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Construct a function which increases faster than its derivative. [closed]
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Is there a function $f$ which is differentiable in $(0, 1)$ and satisfies
beginequation*
begingathered
lim_r to 0+ |f(r)|=infty, \
lim_r to 0+ fracf(r) = infty.
endgathered
endequation*
calculus analysis
asked Aug 23 at 8:36
S. Bryant
36118
closed as off-topic by mfl, user21820, Wouter, Jendrik Stelzner, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mfl, user21820, Wouter, Jendrik Stelzner, user91500
add a comment |Â
up vote
5
down vote
favorite
Is there a function $f$ which is differentiable in $(0, 1)$ and satisfies
beginequation*
begingathered
lim_r to 0+ |f(r)|=infty, \
lim_r to 0+ fracf(r) = infty.
endgathered
endequation*
calculus analysis
asked Aug 23 at 8:36
S. Bryant
36118
closed as off-topic by mfl, user21820, Wouter, Jendrik Stelzner, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mfl, user21820, Wouter, Jendrik Stelzner, user91500
What? Putting this question on hold, after all the stimulating discussion it generated, and the good answers, is ridiculous.
– Giuseppe Negro
Aug 24 at 10:27
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Is there a function $f$ which is differentiable in $(0, 1)$ and satisfies
beginequation*
begingathered
lim_r to 0+ |f(r)|=infty, \
lim_r to 0+ fracf(r) = infty.
endgathered
endequation*
calculus analysis
asked Aug 23 at 8:36
S. Bryant
36118
Is there a function $f$ which is differentiable in $(0, 1)$ and satisfies
beginequation*
begingathered
lim_r to 0+ |f(r)|=infty, \
lim_r to 0+ fracf(r) = infty.
endgathered
endequation*
calculus analysis
asked Aug 23 at 8:36
S. Bryant
36118
edited Aug 23 at 9:30
asked Aug 23 at 8:36
S. Bryant
36118
asked Aug 23 at 8:36
S. Bryant
36118
asked Aug 23 at 8:36
S. Bryant
36118
36118
closed as off-topic by mfl, user21820, Wouter, Jendrik Stelzner, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mfl, user21820, Wouter, Jendrik Stelzner, user91500
closed as off-topic by mfl, user21820, Wouter, Jendrik Stelzner, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mfl, user21820, Wouter, Jendrik Stelzner, user91500
What? Putting this question on hold, after all the stimulating discussion it generated, and the good answers, is ridiculous.
– Giuseppe Negro
Aug 24 at 10:27
add a comment |Â
What? Putting this question on hold, after all the stimulating discussion it generated, and the good answers, is ridiculous.
– Giuseppe Negro
Aug 24 at 10:27
What? Putting this question on hold, after all the stimulating discussion it generated, and the good answers, is ridiculous.
– Giuseppe Negro
Aug 24 at 10:27
What? Putting this question on hold, after all the stimulating discussion it generated, and the good answers, is ridiculous.
– Giuseppe Negro
Aug 24 at 10:27
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
We will deal with $r to 1^-$. The same analysis applies to $r to 0^+$ by replacing $f(r)$ with $f(1-r)$ and $f'(r)$ then with $-f'(1-r)$, which changes the sign of the limit in question.
Suppose that $f(r) to infty$, as $r to 1^-$. Then there is some interval $(1-epsilon, 1)$ on which $log(f(r))$ is defined (because $f(r)$ is eventually positive) and $log(f(r)) to infty$ as $r to 1^-$ as well since $log(s) to infty$ as $s to infty$. Then $$fracddrlog(f(r)) = fracf'(r)f(r)$$ so
Pick a point $t_0 in (1-epsilon, 1)$. Then by the Mean Value Theorem, for each $r in (t_0, 1)$, there is a point $c_r in (t_0, r)$ such that $$fracddrlog(f)(c_r) = fraclog(f(r)) -log(f(t_0))r - t_0$$ and $r-t_0$ remains bounded away from zero as $r$ converges to $1$. Since $log(f(r))$ converges to infinity as $r$ converges to the limit, we see that there is a sequence of points $x_n in (t_0, 1)$ such that $fracddrlog(f)(x_n) to infty$. Because $t_0 in (1-epsilon, 1)$ was arbitrary, this means that $$limsup_rto1^-fracf'(r)f(r) = limsup_rto1^-fracddrlog(f)(r) = infty.$$
This shows the question asked by the OP to be incorrect, because the limsup that we got applies to the reciprocal of the limit wondered about by the OP.
