Converse of Jensen's inequality
up vote
1
down vote
favorite
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 at 18:48
Federico
2,367510
asked Nov 21 at 15:58
D. Brito
345110
add a comment |
up vote
1
down vote
favorite
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 at 18:48
Federico
2,367510
asked Nov 21 at 15:58
D. Brito
345110
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 at 16:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 18:48
Federico
2,367510
asked Nov 21 at 15:58
D. Brito
345110
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 at 18:48
Federico
2,367510
asked Nov 21 at 15:58
D. Brito
345110
edited Nov 21 at 18:48
Federico
2,367510
asked Nov 21 at 15:58
D. Brito
345110
edited Nov 21 at 18:48
Federico
2,367510
edited Nov 21 at 18:48
Federico
2,367510
edited Nov 21 at 18:48
Federico
2,367510
2,367510
asked Nov 21 at 15:58
D. Brito
345110
asked Nov 21 at 15:58
D. Brito
345110
asked Nov 21 at 15:58
D. Brito
345110
345110
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 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 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 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 at 16:13
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
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 at 18:47
Federico
2,367510
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
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 at 18:47
Federico
2,367510
add a comment |
up vote
1
down vote
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 at 18:47
Federico
2,367510
add a comment |
up vote
1
down vote
up vote
1
down vote
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 at 18:47
Federico
2,367510
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 at 18:47
Federico
2,367510
answered Nov 21 at 18:47
Federico
2,367510
answered Nov 21 at 18:47
Federico
2,367510
answered Nov 21 at 18:47
Federico
2,367510
2,367510
add a comment |
add a comment |
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%2f3007917%2fconverse-of-jensens-inequality%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
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 at 16:13