Can Jensen be applied to multivariable functions?
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I was wondering, suppose we have a symmetric two variable function, in my case: $f(x,y)=frac{1}{x+y+1}$ with the restriction that $0le x,y le 1$, I took the second partial derivative with respect to $x$ which is $frac{2}{(x+y+1)^3}$, since the function is symmetric I didn't bother taking the partial derivative with respect to $y$, so we see that $f$ is convex in $(0,+infty)$, now I was wondering, given $0le a,b,c le 1$, can I argue that:$$f(a,b)+f(b,c)+f(c,a)ge 3f(frac{a+b+c}{3},frac{a+b+c}{3})$$
Or is this completely non-sense?
jensen-inequality
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
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add a comment |
$begingroup$
I was wondering, suppose we have a symmetric two variable function, in my case: $f(x,y)=frac{1}{x+y+1}$ with the restriction that $0le x,y le 1$, I took the second partial derivative with respect to $x$ which is $frac{2}{(x+y+1)^3}$, since the function is symmetric I didn't bother taking the partial derivative with respect to $y$, so we see that $f$ is convex in $(0,+infty)$, now I was wondering, given $0le a,b,c le 1$, can I argue that:$$f(a,b)+f(b,c)+f(c,a)ge 3f(frac{a+b+c}{3},frac{a+b+c}{3})$$
Or is this completely non-sense?
jensen-inequality
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
$endgroup$
add a comment |
$begingroup$
I was wondering, suppose we have a symmetric two variable function, in my case: $f(x,y)=frac{1}{x+y+1}$ with the restriction that $0le x,y le 1$, I took the second partial derivative with respect to $x$ which is $frac{2}{(x+y+1)^3}$, since the function is symmetric I didn't bother taking the partial derivative with respect to $y$, so we see that $f$ is convex in $(0,+infty)$, now I was wondering, given $0le a,b,c le 1$, can I argue that:$$f(a,b)+f(b,c)+f(c,a)ge 3f(frac{a+b+c}{3},frac{a+b+c}{3})$$
Or is this completely non-sense?
jensen-inequality
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
$endgroup$
I was wondering, suppose we have a symmetric two variable function, in my case: $f(x,y)=frac{1}{x+y+1}$ with the restriction that $0le x,y le 1$, I took the second partial derivative with respect to $x$ which is $frac{2}{(x+y+1)^3}$, since the function is symmetric I didn't bother taking the partial derivative with respect to $y$, so we see that $f$ is convex in $(0,+infty)$, now I was wondering, given $0le a,b,c le 1$, can I argue that:$$f(a,b)+f(b,c)+f(c,a)ge 3f(frac{a+b+c}{3},frac{a+b+c}{3})$$
Or is this completely non-sense?
jensen-inequality
jensen-inequality
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
asked Dec 23 '18 at 13:41
Spasoje DurovicSpasoje Durovic
36010
36010
add a comment |
add a comment |
1 Answer
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You have the function $$g(rho):={1over 1+rho}qquad(rhogeq0) ,$$
which is certainly convex, and then define $f(x,y):=g(x+y)$. It follows that
$$eqalign{{f(a,b)+f(b,c)+f(c,a)over3}&={g(a+b)+g(b+c)+g(c+a)over3}geq gleft({2a+2b+2cover3}right)cr
&=fleft({a+b+cover3},{a+b+cover3}right) .cr}$$Therefore your final statement is correct.
But you can also prove generally that $f$ is convex. Write $(x,y)=:z$ and $phi(z):=x+y$. Then $f=gcircphi$, and
$$eqalign{(1-lambda)f(z)+lambda f(z')&=(1-lambda)gbigl(phi(z)bigr)+lambda gbigl(phi(z')bigr)cr &geq gbigl((1-lambda)phi(z)+lambdaphi(z')bigr)=gbigl(phi((1-lambda)z+lambda z')bigr)cr &=fbigl((1-lambda)z+lambda z'bigr) .cr}$$
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
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1 Answer
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$begingroup$
You have the function $$g(rho):={1over 1+rho}qquad(rhogeq0) ,$$
which is certainly convex, and then define $f(x,y):=g(x+y)$. It follows that
$$eqalign{{f(a,b)+f(b,c)+f(c,a)over3}&={g(a+b)+g(b+c)+g(c+a)over3}geq gleft({2a+2b+2cover3}right)cr
&=fleft({a+b+cover3},{a+b+cover3}right) .cr}$$Therefore your final statement is correct.