Moreover, we see that this result is the best that we can do, since we cannot guarantee convergence of this ratio to $infty$, because $f'$ can get close to $-infty$ while impacting $f(r)$ very little. For instance, $sqrt-x$ on $(-1, 0)$ has derivative $frac-12sqrt-x to -infty$ as $x to 0$, but the change in the function over this finite interval is finite. Now, it seems that one can combine many of these examples and smoothly connect them to get a function such that $limsup_rto1^-fracf'(r)f(r) = infty$ and $liminf_rto1^-fracf'(r)f(r) = -infty$. In fact, this phenomenon of $f'$ being able to be close to $infty$ in absolute value and $f$ changing very little is another reason that the limit wondered by the OP is false, because $f'$ can suddenly become very large, much larger than $f$, in small spurts and so the limit would not exist.
Also, as a technicality, the denominator might equal zero at times if $f'(r) = 0$ often... but since you are wondering that the limit is infinite, then it really isn't a problem, I suppose.
edited Aug 23 at 12:08
ochsnerd
31
answered Aug 23 at 9:36
4-ier
6089
Thanks a lot for your detailed explanation.
– S. Bryant
Aug 23 at 10:29
"Then there is some interval (1−õ,õ) on which log(f(r)) is defined" -- did you mean: "some interval $(1 - epsilon, 1)$" ?
– CompuChip
Aug 23 at 11:59
Yes, thank you @ochsnerd for the edit.
– 4-ier
Aug 25 at 3:37
add a comment |Â
up vote
6
down vote
The question can be rephrased as:
construct a smooth $f>0$ on $(0,1)$ such that $f(r)to infty$, $(log f(r))'to 0$ as $rto 0$.
Here, of course, the prime denotes derivative. Such a function cannot exist. If it existed, since $(log f)'to 0$ then the integral
$$tag1
int_0^r (log f(s))', ds$$
would be finite for all $rin (0, 1)$. But direct computation shows that (1) equals
$$
log f(r) - lim_sto 0 log f(s) =-infty.$$
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
Strictly speaking, $log(f)$ might not nice enough (absolutely continuous) and if it is strictly increasing and $f > 0$ everywhere, you are only guaranteed that $int_0^r (log(f))'(s)ds leq log(f(r)) - log(f(0))$ and this does not really help you. I am not sure that you can reduce to the case where $f$ is smooth, because you want to control both $f$ and $f'$ in the limit and $f'$ might be erratic. In my response I just do a Mean Value calculation to get a limsup, but I am not sure that one can go from differentiable to smooth. Any ideas?
– 4-ier
Aug 23 at 9:43
@4-ier: I implicitly assume $fin C^1(0,1)$, so that $(log f)'$ is continuous. I think this is not a huge assumption, and it provides a much shorter proof. If you want to remove the hypothesis that $f'$ is continuous then you will have to do some mean values as you did.
– Giuseppe Negro
Aug 23 at 9:50
Thank u very much.
– S. Bryant
Aug 23 at 10:30
add a comment |Â
up vote
4
down vote
The question was modified after I posted this answer. There is no such function: $f(x) >1$ and $frac f(x) f'(x) >1$ for $0<x<r$ for some $r$. In particular $f'(x) >0$ for $0<x<r$ so $f$ is increasing in $(0,r)$. As $x$ decreases $f(x) $ decreases so it cannot have limit $infty $ as $x to 0$.
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
Sorry for missing the absolute value.
– S. Bryant
Aug 23 at 9:33
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
We will deal with $r to 1^-$. The same analysis applies to $r to 0^+$ by replacing $f(r)$ with $f(1-r)$ and $f'(r)$ then with $-f'(1-r)$, which changes the sign of the limit in question.
