convex function $f(x,y)=x^2+y^2-alpha log(x^2y^2)$
up vote
0
down vote
favorite
I have to study where the function $f(x,y)=x^2+y^2-alpha log(x^2y^2)$ is convex or not in its domain. Do I study the Hessian matrix?
real-analysis
edited Nov 27 at 14:19
egreg
176k1384198
asked Nov 27 at 13:58
Giulia B.
413211
add a comment |
up vote
0
down vote
favorite
I have to study where the function $f(x,y)=x^2+y^2-alpha log(x^2y^2)$ is convex or not in its domain. Do I study the Hessian matrix?
real-analysis
edited Nov 27 at 14:19
egreg
176k1384198
asked Nov 27 at 13:58
Giulia B.
413211
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to study where the function $f(x,y)=x^2+y^2-alpha log(x^2y^2)$ is convex or not in its domain. Do I study the Hessian matrix?
real-analysis
edited Nov 27 at 14:19
egreg
176k1384198
asked Nov 27 at 13:58
Giulia B.
413211
I have to study where the function $f(x,y)=x^2+y^2-alpha log(x^2y^2)$ is convex or not in its domain. Do I study the Hessian matrix?
real-analysis
real-analysis
edited Nov 27 at 14:19
egreg
176k1384198
asked Nov 27 at 13:58
Giulia B.
413211
edited Nov 27 at 14:19
egreg
176k1384198
asked Nov 27 at 13:58
Giulia B.
413211
edited Nov 27 at 14:19
egreg
176k1384198
edited Nov 27 at 14:19
egreg
176k1384198
edited Nov 27 at 14:19
egreg
176k1384198
176k1384198
asked Nov 27 at 13:58
Giulia B.
413211
asked Nov 27 at 13:58
Giulia B.
413211
asked Nov 27 at 13:58
Giulia B.
413211
413211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The Hessian matrix is good. For the derivatives, consider
$$
f(x)=x^2+y^2-2alphaloglvert xrvert-2alphaloglvert yrvert
$$
so
begin{align}
frac{partial f}{partial x}&=2x-frac{2alpha}{x} \[4px]
frac{partial f}{partial y}&=2y-frac{2alpha}{y}
end{align}
and therefore the Hessian matrix is
$$
4begin{bmatrix}
1+dfrac{alpha}{x^2} & 0 \
0 & 1+dfrac{alpha}{y^2}
end{bmatrix}
$$
which is positive definite as long as $alphage0$.
If $alpha<0$, there are points where the Hessian matrix is positive definite, points where it's indefinite and points where it's negative definite.
answered Nov 27 at 14:26
egreg
176k1384198
is positive definite as long as $alphage0$ so f is convex or not? f is convex only if is semi-positive definite?
– Giulia B.
Nov 27 at 16:28
@GiuliaB. The positive semidefinite condition is sufficient, in this case the stronger condition of being positive definite holds (when $alphage0$). When $alpha<0$ there are points where the Hessian is even negative definite.
– egreg
Nov 27 at 17:16
a matrix 2x2 is semi positive definite if $tr(A)ge0$ and $det(A)ge 0$?
– Giulia B.
Nov 27 at 17:32
@GiuliaB. Better, when its eigenvalues are $ge0$.
– egreg
Nov 27 at 18:17
add a comment |
up vote
0
down vote
$$
f(x,y) = x^2 + y^2 - 2alphalog(x) - 2alphalog(y)
$$
is convex in $(0,infty)^2$ as long as $alphageq0$ because it's the sum of convex functions.
The function is never convex on its domain of definition, $(mathbb Rsetminus{0})^2$, for any value of $alpha$.
answered Nov 27 at 14:10
Federico
4,178512
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015800%2fconvex-function-fx-y-x2y2-alpha-logx2y2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The Hessian matrix is good. For the derivatives, consider
$$
f(x)=x^2+y^2-2alphaloglvert xrvert-2alphaloglvert yrvert
$$
so
begin{align}
frac{partial f}{partial x}&=2x-frac{2alpha}{x} \[4px]
frac{partial f}{partial y}&=2y-frac{2alpha}{y}
end{align}
and therefore the Hessian matrix is
$$
4begin{bmatrix}
1+dfrac{alpha}{x^2} & 0 \
0 & 1+dfrac{alpha}{y^2}
end{bmatrix}
$$
which is positive definite as long as $alphage0$.
