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Is there any sequence of functions containing functions growing slower than any smooth function?
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The function
$f(x)= begincases
e^-frac1x & x > 0 \
0 & xleq 0 \
endcases$
is a smooth function which grows, at $0$, slower than any function of the form $x^n$.
My question is, does there exist a sequence of functions $(f_n)$, infinitely differentiable at $0$ and with $f_n(0)=0$ and $f_n(x)neq 0$ when $xneq 0$ for all $n$, with the following property?
For any function $f$ infinitely differentiable at 0 where $f(0)=0$ and $f(x)neq 0$ when $xneq 0$ there exists a natural number $N$ such that for all $ngeq N$, we have $lim_xrightarrow 0fracf_n(x)f(x)=0$.
Or is there no such sequence of functions, i.e. for any sequence of functions does there exist a function which grows more slowly than all the functions in the sequence?
calculus real-analysis limits
asked Aug 6 at 6:31
Keshav Srinivasan
1,79111338
add a comment |Â
up vote
6
down vote
favorite
The function
$f(x)= begincases
e^-frac1x & x > 0 \
0 & xleq 0 \
endcases$
is a smooth function which grows, at $0$, slower than any function of the form $x^n$.
My question is, does there exist a sequence of functions $(f_n)$, infinitely differentiable at $0$ and with $f_n(0)=0$ and $f_n(x)neq 0$ when $xneq 0$ for all $n$, with the following property?
For any function $f$ infinitely differentiable at 0 where $f(0)=0$ and $f(x)neq 0$ when $xneq 0$ there exists a natural number $N$ such that for all $ngeq N$, we have $lim_xrightarrow 0fracf_n(x)f(x)=0$.
Or is there no such sequence of functions, i.e. for any sequence of functions does there exist a function which grows more slowly than all the functions in the sequence?
calculus real-analysis limits
asked Aug 6 at 6:31
Keshav Srinivasan
1,79111338
Didn't you mean $lim_xrightarrow0 fracf_n(x)f(x)=0$ and shouldn't there be some restrictions on $f$ to allow the division? E.g., $x^2$ is slower growing than $x$ at $x=0$ and $lim_xrightarrow0 fracx^2x=0$. ... Or maybe I misunderstood the core topic of the question.
– Tobias
Aug 6 at 15:04
@Tobias You’re right, I edited the question.
– Keshav Srinivasan
Aug 6 at 15:29
With the current formulation of the question there is such a sequence, i.e., $f_n(x):=0$ for all admissible $x$ and $n$. That means you need some more restrictions on $f_n$.
– Tobias
Aug 6 at 15:37
@Tobias OK, what about this?
– Keshav Srinivasan
Aug 6 at 15:39
@KeshavSrinivasan I improved my answer to a better result.
– zhw.
Aug 8 at 17:33
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
The function
$f(x)= begincases
e^-frac1x & x > 0 \
0 & xleq 0 \
endcases$
is a smooth function which grows, at $0$, slower than any function of the form $x^n$.
My question is, does there exist a sequence of functions $(f_n)$, infinitely differentiable at $0$ and with $f_n(0)=0$ and $f_n(x)neq 0$ when $xneq 0$ for all $n$, with the following property?
For any function $f$ infinitely differentiable at 0 where $f(0)=0$ and $f(x)neq 0$ when $xneq 0$ there exists a natural number $N$ such that for all $ngeq N$, we have $lim_xrightarrow 0fracf_n(x)f(x)=0$.
Or is there no such sequence of functions, i.e. for any sequence of functions does there exist a function which grows more slowly than all the functions in the sequence?
calculus real-analysis limits
asked Aug 6 at 6:31
Keshav Srinivasan
1,79111338
The function
$f(x)= begincases
e^-frac1x & x > 0 \
0 & xleq 0 \
endcases$
is a smooth function which grows, at $0$, slower than any function of the form $x^n$.
My question is, does there exist a sequence of functions $(f_n)$, infinitely differentiable at $0$ and with $f_n(0)=0$ and $f_n(x)neq 0$ when $xneq 0$ for all $n$, with the following property?
For any function $f$ infinitely differentiable at 0 where $f(0)=0$ and $f(x)neq 0$ when $xneq 0$ there exists a natural number $N$ such that for all $ngeq N$, we have $lim_xrightarrow 0fracf_n(x)f(x)=0$.
Or is there no such sequence of functions, i.e. for any sequence of functions does there exist a function which grows more slowly than all the functions in the sequence?
calculus real-analysis limits
asked Aug 6 at 6:31
Keshav Srinivasan
1,79111338
edited Aug 6 at 15:38
asked Aug 6 at 6:31
Keshav Srinivasan
1,79111338
asked Aug 6 at 6:31
Keshav Srinivasan
1,79111338
asked Aug 6 at 6:31
Keshav Srinivasan
1,79111338
1,79111338
Didn't you mean $lim_xrightarrow0 fracf_n(x)f(x)=0$ and shouldn't there be some restrictions on $f$ to allow the division? E.g., $x^2$ is slower growing than $x$ at $x=0$ and $lim_xrightarrow0 fracx^2x=0$. ... Or maybe I misunderstood the core topic of the question.
– Tobias
Aug 6 at 15:04
@Tobias You’re right, I edited the question.
– Keshav Srinivasan
Aug 6 at 15:29
With the current formulation of the question there is such a sequence, i.e., $f_n(x):=0$ for all admissible $x$ and $n$. That means you need some more restrictions on $f_n$.
– Tobias
Aug 6 at 15:37
@Tobias OK, what about this?
– Keshav Srinivasan
Aug 6 at 15:39
@KeshavSrinivasan I improved my answer to a better result.
– zhw.
Aug 8 at 17:33
add a comment |Â
Didn't you mean $lim_xrightarrow0 fracf_n(x)f(x)=0$ and shouldn't there be some restrictions on $f$ to allow the division? E.g., $x^2$ is slower growing than $x$ at $x=0$ and $lim_xrightarrow0 fracx^2x=0$. ... Or maybe I misunderstood the core topic of the question.
– Tobias
Aug 6 at 15:04
@Tobias You’re right, I edited the question.
– Keshav Srinivasan
Aug 6 at 15:29
With the current formulation of the question there is such a sequence, i.e., $f_n(x):=0$ for all admissible $x$ and $n$. That means you need some more restrictions on $f_n$.
– Tobias
Aug 6 at 15:37
@Tobias OK, what about this?
– Keshav Srinivasan
Aug 6 at 15:39
@KeshavSrinivasan I improved my answer to a better result.
– zhw.