Suppose that $f(r) to infty$, as $r to 1^-$. Then there is some interval $(1-epsilon, 1)$ on which $log(f(r))$ is defined (because $f(r)$ is eventually positive) and $log(f(r)) to infty$ as $r to 1^-$ as well since $log(s) to infty$ as $s to infty$. Then $$fracddrlog(f(r)) = fracf'(r)f(r)$$ so
Pick a point $t_0 in (1-epsilon, 1)$. Then by the Mean Value Theorem, for each $r in (t_0, 1)$, there is a point $c_r in (t_0, r)$ such that $$fracddrlog(f)(c_r) = fraclog(f(r)) -log(f(t_0))r - t_0$$ and $r-t_0$ remains bounded away from zero as $r$ converges to $1$. Since $log(f(r))$ converges to infinity as $r$ converges to the limit, we see that there is a sequence of points $x_n in (t_0, 1)$ such that $fracddrlog(f)(x_n) to infty$. Because $t_0 in (1-epsilon, 1)$ was arbitrary, this means that $$limsup_rto1^-fracf'(r)f(r) = limsup_rto1^-fracddrlog(f)(r) = infty.$$
This shows the question asked by the OP to be incorrect, because the limsup that we got applies to the reciprocal of the limit wondered about by the OP.
Moreover, we see that this result is the best that we can do, since we cannot guarantee convergence of this ratio to $infty$, because $f'$ can get close to $-infty$ while impacting $f(r)$ very little. For instance, $sqrt-x$ on $(-1, 0)$ has derivative $frac-12sqrt-x to -infty$ as $x to 0$, but the change in the function over this finite interval is finite. Now, it seems that one can combine many of these examples and smoothly connect them to get a function such that $limsup_rto1^-fracf'(r)f(r) = infty$ and $liminf_rto1^-fracf'(r)f(r) = -infty$. In fact, this phenomenon of $f'$ being able to be close to $infty$ in absolute value and $f$ changing very little is another reason that the limit wondered by the OP is false, because $f'$ can suddenly become very large, much larger than $f$, in small spurts and so the limit would not exist.
Also, as a technicality, the denominator might equal zero at times if $f'(r) = 0$ often... but since you are wondering that the limit is infinite, then it really isn't a problem, I suppose.
edited Aug 23 at 12:08
ochsnerd
31
answered Aug 23 at 9:36
4-ier
6089
Thanks a lot for your detailed explanation.
– S. Bryant
Aug 23 at 10:29
"Then there is some interval (1−õ,õ) on which log(f(r)) is defined" -- did you mean: "some interval $(1 - epsilon, 1)$" ?
– CompuChip
Aug 23 at 11:59
Yes, thank you @ochsnerd for the edit.
– 4-ier
Aug 25 at 3:37
add a comment |Â
up vote
6
down vote
accepted
We will deal with $r to 1^-$. The same analysis applies to $r to 0^+$ by replacing $f(r)$ with $f(1-r)$ and $f'(r)$ then with $-f'(1-r)$, which changes the sign of the limit in question.
Suppose that $f(r) to infty$, as $r to 1^-$. Then there is some interval $(1-epsilon, 1)$ on which $log(f(r))$ is defined (because $f(r)$ is eventually positive) and $log(f(r)) to infty$ as $r to 1^-$ as well since $log(s) to infty$ as $s to infty$. Then $$fracddrlog(f(r)) = fracf'(r)f(r)$$ so
Pick a point $t_0 in (1-epsilon, 1)$. Then by the Mean Value Theorem, for each $r in (t_0, 1)$, there is a point $c_r in (t_0, r)$ such that $$fracddrlog(f)(c_r) = fraclog(f(r)) -log(f(t_0))r - t_0$$ and $r-t_0$ remains bounded away from zero as $r$ converges to $1$. Since $log(f(r))$ converges to infinity as $r$ converges to the limit, we see that there is a sequence of points $x_n in (t_0, 1)$ such that $fracddrlog(f)(x_n) to infty$. Because $t_0 in (1-epsilon, 1)$ was arbitrary, this means that $$limsup_rto1^-fracf'(r)f(r) = limsup_rto1^-fracddrlog(f)(r) = infty.$$
This shows the question asked by the OP to be incorrect, because the limsup that we got applies to the reciprocal of the limit wondered about by the OP.
Moreover, we see that this result is the best that we can do, since we cannot guarantee convergence of this ratio to $infty$, because $f'$ can get close to $-infty$ while impacting $f(r)$ very little. For instance, $sqrt-x$ on $(-1, 0)$ has derivative $frac-12sqrt-x to -infty$ as $x to 0$, but the change in the function over this finite interval is finite. Now, it seems that one can combine many of these examples and smoothly connect them to get a function such that $limsup_rto1^-fracf'(r)f(r) = infty$ and $liminf_rto1^-fracf'(r)f(r) = -infty$. In fact, this phenomenon of $f'$ being able to be close to $infty$ in absolute value and $f$ changing very little is another reason that the limit wondered by the OP is false, because $f'$ can suddenly become very large, much larger than $f$, in small spurts and so the limit would not exist.