If $alpha<0$, there are points where the Hessian matrix is positive definite, points where it's indefinite and points where it's negative definite.
answered Nov 27 at 14:26
egreg
176k1384198
is positive definite as long as $alphage0$ so f is convex or not? f is convex only if is semi-positive definite?
– Giulia B.
Nov 27 at 16:28
@GiuliaB. The positive semidefinite condition is sufficient, in this case the stronger condition of being positive definite holds (when $alphage0$). When $alpha<0$ there are points where the Hessian is even negative definite.
– egreg
Nov 27 at 17:16
a matrix 2x2 is semi positive definite if $tr(A)ge0$ and $det(A)ge 0$?
– Giulia B.
Nov 27 at 17:32
@GiuliaB. Better, when its eigenvalues are $ge0$.
– egreg
Nov 27 at 18:17
add a comment |
up vote
1
down vote
accepted
The Hessian matrix is good. For the derivatives, consider
$$
f(x)=x^2+y^2-2alphaloglvert xrvert-2alphaloglvert yrvert
$$
so
begin{align}
frac{partial f}{partial x}&=2x-frac{2alpha}{x} \[4px]
frac{partial f}{partial y}&=2y-frac{2alpha}{y}
end{align}
and therefore the Hessian matrix is
$$
4begin{bmatrix}
1+dfrac{alpha}{x^2} & 0 \
0 & 1+dfrac{alpha}{y^2}
end{bmatrix}
$$
which is positive definite as long as $alphage0$.
If $alpha<0$, there are points where the Hessian matrix is positive definite, points where it's indefinite and points where it's negative definite.
answered Nov 27 at 14:26
egreg
176k1384198
is positive definite as long as $alphage0$ so f is convex or not? f is convex only if is semi-positive definite?
– Giulia B.
Nov 27 at 16:28
@GiuliaB. The positive semidefinite condition is sufficient, in this case the stronger condition of being positive definite holds (when $alphage0$). When $alpha<0$ there are points where the Hessian is even negative definite.
– egreg
Nov 27 at 17:16
a matrix 2x2 is semi positive definite if $tr(A)ge0$ and $det(A)ge 0$?
– Giulia B.
Nov 27 at 17:32
@GiuliaB. Better, when its eigenvalues are $ge0$.
– egreg
Nov 27 at 18:17
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The Hessian matrix is good. For the derivatives, consider
$$
f(x)=x^2+y^2-2alphaloglvert xrvert-2alphaloglvert yrvert
$$
so
begin{align}
frac{partial f}{partial x}&=2x-frac{2alpha}{x} \[4px]
frac{partial f}{partial y}&=2y-frac{2alpha}{y}
end{align}
and therefore the Hessian matrix is
$$
4begin{bmatrix}
1+dfrac{alpha}{x^2} & 0 \
0 & 1+dfrac{alpha}{y^2}
end{bmatrix}
$$
which is positive definite as long as $alphage0$.
If $alpha<0$, there are points where the Hessian matrix is positive definite, points where it's indefinite and points where it's negative definite.
answered Nov 27 at 14:26
egreg
176k1384198
The Hessian matrix is good. For the derivatives, consider
$$
f(x)=x^2+y^2-2alphaloglvert xrvert-2alphaloglvert yrvert
$$
so
begin{align}
frac{partial f}{partial x}&=2x-frac{2alpha}{x} \[4px]
frac{partial f}{partial y}&=2y-frac{2alpha}{y}
end{align}
and therefore the Hessian matrix is
$$
4begin{bmatrix}
1+dfrac{alpha}{x^2} & 0 \
0 & 1+dfrac{alpha}{y^2}
end{bmatrix}
$$
which is positive definite as long as $alphage0$.
If $alpha<0$, there are points where the Hessian matrix is positive definite, points where it's indefinite and points where it's negative definite.
answered Nov 27 at 14:26
egreg
176k1384198
answered Nov 27 at 14:26
egreg
176k1384198
answered Nov 27 at 14:26
egreg
176k1384198
answered Nov 27 at 14:26
egreg
176k1384198
176k1384198
is positive definite as long as $alphage0$ so f is convex or not? f is convex only if is semi-positive definite?
– Giulia B.