Aug 8 at 17:33
Didn't you mean $lim_xrightarrow0 fracf_n(x)f(x)=0$ and shouldn't there be some restrictions on $f$ to allow the division? E.g., $x^2$ is slower growing than $x$ at $x=0$ and $lim_xrightarrow0 fracx^2x=0$. ... Or maybe I misunderstood the core topic of the question.
– Tobias
Aug 6 at 15:04
Didn't you mean $lim_xrightarrow0 fracf_n(x)f(x)=0$ and shouldn't there be some restrictions on $f$ to allow the division? E.g., $x^2$ is slower growing than $x$ at $x=0$ and $lim_xrightarrow0 fracx^2x=0$. ... Or maybe I misunderstood the core topic of the question.
– Tobias
Aug 6 at 15:04
@Tobias You’re right, I edited the question.
– Keshav Srinivasan
Aug 6 at 15:29
@Tobias You’re right, I edited the question.
– Keshav Srinivasan
Aug 6 at 15:29
With the current formulation of the question there is such a sequence, i.e., $f_n(x):=0$ for all admissible $x$ and $n$. That means you need some more restrictions on $f_n$.
– Tobias
Aug 6 at 15:37
With the current formulation of the question there is such a sequence, i.e., $f_n(x):=0$ for all admissible $x$ and $n$. That means you need some more restrictions on $f_n$.
– Tobias
Aug 6 at 15:37
@Tobias OK, what about this?
– Keshav Srinivasan
Aug 6 at 15:39
@Tobias OK, what about this?
– Keshav Srinivasan
Aug 6 at 15:39
@KeshavSrinivasan I improved my answer to a better result.
– zhw.
Aug 8 at 17:33
@KeshavSrinivasan I improved my answer to a better result.
– zhw.
Aug 8 at 17:33
add a comment |Â
4 Answers
4
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up vote
1
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accepted
There is no such $C^infty$ function. To prove this, let $f_n$ satisfy the conditions you specify. WLOG, we can assume all $f_n>0$ on $mathbb Rsetminus 0.$ If that is not true, go to the functions $f_n^2.$ We will construct an even $fin C^infty(mathbb R),$ positive on $mathbb Rsetminus 0$ and strictly increasing on $[0,infty),$ such that for every $n,$
$$tag 1lim_xto 0 fracf_n(x)f(x)=infty.$$
For any $gin C^l(mathbb R),$ define
$$|g|_l = sum_j=0^lsup_mathbb R |D^jg|.$$
If $g_kin C^infty(mathbb R), k = 1,2,dots,$ and $sum_k=1^infty |g_k|_l <infty$ for each $l,$ then $sum_k=1^infty g_k in C^infty(mathbb R).$ Furthermore, for each $l,$
$$D^lleft(sum_k=1^infty g_kright ) = sum_k=1^infty D^lg_k.$$
Choose a sequence $a_1>a_2 > cdots to 0.$ For each $k,$ set $E_k=x$ and define $m_k$ to be the smallest of the numbers
$$min_E_k f_1,min_E_k f_2,dots,min_E_k f_k.$$
Now choose a positive sequence $c_kto 0$ such that $c_km_k$ strictly decreases to $0.$
For each $k$ we can choose $g_kin C^infty(mathbb R)$ with $g_k>0$ on $(a_k+1,a_k)$ and $g_k=0$ elsewhere. By multiplying by small enough positive constants we can also require
$$|g_k|_k < 1/2^k text and int_a_k+1^a_k g_k < c_km_k - c_k+1m_k+1.$$
We then define
$$g = sum_k=1^inftyg_k.$$
This $gin C^infty(mathbb R).$
Finally we get to define $f$ (almost): Define
$$f(x) = int_0^x g,,, xin mathbb R.$$
Then $fin C^infty(mathbb R)$ and $f$ is strictly increasing on $[0,infty).$ Fix $n.$ Claim: If $xin [a_k+1,a_k]$ and $kge n,$ then
$$tag 2fracf_n(x)f(x) ge frac1c_k.$$
Since $c_kto 0^+,$ $(2)$ proves $(1),$ at least as $xto 0^+.$ To prove $(2),$ note $f(x)le f(a_k).$ Now
$$f(a_k) = int_0^a_k g =sum_j=k^inftyint_a_j+1^a_j g_j < sum_j=k^infty(c_jm_j - c_j+1m_j+1) = c_km_k.$$
So
$$fracf_n(x)f(x)ge fracf_n(x)f(a_k) > fracm_kc_km_k=frac1c_k,$$
proving $(2).$
We're done except for a minor detail. This $f=0$ on $(-infty,0].$ All we need to do is redefine $f$ for $xle 0$ by setting $f(x)=f(-x).$ Because all $D^lf(0)=0$ for the original $f,$ the redefined $fin C^infty(mathbb R ).$ Since the $m_k$ took into account the behavior of the $f_n$ to the left of $0,$ we have our counterexample as claimed.
answered Aug 8 at 2:07
zhw.
66.1k42870
I think we can do better: There should exist $f$ with $$lim_xto 0 fracf_n(x)f(x)=infty$$ for each $n.$ I'll look into this.
– zhw.
Aug 8 at 15:22
Wow, your edit makes this result even more impressive. This means that the order structure of the set of functions ordered according to how fast or slow they grow has a very interesting structure. It constitutes an ordered set where every countable subset has an upper bound. It may be order-isomorphic to the Long Line or something. I think I’ll post a question on this.
– Keshav Srinivasan
Aug 10 at 1:08
I just posted a follow-up question: math.stackexchange.com/q/2877922/71829
– Keshav Srinivasan
Aug 10 at 2:11
add a comment |Â
up vote
1
down vote
There is no such sequence at least if we don't insist on the requirement that $f$ is infinitely often differentiable.
Assume the contrary.
Construct $a_n$ with $emptysetneqleft(-a_n,a_nright)subseteq f_n^-1left(left(-frac1n,frac1nright)right)$ and $a_n+1leq fraca_n2$ for $n=1,2,ldots$. (Possible since the $f_n$ are continuous at $0$.)
Define $f(x) := f_n(x)$ for $xin (-a_n,a_n)setminus(-a_n+1,a_n+1)$.
Now assume $lim_xrightarrow 0fracf_n(x)f(x)=0$ for $ngeq N$. Then we have some $delta>0$ such that $left|fracf_n(x)f(x)right|< 0.5$ for all $xin(-delta,delta)$ and $ngeq N$. But we also have some $ngeq N$ with $(-a_n,a_n)subset (-delta,delta)$ and $fracf_n(x)f(x)=1$ for $xin(-a_n,a_n)$.
answered Aug 6 at 15:45
Tobias
483314
Thanks for your answer.