Also, as a technicality, the denominator might equal zero at times if $f'(r) = 0$ often... but since you are wondering that the limit is infinite, then it really isn't a problem, I suppose.
edited Aug 23 at 12:08
ochsnerd
31
answered Aug 23 at 9:36
4-ier
6089
Thanks a lot for your detailed explanation.
– S. Bryant
Aug 23 at 10:29
"Then there is some interval (1−õ,õ) on which log(f(r)) is defined" -- did you mean: "some interval $(1 - epsilon, 1)$" ?
– CompuChip
Aug 23 at 11:59
Yes, thank you @ochsnerd for the edit.
– 4-ier
Aug 25 at 3:37
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
We will deal with $r to 1^-$. The same analysis applies to $r to 0^+$ by replacing $f(r)$ with $f(1-r)$ and $f'(r)$ then with $-f'(1-r)$, which changes the sign of the limit in question.
Suppose that $f(r) to infty$, as $r to 1^-$. Then there is some interval $(1-epsilon, 1)$ on which $log(f(r))$ is defined (because $f(r)$ is eventually positive) and $log(f(r)) to infty$ as $r to 1^-$ as well since $log(s) to infty$ as $s to infty$. Then $$fracddrlog(f(r)) = fracf'(r)f(r)$$ so
Pick a point $t_0 in (1-epsilon, 1)$. Then by the Mean Value Theorem, for each $r in (t_0, 1)$, there is a point $c_r in (t_0, r)$ such that $$fracddrlog(f)(c_r) = fraclog(f(r)) -log(f(t_0))r - t_0$$ and $r-t_0$ remains bounded away from zero as $r$ converges to $1$. Since $log(f(r))$ converges to infinity as $r$ converges to the limit, we see that there is a sequence of points $x_n in (t_0, 1)$ such that $fracddrlog(f)(x_n) to infty$. Because $t_0 in (1-epsilon, 1)$ was arbitrary, this means that $$limsup_rto1^-fracf'(r)f(r) = limsup_rto1^-fracddrlog(f)(r) = infty.$$
This shows the question asked by the OP to be incorrect, because the limsup that we got applies to the reciprocal of the limit wondered about by the OP.
Moreover, we see that this result is the best that we can do, since we cannot guarantee convergence of this ratio to $infty$, because $f'$ can get close to $-infty$ while impacting $f(r)$ very little. For instance, $sqrt-x$ on $(-1, 0)$ has derivative $frac-12sqrt-x to -infty$ as $x to 0$, but the change in the function over this finite interval is finite. Now, it seems that one can combine many of these examples and smoothly connect them to get a function such that $limsup_rto1^-fracf'(r)f(r) = infty$ and $liminf_rto1^-fracf'(r)f(r) = -infty$. In fact, this phenomenon of $f'$ being able to be close to $infty$ in absolute value and $f$ changing very little is another reason that the limit wondered by the OP is false, because $f'$ can suddenly become very large, much larger than $f$, in small spurts and so the limit would not exist.
Also, as a technicality, the denominator might equal zero at times if $f'(r) = 0$ often... but since you are wondering that the limit is infinite, then it really isn't a problem, I suppose.
edited Aug 23 at 12:08
ochsnerd
31
answered Aug 23 at 9:36
4-ier
6089
We will deal with $r to 1^-$. The same analysis applies to $r to 0^+$ by replacing $f(r)$ with $f(1-r)$ and $f'(r)$ then with $-f'(1-r)$, which changes the sign of the limit in question.
