Mixing
Converse of Jensen's inequality
Suppose $varphi:mathbb{R}rightarrowmathbb{R}$ and for all bounded measurable $f$,
$$
varphiBig(int_0^1fdlambdaBig) le int_0^1varphi(f)dlambda
$$
I'm asked to prove that $varphi$ is a convex function.
I have no idea how to even begin, only idea I've had is to try to suppose that $varphi''(x)<0$ for some $xin(0,1)$ but then I haven't got a clue.
measure-theory lebesgue-integral
edited Nov 21 '18 at 18:48
Federico
4,799514
asked Nov 21 '18 at 15:58
D. Brito
358111
add a comment |
Suppose $varphi:mathbb{R}rightarrowmathbb{R}$ and for all bounded measurable $f$,
$$
varphiBig(int_0^1fdlambdaBig) le int_0^1varphi(f)dlambda
$$
I'm asked to prove that $varphi$ is a convex function.
I have no idea how to even begin, only idea I've had is to try to suppose that $varphi''(x)<0$ for some $xin(0,1)$ but then I haven't got a clue.
measure-theory lebesgue-integral
edited Nov 21 '18 at 18:48
Federico
4,799514
asked Nov 21 '18 at 15:58
D. Brito
358111
1
This is much simpler than the Jensen's inequality itself. Just recall the definition of convexity, and try to pick your $f$ to arrive at it exactly.
– metamorphy
Nov 21 '18 at 16:13
add a comment |
Suppose $varphi:mathbb{R}rightarrowmathbb{R}$ and for all bounded measurable $f$,
$$
varphiBig(int_0^1fdlambdaBig) le int_0^1varphi(f)dlambda
$$
I'm asked to prove that $varphi$ is a convex function.
I have no idea how to even begin, only idea I've had is to try to suppose that $varphi''(x)<0$ for some $xin(0,1)$ but then I haven't got a clue.
measure-theory lebesgue-integral
edited Nov 21 '18 at 18:48
Federico
4,799514
asked Nov 21 '18 at 15:58
D. Brito
358111
Suppose $varphi:mathbb{R}rightarrowmathbb{R}$ and for all bounded measurable $f$,
$$
varphiBig(int_0^1fdlambdaBig) le int_0^1varphi(f)dlambda
$$
I'm asked to prove that $varphi$ is a convex function.
I have no idea how to even begin, only idea I've had is to try to suppose that $varphi''(x)<0$ for some $xin(0,1)$ but then I haven't got a clue.
measure-theory lebesgue-integral
measure-theory lebesgue-integral
edited Nov 21 '18 at 18:48
Federico
4,799514
asked Nov 21 '18 at 15:58
D. Brito
358111
edited Nov 21 '18 at 18:48
Federico
4,799514
asked Nov 21 '18 at 15:58
D. Brito
358111
edited Nov 21 '18 at 18:48
Federico
4,799514
edited Nov 21 '18 at 18:48
Federico
4,799514
edited Nov 21 '18 at 18:48
Federico
4,799514
4,799514
asked Nov 21 '18 at 15:58
D. Brito
358111
asked Nov 21 '18 at 15:58
D. Brito
358111
asked Nov 21 '18 at 15:58
D. Brito
358111
358111
1
This is much simpler than the Jensen's inequality itself. Just recall the definition of convexity, and try to pick your $f$ to arrive at it exactly.
– metamorphy
Nov 21 '18 at 16:13
add a comment |
1
This is much simpler than the Jensen's inequality itself. Just recall the definition of convexity, and try to pick your $f$ to arrive at it exactly.
– metamorphy
Nov 21 '18 at 16:13
1
1
This is much simpler than the Jensen's inequality itself. Just recall the definition of convexity, and try to pick your $f$ to arrive at it exactly.
– metamorphy
Nov 21 '18 at 16:13
This is much simpler than the Jensen's inequality itself. Just recall the definition of convexity, and try to pick your $f$ to arrive at it exactly.
– metamorphy
Nov 21 '18 at 16:13
add a comment |
1 Answer
1
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oldest
votes
Given $x,yinmathbb R$ and $tin(0,1)$, consider
$$
f(s) = begin{cases} x & sleq t \ y & s>t. end{cases}
$$
Then your inequality tells
$$
phi(tx+(1-t)y) leq tphi(x)+(1-t)phi(y).
$$
answered Nov 21 '18 at 18:47
Federico
4,799514
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Given $x,yinmathbb R$ and $tin(0,1)$, consider
$$
f(s) = begin{cases} x & sleq t \ y & s>t. end{cases}
$$
Then your inequality tells
$$
phi(tx+(1-t)y) leq tphi(x)+(1-t)phi(y).
$$
answered Nov 21 '18 at 18:47
Federico
4,799514
add a comment |
Given $x,yinmathbb R$ and $tin(0,1)$, consider
$$
f(s) = begin{cases} x & sleq t \ y & s>t. end{cases}
$$
Then your inequality tells
$$
phi(tx+(1-t)y) leq tphi(x)+(1-t)phi(y).
$$
answered Nov 21 '18 at 18:47
Federico
4,799514
add a comment |
Given $x,yinmathbb R$ and $tin(0,1)$, consider
$$
f(s) = begin{cases} x & sleq t \ y & s>t. end{cases}
$$
Then your inequality tells
$$
phi(tx+(1-t)y) leq tphi(x)+(1-t)phi(y).
$$
answered Nov 21 '18 at 18:47
Federico
4,799514
Given $x,yinmathbb R$ and $tin(0,1)$, consider
$$
f(s) = begin{cases} x & sleq t \ y & s>t. end{cases}
$$
Then your inequality tells
$$
phi(tx+(1-t)y) leq tphi(x)+(1-t)phi(y).
$$
answered Nov 21 '18 at 18:47
Federico
4,799514
answered Nov 21 '18 at 18:47
Federico
4,799514
answered Nov 21 '18 at 18:47
Federico
4,799514
answered Nov 21 '18 at 18:47
Federico
4,799514
4,799514
add a comment |
add a comment |
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1
This is much simpler than the Jensen's inequality itself. Just recall the definition of convexity, and try to pick your $f$ to arrive at it exactly.
– metamorphy
Nov 21 '18 at 16:13