Bounded function with bounded second derivative imply bounded first derivative
up vote
3
down vote
favorite
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
asked Nov 27 at 15:27
filip
214
add a comment |
up vote
3
down vote
favorite
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
asked Nov 27 at 15:27
filip
214
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 at 16:32
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
asked Nov 27 at 15:27
filip
214
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
real-analysis derivatives taylor-expansion
asked Nov 27 at 15:27
filip
214
asked Nov 27 at 15:27
filip
214
edited Nov 27 at 17:01
asked Nov 27 at 15:27
filip
214
asked Nov 27 at 15:27
filip
214
asked Nov 27 at 15:27
filip
214
214
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 at 16:32
add a comment |
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 at 16:32
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 at 16:30
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 at 16:32
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 at 16:32
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 at 9:13
|
show 4 more comments
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%2f3015909%2fbounded-function-with-bounded-second-derivative-imply-bounded-first-derivative%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
up vote
1
down vote
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 at 9:13
|
show 4 more comments
up vote
1
down vote
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 at 9:13
|
show 4 more comments
up vote
1
down vote
up vote
1
down vote
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
edited Nov 27 at 18:00
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
answered Nov 27 at 16:49
David C. Ullrich
57.7k43891
57.7k43891
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 at 9:13
|
show 4 more comments
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 at 9:13
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 at 17:48
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 at 17:58
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 at 19:31
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 at 20:20
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 at 9:13
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 at 9:13
|
show 4 more comments
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%2f3015909%2fbounded-function-with-bounded-second-derivative-imply-bounded-first-derivative%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
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 at 16:32