Suppose that $f(r) to infty$, as $r to 1^-$. Then there is some interval $(1-epsilon, 1)$ on which $log(f(r))$ is defined (because $f(r)$ is eventually positive) and $log(f(r)) to infty$ as $r to 1^-$ as well since $log(s) to infty$ as $s to infty$. Then $$fracddrlog(f(r)) = fracf'(r)f(r)$$ so
Pick a point $t_0 in (1-epsilon, 1)$. Then by the Mean Value Theorem, for each $r in (t_0, 1)$, there is a point $c_r in (t_0, r)$ such that $$fracddrlog(f)(c_r) = fraclog(f(r)) -log(f(t_0))r - t_0$$ and $r-t_0$ remains bounded away from zero as $r$ converges to $1$. Since $log(f(r))$ converges to infinity as $r$ converges to the limit, we see that there is a sequence of points $x_n in (t_0, 1)$ such that $fracddrlog(f)(x_n) to infty$. Because $t_0 in (1-epsilon, 1)$ was arbitrary, this means that $$limsup_rto1^-fracf'(r)f(r) = limsup_rto1^-fracddrlog(f)(r) = infty.$$
This shows the question asked by the OP to be incorrect, because the limsup that we got applies to the reciprocal of the limit wondered about by the OP.
Moreover, we see that this result is the best that we can do, since we cannot guarantee convergence of this ratio to $infty$, because $f'$ can get close to $-infty$ while impacting $f(r)$ very little. For instance, $sqrt-x$ on $(-1, 0)$ has derivative $frac-12sqrt-x to -infty$ as $x to 0$, but the change in the function over this finite interval is finite. Now, it seems that one can combine many of these examples and smoothly connect them to get a function such that $limsup_rto1^-fracf'(r)f(r) = infty$ and $liminf_rto1^-fracf'(r)f(r) = -infty$. In fact, this phenomenon of $f'$ being able to be close to $infty$ in absolute value and $f$ changing very little is another reason that the limit wondered by the OP is false, because $f'$ can suddenly become very large, much larger than $f$, in small spurts and so the limit would not exist.
Also, as a technicality, the denominator might equal zero at times if $f'(r) = 0$ often... but since you are wondering that the limit is infinite, then it really isn't a problem, I suppose.
edited Aug 23 at 12:08
ochsnerd
31
answered Aug 23 at 9:36
4-ier
6089
edited Aug 23 at 12:08
ochsnerd
31
edited Aug 23 at 12:08
ochsnerd
31
edited Aug 23 at 12:08
ochsnerd
31
31
answered Aug 23 at 9:36
4-ier
6089
answered Aug 23 at 9:36
4-ier
6089
answered Aug 23 at 9:36
4-ier
6089
6089
Thanks a lot for your detailed explanation.
– S. Bryant
Aug 23 at 10:29
"Then there is some interval (1−õ,õ) on which log(f(r)) is defined" -- did you mean: "some interval $(1 - epsilon, 1)$" ?
– CompuChip
Aug 23 at 11:59
Yes, thank you @ochsnerd for the edit.
– 4-ier
Aug 25 at 3:37
add a comment |Â
Thanks a lot for your detailed explanation.
– S. Bryant
Aug 23 at 10:29
"Then there is some interval (1−õ,õ) on which log(f(r)) is defined" -- did you mean: "some interval $(1 - epsilon, 1)$" ?
– CompuChip
Aug 23 at 11:59
Yes, thank you @ochsnerd for the edit.
– 4-ier
Aug 25 at 3:37
Thanks a lot for your detailed explanation.
– S. Bryant
Aug 23 at 10:29
Thanks a lot for your detailed explanation.
– S. Bryant
Aug 23 at 10:29
"Then there is some interval (1−õ,õ) on which log(f(r)) is defined" -- did you mean: "some interval $(1 - epsilon, 1)$" ?
– CompuChip
Aug 23 at 11:59
"Then there is some interval (1−õ,õ) on which log(f(r)) is defined" -- did you mean: "some interval $(1 - epsilon, 1)$" ?
– CompuChip
Aug 23 at 11:59
Yes, thank you @ochsnerd for the edit.
– 4-ier
Aug 25 at 3:37
Yes, thank you @ochsnerd for the edit.
– 4-ier
Aug 25 at 3:37
add a comment |Â
up vote
6
down vote
The question can be rephrased as:
construct a smooth $f>0$ on $(0,1)$ such that $f(r)to infty$, $(log f(r))'to 0$ as $rto 0$.
