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An integral inequality with the Hessian of a convex function defined in the unit disk
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Let $f$ be convex, twice continuously differentible on $D=(x,y);x^2+y^2leq 1$. Suppose that $fgeq 0$ on $(x,y);x^2+y^2=1$. Show that $$f(0)geq -frac1sqrtpileft[iint_D f_xxf_yy-f_xy^2right]^1/2$$ Here $f_xx$ means $partial^2f/partial xpartial y$, and etc.
At first glance, let $g(t)=f(tcos theta,tsintheta)$, and $g(0)geq g(1)+g'(1)(-1)$. What to do next? How to find the $1/2$ in the problem.
real-analysis integration convex-analysis
edited Jul 31 at 17:56
user357151
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asked Jul 31 at 3:22
xldd
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Let $f$ be convex, twice continuously differentible on $D=(x,y);x^2+y^2leq 1$. Suppose that $fgeq 0$ on $(x,y);x^2+y^2=1$. Show that $$f(0)geq -frac1sqrtpileft[iint_D f_xxf_yy-f_xy^2right]^1/2$$ Here $f_xx$ means $partial^2f/partial xpartial y$, and etc.
At first glance, let $g(t)=f(tcos theta,tsintheta)$, and $g(0)geq g(1)+g'(1)(-1)$. What to do next? How to find the $1/2$ in the problem.
real-analysis integration convex-analysis
edited Jul 31 at 17:56
user357151
13.7k31140
asked Jul 31 at 3:22
xldd
1198
unit sphere . ?
– user 108128
Jul 31 at 3:28
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f$ be convex, twice continuously differentible on $D=(x,y);x^2+y^2leq 1$. Suppose that $fgeq 0$ on $(x,y);x^2+y^2=1$. Show that $$f(0)geq -frac1sqrtpileft[iint_D f_xxf_yy-f_xy^2right]^1/2$$ Here $f_xx$ means $partial^2f/partial xpartial y$, and etc.
At first glance, let $g(t)=f(tcos theta,tsintheta)$, and $g(0)geq g(1)+g'(1)(-1)$. What to do next? How to find the $1/2$ in the problem.
real-analysis integration convex-analysis
edited Jul 31 at 17:56
user357151
13.7k31140
asked Jul 31 at 3:22
xldd
1198
Let $f$ be convex, twice continuously differentible on $D=(x,y);x^2+y^2leq 1$. Suppose that $fgeq 0$ on $(x,y);x^2+y^2=1$. Show that $$f(0)geq -frac1sqrtpileft[iint_D f_xxf_yy-f_xy^2right]^1/2$$ Here $f_xx$ means $partial^2f/partial xpartial y$, and etc.
At first glance, let $g(t)=f(tcos theta,tsintheta)$, and $g(0)geq g(1)+g'(1)(-1)$. What to do next? How to find the $1/2$ in the problem.
real-analysis integration convex-analysis
edited Jul 31 at 17:56
user357151
13.7k31140
asked Jul 31 at 3:22
xldd
1198
edited Jul 31 at 17:56
user357151
13.7k31140
edited Jul 31 at 17:56
user357151
13.7k31140
edited Jul 31 at 17:56
user357151
13.7k31140
13.7k31140
asked Jul 31 at 3:22
xldd
1198
asked Jul 31 at 3:22
xldd
1198
asked Jul 31 at 3:22
xldd
1198
1198
unit sphere . ?
– user 108128
Jul 31 at 3:28
add a comment |Â
unit sphere . ?
– user 108128
Jul 31 at 3:28
unit sphere . ?
– user 108128
Jul 31 at 3:28
unit sphere . ?
– user 108128
Jul 31 at 3:28
add a comment |Â
1 Answer
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This is proved by a geometric argument which usually appears along with the names Alexandrov, Bakelman, and Pucci. The idea is to look at the gradient $nabla u$ as a map. Let $Omega$ be the image of $D$ under this map.
Let $M = -f(0)$. For every vector $xi$ with $|xi|<M$ the function $v(x) = f(x)-xicdot x$ is convex in $D$ and satisfies
$$
min_partial D v ge -|xi| > -M = v(0)
$$
Therefore $v$ attains its minimum in $D$, which implies $nabla f(x) = xi$ for some $x$. We conclude that $Omega$ contains an open disk of radius $M$. Thus, the area of $Omega$ is at least $pi M^2$.
On the other hand, this area can be computed by integrating the Jacobian of the gradient map, which is the Hessian $det D^2 f$ (nonnegative by the convexity of $f$):
$$
pi M^2 le |Omega| = iint_D det D^2 f
$$
which proves the claim.