– Keshav Srinivasan
Aug 6 at 15:47
2
How do you know that $f(x) to 0$ as $x to 0$?
– PhoemueX
Aug 6 at 15:51
@PhoemueX Nice catch. We might even have $f(x)=1$ for $xneq 0$ if each of the $f_n$ only gets $neq 1$ for $xinleft(0,frac12(n+1)right)$:-). So some more work is required... I knew it couldn't be that simple.
– Tobias
Aug 6 at 16:06
@KeshavSrinivasan Sorry, I don't have too much time. So I relaxed the strong differentiability conditions for $f$ to make the proof simple. I think it also shows in that form what you essentially wanted to see. Maybe someone other shows the statement for differentiable $f$ and you can accept his answer.
– Tobias
Aug 6 at 17:06
@PhoemueX I dealt with your argument. Hope the proof for the relaxed statement (without differentiability requirement for $f$) is correct now. (Thanks again for your comment.)
– Tobias
Aug 6 at 17:07
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0
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There may be some typos in your question. We can simply put, for each $ninmathbbN$, $f_n(x)=1$.
Then, for any $f$ such that $f'(0)$ exists and $f(0)=0$, we automatically have: For each $n$, $lim_xrightarrow0fracf(x)f_n(x)=lim_xrightarrow0f(x)=0$
because $f$ is continuous at $x=0$.
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
I guess the OP wants $f_n (0)=0$ for all $n$ (although he didn't write it).
– PhoemueX
Aug 6 at 7:15
@PhoemueX Yeah, I edited my question.
– Keshav Srinivasan
Aug 6 at 14:18
add a comment |Â
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0
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You can use
$$f(x)= begincases
e^-frac1x^n & x > 0 \
0 & xleq 0 \
endcases$$
for $n=1,2,3,4,ldots$ as a slower and slower growing sequence.
answered Aug 6 at 14:34
Ross Millikan
276k21187352
Is there no smooth function that grows slower than all the functions in that sequence? Can you prove that?
– Keshav Srinivasan
Aug 6 at 14:35
1
These things always have a next step. The next I would think of is $$e^-frac 1e^-frac 1x$$ but I am not sure it works.
– Ross Millikan
Aug 6 at 14:45
So is there no sequence of functions such that no function grows slower than them all?
– Keshav Srinivasan
Aug 6 at 15:30
I don't believe so
– Ross Millikan
Aug 6 at 15:38
1
Yes, but the statements “For every function there is a slower growing function.†and “There exists a sequence of functions such that no function grows slower than all the functions in the sequence.†are perfectly consistent with one another.
– Keshav Srinivasan
Aug 6 at 16:04
 |Â
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There is no such $C^infty$ function. To prove this, let $f_n$ satisfy the conditions you specify. WLOG, we can assume all $f_n>0$ on $mathbb Rsetminus 0.$ If that is not true, go to the functions $f_n^2.$ We will construct an even $fin C^infty(mathbb R),$ positive on $mathbb Rsetminus 0$ and strictly increasing on $[0,infty),$ such that for every $n,$
$$tag 1lim_xto 0 fracf_n(x)f(x)=infty.$$
For any $gin C^l(mathbb R),$ define
$$|g|_l = sum_j=0^lsup_mathbb R |D^jg|.$$
If $g_kin C^infty(mathbb R), k = 1,2,dots,$ and $sum_k=1^infty |g_k|_l <infty$ for each $l,$ then $sum_k=1^infty g_k in C^infty(mathbb R).$ Furthermore, for each $l,$
$$D^lleft(sum_k=1^infty g_kright ) = sum_k=1^infty D^lg_k.$$
Choose a sequence $a_1>a_2 > cdots to 0.$ For each $k,$ set $E_k=x$ and define $m_k$ to be the smallest of the numbers
$$min_E_k f_1,min_E_k f_2,dots,min_E_k f_k.$$
Now choose a positive sequence $c_kto 0$ such that $c_km_k$ strictly decreases to $0.$
For each $k$ we can choose $g_kin C^infty(mathbb R)$ with $g_k>0$ on $(a_k+1,a_k)$ and $g_k=0$ elsewhere. By multiplying by small enough positive constants we can also require
$$|g_k|_k < 1/2^k text and int_a_k+1^a_k g_k < c_km_k - c_k+1m_k+1.$$
We then define
$$g = sum_k=1^inftyg_k.$$
This $gin C^infty(mathbb R).$
Finally we get to define $f$ (almost): Define
$$f(x) = int_0^x g,,, xin mathbb R.$$
Then $fin C^infty(mathbb R)$ and $f$ is strictly increasing on $[0,infty).$ Fix $n.$ Claim: If $xin [a_k+1,a_k]$ and $kge n,$ then
$$tag 2fracf_n(x)f(x) ge frac1c_k.$$
Since $c_kto 0^+,$ $(2)$ proves $(1),$ at least as $xto 0^+.$ To prove $(2),$ note $f(x)le f(a_k).$ Now
$$f(a_k) = int_0^a_k g =sum_j=k^inftyint_a_j+1^a_j g_j < sum_j=k^infty(c_jm_j - c_j+1m_j+1) = c_km_k.$$
So
$$fracf_n(x)f(x)ge fracf_n(x)f(a_k) > fracm_kc_km_k=frac1c_k,$$
proving $(2).$
We're done except for a minor detail. This $f=0$ on $(-infty,0].$ All we need to do is redefine $f$ for $xle 0$ by setting $f(x)=f(-x).$ Because all $D^lf(0)=0$ for the original $f,$ the redefined $fin C^infty(mathbb R ).$ Since the $m_k$ took into account the behavior of the $f_n$ to the left of $0,$ we have our counterexample as claimed.
answered Aug 8 at 2:07
zhw.
66.1k42870
I think we can do better: There should exist $f$ with $$lim_xto 0 fracf_n(x)f(x)=infty$$ for each $n.$ I'll look into this.
– zhw.
Aug 8 at 15:22
Wow, your edit makes this result even more impressive. This means that the order structure of the set of functions ordered according to how fast or slow they grow has a very interesting structure. It constitutes an ordered set where every countable subset has an upper bound. It may be order-isomorphic to the Long Line or something. I think I’ll post a question on this.
