Smooth function with equibounded family of derivatives
$begingroup$
By $mathcal{C}^{infty}(mathbb{R})$ we denote the space of smooth functions $mathbb{R}rightarrow mathbb{R}$. Also, by $mathrm{supp}(f)$ we denote the closure $mathrm{Cl}(f^{-1}(mathbb{R}_{ne 0}))$. I'm trying to build a function $fin mathcal{C}^{infty}(mathbb{R})$ such that the family of higher derivatives ${f^{(k)}: kin mathbb{N}}$ is equibounded, i.e. there exists $c>0$ such that for each $k$ the image $f^{(k)}(mathbb{R})subset (-c,c)$. Furthermore, I want $mathrm{supp}(f)$ to be a compact set. Is it possible?
Thus I want to give an important example in functional analysis. I'll be grateful for any hints!
I know the example of a smooth function $g$ such that $mathrm{supp}(f)=[-1,1]$ but the family of derivatives is not equibounded! (Really, let $f(x)=e^{-frac{1}{x^2}}$ for $x>0$ and $f(x)=0$ for $xle 0$. Let $g(x)=f(1-x^2)$. Well, $gin mathcal{C}^{infty}(mathbb{R})$ and $mathrm{supp}(g)=[-1,1]$).
calculus functional-analysis differential-topology
$endgroup$
add a comment |
$begingroup$
By $mathcal{C}^{infty}(mathbb{R})$ we denote the space of smooth functions $mathbb{R}rightarrow mathbb{R}$. Also, by $mathrm{supp}(f)$ we denote the closure $mathrm{Cl}(f^{-1}(mathbb{R}_{ne 0}))$. I'm trying to build a function $fin mathcal{C}^{infty}(mathbb{R})$ such that the family of higher derivatives ${f^{(k)}: kin mathbb{N}}$ is equibounded, i.e. there exists $c>0$ such that for each $k$ the image $f^{(k)}(mathbb{R})subset (-c,c)$. Furthermore, I want $mathrm{supp}(f)$ to be a compact set. Is it possible?
Thus I want to give an important example in functional analysis. I'll be grateful for any hints!
I know the example of a smooth function $g$ such that $mathrm{supp}(f)=[-1,1]$ but the family of derivatives is not equibounded! (Really, let $f(x)=e^{-frac{1}{x^2}}$ for $x>0$ and $f(x)=0$ for $xle 0$. Let $g(x)=f(1-x^2)$. Well, $gin mathcal{C}^{infty}(mathbb{R})$ and $mathrm{supp}(g)=[-1,1]$).
calculus functional-analysis differential-topology
$endgroup$
$begingroup$
Well, there's $f equiv 0$. I think that's the only one.
$endgroup$
– Daniel Fischer♦
Oct 25 '13 at 17:24
$begingroup$
Of course, I meant $fne 0$
$endgroup$
– user74574
Oct 25 '13 at 17:24
add a comment |
$begingroup$
By $mathcal{C}^{infty}(mathbb{R})$ we denote the space of smooth functions $mathbb{R}rightarrow mathbb{R}$. Also, by $mathrm{supp}(f)$ we denote the closure $mathrm{Cl}(f^{-1}(mathbb{R}_{ne 0}))$. I'm trying to build a function $fin mathcal{C}^{infty}(mathbb{R})$ such that the family of higher derivatives ${f^{(k)}: kin mathbb{N}}$ is equibounded, i.e. there exists $c>0$ such that for each $k$ the image $f^{(k)}(mathbb{R})subset (-c,c)$. Furthermore, I want $mathrm{supp}(f)$ to be a compact set. Is it possible?
Thus I want to give an important example in functional analysis. I'll be grateful for any hints!
I know the example of a smooth function $g$ such that $mathrm{supp}(f)=[-1,1]$ but the family of derivatives is not equibounded! (Really, let $f(x)=e^{-frac{1}{x^2}}$ for $x>0$ and $f(x)=0$ for $xle 0$. Let $g(x)=f(1-x^2)$. Well, $gin mathcal{C}^{infty}(mathbb{R})$ and $mathrm{supp}(g)=[-1,1]$).
calculus functional-analysis differential-topology
$endgroup$
By $mathcal{C}^{infty}(mathbb{R})$ we denote the space of smooth functions $mathbb{R}rightarrow mathbb{R}$. Also, by $mathrm{supp}(f)$ we denote the closure $mathrm{Cl}(f^{-1}(mathbb{R}_{ne 0}))$. I'm trying to build a function $fin mathcal{C}^{infty}(mathbb{R})$ such that the family of higher derivatives ${f^{(k)}: kin mathbb{N}}$ is equibounded, i.e. there exists $c>0$ such that for each $k$ the image $f^{(k)}(mathbb{R})subset (-c,c)$. Furthermore, I want $mathrm{supp}(f)$ to be a compact set. Is it possible?