Section 9.1 of Gilbarg-Trudinger has more general results.
answered Jul 31 at 17:55
user357151
13.7k31140
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This is proved by a geometric argument which usually appears along with the names Alexandrov, Bakelman, and Pucci. The idea is to look at the gradient $nabla u$ as a map. Let $Omega$ be the image of $D$ under this map.
Let $M = -f(0)$. For every vector $xi$ with $|xi|<M$ the function $v(x) = f(x)-xicdot x$ is convex in $D$ and satisfies
$$
min_partial D v ge -|xi| > -M = v(0)
$$
Therefore $v$ attains its minimum in $D$, which implies $nabla f(x) = xi$ for some $x$. We conclude that $Omega$ contains an open disk of radius $M$. Thus, the area of $Omega$ is at least $pi M^2$.
On the other hand, this area can be computed by integrating the Jacobian of the gradient map, which is the Hessian $det D^2 f$ (nonnegative by the convexity of $f$):
$$
pi M^2 le |Omega| = iint_D det D^2 f
$$
which proves the claim.
Section 9.1 of Gilbarg-Trudinger has more general results.
answered Jul 31 at 17:55
user357151
13.7k31140
add a comment |Â
up vote
0
down vote
accepted
This is proved by a geometric argument which usually appears along with the names Alexandrov, Bakelman, and Pucci. The idea is to look at the gradient $nabla u$ as a map. Let $Omega$ be the image of $D$ under this map.
Let $M = -f(0)$. For every vector $xi$ with $|xi|<M$ the function $v(x) = f(x)-xicdot x$ is convex in $D$ and satisfies
$$
min_partial D v ge -|xi| > -M = v(0)
$$
Therefore $v$ attains its minimum in $D$, which implies $nabla f(x) = xi$ for some $x$. We conclude that $Omega$ contains an open disk of radius $M$. Thus, the area of $Omega$ is at least $pi M^2$.
On the other hand, this area can be computed by integrating the Jacobian of the gradient map, which is the Hessian $det D^2 f$ (nonnegative by the convexity of $f$):
$$
pi M^2 le |Omega| = iint_D det D^2 f
$$
which proves the claim.
Section 9.1 of Gilbarg-Trudinger has more general results.
answered Jul 31 at 17:55
user357151
13.7k31140
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This is proved by a geometric argument which usually appears along with the names Alexandrov, Bakelman, and Pucci. The idea is to look at the gradient $nabla u$ as a map. Let $Omega$ be the image of $D$ under this map.
Let $M = -f(0)$. For every vector $xi$ with $|xi|<M$ the function $v(x) = f(x)-xicdot x$ is convex in $D$ and satisfies
$$
min_partial D v ge -|xi| > -M = v(0)
$$
Therefore $v$ attains its minimum in $D$, which implies $nabla f(x) = xi$ for some $x$. We conclude that $Omega$ contains an open disk of radius $M$. Thus, the area of $Omega$ is at least $pi M^2$.
On the other hand, this area can be computed by integrating the Jacobian of the gradient map, which is the Hessian $det D^2 f$ (nonnegative by the convexity of $f$):
$$
pi M^2 le |Omega| = iint_D det D^2 f
$$
which proves the claim.
Section 9.1 of Gilbarg-Trudinger has more general results.
answered Jul 31 at 17:55
user357151
13.7k31140
This is proved by a geometric argument which usually appears along with the names Alexandrov, Bakelman, and Pucci. The idea is to look at the gradient $nabla u$ as a map. Let $Omega$ be the image of $D$ under this map.
Let $M = -f(0)$. For every vector $xi$ with $|xi|<M$ the function $v(x) = f(x)-xicdot x$ is convex in $D$ and satisfies
$$
min_partial D v ge -|xi| > -M = v(0)
$$
Therefore $v$ attains its minimum in $D$, which implies $nabla f(x) = xi$ for some $x$. We conclude that $Omega$ contains an open disk of radius $M$. Thus, the area of $Omega$ is at least $pi M^2$.
On the other hand, this area can be computed by integrating the Jacobian of the gradient map, which is the Hessian $det D^2 f$ (nonnegative by the convexity of $f$):
$$
pi M^2 le |Omega| = iint_D det D^2 f
$$
which proves the claim.
Section 9.1 of Gilbarg-Trudinger has more general results.
answered Jul 31 at 17:55
user357151
13.7k31140
answered Jul 31 at 17:55
user357151
13.7k31140
answered Jul 31 at 17:55
user357151
13.7k31140
answered Jul 31 at 17:55
user357151
13.7k31140
13.7k31140
add a comment |Â
add a comment |Â
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unit sphere . ?
– user 108128
Jul 31 at 3:28