If $f(y)=limsup_k|y-x_k|^2$ then $f$ is a strictly convex function












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f(y)$ 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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f(y)$ 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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 19:41









Masacroso

13.1k41747




13.1k41747










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




















  • $begingroup$
    $f(y)$ 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












2 Answers
2






active

oldest

votes


















2












$begingroup$

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
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$






share|cite|improve this answer











$endgroup$



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
$$
f(lambda y + (1-lambda) z) le lambda f(y) + (1-lambda) f(z) - lambda (1-lambda) Vert y-z Vert^2 \
< lambda f(y) + (1-lambda) f(z), .
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 20:49

























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


















  • $begingroup$
    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











0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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