– Keshav Srinivasan
Aug 10 at 1:08
I just posted a follow-up question: math.stackexchange.com/q/2877922/71829
– Keshav Srinivasan
Aug 10 at 2:11
add a comment |Â
up vote
1
down vote
accepted
There is no such $C^infty$ function. To prove this, let $f_n$ satisfy the conditions you specify. WLOG, we can assume all $f_n>0$ on $mathbb Rsetminus 0.$ If that is not true, go to the functions $f_n^2.$ We will construct an even $fin C^infty(mathbb R),$ positive on $mathbb Rsetminus 0$ and strictly increasing on $[0,infty),$ such that for every $n,$
$$tag 1lim_xto 0 fracf_n(x)f(x)=infty.$$
For any $gin C^l(mathbb R),$ define
$$|g|_l = sum_j=0^lsup_mathbb R |D^jg|.$$
If $g_kin C^infty(mathbb R), k = 1,2,dots,$ and $sum_k=1^infty |g_k|_l <infty$ for each $l,$ then $sum_k=1^infty g_k in C^infty(mathbb R).$ Furthermore, for each $l,$
$$D^lleft(sum_k=1^infty g_kright ) = sum_k=1^infty D^lg_k.$$
Choose a sequence $a_1>a_2 > cdots to 0.$ For each $k,$ set $E_k=x$ and define $m_k$ to be the smallest of the numbers
$$min_E_k f_1,min_E_k f_2,dots,min_E_k f_k.$$
Now choose a positive sequence $c_kto 0$ such that $c_km_k$ strictly decreases to $0.$
For each $k$ we can choose $g_kin C^infty(mathbb R)$ with $g_k>0$ on $(a_k+1,a_k)$ and $g_k=0$ elsewhere. By multiplying by small enough positive constants we can also require
$$|g_k|_k < 1/2^k text and int_a_k+1^a_k g_k < c_km_k - c_k+1m_k+1.$$
We then define
$$g = sum_k=1^inftyg_k.$$
This $gin C^infty(mathbb R).$
Finally we get to define $f$ (almost): Define
$$f(x) = int_0^x g,,, xin mathbb R.$$
Then $fin C^infty(mathbb R)$ and $f$ is strictly increasing on $[0,infty).$ Fix $n.$ Claim: If $xin [a_k+1,a_k]$ and $kge n,$ then
$$tag 2fracf_n(x)f(x) ge frac1c_k.$$
Since $c_kto 0^+,$ $(2)$ proves $(1),$ at least as $xto 0^+.$ To prove $(2),$ note $f(x)le f(a_k).$ Now
$$f(a_k) = int_0^a_k g =sum_j=k^inftyint_a_j+1^a_j g_j < sum_j=k^infty(c_jm_j - c_j+1m_j+1) = c_km_k.$$
So
$$fracf_n(x)f(x)ge fracf_n(x)f(a_k) > fracm_kc_km_k=frac1c_k,$$
proving $(2).$
We're done except for a minor detail. This $f=0$ on $(-infty,0].$ All we need to do is redefine $f$ for $xle 0$ by setting $f(x)=f(-x).$ Because all $D^lf(0)=0$ for the original $f,$ the redefined $fin C^infty(mathbb R ).$ Since the $m_k$ took into account the behavior of the $f_n$ to the left of $0,$ we have our counterexample as claimed.
answered Aug 8 at 2:07
zhw.
66.1k42870
I think we can do better: There should exist $f$ with $$lim_xto 0 fracf_n(x)f(x)=infty$$ for each $n.$ I'll look into this.
– zhw.
Aug 8 at 15:22
Wow, your edit makes this result even more impressive. This means that the order structure of the set of functions ordered according to how fast or slow they grow has a very interesting structure. It constitutes an ordered set where every countable subset has an upper bound. It may be order-isomorphic to the Long Line or something. I think I’ll post a question on this.
– Keshav Srinivasan
Aug 10 at 1:08
I just posted a follow-up question: math.stackexchange.com/q/2877922/71829
– Keshav Srinivasan
Aug 10 at 2:11
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is no such $C^infty$ function. To prove this, let $f_n$ satisfy the conditions you specify. WLOG, we can assume all $f_n>0$ on $mathbb Rsetminus 0.$ If that is not true, go to the functions $f_n^2.$ We will construct an even $fin C^infty(mathbb R),$ positive on $mathbb Rsetminus 0$ and strictly increasing on $[0,infty),$ such that for every $n,$
$$tag 1lim_xto 0 fracf_n(x)f(x)=infty.$$
For any $gin C^l(mathbb R),$ define
$$|g|_l = sum_j=0^lsup_mathbb R |D^jg|.$$
If $g_kin C^infty(mathbb R), k = 1,2,dots,$ and $sum_k=1^infty |g_k|_l <infty$ for each $l,$ then $sum_k=1^infty g_k in C^infty(mathbb R).$ Furthermore, for each $l,$
$$D^lleft(sum_k=1^infty g_kright ) = sum_k=1^infty D^lg_k.$$
Choose a sequence $a_1>a_2 > cdots to 0.$ For each $k,$ set $E_k=x$ and define $m_k$ to be the smallest of the numbers
$$min_E_k f_1,min_E_k f_2,dots,min_E_k f_k.$$
Now choose a positive sequence $c_kto 0$ such that $c_km_k$ strictly decreases to $0.$
For each $k$ we can choose $g_kin C^infty(mathbb R)$ with $g_k>0$ on $(a_k+1,a_k)$ and $g_k=0$ elsewhere. By multiplying by small enough positive constants we can also require
$$|g_k|_k < 1/2^k text and int_a_k+1^a_k g_k < c_km_k - c_k+1m_k+1.$$
We then define
$$g = sum_k=1^inftyg_k.$$
This $gin C^infty(mathbb R).$
Finally we get to define $f$ (almost): Define
$$f(x) = int_0^x g,,, xin mathbb R.$$
Then $fin C^infty(mathbb R)$ and $f$ is strictly increasing on $[0,infty).$ Fix $n.$ Claim: If $xin [a_k+1,a_k]$ and $kge n,$ then
$$tag 2fracf_n(x)f(x) ge frac1c_k.$$
Since $c_kto 0^+,$ $(2)$ proves $(1),$ at least as $xto 0^+.$ To prove $(2),$ note $f(x)le f(a_k).$ Now
$$f(a_k) = int_0^a_k g =sum_j=k^inftyint_a_j+1^a_j g_j < sum_j=k^infty(c_jm_j - c_j+1m_j+1) = c_km_k.$$
So
$$fracf_n(x)f(x)ge fracf_n(x)f(a_k) > fracm_kc_km_k=frac1c_k,$$
proving $(2).$
We're done except for a minor detail. This $f=0$ on $(-infty,0].$ All we need to do is redefine $f$ for $xle 0$ by setting $f(x)=f(-x).$ Because all $D^lf(0)=0$ for the original $f,$ the redefined $fin C^infty(mathbb R ).$ Since the $m_k$ took into account the behavior of the $f_n$ to the left of $0,$ we have our counterexample as claimed.
answered Aug 8 at 2:07
zhw.
