BOUALII
Dec 19 '18 at 15:35
$begingroup$
@AnasBOUALII: I have applied the first identity with $a = y-x_k$ and $b=z-x_k$.
$endgroup$
– Martin R
Dec 19 '18 at 15:58
$begingroup$
yeah that s obvious, but I asked how we prove it wasnt clair at first(the identity).
$endgroup$
– Anas BOUALII
Dec 19 '18 at 18:34
|
show 4 more comments
$begingroup$
Why $f(y)=f(z)=0$ otherwise ?
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:18
$begingroup$
@AnasBOUALII: My original answer was indeed wrong. – Have a look at the new version, it should be correct now.
$endgroup$
– Martin R
Dec 19 '18 at 2:40
$begingroup$
yeah, well i didnt get the second identity used, where u took $z$ diffrent from $y$.
$endgroup$
– Anas BOUALII
Dec 19 '18 at 15:35
$begingroup$
@AnasBOUALII: I have applied the first identity with $a = y-x_k$ and $b=z-x_k$.
$endgroup$
– Martin R
Dec 19 '18 at 15:58
$begingroup$
yeah that s obvious, but I asked how we prove it wasnt clair at first(the identity).
$endgroup$
– Anas BOUALII
Dec 19 '18 at 18:34
$begingroup$
Why $f(y)=f(z)=0$ otherwise ?
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:18
$begingroup$
Why $f(y)=f(z)=0$ otherwise ?
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:18
$begingroup$
@AnasBOUALII: My original answer was indeed wrong. – Have a look at the new version, it should be correct now.
$endgroup$
– Martin R
Dec 19 '18 at 2:40
$begingroup$
@AnasBOUALII: My original answer was indeed wrong. – Have a look at the new version, it should be correct now.
$endgroup$
– Martin R
Dec 19 '18 at 2:40
$begingroup$
yeah, well i didnt get the second identity used, where u took $z$ diffrent from $y$.
$endgroup$
– Anas BOUALII
Dec 19 '18 at 15:35
$begingroup$
yeah, well i didnt get the second identity used, where u took $z$ diffrent from $y$.
$endgroup$
– Anas BOUALII
Dec 19 '18 at 15:35
$begingroup$
@AnasBOUALII: I have applied the first identity with $a = y-x_k$ and $b=z-x_k$.
$endgroup$
– Martin R
Dec 19 '18 at 15:58
$begingroup$
@AnasBOUALII: I have applied the first identity with $a = y-x_k$ and $b=z-x_k$.
$endgroup$
– Martin R
Dec 19 '18 at 15:58
$begingroup$
yeah that s obvious, but I asked how we prove it wasnt clair at first(the identity).
$endgroup$
– Anas BOUALII
Dec 19 '18 at 18:34
$begingroup$
yeah that s obvious, but I asked how we prove it wasnt clair at first(the identity).
$endgroup$
– Anas BOUALII
Dec 19 '18 at 18:34
|
show 4 more comments
$begingroup$
we first prove that $g_k(x)=||x-x_k||^2$ is a convex function.
Then we will have $g_k(tx+(1-t)y)leq tg_k(x)+(1-t)g_k(y)$ for all $tin [0,1] $ and $x,yin mathbb{R^n}.$
which means $$||tx+(1-t)y-x_k|| leq ||x-x_k||+ ||y-x_k||$$
then $$limsup_k||tx+(1-t)y-x_k|| leq limsup_k||x-x_k||+ limsup_k||y-x_k||$$
Hence, $f$ is convex.
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
$endgroup$
$begingroup$
But after we pass to the limit all strict inequality become $leq$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:03
1
$begingroup$
@Masacroso: Anas is the OP. This might be a misunderstanding from my above comment. What I meant is that Anas adds his/her proof of weak convexity to the question.
$endgroup$
– Martin R
Dec 18 '18 at 20:05
$begingroup$
@Martin lol, thanks, I didnt noticed. Yes, this must be written in the text body of the original question
$endgroup$
– Masacroso
Dec 18 '18 at 20:07
$begingroup$
This is not a complete ''answer", just the convexity of $f$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:09
add a comment |
$begingroup$
we first prove that $g_k(x)=||x-x_k||^2$ is a convex function.
Then we will have $g_k(tx+(1-t)y)leq tg_k(x)+(1-t)g_k(y)$ for all $tin [0,1] $ and $x,yin mathbb{R^n}.$
which means $$||tx+(1-t)y-x_k|| leq ||x-x_k||+ ||y-x_k||$$
then $$limsup_k||tx+(1-t)y-x_k|| leq limsup_k||x-x_k||+ limsup_k||y-x_k||$$
Hence, $f$ is convex.