Nov 27 at 16:28
@GiuliaB. The positive semidefinite condition is sufficient, in this case the stronger condition of being positive definite holds (when $alphage0$). When $alpha<0$ there are points where the Hessian is even negative definite.
– egreg
Nov 27 at 17:16
a matrix 2x2 is semi positive definite if $tr(A)ge0$ and $det(A)ge 0$?
– Giulia B.
Nov 27 at 17:32
@GiuliaB. Better, when its eigenvalues are $ge0$.
– egreg
Nov 27 at 18:17
add a comment |
is positive definite as long as $alphage0$ so f is convex or not? f is convex only if is semi-positive definite?
– Giulia B.
Nov 27 at 16:28
@GiuliaB. The positive semidefinite condition is sufficient, in this case the stronger condition of being positive definite holds (when $alphage0$). When $alpha<0$ there are points where the Hessian is even negative definite.
– egreg
Nov 27 at 17:16
a matrix 2x2 is semi positive definite if $tr(A)ge0$ and $det(A)ge 0$?
– Giulia B.
Nov 27 at 17:32
@GiuliaB. Better, when its eigenvalues are $ge0$.
– egreg
Nov 27 at 18:17
is positive definite as long as $alphage0$ so f is convex or not? f is convex only if is semi-positive definite?
– Giulia B.
Nov 27 at 16:28
is positive definite as long as $alphage0$ so f is convex or not? f is convex only if is semi-positive definite?
– Giulia B.
Nov 27 at 16:28
@GiuliaB. The positive semidefinite condition is sufficient, in this case the stronger condition of being positive definite holds (when $alphage0$). When $alpha<0$ there are points where the Hessian is even negative definite.
– egreg
Nov 27 at 17:16
@GiuliaB. The positive semidefinite condition is sufficient, in this case the stronger condition of being positive definite holds (when $alphage0$). When $alpha<0$ there are points where the Hessian is even negative definite.
– egreg
Nov 27 at 17:16
a matrix 2x2 is semi positive definite if $tr(A)ge0$ and $det(A)ge 0$?
– Giulia B.
Nov 27 at 17:32
a matrix 2x2 is semi positive definite if $tr(A)ge0$ and $det(A)ge 0$?
– Giulia B.
Nov 27 at 17:32
@GiuliaB. Better, when its eigenvalues are $ge0$.
– egreg
Nov 27 at 18:17
@GiuliaB. Better, when its eigenvalues are $ge0$.
– egreg
Nov 27 at 18:17
add a comment |
up vote
0
down vote
$$
f(x,y) = x^2 + y^2 - 2alphalog(x) - 2alphalog(y)
$$
is convex in $(0,infty)^2$ as long as $alphageq0$ because it's the sum of convex functions.
The function is never convex on its domain of definition, $(mathbb Rsetminus{0})^2$, for any value of $alpha$.
answered Nov 27 at 14:10
Federico
4,178512
add a comment |
up vote
0
down vote
$$
f(x,y) = x^2 + y^2 - 2alphalog(x) - 2alphalog(y)
$$
is convex in $(0,infty)^2$ as long as $alphageq0$ because it's the sum of convex functions.
The function is never convex on its domain of definition, $(mathbb Rsetminus{0})^2$, for any value of $alpha$.
answered Nov 27 at 14:10
Federico
4,178512
add a comment |
up vote
0
down vote
up vote
0
down vote
$$
f(x,y) = x^2 + y^2 - 2alphalog(x) - 2alphalog(y)
$$
is convex in $(0,infty)^2$ as long as $alphageq0$ because it's the sum of convex functions.
The function is never convex on its domain of definition, $(mathbb Rsetminus{0})^2$, for any value of $alpha$.
answered Nov 27 at 14:10
Federico
4,178512
$$
f(x,y) = x^2 + y^2 - 2alphalog(x) - 2alphalog(y)
$$
is convex in $(0,infty)^2$ as long as $alphageq0$ because it's the sum of convex functions.
The function is never convex on its domain of definition, $(mathbb Rsetminus{0})^2$, for any value of $alpha$.
answered Nov 27 at 14:10
Federico
4,178512
answered Nov 27 at 14:10
Federico
4,178512
answered Nov 27 at 14:10
Federico
4,178512
answered Nov 27 at 14:10
Federico
4,178512
4,178512
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015800%2fconvex-function-fx-y-x2y2-alpha-logx2y2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
