Dominated convergence 2.1?

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After this question : Dominated convergence 2.0?



I want to know, what about the case when $hin L^1([0,1])$.



The completed question :



Let $(f_n)_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g in L^1([0,1])$ and $forall x in [0,1], g(x)in mathbb R$.



Assume that:



$forall ninmathbb N, f_n''<h$, where $h in L^1([0,1])$.



Is it true that $lim int_0^1 f_n=int_0^1 g$ ?










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  • +1. Maybe it is worthwhile mentioning in the post that you assume $g$ to take values in $mathbbR$ rather than in $[-infty, infty]$?
    – Jochen Glueck
    Aug 19 at 18:49














up vote
5
down vote

favorite
3












After this question : Dominated convergence 2.0?



I want to know, what about the case when $hin L^1([0,1])$.



The completed question :



Let $(f_n)_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g in L^1([0,1])$ and $forall x in [0,1], g(x)in mathbb R$.



Assume that:



$forall ninmathbb N, f_n''<h$, where $h in L^1([0,1])$.



Is it true that $lim int_0^1 f_n=int_0^1 g$ ?










share|cite|improve this question























  • +1. Maybe it is worthwhile mentioning in the post that you assume $g$ to take values in $mathbbR$ rather than in $[-infty, infty]$?
    – Jochen Glueck
    Aug 19 at 18:49












up vote
5
down vote

favorite
3









up vote
5
down vote

favorite
3






3





After this question : Dominated convergence 2.0?



I want to know, what about the case when $hin L^1([0,1])$.



The completed question :



Let $(f_n)_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g in L^1([0,1])$ and $forall x in [0,1], g(x)in mathbb R$.



Assume that:



$forall ninmathbb N, f_n''<h$, where $h in L^1([0,1])$.



Is it true that $lim int_0^1 f_n=int_0^1 g$ ?










share|cite|improve this question















After this question : Dominated convergence 2.0?



I want to know, what about the case when $hin L^1([0,1])$.



The completed question :



Let $(f_n)_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g in L^1([0,1])$ and $forall x in [0,1], g(x)in mathbb R$.



Assume that:



$forall ninmathbb N, f_n''<h$, where $h in L^1([0,1])$.



Is it true that $lim int_0^1 f_n=int_0^1 g$ ?







real-analysis integration






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edited Aug 19 at 19:13

























asked Aug 19 at 18:40









Dattier

612314




612314











  • +1. Maybe it is worthwhile mentioning in the post that you assume $g$ to take values in $mathbbR$ rather than in $[-infty, infty]$?
    – Jochen Glueck
    Aug 19 at 18:49
















  • +1. Maybe it is worthwhile mentioning in the post that you assume $g$ to take values in $mathbbR$ rather than in $[-infty, infty]$?
    – Jochen Glueck
    Aug 19 at 18:49















+1. Maybe it is worthwhile mentioning in the post that you assume $g$ to take values in $mathbbR$ rather than in $[-infty, infty]$?
– Jochen Glueck
Aug 19 at 18:49




+1. Maybe it is worthwhile mentioning in the post that you assume $g$ to take values in $mathbbR$ rather than in $[-infty, infty]$?
– Jochen Glueck
Aug 19 at 18:49










2 Answers
2






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up vote
3
down vote



accepted










I claim that under these assumptions, the functions $f_n$ are uniformly bounded. Then the conclusion follows from the dominated convergence theorem.



First set $H(x) = int_0^x int_0^t h(s),ds$, which is $C^1$. Letting $u_n = f_n-H$, we have that $u_n$ is concave (i.e. $-u_n$ is convex), continuous, and still converges pointwise.



Now let $v_n(x) = u_n(x) - (x u_n(1) + (1-x) u_n(0))$. Now $v_n$ is again concave and continuous, $v_n(0)= v_n(1)=0$, and $v_n$ still converges pointwise (note that $u_n(0), u_n(1)$ both converge to finite limits). In particular we have $v_n ge 0$ everywhere.



Let $M_n$ be the maximum value of $v_n$, and let $x_n$ be the point where it is attained. Suppose first that $x_n ge 1/2$. By concavity we have $v_n(x) ge fracxx_n v(x_n) ge x v(x_n) = x M_n$ for all $0 le x le x_n$. In particular, we have $v_n(1/2) ge frac12 M_n$. If $x_n le 1/2$, we can get the same result by a similar argument (or by replacing $v_n(x)$ with $v_n(1-x)$).