66.1k42870
There is no such $C^infty$ function. To prove this, let $f_n$ satisfy the conditions you specify. WLOG, we can assume all $f_n>0$ on $mathbb Rsetminus 0.$ If that is not true, go to the functions $f_n^2.$ We will construct an even $fin C^infty(mathbb R),$ positive on $mathbb Rsetminus 0$ and strictly increasing on $[0,infty),$ such that for every $n,$
$$tag 1lim_xto 0 fracf_n(x)f(x)=infty.$$
For any $gin C^l(mathbb R),$ define
$$|g|_l = sum_j=0^lsup_mathbb R |D^jg|.$$
If $g_kin C^infty(mathbb R), k = 1,2,dots,$ and $sum_k=1^infty |g_k|_l <infty$ for each $l,$ then $sum_k=1^infty g_k in C^infty(mathbb R).$ Furthermore, for each $l,$
$$D^lleft(sum_k=1^infty g_kright ) = sum_k=1^infty D^lg_k.$$
Choose a sequence $a_1>a_2 > cdots to 0.$ For each $k,$ set $E_k=x$ and define $m_k$ to be the smallest of the numbers
$$min_E_k f_1,min_E_k f_2,dots,min_E_k f_k.$$
Now choose a positive sequence $c_kto 0$ such that $c_km_k$ strictly decreases to $0.$
For each $k$ we can choose $g_kin C^infty(mathbb R)$ with $g_k>0$ on $(a_k+1,a_k)$ and $g_k=0$ elsewhere. By multiplying by small enough positive constants we can also require
$$|g_k|_k < 1/2^k text and int_a_k+1^a_k g_k < c_km_k - c_k+1m_k+1.$$
We then define
$$g = sum_k=1^inftyg_k.$$
This $gin C^infty(mathbb R).$
Finally we get to define $f$ (almost): Define
$$f(x) = int_0^x g,,, xin mathbb R.$$
Then $fin C^infty(mathbb R)$ and $f$ is strictly increasing on $[0,infty).$ Fix $n.$ Claim: If $xin [a_k+1,a_k]$ and $kge n,$ then
$$tag 2fracf_n(x)f(x) ge frac1c_k.$$
Since $c_kto 0^+,$ $(2)$ proves $(1),$ at least as $xto 0^+.$ To prove $(2),$ note $f(x)le f(a_k).$ Now
$$f(a_k) = int_0^a_k g =sum_j=k^inftyint_a_j+1^a_j g_j < sum_j=k^infty(c_jm_j - c_j+1m_j+1) = c_km_k.$$
So
$$fracf_n(x)f(x)ge fracf_n(x)f(a_k) > fracm_kc_km_k=frac1c_k,$$
proving $(2).$
We're done except for a minor detail. This $f=0$ on $(-infty,0].$ All we need to do is redefine $f$ for $xle 0$ by setting $f(x)=f(-x).$ Because all $D^lf(0)=0$ for the original $f,$ the redefined $fin C^infty(mathbb R ).$ Since the $m_k$ took into account the behavior of the $f_n$ to the left of $0,$ we have our counterexample as claimed.
answered Aug 8 at 2:07
zhw.
66.1k42870
edited Aug 8 at 17:32
answered Aug 8 at 2:07
zhw.
66.1k42870
answered Aug 8 at 2:07
zhw.
66.1k42870
answered Aug 8 at 2:07
zhw.
66.1k42870
66.1k42870
I think we can do better: There should exist $f$ with $$lim_xto 0 fracf_n(x)f(x)=infty$$ for each $n.$ I'll look into this.
– zhw.
Aug 8 at 15:22
Wow, your edit makes this result even more impressive. This means that the order structure of the set of functions ordered according to how fast or slow they grow has a very interesting structure. It constitutes an ordered set where every countable subset has an upper bound. It may be order-isomorphic to the Long Line or something. I think I’ll post a question on this.
– Keshav Srinivasan
Aug 10 at 1:08
I just posted a follow-up question: math.stackexchange.com/q/2877922/71829
– Keshav Srinivasan
Aug 10 at 2:11
add a comment |Â
I think we can do better: There should exist $f$ with $$lim_xto 0 fracf_n(x)f(x)=infty$$ for each $n.$ I'll look into this.
– zhw.
Aug 8 at 15:22
Wow, your edit makes this result even more impressive. This means that the order structure of the set of functions ordered according to how fast or slow they grow has a very interesting structure. It constitutes an ordered set where every countable subset has an upper bound. It may be order-isomorphic to the Long Line or something. I think I’ll post a question on this.
– Keshav Srinivasan
Aug 10 at 1:08
I just posted a follow-up question: math.stackexchange.com/q/2877922/71829
– Keshav Srinivasan
Aug 10 at 2:11
I think we can do better: There should exist $f$ with $$lim_xto 0 fracf_n(x)f(x)=infty$$ for each $n.$ I'll look into this.
– zhw.
Aug 8 at 15:22
I think we can do better: There should exist $f$ with $$lim_xto 0 fracf_n(x)f(x)=infty$$ for each $n.$ I'll look into this.
– zhw.
Aug 8 at 15:22
Wow, your edit makes this result even more impressive. This means that the order structure of the set of functions ordered according to how fast or slow they grow has a very interesting structure. It constitutes an ordered set where every countable subset has an upper bound. It may be order-isomorphic to the Long Line or something. I think I’ll post a question on this.
– Keshav Srinivasan
Aug 10 at 1:08
Wow, your edit makes this result even more impressive. This means that the order structure of the set of functions ordered according to how fast or slow they grow has a very interesting structure. It constitutes an ordered set where every countable subset has an upper bound. It may be order-isomorphic to the Long Line or something. I think I’ll post a question on this.
– Keshav Srinivasan
Aug 10 at 1:08
I just posted a follow-up question: math.stackexchange.com/q/2877922/71829
– Keshav Srinivasan
Aug 10 at 2:11
I just posted a follow-up question: math.stackexchange.com/q/2877922/71829
– Keshav Srinivasan
Aug 10 at 2:11
add a comment |Â
up vote
1
down vote
There is no such sequence at least if we don't insist on the requirement that $f$ is infinitely often differentiable.
Assume the contrary.
Construct $a_n$ with $emptysetneqleft(-a_n,a_nright)subseteq f_n^-1left(left(-frac1n,frac1nright)right)$ and $a_n+1leq fraca_n2$ for $n=1,2,ldots$. (Possible since the $f_n$ are continuous at $0$.)