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
$endgroup$
$begingroup$
But after we pass to the limit all strict inequality become $leq$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:03
1
$begingroup$
@Masacroso: Anas is the OP. This might be a misunderstanding from my above comment. What I meant is that Anas adds his/her proof of weak convexity to the question.
$endgroup$
– Martin R
Dec 18 '18 at 20:05
$begingroup$
@Martin lol, thanks, I didnt noticed. Yes, this must be written in the text body of the original question
$endgroup$
– Masacroso
Dec 18 '18 at 20:07
$begingroup$
This is not a complete ''answer", just the convexity of $f$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:09
add a comment |
$begingroup$
we first prove that $g_k(x)=||x-x_k||^2$ is a convex function.
Then we will have $g_k(tx+(1-t)y)leq tg_k(x)+(1-t)g_k(y)$ for all $tin [0,1] $ and $x,yin mathbb{R^n}.$
which means $$||tx+(1-t)y-x_k|| leq ||x-x_k||+ ||y-x_k||$$
then $$limsup_k||tx+(1-t)y-x_k|| leq limsup_k||x-x_k||+ limsup_k||y-x_k||$$
Hence, $f$ is convex.
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
$endgroup$
we first prove that $g_k(x)=||x-x_k||^2$ is a convex function.
Then we will have $g_k(tx+(1-t)y)leq tg_k(x)+(1-t)g_k(y)$ for all $tin [0,1] $ and $x,yin mathbb{R^n}.$
which means $$||tx+(1-t)y-x_k|| leq ||x-x_k||+ ||y-x_k||$$
then $$limsup_k||tx+(1-t)y-x_k|| leq limsup_k||x-x_k||+ limsup_k||y-x_k||$$
Hence, $f$ is convex.
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
answered Dec 18 '18 at 19:59
Anas BOUALIIAnas BOUALII
1388
1388
$begingroup$
But after we pass to the limit all strict inequality become $leq$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:03
1
$begingroup$
@Masacroso: Anas is the OP. This might be a misunderstanding from my above comment. What I meant is that Anas adds his/her proof of weak convexity to the question.
$endgroup$
– Martin R
Dec 18 '18 at 20:05
$begingroup$
@Martin lol, thanks, I didnt noticed. Yes, this must be written in the text body of the original question
$endgroup$
– Masacroso
Dec 18 '18 at 20:07
$begingroup$
This is not a complete ''answer", just the convexity of $f$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:09
add a comment |
$begingroup$
But after we pass to the limit all strict inequality become $leq$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:03
1
$begingroup$
@Masacroso: Anas is the OP. This might be a misunderstanding from my above comment. What I meant is that Anas adds his/her proof of weak convexity to the question.
$endgroup$
– Martin R
Dec 18 '18 at 20:05
$begingroup$
@Martin lol, thanks, I didnt noticed. Yes, this must be written in the text body of the original question
$endgroup$
– Masacroso
Dec 18 '18 at 20:07
$begingroup$
This is not a complete ''answer", just the convexity of $f$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:09
$begingroup$
But after we pass to the limit all strict inequality become $leq$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:03
$begingroup$
But after we pass to the limit all strict inequality become $leq$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:03
1
1
$begingroup$
@Masacroso: Anas is the OP. This might be a misunderstanding from my above comment. What I meant is that Anas adds his/her proof of weak convexity to the question.
$endgroup$
– Martin R
Dec 18 '18 at 20:05
$begingroup$
@Masacroso: Anas is the OP. This might be a misunderstanding from my above comment. What I meant is that Anas adds his/her proof of weak convexity to the question.
$endgroup$
– Martin R
Dec 18 '18 at 20:05
$begingroup$
@Martin lol, thanks, I didnt noticed. Yes, this must be written in the text body of the original question
$endgroup$
– Masacroso
Dec 18 '18 at 20:07
$begingroup$
@Martin lol, thanks, I didnt noticed. Yes, this must be written in the text body of the original question
$endgroup$
– Masacroso
Dec 18 '18 at 20:07
$begingroup$
This is not a complete ''answer", just the convexity of $f$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:09
$begingroup$
This is not a complete ''answer", just the convexity of $f$.
$endgroup$
– Anas BOUALII
Dec 18 '18 at 20:09
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%2f3045601%2fif-fy-limsup-k-y-x-k-2-then-f-is-a-strictly-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
$begingroup$
$f(y)$ could be $infty$ everywhere
$endgroup$
– LinAlg
Dec 18 '18 at 19:49
$begingroup$
Why don't you show your proof for (weak) convexity? Then we can try to point out where you can replace $le$ by $<$.
$endgroup$
– Martin R
Dec 18 '18 at 19:49