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












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1 Answer
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1 Answer
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active

oldest

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oldest

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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


















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