Basic question: why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to $g(x):=...
$begingroup$
I am so sorry to ask probably the most trivial and fundamental question. But it is just bothering me and not able to understand,
why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to another convex function $g(x):= frac{L}{2} x^T x - f(x)$?
(reference: page 1-11 of http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf)
I am so confused. If I minimize the function $f(x)$, then I need to maximize the function $g(x)$, right? or I have confused myself more.
convex-analysis
asked Jan 24 at 17:00
learninglearning
566
$endgroup$
add a comment |
$begingroup$
I am so sorry to ask probably the most trivial and fundamental question. But it is just bothering me and not able to understand,
why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to another convex function $g(x):= frac{L}{2} x^T x - f(x)$?
(reference: page 1-11 of http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf)
I am so confused. If I minimize the function $f(x)$, then I need to maximize the function $g(x)$, right? or I have confused myself more.
convex-analysis
asked Jan 24 at 17:00
learninglearning
566
$endgroup$
add a comment |
$begingroup$
I am so sorry to ask probably the most trivial and fundamental question. But it is just bothering me and not able to understand,
why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to another convex function $g(x):= frac{L}{2} x^T x - f(x)$?
(reference: page 1-11 of http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf)
I am so confused. If I minimize the function $f(x)$, then I need to maximize the function $g(x)$, right? or I have confused myself more.
convex-analysis
asked Jan 24 at 17:00
learninglearning
566
$endgroup$
I am so sorry to ask probably the most trivial and fundamental question. But it is just bothering me and not able to understand,
why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to another convex function $g(x):= frac{L}{2} x^T x - f(x)$?
(reference: page 1-11 of http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf)
I am so confused. If I minimize the function $f(x)$, then I need to maximize the function $g(x)$, right? or I have confused myself more.
convex-analysis
convex-analysis
asked Jan 24 at 17:00
learninglearning
566
asked Jan 24 at 17:00
learninglearning
566
edited Jan 24 at 17:07
learning
asked Jan 24 at 17:00
learninglearning
566
asked Jan 24 at 17:00
learninglearning
566
asked Jan 24 at 17:00
learninglearning
566
566
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No wonder you're confused, since you're reading it all wrong.
The claim that they make is that “this” (which refers to the inequality at the top of the page, not to the function $f$) is equivalent to the condition that $frac{L}{2} x^T x - f(x)$ is convex, provided that $f$ is convex with domain $mathbf{R}^n$.
(What would it even mean for a function $f$ to be equivalent to another function $g$ in this context?)
And it also says “we will see that”, meaning (I suppose) that you will get an explanation of why this is true later in the text, or in a later lecture perhaps (since this seems to be lecture notes). So just keep reading on and see if it gets clearer.
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
$begingroup$
Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ...
$endgroup$
– learning
Jan 24 at 18:30
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',
autoActivateHeartbeat: false,
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%2f3086120%2fbasic-question-why-and-how-smooth-convex-function-fx-with-domain-rn-is-e%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No wonder you're confused, since you're reading it all wrong.
The claim that they make is that “this” (which refers to the inequality at the top of the page, not to the function $f$) is equivalent to the condition that $frac{L}{2} x^T x - f(x)$ is convex, provided that $f$ is convex with domain $mathbf{R}^n$.
(What would it even mean for a function $f$ to be equivalent to another function $g$ in this context?)
And it also says “we will see that”, meaning (I suppose) that you will get an explanation of why this is true later in the text, or in a later lecture perhaps (since this seems to be lecture notes). So just keep reading on and see if it gets clearer.
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
$begingroup$
Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ...
$endgroup$
– learning
Jan 24 at 18:30
add a comment |
$begingroup$
No wonder you're confused, since you're reading it all wrong.
The claim that they make is that “this” (which refers to the inequality at the top of the page, not to the function $f$) is equivalent to the condition that $frac{L}{2} x^T x - f(x)$ is convex, provided that $f$ is convex with domain $mathbf{R}^n$.
(What would it even mean for a function $f$ to be equivalent to another function $g$ in this context?)
And it also says “we will see that”, meaning (I suppose) that you will get an explanation of why this is true later in the text, or in a later lecture perhaps (since this seems to be lecture notes). So just keep reading on and see if it gets clearer.
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
$begingroup$
Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ...
$endgroup$
– learning
Jan 24 at 18:30
add a comment |
$begingroup$
No wonder you're confused, since you're reading it all wrong.
The claim that they make is that “this” (which refers to the inequality at the top of the page, not to the function $f$) is equivalent to the condition that $frac{L}{2} x^T x - f(x)$ is convex, provided that $f$ is convex with domain $mathbf{R}^n$.
(What would it even mean for a function $f$ to be equivalent to another function $g$ in this context?)
And it also says “we will see that”, meaning (I suppose) that you will get an explanation of why this is true later in the text, or in a later lecture perhaps (since this seems to be lecture notes). So just keep reading on and see if it gets clearer.
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
No wonder you're confused, since you're reading it all wrong.
The claim that they make is that “this” (which refers to the inequality at the top of the page, not to the function $f$) is equivalent to the condition that $frac{L}{2} x^T x - f(x)$ is convex, provided that $f$ is convex with domain $mathbf{R}^n$.
(What would it even mean for a function $f$ to be equivalent to another function $g$ in this context?)
And it also says “we will see that”, meaning (I suppose) that you will get an explanation of why this is true later in the text, or in a later lecture perhaps (since this seems to be lecture notes). So just keep reading on and see if it gets clearer.
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
answered Jan 24 at 17:55
Hans LundmarkHans Lundmark
35.8k564115
35.8k564115
$begingroup$
Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ...
$endgroup$
– learning
Jan 24 at 18:30
add a comment |
$begingroup$
Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ...
$endgroup$
– learning
Jan 24 at 18:30
$begingroup$
Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ...
$endgroup$
– learning
Jan 24 at 18:30
$begingroup$
Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ...
$endgroup$
– learning
Jan 24 at 18:30
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.
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%2f3086120%2fbasic-question-why-and-how-smooth-convex-function-fx-with-domain-rn-is-e%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