So $M := sup_n M_n le 2 sup_n v_n(1/2)$ which is finite because $v_n$ converges pointwise. So we have $0 le v_n(x) le M$ for all $x,n$. It follows easily that $f_n$ is uniformly bounded as well (by, say, $M + sup_n |u_n(0)| + sup_n |u_n(1)| + sup_x |H(x)|$).






share|cite|improve this answer






















  • change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge
    – Dattier
    Aug 20 at 8:41











  • you should corriged your tipos
    – Dattier
    Aug 20 at 8:47

















up vote
3
down vote













The answer is yes. Indeed, let
$u_n:=f_n-H$ and $v:=g-H$, where
beginequation
H(x):=int_0^x(x-t)h(t),dt
endequation
for $xin[0,1]$.
Then $u_nto v$ pointwise and $u_n$ is concave for each $n$ (see the Detail below). So, by Lemma 1 below, $u_nto v$ in $L^1[0,1]$ and hence $lim int_0^1 f_n=int_0^1 g$.



Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $xin[0,1]$,

beginequation
u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+int_0^1(x-t)_+[f''_n(t)-h(t)],dt.
endequation
Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.



Lemma 1. Suppose that $f_n$ are convex real-valued functions on $[0,1]$ converging pointwise to a real-valued function $f$. Then $f_nto f$ in $L^1[0,1]$.



Proof. The function $f$ is real-valued and convex and hence bounded from below. So, by Corollary 3, all the functions $f_n$ are uniformly bounded from below. On the other hand, all the convex functions $f_n$ are uniformly bounded from above by $sup_n(f_n(0)vee f_n(1))$. So, Lemma 1 follows by dominated convergence.






share|cite|improve this answer






















  • why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis
    – Dattier
    Aug 19 at 19:55











  • @Dattier : see the edited version.
    – Iosif Pinelis
    Aug 19 at 20:26










  • @NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true.
    – Iosif Pinelis
    Aug 19 at 20:49











  • @NateEldredge : Please see the latest edit.
    – Iosif Pinelis
    Aug 19 at 21:14










  • @IosifPinelis: Yep, I think we basically found the same argument :-)
    – Nate Eldredge
    Aug 19 at 21:16










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










I claim that under these assumptions, the functions $f_n$ are uniformly bounded. Then the conclusion follows from the dominated convergence theorem.



First set $H(x) = int_0^x int_0^t h(s),ds$, which is $C^1$. Letting $u_n = f_n-H$, we have that $u_n$ is concave (i.e. $-u_n$ is convex), continuous, and still converges pointwise.



Now let $v_n(x) = u_n(x) - (x u_n(1) + (1-x) u_n(0))$. Now $v_n$ is again concave and continuous, $v_n(0)= v_n(1)=0$, and $v_n$ still converges pointwise (note that $u_n(0), u_n(1)$ both converge to finite limits). In particular we have $v_n ge 0$ everywhere.



Let $M_n$ be the maximum value of $v_n$, and let $x_n$ be the point where it is attained. Suppose first that $x_n ge 1/2$. By concavity we have $v_n(x) ge fracxx_n v(x_n) ge x v(x_n) = x M_n$ for all $0 le x le x_n$. In particular, we have $v_n(1/2) ge frac12 M_n$. If $x_n le 1/2$, we can get the same result by a similar argument (or by replacing $v_n(x)$ with $v_n(1-x)$).



So $M := sup_n M_n le 2 sup_n v_n(1/2)$ which is finite because $v_n$ converges pointwise. So we have $0 le v_n(x) le M$ for all $x,n$. It follows easily that $f_n$ is uniformly bounded as well (by, say, $M + sup_n |u_n(0)| + sup_n |u_n(1)| + sup_x |H(x)|$).






share|cite|improve this answer






















  • change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge
    – Dattier
    Aug 20 at 8:41











  • you should corriged your tipos
    – Dattier
    Aug 20 at 8:47














up vote
3
down vote



accepted










I claim that under these assumptions, the functions $f_n$ are uniformly bounded. Then the conclusion follows from the dominated convergence theorem.