Define $f(x) := f_n(x)$ for $xin (-a_n,a_n)setminus(-a_n+1,a_n+1)$.
Now assume $lim_xrightarrow 0fracf_n(x)f(x)=0$ for $ngeq N$. Then we have some $delta>0$ such that $left|fracf_n(x)f(x)right|< 0.5$ for all $xin(-delta,delta)$ and $ngeq N$. But we also have some $ngeq N$ with $(-a_n,a_n)subset (-delta,delta)$ and $fracf_n(x)f(x)=1$ for $xin(-a_n,a_n)$.
answered Aug 6 at 15:45
Tobias
483314
Thanks for your answer.
– Keshav Srinivasan
Aug 6 at 15:47
2
How do you know that $f(x) to 0$ as $x to 0$?
– PhoemueX
Aug 6 at 15:51
@PhoemueX Nice catch. We might even have $f(x)=1$ for $xneq 0$ if each of the $f_n$ only gets $neq 1$ for $xinleft(0,frac12(n+1)right)$:-). So some more work is required... I knew it couldn't be that simple.
– Tobias
Aug 6 at 16:06
@KeshavSrinivasan Sorry, I don't have too much time. So I relaxed the strong differentiability conditions for $f$ to make the proof simple. I think it also shows in that form what you essentially wanted to see. Maybe someone other shows the statement for differentiable $f$ and you can accept his answer.
– Tobias
Aug 6 at 17:06
@PhoemueX I dealt with your argument. Hope the proof for the relaxed statement (without differentiability requirement for $f$) is correct now. (Thanks again for your comment.)
– Tobias
Aug 6 at 17:07
 |Â
show 2 more comments
up vote
1
down vote
There is no such sequence at least if we don't insist on the requirement that $f$ is infinitely often differentiable.
Assume the contrary.
Construct $a_n$ with $emptysetneqleft(-a_n,a_nright)subseteq f_n^-1left(left(-frac1n,frac1nright)right)$ and $a_n+1leq fraca_n2$ for $n=1,2,ldots$. (Possible since the $f_n$ are continuous at $0$.)
Define $f(x) := f_n(x)$ for $xin (-a_n,a_n)setminus(-a_n+1,a_n+1)$.
Now assume $lim_xrightarrow 0fracf_n(x)f(x)=0$ for $ngeq N$. Then we have some $delta>0$ such that $left|fracf_n(x)f(x)right|< 0.5$ for all $xin(-delta,delta)$ and $ngeq N$. But we also have some $ngeq N$ with $(-a_n,a_n)subset (-delta,delta)$ and $fracf_n(x)f(x)=1$ for $xin(-a_n,a_n)$.
answered Aug 6 at 15:45
Tobias
483314
Thanks for your answer.
– Keshav Srinivasan
Aug 6 at 15:47
2
How do you know that $f(x) to 0$ as $x to 0$?
– PhoemueX
Aug 6 at 15:51
@PhoemueX Nice catch. We might even have $f(x)=1$ for $xneq 0$ if each of the $f_n$ only gets $neq 1$ for $xinleft(0,frac12(n+1)right)$:-). So some more work is required... I knew it couldn't be that simple.
– Tobias
Aug 6 at 16:06
@KeshavSrinivasan Sorry, I don't have too much time. So I relaxed the strong differentiability conditions for $f$ to make the proof simple. I think it also shows in that form what you essentially wanted to see. Maybe someone other shows the statement for differentiable $f$ and you can accept his answer.
– Tobias
Aug 6 at 17:06
@PhoemueX I dealt with your argument. Hope the proof for the relaxed statement (without differentiability requirement for $f$) is correct now. (Thanks again for your comment.)
– Tobias
Aug 6 at 17:07
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
There is no such sequence at least if we don't insist on the requirement that $f$ is infinitely often differentiable.
Assume the contrary.
Construct $a_n$ with $emptysetneqleft(-a_n,a_nright)subseteq f_n^-1left(left(-frac1n,frac1nright)right)$ and $a_n+1leq fraca_n2$ for $n=1,2,ldots$. (Possible since the $f_n$ are continuous at $0$.)
Define $f(x) := f_n(x)$ for $xin (-a_n,a_n)setminus(-a_n+1,a_n+1)$.
Now assume $lim_xrightarrow 0fracf_n(x)f(x)=0$ for $ngeq N$. Then we have some $delta>0$ such that $left|fracf_n(x)f(x)right|< 0.5$ for all $xin(-delta,delta)$ and $ngeq N$. But we also have some $ngeq N$ with $(-a_n,a_n)subset (-delta,delta)$ and $fracf_n(x)f(x)=1$ for $xin(-a_n,a_n)$.
answered Aug 6 at 15:45
Tobias
483314
There is no such sequence at least if we don't insist on the requirement that $f$ is infinitely often differentiable.
Assume the contrary.
Construct $a_n$ with $emptysetneqleft(-a_n,a_nright)subseteq f_n^-1left(left(-frac1n,frac1nright)right)$ and $a_n+1leq fraca_n2$ for $n=1,2,ldots$. (Possible since the $f_n$ are continuous at $0$.)
Define $f(x) := f_n(x)$ for $xin (-a_n,a_n)setminus(-a_n+1,a_n+1)$.
Now assume $lim_xrightarrow 0fracf_n(x)f(x)=0$ for $ngeq N$. Then we have some $delta>0$ such that $left|fracf_n(x)f(x)right|< 0.5$ for all $xin(-delta,delta)$ and $ngeq N$. But we also have some $ngeq N$ with $(-a_n,a_n)subset (-delta,delta)$ and $fracf_n(x)f(x)=1$ for $xin(-a_n,a_n)$.
answered Aug 6 at 15:45
Tobias
483314
edited Aug 6 at 16:57
answered Aug 6 at 15:45
Tobias
483314
answered Aug 6 at 15:45
Tobias
483314
answered Aug 6 at 15:45
Tobias
483314
483314
Thanks for your answer.
– Keshav Srinivasan
Aug 6 at 15:47
2
How do you know that $f(x) to 0$ as $x to 0$?
– PhoemueX
Aug 6 at 15:51
@PhoemueX Nice catch. We might even have $f(x)=1$ for $xneq 0$ if each of the $f_n$ only gets $neq 1$ for $xinleft(0,frac12(n+1)right)$:-). So some more work is required... I knew it couldn't be that simple.
– Tobias
Aug 6 at 16:06
@KeshavSrinivasan Sorry, I don't have too much time. So I relaxed the strong differentiability conditions for $f$ to make the proof simple. I think it also shows in that form what you essentially wanted to see. Maybe someone other shows the statement for differentiable $f$ and you can accept his answer.