Thus I want to give an important example in functional analysis. I'll be grateful for any hints!
I know the example of a smooth function $g$ such that $mathrm{supp}(f)=[-1,1]$ but the family of derivatives is not equibounded! (Really, let $f(x)=e^{-frac{1}{x^2}}$ for $x>0$ and $f(x)=0$ for $xle 0$. Let $g(x)=f(1-x^2)$. Well, $gin mathcal{C}^{infty}(mathbb{R})$ and $mathrm{supp}(g)=[-1,1]$).
calculus functional-analysis differential-topology
calculus functional-analysis differential-topology
asked Oct 25 '13 at 17:20
user74574
$begingroup$
Well, there's $f equiv 0$. I think that's the only one.
$endgroup$
– Daniel Fischer♦
Oct 25 '13 at 17:24
$begingroup$
Of course, I meant $fne 0$
$endgroup$
– user74574
Oct 25 '13 at 17:24
add a comment |
$begingroup$
Well, there's $f equiv 0$. I think that's the only one.
$endgroup$
– Daniel Fischer♦
Oct 25 '13 at 17:24
$begingroup$
Of course, I meant $fne 0$
$endgroup$
– user74574
Oct 25 '13 at 17:24
$begingroup$
Well, there's $f equiv 0$. I think that's the only one.
$endgroup$
– Daniel Fischer♦
Oct 25 '13 at 17:24
$begingroup$
Well, there's $f equiv 0$. I think that's the only one.
$endgroup$
– Daniel Fischer♦
Oct 25 '13 at 17:24
$begingroup$
Of course, I meant $fne 0$
$endgroup$
– user74574
Oct 25 '13 at 17:24
$begingroup$
Of course, I meant $fne 0$
$endgroup$
– user74574
Oct 25 '13 at 17:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A nonzero function in $C_c^infty(mathbb{R})$ cannot have equibounded derivatives. Without loss of generality, suppose $f$ has support in $[-1,1]$. Suppose you have $lvert f^{(k)}(t)rvert leqslant C$ for some $k > 0$.
Then you have
$$lvert f^{(k-1)}(t)rvert = leftlvert int_{-1}^t f^{(k)}(s),dsrightrvert leqslant Ccdot (t+1)$$
for $-1 leqslant t leqslant 0$, and similarly $lvert f^{(k-1)}(t)rvert leqslant C(1-t)$ for $0leqslant t leqslant 1$. And further
$$lvert f^{(k-2)}(t)rvert leqslant int_{-1}^t C(s+1),ds leqslant frac{C}{2}(t+1)^2$$
for $-1 leqslant t leqslant 0$, continuing, you get
$$lvert f^{(k-m)}(t)rvert leqslant frac{C}{m!}(t+1)^m,$$
and similarly for $0 leqslant t leqslant 1$. Altogether,
$$lvert f(t)rvert leqslant frac{C}{k!}$$
for all $t$. If the $f^{(k)}$ are equibounded, that estimate forces $f equiv 0$ by letting $ktoinfty$.
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
$endgroup$
add a comment |
$begingroup$
This is a question that was considered by Denjoy back in the 1920's in the study of quasi-analytic functions. Thanks to the Denjoy-Carleman theorem (Theorem 19.11 in Rudin's Real and Complex Analysis), we actually have a precise answer.
Take $M_1$, $M_2$, ..., $M_k$, ... to be an infinite sequence of positive real numbers satisfying
$$M_k^2 le M_{k-1}M_{k+1}$$
for $k > 1$. Then there is a smooth nonzero function with compact support whose $k^{th}$ derivative is bounded everywhere by $M_k$ for all $k > 0$ if and only if
$$sum_{k ge 1} (M_k)^{-1/k}$$
converges.
In particular, any constant positive sequence will give a divergent sum.