First set $H(x) = int_0^x int_0^t h(s),ds$, which is $C^1$. Letting $u_n = f_n-H$, we have that $u_n$ is concave (i.e. $-u_n$ is convex), continuous, and still converges pointwise.



Now let $v_n(x) = u_n(x) - (x u_n(1) + (1-x) u_n(0))$. Now $v_n$ is again concave and continuous, $v_n(0)= v_n(1)=0$, and $v_n$ still converges pointwise (note that $u_n(0), u_n(1)$ both converge to finite limits). In particular we have $v_n ge 0$ everywhere.



Let $M_n$ be the maximum value of $v_n$, and let $x_n$ be the point where it is attained. Suppose first that $x_n ge 1/2$. By concavity we have $v_n(x) ge fracxx_n v(x_n) ge x v(x_n) = x M_n$ for all $0 le x le x_n$. In particular, we have $v_n(1/2) ge frac12 M_n$. If $x_n le 1/2$, we can get the same result by a similar argument (or by replacing $v_n(x)$ with $v_n(1-x)$).



So $M := sup_n M_n le 2 sup_n v_n(1/2)$ which is finite because $v_n$ converges pointwise. So we have $0 le v_n(x) le M$ for all $x,n$. It follows easily that $f_n$ is uniformly bounded as well (by, say, $M + sup_n |u_n(0)| + sup_n |u_n(1)| + sup_x |H(x)|$).






share|cite|improve this answer






















  • change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge
    – Dattier
    Aug 20 at 8:41











  • you should corriged your tipos
    – Dattier
    Aug 20 at 8:47












up vote
3
down vote



accepted







up vote
3
down vote



accepted






I claim that under these assumptions, the functions $f_n$ are uniformly bounded. Then the conclusion follows from the dominated convergence theorem.



First set $H(x) = int_0^x int_0^t h(s),ds$, which is $C^1$. Letting $u_n = f_n-H$, we have that $u_n$ is concave (i.e. $-u_n$ is convex), continuous, and still converges pointwise.



Now let $v_n(x) = u_n(x) - (x u_n(1) + (1-x) u_n(0))$. Now $v_n$ is again concave and continuous, $v_n(0)= v_n(1)=0$, and $v_n$ still converges pointwise (note that $u_n(0), u_n(1)$ both converge to finite limits). In particular we have $v_n ge 0$ everywhere.



Let $M_n$ be the maximum value of $v_n$, and let $x_n$ be the point where it is attained. Suppose first that $x_n ge 1/2$. By concavity we have $v_n(x) ge fracxx_n v(x_n) ge x v(x_n) = x M_n$ for all $0 le x le x_n$. In particular, we have $v_n(1/2) ge frac12 M_n$. If $x_n le 1/2$, we can get the same result by a similar argument (or by replacing $v_n(x)$ with $v_n(1-x)$).



So $M := sup_n M_n le 2 sup_n v_n(1/2)$ which is finite because $v_n$ converges pointwise. So we have $0 le v_n(x) le M$ for all $x,n$. It follows easily that $f_n$ is uniformly bounded as well (by, say, $M + sup_n |u_n(0)| + sup_n |u_n(1)| + sup_x |H(x)|$).






share|cite|improve this answer














I claim that under these assumptions, the functions $f_n$ are uniformly bounded. Then the conclusion follows from the dominated convergence theorem.



First set $H(x) = int_0^x int_0^t h(s),ds$, which is $C^1$. Letting $u_n = f_n-H$, we have that $u_n$ is concave (i.e. $-u_n$ is convex), continuous, and still converges pointwise.



Now let $v_n(x) = u_n(x) - (x u_n(1) + (1-x) u_n(0))$. Now $v_n$ is again concave and continuous, $v_n(0)= v_n(1)=0$, and $v_n$ still converges pointwise (note that $u_n(0), u_n(1)$ both converge to finite limits). In particular we have $v_n ge 0$ everywhere.



Let $M_n$ be the maximum value of $v_n$, and let $x_n$ be the point where it is attained. Suppose first that $x_n ge 1/2$. By concavity we have $v_n(x) ge fracxx_n v(x_n) ge x v(x_n) = x M_n$ for all $0 le x le x_n$. In particular, we have $v_n(1/2) ge frac12 M_n$. If $x_n le 1/2$, we can get the same result by a similar argument (or by replacing $v_n(x)$ with $v_n(1-x)$).



