Mixing
How does $f_0(z(0))' < 0 Rightarrow f_0(z(t)) lt f_0(x)$?
$begingroup$
From Convex optimization by Boyd and Vanderberghe:
In the red box in the proof below:
How does the fact that the derivative of $f_0(z(t))$ at $t=0$ show that $f_0(z(t)) lt f_0(x)$?
proof-explanation convex-optimization
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
$endgroup$
add a comment |
$begingroup$
From Convex optimization by Boyd and Vanderberghe:
In the red box in the proof below:
How does the fact that the derivative of $f_0(z(t))$ at $t=0$ show that $f_0(z(t)) lt f_0(x)$?
proof-explanation convex-optimization
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
$endgroup$
add a comment |
$begingroup$
From Convex optimization by Boyd and Vanderberghe:
In the red box in the proof below:
How does the fact that the derivative of $f_0(z(t))$ at $t=0$ show that $f_0(z(t)) lt f_0(x)$?
proof-explanation convex-optimization
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
$endgroup$
From Convex optimization by Boyd and Vanderberghe:
In the red box in the proof below:
How does the fact that the derivative of $f_0(z(t))$ at $t=0$ show that $f_0(z(t)) lt f_0(x)$?
proof-explanation convex-optimization
proof-explanation convex-optimization
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
asked Feb 2 at 12:17
Oliver GOliver G
1,2621634
1,2621634
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the function $g(t) = f_0(z(t))$. It is given that
$$
g'(0) = lim_{t to 0} frac{g(t) - g(0)}{t-0}
$$
exists and is negative. From the definition of the limit it follows that
there is a $delta > 0$ such that
$$
frac{g(t) - g(0)}{t-0} < frac 12 g'(0)
$$
for $|t| < delta$. Consequently, for $0 < t < delta$,
$$
g(t) < g(0) + frac t2 g'(0) < g(0) , .
$$
answered Feb 2 at 12:31
Martin RMartin R
31k33561
$endgroup$
$begingroup$
How does it follow that for small $t$ we have $g(t) - g(0) lt 0$? Where is $t$ sufficient?
$endgroup$
– Oliver G
Feb 2 at 12:35
$begingroup$
And why can you be sure that it's not $0$ or positive for small positive $t$?
$endgroup$
– Oliver G
Feb 2 at 12:54
$begingroup$
@OliverG: I have added more details.
$endgroup$
– Martin R
Feb 2 at 13:22
add a comment |
Your Answer
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%2f3097233%2fhow-does-f-0z0-0-rightarrow-f-0zt-lt-f-0x%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
$begingroup$
Consider the function $g(t) = f_0(z(t))$. It is given that
$$
g'(0) = lim_{t to 0} frac{g(t) - g(0)}{t-0}
$$
exists and is negative. From the definition of the limit it follows that
there is a $delta > 0$ such that
$$
frac{g(t) - g(0)}{t-0} < frac 12 g'(0)
$$
for $|t| < delta$. Consequently, for $0 < t < delta$,
$$
g(t) < g(0) + frac t2 g'(0) < g(0) , .
$$
answered Feb 2 at 12:31
Martin RMartin R
31k33561
$endgroup$
$begingroup$
How does it follow that for small $t$ we have $g(t) - g(0) lt 0$? Where is $t$ sufficient?
$endgroup$
– Oliver G
Feb 2 at 12:35
$begingroup$
And why can you be sure that it's not $0$ or positive for small positive $t$?
$endgroup$
– Oliver G
Feb 2 at 12:54
$begingroup$
@OliverG: I have added more details.
$endgroup$
– Martin R
Feb 2 at 13:22
add a comment |
$begingroup$
Consider the function $g(t) = f_0(z(t))$. It is given that
$$
g'(0) = lim_{t to 0} frac{g(t) - g(0)}{t-0}
$$
exists and is negative. From the definition of the limit it follows that
there is a $delta > 0$ such that
$$
frac{g(t) - g(0)}{t-0} < frac 12 g'(0)
$$
for $|t| < delta$. Consequently, for $0 < t < delta$,
$$
g(t) < g(0) + frac t2 g'(0) < g(0) , .
$$
answered Feb 2 at 12:31
Martin RMartin R
31k33561
$endgroup$
$begingroup$
How does it follow that for small $t$ we have $g(t) - g(0) lt 0$? Where is $t$ sufficient?