Here, of course, the prime denotes derivative. Such a function cannot exist. If it existed, since $(log f)'to 0$ then the integral
$$tag1
int_0^r (log f(s))', ds$$
would be finite for all $rin (0, 1)$. But direct computation shows that (1) equals
$$
log f(r) - lim_sto 0 log f(s) =-infty.$$
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
Strictly speaking, $log(f)$ might not nice enough (absolutely continuous) and if it is strictly increasing and $f > 0$ everywhere, you are only guaranteed that $int_0^r (log(f))'(s)ds leq log(f(r)) - log(f(0))$ and this does not really help you. I am not sure that you can reduce to the case where $f$ is smooth, because you want to control both $f$ and $f'$ in the limit and $f'$ might be erratic. In my response I just do a Mean Value calculation to get a limsup, but I am not sure that one can go from differentiable to smooth. Any ideas?
– 4-ier
Aug 23 at 9:43
@4-ier: I implicitly assume $fin C^1(0,1)$, so that $(log f)'$ is continuous. I think this is not a huge assumption, and it provides a much shorter proof. If you want to remove the hypothesis that $f'$ is continuous then you will have to do some mean values as you did.
– Giuseppe Negro
Aug 23 at 9:50
Thank u very much.
– S. Bryant
Aug 23 at 10:30
add a comment |Â
up vote
6
down vote
The question can be rephrased as:
construct a smooth $f>0$ on $(0,1)$ such that $f(r)to infty$, $(log f(r))'to 0$ as $rto 0$.
Here, of course, the prime denotes derivative. Such a function cannot exist. If it existed, since $(log f)'to 0$ then the integral
$$tag1
int_0^r (log f(s))', ds$$
would be finite for all $rin (0, 1)$. But direct computation shows that (1) equals
$$
log f(r) - lim_sto 0 log f(s) =-infty.$$
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
Strictly speaking, $log(f)$ might not nice enough (absolutely continuous) and if it is strictly increasing and $f > 0$ everywhere, you are only guaranteed that $int_0^r (log(f))'(s)ds leq log(f(r)) - log(f(0))$ and this does not really help you. I am not sure that you can reduce to the case where $f$ is smooth, because you want to control both $f$ and $f'$ in the limit and $f'$ might be erratic. In my response I just do a Mean Value calculation to get a limsup, but I am not sure that one can go from differentiable to smooth. Any ideas?
– 4-ier
Aug 23 at 9:43
@4-ier: I implicitly assume $fin C^1(0,1)$, so that $(log f)'$ is continuous. I think this is not a huge assumption, and it provides a much shorter proof. If you want to remove the hypothesis that $f'$ is continuous then you will have to do some mean values as you did.
– Giuseppe Negro
Aug 23 at 9:50
Thank u very much.
– S. Bryant
Aug 23 at 10:30
add a comment |Â
up vote
6
down vote
up vote
6
down vote
The question can be rephrased as:
construct a smooth $f>0$ on $(0,1)$ such that $f(r)to infty$, $(log f(r))'to 0$ as $rto 0$.
Here, of course, the prime denotes derivative. Such a function cannot exist. If it existed, since $(log f)'to 0$ then the integral
$$tag1
int_0^r (log f(s))', ds$$
would be finite for all $rin (0, 1)$. But direct computation shows that (1) equals
$$
log f(r) - lim_sto 0 log f(s) =-infty.$$
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
The question can be rephrased as:
construct a smooth $f>0$ on $(0,1)$ such that $f(r)to infty$, $(log f(r))'to 0$ as $rto 0$.
Here, of course, the prime denotes derivative. Such a function cannot exist. If it existed, since $(log f)'to 0$ then the integral
$$tag1
int_0^r (log f(s))', ds$$
would be finite for all $rin (0, 1)$. But direct computation shows that (1) equals
$$
log f(r) - lim_sto 0 log f(s) =-infty.$$
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
answered Aug 23 at 9:35
Giuseppe Negro
16k328117
16k328117
Strictly speaking, $log(f)$ might not nice enough (absolutely continuous) and if it is strictly increasing and $f > 0$ everywhere, you are only guaranteed that $int_0^r (log(f))'(s)ds leq log(f(r)) - log(f(0))$ and this does not really help you. I am not sure that you can reduce to the case where $f$ is smooth, because you want to control both $f$ and $f'$ in the limit and $f'$ might be erratic. In my response I just do a Mean Value calculation to get a limsup, but I am not sure that one can go from differentiable to smooth. Any ideas?
– 4-ier
Aug 23 at 9:43
@4-ier: I implicitly assume $fin C^1(0,1)$, so that $(log f)'$ is continuous. I think this is not a huge assumption, and it provides a much shorter proof. If you want to remove the hypothesis that $f'$ is continuous then you will have to do some mean values as you did.
