Showing that the quadratic over linear function is convex?
$begingroup$
Let $f(x,y) = frac{x^2}{y}$ be the quadratic-over-linear function.
To show it's convex over $Bbb R times Bbb R_{++}$, we have to show that $nabla^2f$ is positive semi definite.I see that
$$nabla f(x,y) =
begin{bmatrix}2x/y\-x^2/y^2end{bmatrix},
$$
but the book is saying that
$$
nabla^2 f(x,y)=frac{2}{y^3}begin{bmatrix}y^2 &-xy\-xy & x^2end{bmatrix}.
$$
But isn't it the case that $nabla^2f = nabla * nabla f$ where $*$ means dot product and therefore $nabla^2f = 2/y + 2x^2 / y^3$?
What have I misunderstood here?
convex-analysis proof-explanation
edited Jan 22 at 16:22
A.Γ.
22.8k32656
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
$endgroup$
add a comment |
$begingroup$
Let $f(x,y) = frac{x^2}{y}$ be the quadratic-over-linear function.
To show it's convex over $Bbb R times Bbb R_{++}$, we have to show that $nabla^2f$ is positive semi definite.I see that
$$nabla f(x,y) =
begin{bmatrix}2x/y\-x^2/y^2end{bmatrix},
$$
but the book is saying that
$$
nabla^2 f(x,y)=frac{2}{y^3}begin{bmatrix}y^2 &-xy\-xy & x^2end{bmatrix}.
$$
But isn't it the case that $nabla^2f = nabla * nabla f$ where $*$ means dot product and therefore $nabla^2f = 2/y + 2x^2 / y^3$?
What have I misunderstood here?
convex-analysis proof-explanation
edited Jan 22 at 16:22
A.Γ.
22.8k32656
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
$endgroup$
1
$begingroup$
There seems to be some ambiguity of notation for $nabla^2$ as either the "Laplace operator" or the "Hessian matrix," see here: en.wikipedia.org/wiki/Del_squared [In your case you want the Hessian matrix.] Note that it only makes sense for a square matrix to be "positive semi-definite."
$endgroup$
– Michael
Jan 20 at 16:05
add a comment |
$begingroup$
Let $f(x,y) = frac{x^2}{y}$ be the quadratic-over-linear function.
To show it's convex over $Bbb R times Bbb R_{++}$, we have to show that $nabla^2f$ is positive semi definite.I see that
$$nabla f(x,y) =
begin{bmatrix}2x/y\-x^2/y^2end{bmatrix},
$$
but the book is saying that
$$
nabla^2 f(x,y)=frac{2}{y^3}begin{bmatrix}y^2 &-xy\-xy & x^2end{bmatrix}.
$$
But isn't it the case that $nabla^2f = nabla * nabla f$ where $*$ means dot product and therefore $nabla^2f = 2/y + 2x^2 / y^3$?
What have I misunderstood here?
convex-analysis proof-explanation
edited Jan 22 at 16:22
A.Γ.
22.8k32656
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
$endgroup$
Let $f(x,y) = frac{x^2}{y}$ be the quadratic-over-linear function.
To show it's convex over $Bbb R times Bbb R_{++}$, we have to show that $nabla^2f$ is positive semi definite.I see that
$$nabla f(x,y) =
begin{bmatrix}2x/y\-x^2/y^2end{bmatrix},
$$
but the book is saying that
$$
nabla^2 f(x,y)=frac{2}{y^3}begin{bmatrix}y^2 &-xy\-xy & x^2end{bmatrix}.
$$
But isn't it the case that $nabla^2f = nabla * nabla f$ where $*$ means dot product and therefore $nabla^2f = 2/y + 2x^2 / y^3$?
What have I misunderstood here?
convex-analysis proof-explanation
convex-analysis proof-explanation
edited Jan 22 at 16:22
A.Γ.
22.8k32656
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
edited Jan 22 at 16:22
A.Γ.
22.8k32656
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
edited Jan 22 at 16:22
A.Γ.
22.8k32656
edited Jan 22 at 16:22
A.Γ.
22.8k32656
edited Jan 22 at 16:22
A.Γ.
22.8k32656
22.8k32656
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
asked Jan 20 at 15:04
Oliver GOliver G
1,4471532
1,4471532
1
$begingroup$
There seems to be some ambiguity of notation for $nabla^2$ as either the "Laplace operator" or the "Hessian matrix," see here: en.wikipedia.org/wiki/Del_squared [In your case you want the Hessian matrix.] Note that it only makes sense for a square matrix to be "positive semi-definite."
$endgroup$
– Michael
Jan 20 at 16:05
add a comment |
1
$begingroup$
There seems to be some ambiguity of notation for $nabla^2$ as either the "Laplace operator" or the "Hessian matrix," see here: en.wikipedia.org/wiki/Del_squared [In your case you want the Hessian matrix.] Note that it only makes sense for a square matrix to be "positive semi-definite."
$endgroup$
– Michael
Jan 20 at 16:05
1
1
$begingroup$
There seems to be some ambiguity of notation for $nabla^2$ as either the "Laplace operator" or the "Hessian matrix," see here: en.wikipedia.org/wiki/Del_squared [In your case you want the Hessian matrix.] Note that it only makes sense for a square matrix to be "positive semi-definite."
$endgroup$
– Michael
Jan 20 at 16:05
$begingroup$
There seems to be some ambiguity of notation for $nabla^2$ as either the "Laplace operator" or the "Hessian matrix," see here: en.wikipedia.org/wiki/Del_squared [In your case you want the Hessian matrix.] Note that it only makes sense for a square matrix to be "positive semi-definite."
$endgroup$
– Michael
Jan 20 at 16:05
add a comment |
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$begingroup$
There seems to be some ambiguity of notation for $nabla^2$ as either the "Laplace operator" or the "Hessian matrix," see here: en.wikipedia.org/wiki/Del_squared [In your case you want the Hessian matrix.] Note that it only makes sense for a square matrix to be "positive semi-definite."
$endgroup$
– Michael
Jan 20 at 16:05