$endgroup$
– Oliver G
Feb 2 at 12:35
$begingroup$
And why can you be sure that it's not $0$ or positive for small positive $t$?
$endgroup$
– Oliver G
Feb 2 at 12:54
$begingroup$
@OliverG: I have added more details.
$endgroup$
– Martin R
Feb 2 at 13:22
add a comment |
$begingroup$
Consider the function $g(t) = f_0(z(t))$. It is given that
$$
g'(0) = lim_{t to 0} frac{g(t) - g(0)}{t-0}
$$
exists and is negative. From the definition of the limit it follows that
there is a $delta > 0$ such that
$$
frac{g(t) - g(0)}{t-0} < frac 12 g'(0)
$$
for $|t| < delta$. Consequently, for $0 < t < delta$,
$$
g(t) < g(0) + frac t2 g'(0) < g(0) , .
$$
answered Feb 2 at 12:31
Martin RMartin R
31k33561
$endgroup$
Consider the function $g(t) = f_0(z(t))$. It is given that
$$
g'(0) = lim_{t to 0} frac{g(t) - g(0)}{t-0}
$$
exists and is negative. From the definition of the limit it follows that
there is a $delta > 0$ such that
$$
frac{g(t) - g(0)}{t-0} < frac 12 g'(0)
$$
for $|t| < delta$. Consequently, for $0 < t < delta$,
$$
g(t) < g(0) + frac t2 g'(0) < g(0) , .
$$
answered Feb 2 at 12:31
Martin RMartin R
31k33561
edited Feb 2 at 13:22
answered Feb 2 at 12:31
Martin RMartin R
31k33561
answered Feb 2 at 12:31
Martin RMartin R
31k33561
answered Feb 2 at 12:31
Martin RMartin R
31k33561
31k33561
$begingroup$
How does it follow that for small $t$ we have $g(t) - g(0) lt 0$? Where is $t$ sufficient?
$endgroup$
– Oliver G
Feb 2 at 12:35
$begingroup$
And why can you be sure that it's not $0$ or positive for small positive $t$?
$endgroup$
– Oliver G
Feb 2 at 12:54
$begingroup$
@OliverG: I have added more details.
$endgroup$
– Martin R
Feb 2 at 13:22
add a comment |
$begingroup$
How does it follow that for small $t$ we have $g(t) - g(0) lt 0$? Where is $t$ sufficient?
$endgroup$
– Oliver G
Feb 2 at 12:35
$begingroup$
And why can you be sure that it's not $0$ or positive for small positive $t$?
$endgroup$
– Oliver G
Feb 2 at 12:54
$begingroup$
@OliverG: I have added more details.
$endgroup$
– Martin R
Feb 2 at 13:22
$begingroup$
How does it follow that for small $t$ we have $g(t) - g(0) lt 0$? Where is $t$ sufficient?
$endgroup$
– Oliver G
Feb 2 at 12:35
$begingroup$
How does it follow that for small $t$ we have $g(t) - g(0) lt 0$? Where is $t$ sufficient?
$endgroup$
– Oliver G
Feb 2 at 12:35
$begingroup$
And why can you be sure that it's not $0$ or positive for small positive $t$?
$endgroup$
– Oliver G
Feb 2 at 12:54
$begingroup$
And why can you be sure that it's not $0$ or positive for small positive $t$?
$endgroup$
– Oliver G
Feb 2 at 12:54
$begingroup$
@OliverG: I have added more details.
$endgroup$
– Martin R
Feb 2 at 13:22
$begingroup$
@OliverG: I have added more details.
$endgroup$
– Martin R
Feb 2 at 13:22
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%2f3097233%2fhow-does-f-0z0-0-rightarrow-f-0zt-lt-f-0x%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