– Tobias
Aug 6 at 17:06
@PhoemueX I dealt with your argument. Hope the proof for the relaxed statement (without differentiability requirement for $f$) is correct now. (Thanks again for your comment.)
– Tobias
Aug 6 at 17:07
 |Â
show 2 more comments
Thanks for your answer.
– Keshav Srinivasan
Aug 6 at 15:47
2
How do you know that $f(x) to 0$ as $x to 0$?
– PhoemueX
Aug 6 at 15:51
@PhoemueX Nice catch. We might even have $f(x)=1$ for $xneq 0$ if each of the $f_n$ only gets $neq 1$ for $xinleft(0,frac12(n+1)right)$:-). So some more work is required... I knew it couldn't be that simple.
– Tobias
Aug 6 at 16:06
@KeshavSrinivasan Sorry, I don't have too much time. So I relaxed the strong differentiability conditions for $f$ to make the proof simple. I think it also shows in that form what you essentially wanted to see. Maybe someone other shows the statement for differentiable $f$ and you can accept his answer.
– Tobias
Aug 6 at 17:06
@PhoemueX I dealt with your argument. Hope the proof for the relaxed statement (without differentiability requirement for $f$) is correct now. (Thanks again for your comment.)
– Tobias
Aug 6 at 17:07
Thanks for your answer.
– Keshav Srinivasan
Aug 6 at 15:47
Thanks for your answer.
– Keshav Srinivasan
Aug 6 at 15:47
2
2
How do you know that $f(x) to 0$ as $x to 0$?
– PhoemueX
Aug 6 at 15:51
How do you know that $f(x) to 0$ as $x to 0$?
– PhoemueX
Aug 6 at 15:51
@PhoemueX Nice catch. We might even have $f(x)=1$ for $xneq 0$ if each of the $f_n$ only gets $neq 1$ for $xinleft(0,frac12(n+1)right)$:-). So some more work is required... I knew it couldn't be that simple.
– Tobias
Aug 6 at 16:06
@PhoemueX Nice catch. We might even have $f(x)=1$ for $xneq 0$ if each of the $f_n$ only gets $neq 1$ for $xinleft(0,frac12(n+1)right)$:-). So some more work is required... I knew it couldn't be that simple.
– Tobias
Aug 6 at 16:06
@KeshavSrinivasan Sorry, I don't have too much time. So I relaxed the strong differentiability conditions for $f$ to make the proof simple. I think it also shows in that form what you essentially wanted to see. Maybe someone other shows the statement for differentiable $f$ and you can accept his answer.
– Tobias
Aug 6 at 17:06
@KeshavSrinivasan Sorry, I don't have too much time. So I relaxed the strong differentiability conditions for $f$ to make the proof simple. I think it also shows in that form what you essentially wanted to see. Maybe someone other shows the statement for differentiable $f$ and you can accept his answer.
– Tobias
Aug 6 at 17:06
@PhoemueX I dealt with your argument. Hope the proof for the relaxed statement (without differentiability requirement for $f$) is correct now. (Thanks again for your comment.)
– Tobias
Aug 6 at 17:07
@PhoemueX I dealt with your argument. Hope the proof for the relaxed statement (without differentiability requirement for $f$) is correct now. (Thanks again for your comment.)
– Tobias
Aug 6 at 17:07
 |Â
show 2 more comments
up vote
0
down vote
There may be some typos in your question. We can simply put, for each $ninmathbbN$, $f_n(x)=1$.
Then, for any $f$ such that $f'(0)$ exists and $f(0)=0$, we automatically have: For each $n$, $lim_xrightarrow0fracf(x)f_n(x)=lim_xrightarrow0f(x)=0$
because $f$ is continuous at $x=0$.
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
I guess the OP wants $f_n (0)=0$ for all $n$ (although he didn't write it).
– PhoemueX
Aug 6 at 7:15
@PhoemueX Yeah, I edited my question.
– Keshav Srinivasan
Aug 6 at 14:18
add a comment |Â
up vote
0
down vote
There may be some typos in your question. We can simply put, for each $ninmathbbN$, $f_n(x)=1$.
Then, for any $f$ such that $f'(0)$ exists and $f(0)=0$, we automatically have: For each $n$, $lim_xrightarrow0fracf(x)f_n(x)=lim_xrightarrow0f(x)=0$
because $f$ is continuous at $x=0$.
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
I guess the OP wants $f_n (0)=0$ for all $n$ (although he didn't write it).
– PhoemueX
Aug 6 at 7:15
@PhoemueX Yeah, I edited my question.
– Keshav Srinivasan
Aug 6 at 14:18
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There may be some typos in your question. We can simply put, for each $ninmathbbN$, $f_n(x)=1$.
Then, for any $f$ such that $f'(0)$ exists and $f(0)=0$, we automatically have: For each $n$, $lim_xrightarrow0fracf(x)f_n(x)=lim_xrightarrow0f(x)=0$
because $f$ is continuous at $x=0$.
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
There may be some typos in your question. We can simply put, for each $ninmathbbN$, $f_n(x)=1$.
Then, for any $f$ such that $f'(0)$ exists and $f(0)=0$, we automatically have: For each $n$, $lim_xrightarrow0fracf(x)f_n(x)=lim_xrightarrow0f(x)=0$
because $f$ is continuous at $x=0$.
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 6:47
Danny Pak-Keung Chan
1,87628
1,87628
I guess the OP wants $f_n (0)=0$ for all $n$ (although he didn't write it).
– PhoemueX
Aug 6 at 7:15
@PhoemueX Yeah, I edited my question.
– Keshav Srinivasan
Aug 6 at 14:18
add a comment |Â
I guess the OP wants $f_n (0)=0$ for all $n$ (although he didn't write it).
– PhoemueX
Aug 6 at 7:15
@PhoemueX Yeah, I edited my question.
– Keshav Srinivasan
Aug 6 at 14:18
I guess the OP wants $f_n (0)=0$ for all $n$ (although he didn't write it).
– PhoemueX
Aug 6 at 7:15
I guess the OP wants $f_n (0)=0$ for all $n$ (although he didn't write it).
– PhoemueX
Aug 6 at 7:15
@PhoemueX Yeah, I edited my question.
– Keshav Srinivasan
Aug 6 at 14:18
@PhoemueX Yeah, I edited my question.
– Keshav Srinivasan
Aug 6 at 14:18
add a comment |Â
up vote
0
down vote
You can use
$$f(x)= begincases
e^-frac1x^n & x > 0 \
0 & xleq 0 \
endcases$$
for $n=1,2,3,4,ldots$ as a slower and slower growing sequence.
answered Aug 6 at 14:34
Ross Millikan
276k21187352
Is there no smooth function that grows slower than all the functions in that sequence? Can you prove that?