But you can also prove generally that $f$ is convex. Write $(x,y)=:z$ and $phi(z):=x+y$. Then $f=gcircphi$, and
$$eqalign{(1-lambda)f(z)+lambda f(z')&=(1-lambda)gbigl(phi(z)bigr)+lambda gbigl(phi(z')bigr)cr &geq gbigl((1-lambda)phi(z)+lambdaphi(z')bigr)=gbigl(phi((1-lambda)z+lambda z')bigr)cr &=fbigl((1-lambda)z+lambda z'bigr) .cr}$$
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
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add a comment |
$begingroup$
You have the function $$g(rho):={1over 1+rho}qquad(rhogeq0) ,$$
which is certainly convex, and then define $f(x,y):=g(x+y)$. It follows that
$$eqalign{{f(a,b)+f(b,c)+f(c,a)over3}&={g(a+b)+g(b+c)+g(c+a)over3}geq gleft({2a+2b+2cover3}right)cr
&=fleft({a+b+cover3},{a+b+cover3}right) .cr}$$Therefore your final statement is correct.
But you can also prove generally that $f$ is convex. Write $(x,y)=:z$ and $phi(z):=x+y$. Then $f=gcircphi$, and
$$eqalign{(1-lambda)f(z)+lambda f(z')&=(1-lambda)gbigl(phi(z)bigr)+lambda gbigl(phi(z')bigr)cr &geq gbigl((1-lambda)phi(z)+lambdaphi(z')bigr)=gbigl(phi((1-lambda)z+lambda z')bigr)cr &=fbigl((1-lambda)z+lambda z'bigr) .cr}$$
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
You have the function $$g(rho):={1over 1+rho}qquad(rhogeq0) ,$$
which is certainly convex, and then define $f(x,y):=g(x+y)$. It follows that
$$eqalign{{f(a,b)+f(b,c)+f(c,a)over3}&={g(a+b)+g(b+c)+g(c+a)over3}geq gleft({2a+2b+2cover3}right)cr
&=fleft({a+b+cover3},{a+b+cover3}right) .cr}$$Therefore your final statement is correct.
But you can also prove generally that $f$ is convex. Write $(x,y)=:z$ and $phi(z):=x+y$. Then $f=gcircphi$, and
$$eqalign{(1-lambda)f(z)+lambda f(z')&=(1-lambda)gbigl(phi(z)bigr)+lambda gbigl(phi(z')bigr)cr &geq gbigl((1-lambda)phi(z)+lambdaphi(z')bigr)=gbigl(phi((1-lambda)z+lambda z')bigr)cr &=fbigl((1-lambda)z+lambda z'bigr) .cr}$$
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
$endgroup$
You have the function $$g(rho):={1over 1+rho}qquad(rhogeq0) ,$$
which is certainly convex, and then define $f(x,y):=g(x+y)$. It follows that
$$eqalign{{f(a,b)+f(b,c)+f(c,a)over3}&={g(a+b)+g(b+c)+g(c+a)over3}geq gleft({2a+2b+2cover3}right)cr
&=fleft({a+b+cover3},{a+b+cover3}right) .cr}$$Therefore your final statement is correct.
But you can also prove generally that $f$ is convex. Write $(x,y)=:z$ and $phi(z):=x+y$. Then $f=gcircphi$, and
$$eqalign{(1-lambda)f(z)+lambda f(z')&=(1-lambda)gbigl(phi(z)bigr)+lambda gbigl(phi(z')bigr)cr &geq gbigl((1-lambda)phi(z)+lambdaphi(z')bigr)=gbigl(phi((1-lambda)z+lambda z')bigr)cr &=fbigl((1-lambda)z+lambda z'bigr) .cr}$$
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
answered Dec 23 '18 at 16:35
Christian BlatterChristian Blatter
173k7113326
173k7113326
add a comment |
add a comment |
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