Showing a function is convex by looking at the hessian.
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Let $f: mathbb{R}^n rightarrow mathbb{R}$ be a function that is twice-differentiable almost everywhere. That is, $f$ is differentiable almost everywhere, and its $nabla f$ is also differentiable almost everywhere. Furthermore $nabla^2 f = 0$ where defined, and so the derivatives of $nabla f$ agrees as you approach the non-differential measure zero set from any direction.
My end goal is to show that $f$ is convex. The approach I'm taking is showing that the Hessian is PSD. If $f$ were twice differentiable everywhere, this would be sufficient. However $f$ is not even differentiable everywhere, it is differentiable almost everywhere, and where its Hessian is defined $nabla^2 f = 0$. Does the conclusion that $f$ is convex still hold?
multivariable-calculus derivatives convex-analysis convex-optimization
asked Dec 16 '18 at 1:31
user395788user395788
405
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^n rightarrow mathbb{R}$ be a function that is twice-differentiable almost everywhere. That is, $f$ is differentiable almost everywhere, and its $nabla f$ is also differentiable almost everywhere. Furthermore $nabla^2 f = 0$ where defined, and so the derivatives of $nabla f$ agrees as you approach the non-differential measure zero set from any direction.
My end goal is to show that $f$ is convex. The approach I'm taking is showing that the Hessian is PSD. If $f$ were twice differentiable everywhere, this would be sufficient. However $f$ is not even differentiable everywhere, it is differentiable almost everywhere, and where its Hessian is defined $nabla^2 f = 0$. Does the conclusion that $f$ is convex still hold?
multivariable-calculus derivatives convex-analysis convex-optimization
asked Dec 16 '18 at 1:31
user395788user395788
405
$endgroup$
1
$begingroup$
is the Heaviside step function a counterexample?
$endgroup$
– LinAlg
Dec 16 '18 at 3:02
$begingroup$
Isn't $f:xmapsto -|x|$ a concave non-convex function $mathbb Rtomathbb R$ that satisfies your conditions?
$endgroup$
– kimchi lover
Dec 16 '18 at 3:41
add a comment |
$begingroup$
Let $f: mathbb{R}^n rightarrow mathbb{R}$ be a function that is twice-differentiable almost everywhere. That is, $f$ is differentiable almost everywhere, and its $nabla f$ is also differentiable almost everywhere. Furthermore $nabla^2 f = 0$ where defined, and so the derivatives of $nabla f$ agrees as you approach the non-differential measure zero set from any direction.
My end goal is to show that $f$ is convex. The approach I'm taking is showing that the Hessian is PSD. If $f$ were twice differentiable everywhere, this would be sufficient. However $f$ is not even differentiable everywhere, it is differentiable almost everywhere, and where its Hessian is defined $nabla^2 f = 0$. Does the conclusion that $f$ is convex still hold?
multivariable-calculus derivatives convex-analysis convex-optimization
asked Dec 16 '18 at 1:31
user395788user395788
405
$endgroup$
Let $f: mathbb{R}^n rightarrow mathbb{R}$ be a function that is twice-differentiable almost everywhere. That is, $f$ is differentiable almost everywhere, and its $nabla f$ is also differentiable almost everywhere. Furthermore $nabla^2 f = 0$ where defined, and so the derivatives of $nabla f$ agrees as you approach the non-differential measure zero set from any direction.
My end goal is to show that $f$ is convex. The approach I'm taking is showing that the Hessian is PSD. If $f$ were twice differentiable everywhere, this would be sufficient. However $f$ is not even differentiable everywhere, it is differentiable almost everywhere, and where its Hessian is defined $nabla^2 f = 0$. Does the conclusion that $f$ is convex still hold?
multivariable-calculus derivatives convex-analysis convex-optimization
multivariable-calculus derivatives convex-analysis convex-optimization
asked Dec 16 '18 at 1:31
user395788user395788
405
asked Dec 16 '18 at 1:31
user395788user395788
405
asked Dec 16 '18 at 1:31
user395788user395788
405
asked Dec 16 '18 at 1:31
user395788user395788
405
asked Dec 16 '18 at 1:31
user395788user395788
405
405
1
$begingroup$
is the Heaviside step function a counterexample?
$endgroup$
– LinAlg
Dec 16 '18 at 3:02
$begingroup$
Isn't $f:xmapsto -|x|$ a concave non-convex function $mathbb Rtomathbb R$ that satisfies your conditions?
$endgroup$
– kimchi lover
Dec 16 '18 at 3:41
add a comment |
1
$begingroup$
is the Heaviside step function a counterexample?
$endgroup$
– LinAlg
Dec 16 '18 at 3:02
$begingroup$
Isn't $f:xmapsto -|x|$ a concave non-convex function $mathbb Rtomathbb R$ that satisfies your conditions?
$endgroup$
– kimchi lover
Dec 16 '18 at 3:41
1
1
$begingroup$
is the Heaviside step function a counterexample?
$endgroup$
– LinAlg
Dec 16 '18 at 3:02
$begingroup$
is the Heaviside step function a counterexample?
$endgroup$
– LinAlg
Dec 16 '18 at 3:02
$begingroup$
Isn't $f:xmapsto -|x|$ a concave non-convex function $mathbb Rtomathbb R$ that satisfies your conditions?
$endgroup$
– kimchi lover
Dec 16 '18 at 3:41
$begingroup$
Isn't $f:xmapsto -|x|$ a concave non-convex function $mathbb Rtomathbb R$ that satisfies your conditions?
$endgroup$
– kimchi lover
Dec 16 '18 at 3:41
add a comment |
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1
$begingroup$
is the Heaviside step function a counterexample?
$endgroup$
– LinAlg
Dec 16 '18 at 3:02
$begingroup$
Isn't $f:xmapsto -|x|$ a concave non-convex function $mathbb Rtomathbb R$ that satisfies your conditions?
$endgroup$
– kimchi lover
Dec 16 '18 at 3:41