– Keshav Srinivasan
Aug 6 at 14:35
1
These things always have a next step. The next I would think of is $$e^-frac 1e^-frac 1x$$ but I am not sure it works.
– Ross Millikan
Aug 6 at 14:45
So is there no sequence of functions such that no function grows slower than them all?
– Keshav Srinivasan
Aug 6 at 15:30
I don't believe so
– Ross Millikan
Aug 6 at 15:38
1
Yes, but the statements “For every function there is a slower growing function.†and “There exists a sequence of functions such that no function grows slower than all the functions in the sequence.†are perfectly consistent with one another.
– Keshav Srinivasan
Aug 6 at 16:04
 |Â
show 2 more comments
up vote
0
down vote
You can use
$$f(x)= begincases
e^-frac1x^n & x > 0 \
0 & xleq 0 \
endcases$$
for $n=1,2,3,4,ldots$ as a slower and slower growing sequence.
answered Aug 6 at 14:34
Ross Millikan
276k21187352
Is there no smooth function that grows slower than all the functions in that sequence? Can you prove that?
– Keshav Srinivasan
Aug 6 at 14:35
1
These things always have a next step. The next I would think of is $$e^-frac 1e^-frac 1x$$ but I am not sure it works.
– Ross Millikan
Aug 6 at 14:45
So is there no sequence of functions such that no function grows slower than them all?
– Keshav Srinivasan
Aug 6 at 15:30
I don't believe so
– Ross Millikan
Aug 6 at 15:38
1
Yes, but the statements “For every function there is a slower growing function.†and “There exists a sequence of functions such that no function grows slower than all the functions in the sequence.†are perfectly consistent with one another.
– Keshav Srinivasan
Aug 6 at 16:04
 |Â
show 2 more comments
up vote
0
down vote
up vote
0
down vote
You can use
$$f(x)= begincases
e^-frac1x^n & x > 0 \
0 & xleq 0 \
endcases$$
for $n=1,2,3,4,ldots$ as a slower and slower growing sequence.
answered Aug 6 at 14:34
Ross Millikan
276k21187352
You can use
$$f(x)= begincases
e^-frac1x^n & x > 0 \
0 & xleq 0 \
endcases$$
for $n=1,2,3,4,ldots$ as a slower and slower growing sequence.
answered Aug 6 at 14:34
Ross Millikan
276k21187352
edited Aug 6 at 14:39
answered Aug 6 at 14:34
Ross Millikan
276k21187352
answered Aug 6 at 14:34
Ross Millikan
276k21187352
answered Aug 6 at 14:34
Ross Millikan
276k21187352
276k21187352
Is there no smooth function that grows slower than all the functions in that sequence? Can you prove that?
– Keshav Srinivasan
Aug 6 at 14:35
1
These things always have a next step. The next I would think of is $$e^-frac 1e^-frac 1x$$ but I am not sure it works.
– Ross Millikan
Aug 6 at 14:45
So is there no sequence of functions such that no function grows slower than them all?
– Keshav Srinivasan
Aug 6 at 15:30
I don't believe so
– Ross Millikan
Aug 6 at 15:38
1
Yes, but the statements “For every function there is a slower growing function.†and “There exists a sequence of functions such that no function grows slower than all the functions in the sequence.†are perfectly consistent with one another.
– Keshav Srinivasan
Aug 6 at 16:04
 |Â
show 2 more comments
Is there no smooth function that grows slower than all the functions in that sequence? Can you prove that?
– Keshav Srinivasan
Aug 6 at 14:35
1
These things always have a next step. The next I would think of is $$e^-frac 1e^-frac 1x$$ but I am not sure it works.
– Ross Millikan
Aug 6 at 14:45
So is there no sequence of functions such that no function grows slower than them all?
– Keshav Srinivasan
Aug 6 at 15:30
I don't believe so
– Ross Millikan
Aug 6 at 15:38
1
Yes, but the statements “For every function there is a slower growing function.†and “There exists a sequence of functions such that no function grows slower than all the functions in the sequence.†are perfectly consistent with one another.
– Keshav Srinivasan
Aug 6 at 16:04
Is there no smooth function that grows slower than all the functions in that sequence? Can you prove that?
– Keshav Srinivasan
Aug 6 at 14:35
Is there no smooth function that grows slower than all the functions in that sequence? Can you prove that?
– Keshav Srinivasan
Aug 6 at 14:35
1
1
These things always have a next step. The next I would think of is $$e^-frac 1e^-frac 1x$$ but I am not sure it works.
– Ross Millikan
Aug 6 at 14:45
These things always have a next step. The next I would think of is $$e^-frac 1e^-frac 1x$$ but I am not sure it works.
– Ross Millikan
Aug 6 at 14:45
So is there no sequence of functions such that no function grows slower than them all?
– Keshav Srinivasan
Aug 6 at 15:30
So is there no sequence of functions such that no function grows slower than them all?
– Keshav Srinivasan
Aug 6 at 15:30
I don't believe so
– Ross Millikan
Aug 6 at 15:38
I don't believe so
– Ross Millikan
Aug 6 at 15:38
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Yes, but the statements “For every function there is a slower growing function.†and “There exists a sequence of functions such that no function grows slower than all the functions in the sequence.†are perfectly consistent with one another.
– Keshav Srinivasan
Aug 6 at 16:04
Yes, but the statements “For every function there is a slower growing function.†and “There exists a sequence of functions such that no function grows slower than all the functions in the sequence.†are perfectly consistent with one another.
– Keshav Srinivasan
Aug 6 at 16:04
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Didn't you mean $lim_xrightarrow0 fracf_n(x)f(x)=0$ and shouldn't there be some restrictions on $f$ to allow the division? E.g., $x^2$ is slower growing than $x$ at $x=0$ and $lim_xrightarrow0 fracx^2x=0$. ... Or maybe I misunderstood the core topic of the question.
– Tobias
Aug 6 at 15:04
@Tobias You’re right, I edited the question.
– Keshav Srinivasan
Aug 6 at 15:29
With the current formulation of the question there is such a sequence, i.e., $f_n(x):=0$ for all admissible $x$ and $n$. That means you need some more restrictions on $f_n$.
– Tobias
Aug 6 at 15:37
@Tobias OK, what about this?
– Keshav Srinivasan
Aug 6 at 15:39
@KeshavSrinivasan I improved my answer to a better result.
– zhw.
Aug 8 at 17:33