Can $f^{(infty)}(a)=0$ for almost all $a$?












2












$begingroup$


My question is:




Does there exist an infinitely differentiable function $f$ such that
$$lim_{ntoinfty} f^{(n)}(a)=0qquad{text{for almost all } ain[0,infty)}$$
?




($f$ cannot be a constant function or a polynomial.)




If we restrict us to $C^{omega}$ (i.e. assuming $f$ is holomorphic), it is likely that the answer is no, since
$$f^{(n)}(a)=frac{n!}{2pi i}oint_{gamma}frac{f(z)}{(z-a)^{n+1}}dz$$
the integrand only decays exponentially in $n$ while there is a $n!$ factor there.




However, I am not quite sure if the above argument is correct.



For smooth functions, I have no ideas.



Any help will be appreciated.



Thanks in advance.






real-analysis complex-analysis derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not clear to me what "almost all" means.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 3:10








  • 1




    $begingroup$
    @JohnOmielan it is a term used in Lebesgue measure theory. You can find the explanation of this term on Wikipedia.
    $endgroup$
    – Szeto
    Dec 24 '18 at 3:11










  • $begingroup$
    fundamental theorem of calculus applied to the derivatives of $f$ along with the almost everywhere assumption and dominated convergence may lead to something (?). not entirely sure, just throwing out an idea
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 3:11
















2












$begingroup$


My question is:




Does there exist an infinitely differentiable function $f$ such that
$$lim_{ntoinfty} f^{(n)}(a)=0qquad{text{for almost all } ain[0,infty)}$$
?




($f$ cannot be a constant function or a polynomial.)




If we restrict us to $C^{omega}$ (i.e. assuming $f$ is holomorphic), it is likely that the answer is no, since
$$f^{(n)}(a)=frac{n!}{2pi i}oint_{gamma}frac{f(z)}{(z-a)^{n+1}}dz$$
the integrand only decays exponentially in $n$ while there is a $n!$ factor there.




However, I am not quite sure if the above argument is correct.



For smooth functions, I have no ideas.



Any help will be appreciated.



Thanks in advance.






real-analysis complex-analysis derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not clear to me what "almost all" means.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 3:10








  • 1




    $begingroup$
    @JohnOmielan it is a term used in Lebesgue measure theory. You can find the explanation of this term on Wikipedia.
    $endgroup$
    – Szeto
    Dec 24 '18 at 3:11










  • $begingroup$
    fundamental theorem of calculus applied to the derivatives of $f$ along with the almost everywhere assumption and dominated convergence may lead to something (?). not entirely sure, just throwing out an idea
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 3:11














2












2








2


1



$begingroup$


My question is:




Does there exist an infinitely differentiable function $f$ such that
$$lim_{ntoinfty} f^{(n)}(a)=0qquad{text{for almost all } ain[0,infty)}$$
?




($f$ cannot be a constant function or a polynomial.)




If we restrict us to $C^{omega}$ (i.e. assuming $f$ is holomorphic), it is likely that the answer is no, since
$$f^{(n)}(a)=frac{n!}{2pi i}oint_{gamma}frac{f(z)}{(z-a)^{n+1}}dz$$
the integrand only decays exponentially in $n$ while there is a $n!$ factor there.




However, I am not quite sure if the above argument is correct.



For smooth functions, I have no ideas.



Any help will be appreciated.



Thanks in advance.






real-analysis complex-analysis derivatives







share|cite|improve this question











$endgroup$




My question is:




Does there exist an infinitely differentiable function $f$ such that
$$lim_{ntoinfty} f^{(n)}(a)=0qquad{text{for almost all } ain[0,infty)}$$
?




($f$ cannot be a constant function or a polynomial.)




If we restrict us to $C^{omega}$ (i.e. assuming $f$ is holomorphic), it is likely that the answer is no, since
$$f^{(n)}(a)=frac{n!}{2pi i}oint_{gamma}frac{f(z)}{(z-a)^{n+1}}dz$$
the integrand only decays exponentially in $n$ while there is a $n!$ factor there.




However, I am not quite sure if the above argument is correct.



For smooth functions, I have no ideas.