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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%2f539523%2fsmooth-function-with-equibounded-family-of-derivatives%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
$begingroup$
A nonzero function in $C_c^infty(mathbb{R})$ cannot have equibounded derivatives. Without loss of generality, suppose $f$ has support in $[-1,1]$. Suppose you have $lvert f^{(k)}(t)rvert leqslant C$ for some $k > 0$.
Then you have
$$lvert f^{(k-1)}(t)rvert = leftlvert int_{-1}^t f^{(k)}(s),dsrightrvert leqslant Ccdot (t+1)$$
for $-1 leqslant t leqslant 0$, and similarly $lvert f^{(k-1)}(t)rvert leqslant C(1-t)$ for $0leqslant t leqslant 1$. And further
$$lvert f^{(k-2)}(t)rvert leqslant int_{-1}^t C(s+1),ds leqslant frac{C}{2}(t+1)^2$$
for $-1 leqslant t leqslant 0$, continuing, you get
$$lvert f^{(k-m)}(t)rvert leqslant frac{C}{m!}(t+1)^m,$$
and similarly for $0 leqslant t leqslant 1$. Altogether,
$$lvert f(t)rvert leqslant frac{C}{k!}$$
for all $t$. If the $f^{(k)}$ are equibounded, that estimate forces $f equiv 0$ by letting $ktoinfty$.
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
$endgroup$
add a comment |
$begingroup$
A nonzero function in $C_c^infty(mathbb{R})$ cannot have equibounded derivatives. Without loss of generality, suppose $f$ has support in $[-1,1]$. Suppose you have $lvert f^{(k)}(t)rvert leqslant C$ for some $k > 0$.
Then you have
$$lvert f^{(k-1)}(t)rvert = leftlvert int_{-1}^t f^{(k)}(s),dsrightrvert leqslant Ccdot (t+1)$$
for $-1 leqslant t leqslant 0$, and similarly $lvert f^{(k-1)}(t)rvert leqslant C(1-t)$ for $0leqslant t leqslant 1$. And further
$$lvert f^{(k-2)}(t)rvert leqslant int_{-1}^t C(s+1),ds leqslant frac{C}{2}(t+1)^2$$
for $-1 leqslant t leqslant 0$, continuing, you get
$$lvert f^{(k-m)}(t)rvert leqslant frac{C}{m!}(t+1)^m,$$
and similarly for $0 leqslant t leqslant 1$. Altogether,
$$lvert f(t)rvert leqslant frac{C}{k!}$$
for all $t$. If the $f^{(k)}$ are equibounded, that estimate forces $f equiv 0$ by letting $ktoinfty$.
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
$endgroup$
add a comment |
$begingroup$
A nonzero function in $C_c^infty(mathbb{R})$ cannot have equibounded derivatives. Without loss of generality, suppose $f$ has support in $[-1,1]$. Suppose you have $lvert f^{(k)}(t)rvert leqslant C$ for some $k > 0$.
Then you have
$$lvert f^{(k-1)}(t)rvert = leftlvert int_{-1}^t f^{(k)}(s),dsrightrvert leqslant Ccdot (t+1)$$
for $-1 leqslant t leqslant 0$, and similarly $lvert f^{(k-1)}(t)rvert leqslant C(1-t)$ for $0leqslant t leqslant 1$. And further
$$lvert f^{(k-2)}(t)rvert leqslant int_{-1}^t C(s+1),ds leqslant frac{C}{2}(t+1)^2$$
for $-1 leqslant t leqslant 0$, continuing, you get
$$lvert f^{(k-m)}(t)rvert leqslant frac{C}{m!}(t+1)^m,$$
and similarly for $0 leqslant t leqslant 1$. Altogether,
$$lvert f(t)rvert leqslant frac{C}{k!}$$
for all $t$. If the $f^{(k)}$ are equibounded, that estimate forces $f equiv 0$ by letting $ktoinfty$.
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
$endgroup$
A nonzero function in $C_c^infty(mathbb{R})$ cannot have equibounded derivatives. Without loss of generality, suppose $f$ has support in $[-1,1]$. Suppose you have $lvert f^{(k)}(t)rvert leqslant C$ for some $k > 0$.