So $M := sup_n M_n le 2 sup_n v_n(1/2)$ which is finite because $v_n$ converges pointwise. So we have $0 le v_n(x) le M$ for all $x,n$. It follows easily that $f_n$ is uniformly bounded as well (by, say, $M + sup_n |u_n(0)| + sup_n |u_n(1)| + sup_x |H(x)|$).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 20 at 13:19

























answered Aug 19 at 21:09









Nate Eldredge

18.8k362108




18.8k362108











  • change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge
    – Dattier
    Aug 20 at 8:41











  • you should corriged your tipos
    – Dattier
    Aug 20 at 8:47
















  • change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge
    – Dattier
    Aug 20 at 8:41











  • you should corriged your tipos
    – Dattier
    Aug 20 at 8:47















change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge
– Dattier
Aug 20 at 8:41





change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge
– Dattier
Aug 20 at 8:41













you should corriged your tipos
– Dattier
Aug 20 at 8:47




you should corriged your tipos
– Dattier
Aug 20 at 8:47










up vote
3
down vote













The answer is yes. Indeed, let
$u_n:=f_n-H$ and $v:=g-H$, where
beginequation
H(x):=int_0^x(x-t)h(t),dt
endequation
for $xin[0,1]$.
Then $u_nto v$ pointwise and $u_n$ is concave for each $n$ (see the Detail below). So, by Lemma 1 below, $u_nto v$ in $L^1[0,1]$ and hence $lim int_0^1 f_n=int_0^1 g$.



Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $xin[0,1]$,

beginequation
u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+int_0^1(x-t)_+[f''_n(t)-h(t)],dt.
endequation
Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.



Lemma 1. Suppose that $f_n$ are convex real-valued functions on $[0,1]$ converging pointwise to a real-valued function $f$. Then $f_nto f$ in $L^1[0,1]$.



Proof. The function $f$ is real-valued and convex and hence bounded from below. So, by Corollary 3, all the functions $f_n$ are uniformly bounded from below. On the other hand, all the convex functions $f_n$ are uniformly bounded from above by $sup_n(f_n(0)vee f_n(1))$. So, Lemma 1 follows by dominated convergence.






share|cite|improve this answer






















  • why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis
    – Dattier
    Aug 19 at 19:55











  • @Dattier : see the edited version.
    – Iosif Pinelis
    Aug 19 at 20:26










  • @NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true.
    – Iosif Pinelis
    Aug 19 at 20:49











  • @NateEldredge : Please see the latest edit.
    – Iosif Pinelis
    Aug 19 at 21:14










  • @IosifPinelis: Yep, I think we basically found the same argument :-)
    – Nate Eldredge
    Aug 19 at 21:16














up vote
3
down vote













The answer is yes. Indeed, let
$u_n:=f_n-H$ and $v:=g-H$, where
beginequation
H(x):=int_0^x(x-t)h(t),dt
endequation
for $xin[0,1]$.
Then $u_nto v$ pointwise and $u_n$ is concave for each $n$ (see the Detail below). So, by Lemma 1 below, $u_nto v$ in $L^1[0,1]$ and hence $lim int_0^1 f_n=int_0^1 g$.



Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $xin[0,1]$,

beginequation
u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+int_0^1(x-t)_+[f''_n(t)-h(t)],dt.
endequation
Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.



Lemma 1. Suppose that $f_n$ are convex real-valued functions on $[0,1]$ converging pointwise to a real-valued function $f$. Then $f_nto f$ in $L^1[0,1]$.



Proof. The function $f$ is real-valued and convex and hence bounded from below. So, by Corollary 3, all the functions $f_n$ are uniformly bounded from below. On the other hand, all the convex functions $f_n$ are uniformly bounded from above by $sup_n(f_n(0)vee f_n(1))$. So, Lemma 1 follows by dominated convergence.






share|cite|improve this answer






















  • why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis
    – Dattier
    Aug 19 at 19:55











  • @Dattier : see the edited version.
    – Iosif Pinelis
    Aug 19 at 20:26










  • @NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true.
    – Iosif Pinelis
    Aug 19 at 20:49











  • @NateEldredge : Please see the latest edit.
    – Iosif Pinelis
    Aug 19 at 21:14










  • @IosifPinelis: Yep, I think we basically found the same argument :-)
    – Nate Eldredge
    Aug 19 at 21:16












up vote
3
down vote










up vote
3
down vote









The answer is yes. Indeed, let
$u_n:=f_n-H$ and $v:=g-H$, where
beginequation
H(x):=int_0^x(x-t)h(t),dt
endequation
for $xin[0,1]$.
Then $u_nto v$ pointwise and $u_n$ is concave for each $n$ (see the Detail below). So, by Lemma 1 below, $u_nto v$ in $L^1[0,1]$ and hence $lim int_0^1 f_n=int_0^1 g$.



Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $xin[0,1]$,

beginequation
u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+int_0^1(x-t)_+[f''_n(t)-h(t)],dt.
endequation
Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.



Lemma 1. Suppose that $f_n$ are convex real-valued functions on $[0,1]$ converging pointwise to a real-valued function $f$. Then $f_nto f$ in $L^1[0,1]$.



Proof. The function $f$ is real-valued and convex and hence bounded from below. So, by Corollary 3, all the functions $f_n$ are uniformly bounded from below. On the other hand, all the convex functions $f_n$ are uniformly bounded from above by $sup_n(f_n(0)vee f_n(1))$. So, Lemma 1 follows by dominated convergence.






share|cite|improve this answer














The answer is yes. Indeed, let
$u_n:=f_n-H$ and $v:=g-H$, where
beginequation
H(x):=int_0^x(x-t)h(t),dt
endequation
for $xin[0,1]$.
Then $u_nto v$ pointwise and $u_n$ is concave for each $n$ (see the Detail below). So, by Lemma 1 below, $u_nto v$ in $L^1[0,1]$ and hence $lim int_0^1 f_n=int_0^1 g$.



Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $xin[0,1]$,

beginequation
u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+int_0^1(x-t)_+[f''_n(t)-h(t)],dt.
endequation
Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.



Lemma 1. Suppose that $f_n$ are convex real-valued functions on $[0,1]$ converging pointwise to a real-valued function $f$. Then $f_nto f$ in $L^1[0,1]$.



Proof. The function $f$ is real-valued and convex and hence bounded from below. So, by Corollary 3, all the functions $f_n$ are uniformly bounded from below. On the other hand, all the convex functions $f_n$ are uniformly bounded from above by $sup_n(f_n(0)vee f_n(1))$. So, Lemma 1 follows by dominated convergence.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 19 at 21:13

























answered Aug 19 at 19:17









Iosif Pinelis

14.9k12154




14.9k12154











  • why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis
    – Dattier
    Aug 19 at 19:55











  • @Dattier : see the edited version.
    – Iosif Pinelis
    Aug 19 at 20:26










  • @NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true.
    – Iosif Pinelis
    Aug 19 at 20:49











  • @NateEldredge : Please see the latest edit.
    – Iosif Pinelis
    Aug 19 at 21:14










  • @IosifPinelis: Yep, I think we basically found the same argument :-)
    – Nate Eldredge
    Aug 19 at 21:16
















  • why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis
    – Dattier
    Aug 19 at 19:55











  • @Dattier : see the edited version.
    – Iosif Pinelis
    Aug 19 at 20:26










  • @NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true.
    – Iosif Pinelis
    Aug 19 at 20:49











  • @NateEldredge : Please see the latest edit.
    – Iosif Pinelis
    Aug 19 at 21:14










  • @IosifPinelis: Yep, I think we basically found the same argument :-)
    – Nate Eldredge
    Aug 19 at 21:16















why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis
– Dattier
Aug 19 at 19:55





why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis
– Dattier
Aug 19 at 19:55













@Dattier : see the edited version.
– Iosif Pinelis
Aug 19 at 20:26




@Dattier : see the edited version.
– Iosif Pinelis
Aug 19 at 20:26












@NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true.
– Iosif Pinelis
Aug 19 at 20:49





@NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true.
– Iosif Pinelis
Aug 19 at 20:49













@NateEldredge : Please see the latest edit.
– Iosif Pinelis
Aug 19 at 21:14




@NateEldredge : Please see the latest edit.
– Iosif Pinelis
Aug 19 at 21:14












@IosifPinelis: Yep, I think we basically found the same argument :-)
– Nate Eldredge
Aug 19 at 21:16




@IosifPinelis: Yep, I think we basically found the same argument :-)
– Nate Eldredge
Aug 19 at 21:16

















 

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