Any help will be appreciated.



Thanks in advance.






real-analysis complex-analysis derivatives




real-analysis complex-analysis derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 3:11







Szeto

















asked Dec 24 '18 at 3:03









SzetoSzeto

6,7162927




6,7162927












  • $begingroup$
    It's not clear to me what "almost all" means.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 3:10








  • 1




    $begingroup$
    @JohnOmielan it is a term used in Lebesgue measure theory. You can find the explanation of this term on Wikipedia.
    $endgroup$
    – Szeto
    Dec 24 '18 at 3:11










  • $begingroup$
    fundamental theorem of calculus applied to the derivatives of $f$ along with the almost everywhere assumption and dominated convergence may lead to something (?). not entirely sure, just throwing out an idea
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 3:11


















  • $begingroup$
    It's not clear to me what "almost all" means.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 3:10








  • 1




    $begingroup$
    @JohnOmielan it is a term used in Lebesgue measure theory. You can find the explanation of this term on Wikipedia.
    $endgroup$
    – Szeto
    Dec 24 '18 at 3:11










  • $begingroup$
    fundamental theorem of calculus applied to the derivatives of $f$ along with the almost everywhere assumption and dominated convergence may lead to something (?). not entirely sure, just throwing out an idea
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 3:11
















$begingroup$
It's not clear to me what "almost all" means.
$endgroup$
– John Omielan
Dec 24 '18 at 3:10






$begingroup$
It's not clear to me what "almost all" means.
$endgroup$
– John Omielan
Dec 24 '18 at 3:10






1




1




$begingroup$
@JohnOmielan it is a term used in Lebesgue measure theory. You can find the explanation of this term on Wikipedia.
$endgroup$
– Szeto
Dec 24 '18 at 3:11




$begingroup$
@JohnOmielan it is a term used in Lebesgue measure theory. You can find the explanation of this term on Wikipedia.
$endgroup$
– Szeto
Dec 24 '18 at 3:11












$begingroup$
fundamental theorem of calculus applied to the derivatives of $f$ along with the almost everywhere assumption and dominated convergence may lead to something (?). not entirely sure, just throwing out an idea
$endgroup$
– rubikscube09
Dec 24 '18 at 3:11




$begingroup$
fundamental theorem of calculus applied to the derivatives of $f$ along with the almost everywhere assumption and dominated convergence may lead to something (?). not entirely sure, just throwing out an idea
$endgroup$
– rubikscube09
Dec 24 '18 at 3:11










1 Answer
1






active

oldest

votes


















8












$begingroup$

Try $$f(x) = sum_{k=0}^infty frac{x^k}{(k!)^2} = I_0(2 sqrt{x})$$
This is an entire function, and
$$ f^{(n)}(x) = sum_{k=n}^infty frac{x^{k-n}}{k! (k-n)!} = sum_{j=0}^infty frac{x^j}{(j+n)! j!} $$
so that for all $x$
$$ |f^{(n)}(x)| le sum_{j=0}^infty frac{|x|^j}{(j+n)! j!} le frac{1}{n!} sum_{j=0}^infty frac{x^j}{j!} = frac{exp(|x|)}{n!}$$
and thus $f^{(n)}(x) to 0$ for all $x$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    What about $sin(x/2)$ or $e^{x/2}$.
    $endgroup$
    – Shalop
    Dec 24 '18 at 3:43










  • $begingroup$
    Those work too, and much simpler.
    $endgroup$
    – Robert Israel
    Dec 24 '18 at 5:07












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050894%2fcan-f-inftya-0-for-almost-all-a%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









8












$begingroup$

Try $$f(x) = sum_{k=0}^infty frac{x^k}{(k!)^2} = I_0(2 sqrt{x})$$
This is an entire function, and
$$ f^{(n)}(x) = sum_{k=n}^infty frac{x^{k-n}}{k! (k-n)!} = sum_{j=0}^infty frac{x^j}{(j+n)! j!} $$
so that for all $x$
$$ |f^{(n)}(x)| le sum_{j=0}^infty frac{|x|^j}{(j+n)! j!} le frac{1}{n!} sum_{j=0}^infty frac{x^j}{j!} = frac{exp(|x|)}{n!}$$
and thus $f^{(n)}(x) to 0$ for all $x$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    What about $sin(x/2)$ or $e^{x/2}$.
    $endgroup$
    – Shalop
    Dec 24 '18 at 3:43