Then you have
$$lvert f^{(k-1)}(t)rvert = leftlvert int_{-1}^t f^{(k)}(s),dsrightrvert leqslant Ccdot (t+1)$$
for $-1 leqslant t leqslant 0$, and similarly $lvert f^{(k-1)}(t)rvert leqslant C(1-t)$ for $0leqslant t leqslant 1$. And further
$$lvert f^{(k-2)}(t)rvert leqslant int_{-1}^t C(s+1),ds leqslant frac{C}{2}(t+1)^2$$
for $-1 leqslant t leqslant 0$, continuing, you get
$$lvert f^{(k-m)}(t)rvert leqslant frac{C}{m!}(t+1)^m,$$
and similarly for $0 leqslant t leqslant 1$. Altogether,
$$lvert f(t)rvert leqslant frac{C}{k!}$$
for all $t$. If the $f^{(k)}$ are equibounded, that estimate forces $f equiv 0$ by letting $ktoinfty$.
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
answered Oct 25 '13 at 17:38
Daniel Fischer♦Daniel Fischer
173k16163285
173k16163285
add a comment |
add a comment |
$begingroup$
This is a question that was considered by Denjoy back in the 1920's in the study of quasi-analytic functions. Thanks to the Denjoy-Carleman theorem (Theorem 19.11 in Rudin's Real and Complex Analysis), we actually have a precise answer.
Take $M_1$, $M_2$, ..., $M_k$, ... to be an infinite sequence of positive real numbers satisfying
$$M_k^2 le M_{k-1}M_{k+1}$$
for $k > 1$. Then there is a smooth nonzero function with compact support whose $k^{th}$ derivative is bounded everywhere by $M_k$ for all $k > 0$ if and only if
$$sum_{k ge 1} (M_k)^{-1/k}$$
converges.
In particular, any constant positive sequence will give a divergent sum.
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This is a question that was considered by Denjoy back in the 1920's in the study of quasi-analytic functions. Thanks to the Denjoy-Carleman theorem (Theorem 19.11 in Rudin's Real and Complex Analysis), we actually have a precise answer.
Take $M_1$, $M_2$, ..., $M_k$, ... to be an infinite sequence of positive real numbers satisfying
$$M_k^2 le M_{k-1}M_{k+1}$$
for $k > 1$. Then there is a smooth nonzero function with compact support whose $k^{th}$ derivative is bounded everywhere by $M_k$ for all $k > 0$ if and only if
$$sum_{k ge 1} (M_k)^{-1/k}$$
converges.
In particular, any constant positive sequence will give a divergent sum.
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This is a question that was considered by Denjoy back in the 1920's in the study of quasi-analytic functions. Thanks to the Denjoy-Carleman theorem (Theorem 19.11 in Rudin's Real and Complex Analysis), we actually have a precise answer.
Take $M_1$, $M_2$, ..., $M_k$, ... to be an infinite sequence of positive real numbers satisfying
$$M_k^2 le M_{k-1}M_{k+1}$$
for $k > 1$. Then there is a smooth nonzero function with compact support whose $k^{th}$ derivative is bounded everywhere by $M_k$ for all $k > 0$ if and only if
$$sum_{k ge 1} (M_k)^{-1/k}$$
converges.
In particular, any constant positive sequence will give a divergent sum.
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This is a question that was considered by Denjoy back in the 1920's in the study of quasi-analytic functions. Thanks to the Denjoy-Carleman theorem (Theorem 19.11 in Rudin's Real and Complex Analysis), we actually have a precise answer.
Take $M_1$, $M_2$, ..., $M_k$, ... to be an infinite sequence of positive real numbers satisfying
$$M_k^2 le M_{k-1}M_{k+1}$$
for $k > 1$. Then there is a smooth nonzero function with compact support whose $k^{th}$ derivative is bounded everywhere by $M_k$ for all $k > 0$ if and only if
$$sum_{k ge 1} (M_k)^{-1/k}$$
converges.
In particular, any constant positive sequence will give a divergent sum.
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
answered Jan 8 at 0:46
Alexander SmithAlexander Smith
1
1
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alexander Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
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%2f539523%2fsmooth-function-with-equibounded-family-of-derivatives%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
$begingroup$
Well, there's $f equiv 0$. I think that's the only one.
$endgroup$
– Daniel Fischer♦
Oct 25 '13 at 17:24
$begingroup$
Of course, I meant $fne 0$
$endgroup$
– user74574
Oct 25 '13 at 17:24