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Issue with proof by contradiction of necessary second-order condition for convexity
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We wish to prove that if $f : K to Bbb R$ is a $mathcal C^2$, convex function on a convex domain $K subset Bbb R^n$, then it has a positive semidefinite Hessian matrix on every point $x in K$.
To do this, I have a proof by contradiction: we assume there's a point $bar x in K$ where the Hessian is not positive semidefinite. That should mean
$$
exists y in Bbb R^n : y^T nabla^2f(bar x) y < 0
$$
but then instead the proof seems to go on implying that $bar x^Tnabla^2f(bar x)bar x < 0$, while this does not seem true in general.
Or is it?
And hence how would you write such a proof (by contradiction)?
multivariable-calculus convex-analysis positive-semidefinite
asked Jul 28 at 12:16
mattecapu
667617
add a comment |Â
up vote
-1
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favorite
We wish to prove that if $f : K to Bbb R$ is a $mathcal C^2$, convex function on a convex domain $K subset Bbb R^n$, then it has a positive semidefinite Hessian matrix on every point $x in K$.
To do this, I have a proof by contradiction: we assume there's a point $bar x in K$ where the Hessian is not positive semidefinite. That should mean
$$
exists y in Bbb R^n : y^T nabla^2f(bar x) y < 0
$$
but then instead the proof seems to go on implying that $bar x^Tnabla^2f(bar x)bar x < 0$, while this does not seem true in general.
Or is it?
And hence how would you write such a proof (by contradiction)?
multivariable-calculus convex-analysis positive-semidefinite
asked Jul 28 at 12:16
mattecapu
667617
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
We wish to prove that if $f : K to Bbb R$ is a $mathcal C^2$, convex function on a convex domain $K subset Bbb R^n$, then it has a positive semidefinite Hessian matrix on every point $x in K$.
To do this, I have a proof by contradiction: we assume there's a point $bar x in K$ where the Hessian is not positive semidefinite. That should mean
$$
exists y in Bbb R^n : y^T nabla^2f(bar x) y < 0
$$
but then instead the proof seems to go on implying that $bar x^Tnabla^2f(bar x)bar x < 0$, while this does not seem true in general.
Or is it?
And hence how would you write such a proof (by contradiction)?
multivariable-calculus convex-analysis positive-semidefinite
asked Jul 28 at 12:16
mattecapu
667617
We wish to prove that if $f : K to Bbb R$ is a $mathcal C^2$, convex function on a convex domain $K subset Bbb R^n$, then it has a positive semidefinite Hessian matrix on every point $x in K$.
To do this, I have a proof by contradiction: we assume there's a point $bar x in K$ where the Hessian is not positive semidefinite. That should mean
$$
exists y in Bbb R^n : y^T nabla^2f(bar x) y < 0
$$
but then instead the proof seems to go on implying that $bar x^Tnabla^2f(bar x)bar x < 0$, while this does not seem true in general.
Or is it?
And hence how would you write such a proof (by contradiction)?
multivariable-calculus convex-analysis positive-semidefinite
asked Jul 28 at 12:16
mattecapu
667617
edited Jul 28 at 12:26
asked Jul 28 at 12:16
mattecapu
667617
asked Jul 28 at 12:16
mattecapu
667617
asked Jul 28 at 12:16
mattecapu
667617
667617
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
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up vote
-1
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I think the problem is a subtle issue with the meaning of "positive semidefinite Hessian matrix on every point $x in K$".
The actual theorem should be:
Theorem. If $f : K to Bbb R$ is twice differentiable, convex function on the convex domain $K subseteq Bbb R^n$, then for every point $x in K$, the Hessian matrix $nabla^2 f(x)$ is positive semidefinite on every point $x in K$:
$$
forall y in K, quad y^Tnabla^2f(x)y geq 0
$$
So that the definiteness condition is stated for $K$. Now the negation of this fact is precisely that there exists an $bar x in K$ such that
$$
exists y in K, quad y^Tnabla^2(bar x)y < 0.
$$
answered Jul 28 at 12:26
mattecapu
667617
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
I think the problem is a subtle issue with the meaning of "positive semidefinite Hessian matrix on every point $x in K$".
The actual theorem should be:
Theorem. If $f : K to Bbb R$ is twice differentiable, convex function on the convex domain $K subseteq Bbb R^n$, then for every point $x in K$, the Hessian matrix $nabla^2 f(x)$ is positive semidefinite on every point $x in K$:
$$
forall y in K, quad y^Tnabla^2f(x)y geq 0
$$
So that the definiteness condition is stated for $K$. Now the negation of this fact is precisely that there exists an $bar x in K$ such that
$$
exists y in K, quad y^Tnabla^2(bar x)y < 0.
$$
answered Jul 28 at 12:26
mattecapu
667617
add a comment |Â
up vote
-1
down vote
I think the problem is a subtle issue with the meaning of "positive semidefinite Hessian matrix on every point $x in K$".
The actual theorem should be:
Theorem. If $f : K to Bbb R$ is twice differentiable, convex function on the convex domain $K subseteq Bbb R^n$, then for every point $x in K$, the Hessian matrix $nabla^2 f(x)$ is positive semidefinite on every point $x in K$:
$$
forall y in K, quad y^Tnabla^2f(x)y geq 0
$$
So that the definiteness condition is stated for $K$. Now the negation of this fact is precisely that there exists an $bar x in K$ such that
$$
exists y in K, quad y^Tnabla^2(bar x)y < 0.
$$
answered Jul 28 at 12:26
mattecapu
667617
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
I think the problem is a subtle issue with the meaning of "positive semidefinite Hessian matrix on every point $x in K$".
The actual theorem should be:
Theorem. If $f : K to Bbb R$ is twice differentiable, convex function on the convex domain $K subseteq Bbb R^n$, then for every point $x in K$, the Hessian matrix $nabla^2 f(x)$ is positive semidefinite on every point $x in K$:
$$
forall y in K, quad y^Tnabla^2f(x)y geq 0
$$
So that the definiteness condition is stated for $K$. Now the negation of this fact is precisely that there exists an $bar x in K$ such that
$$
exists y in K, quad y^Tnabla^2(bar x)y < 0.
$$
answered Jul 28 at 12:26
mattecapu
667617
I think the problem is a subtle issue with the meaning of "positive semidefinite Hessian matrix on every point $x in K$".
The actual theorem should be:
Theorem. If $f : K to Bbb R$ is twice differentiable, convex function on the convex domain $K subseteq Bbb R^n$, then for every point $x in K$, the Hessian matrix $nabla^2 f(x)$ is positive semidefinite on every point $x in K$:
$$
forall y in K, quad y^Tnabla^2f(x)y geq 0
$$
So that the definiteness condition is stated for $K$. Now the negation of this fact is precisely that there exists an $bar x in K$ such that
$$
exists y in K, quad y^Tnabla^2(bar x)y < 0.
$$
answered Jul 28 at 12:26
mattecapu
667617
answered Jul 28 at 12:26
mattecapu
667617
answered Jul 28 at 12:26
mattecapu
667617
answered Jul 28 at 12:26
mattecapu
667617
667617
add a comment |Â
add a comment |Â
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