  • $begingroup$
    Those work too, and much simpler.
    $endgroup$
    – Robert Israel
    Dec 24 '18 at 5:07
















8












$begingroup$

Try $$f(x) = sum_{k=0}^infty frac{x^k}{(k!)^2} = I_0(2 sqrt{x})$$
This is an entire function, and
$$ f^{(n)}(x) = sum_{k=n}^infty frac{x^{k-n}}{k! (k-n)!} = sum_{j=0}^infty frac{x^j}{(j+n)! j!} $$
so that for all $x$
$$ |f^{(n)}(x)| le sum_{j=0}^infty frac{|x|^j}{(j+n)! j!} le frac{1}{n!} sum_{j=0}^infty frac{x^j}{j!} = frac{exp(|x|)}{n!}$$
and thus $f^{(n)}(x) to 0$ for all $x$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    What about $sin(x/2)$ or $e^{x/2}$.
    $endgroup$
    – Shalop
    Dec 24 '18 at 3:43










  • $begingroup$
    Those work too, and much simpler.
    $endgroup$
    – Robert Israel
    Dec 24 '18 at 5:07














8












8








8





$begingroup$

Try $$f(x) = sum_{k=0}^infty frac{x^k}{(k!)^2} = I_0(2 sqrt{x})$$
This is an entire function, and
$$ f^{(n)}(x) = sum_{k=n}^infty frac{x^{k-n}}{k! (k-n)!} = sum_{j=0}^infty frac{x^j}{(j+n)! j!} $$
so that for all $x$
$$ |f^{(n)}(x)| le sum_{j=0}^infty frac{|x|^j}{(j+n)! j!} le frac{1}{n!} sum_{j=0}^infty frac{x^j}{j!} = frac{exp(|x|)}{n!}$$
and thus $f^{(n)}(x) to 0$ for all $x$.






share|cite|improve this answer









$endgroup$



Try $$f(x) = sum_{k=0}^infty frac{x^k}{(k!)^2} = I_0(2 sqrt{x})$$
This is an entire function, and
$$ f^{(n)}(x) = sum_{k=n}^infty frac{x^{k-n}}{k! (k-n)!} = sum_{j=0}^infty frac{x^j}{(j+n)! j!} $$
so that for all $x$
$$ |f^{(n)}(x)| le sum_{j=0}^infty frac{|x|^j}{(j+n)! j!} le frac{1}{n!} sum_{j=0}^infty frac{x^j}{j!} = frac{exp(|x|)}{n!}$$
and thus $f^{(n)}(x) to 0$ for all $x$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 3:15









Robert IsraelRobert Israel

332k23222482




332k23222482








  • 2




    $begingroup$
    What about $sin(x/2)$ or $e^{x/2}$.
    $endgroup$
    – Shalop
    Dec 24 '18 at 3:43










  • $begingroup$
    Those work too, and much simpler.
    $endgroup$
    – Robert Israel
    Dec 24 '18 at 5:07














  • 2




    $begingroup$
    What about $sin(x/2)$ or $e^{x/2}$.
    $endgroup$
    – Shalop
    Dec 24 '18 at 3:43










  • $begingroup$
    Those work too, and much simpler.
    $endgroup$
    – Robert Israel
    Dec 24 '18 at 5:07








2




2




$begingroup$
What about $sin(x/2)$ or $e^{x/2}$.
$endgroup$
– Shalop
Dec 24 '18 at 3:43




$begingroup$
What about $sin(x/2)$ or $e^{x/2}$.
$endgroup$
– Shalop
Dec 24 '18 at 3:43












$begingroup$
Those work too, and much simpler.
$endgroup$
– Robert Israel
Dec 24 '18 at 5:07




$begingroup$
Those work too, and much simpler.
$endgroup$
– Robert Israel
Dec 24 '18 at 5:07


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050894%2fcan-f-inftya-0-for-almost-all-a%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more