– Giuseppe Negro
Aug 23 at 9:50
Thank u very much.
– S. Bryant
Aug 23 at 10:30
add a comment |Â
Strictly speaking, $log(f)$ might not nice enough (absolutely continuous) and if it is strictly increasing and $f > 0$ everywhere, you are only guaranteed that $int_0^r (log(f))'(s)ds leq log(f(r)) - log(f(0))$ and this does not really help you. I am not sure that you can reduce to the case where $f$ is smooth, because you want to control both $f$ and $f'$ in the limit and $f'$ might be erratic. In my response I just do a Mean Value calculation to get a limsup, but I am not sure that one can go from differentiable to smooth. Any ideas?
– 4-ier
Aug 23 at 9:43
@4-ier: I implicitly assume $fin C^1(0,1)$, so that $(log f)'$ is continuous. I think this is not a huge assumption, and it provides a much shorter proof. If you want to remove the hypothesis that $f'$ is continuous then you will have to do some mean values as you did.
– Giuseppe Negro
Aug 23 at 9:50
Thank u very much.
– S. Bryant
Aug 23 at 10:30
Strictly speaking, $log(f)$ might not nice enough (absolutely continuous) and if it is strictly increasing and $f > 0$ everywhere, you are only guaranteed that $int_0^r (log(f))'(s)ds leq log(f(r)) - log(f(0))$ and this does not really help you. I am not sure that you can reduce to the case where $f$ is smooth, because you want to control both $f$ and $f'$ in the limit and $f'$ might be erratic. In my response I just do a Mean Value calculation to get a limsup, but I am not sure that one can go from differentiable to smooth. Any ideas?
– 4-ier
Aug 23 at 9:43
Strictly speaking, $log(f)$ might not nice enough (absolutely continuous) and if it is strictly increasing and $f > 0$ everywhere, you are only guaranteed that $int_0^r (log(f))'(s)ds leq log(f(r)) - log(f(0))$ and this does not really help you. I am not sure that you can reduce to the case where $f$ is smooth, because you want to control both $f$ and $f'$ in the limit and $f'$ might be erratic. In my response I just do a Mean Value calculation to get a limsup, but I am not sure that one can go from differentiable to smooth. Any ideas?
– 4-ier
Aug 23 at 9:43
@4-ier: I implicitly assume $fin C^1(0,1)$, so that $(log f)'$ is continuous. I think this is not a huge assumption, and it provides a much shorter proof. If you want to remove the hypothesis that $f'$ is continuous then you will have to do some mean values as you did.
– Giuseppe Negro
Aug 23 at 9:50
@4-ier: I implicitly assume $fin C^1(0,1)$, so that $(log f)'$ is continuous. I think this is not a huge assumption, and it provides a much shorter proof. If you want to remove the hypothesis that $f'$ is continuous then you will have to do some mean values as you did.
– Giuseppe Negro
Aug 23 at 9:50
Thank u very much.
– S. Bryant
Aug 23 at 10:30
Thank u very much.
– S. Bryant
Aug 23 at 10:30
add a comment |Â
up vote
4
down vote
The question was modified after I posted this answer. There is no such function: $f(x) >1$ and $frac f(x) f'(x) >1$ for $0<x<r$ for some $r$. In particular $f'(x) >0$ for $0<x<r$ so $f$ is increasing in $(0,r)$. As $x$ decreases $f(x) $ decreases so it cannot have limit $infty $ as $x to 0$.
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
Sorry for missing the absolute value.
– S. Bryant
Aug 23 at 9:33
add a comment |Â
up vote
4
down vote
The question was modified after I posted this answer. There is no such function: $f(x) >1$ and $frac f(x) f'(x) >1$ for $0<x<r$ for some $r$. In particular $f'(x) >0$ for $0<x<r$ so $f$ is increasing in $(0,r)$. As $x$ decreases $f(x) $ decreases so it cannot have limit $infty $ as $x to 0$.
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
Sorry for missing the absolute value.
– S. Bryant
Aug 23 at 9:33
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The question was modified after I posted this answer. There is no such function: $f(x) >1$ and $frac f(x) f'(x) >1$ for $0<x<r$ for some $r$. In particular $f'(x) >0$ for $0<x<r$ so $f$ is increasing in $(0,r)$. As $x$ decreases $f(x) $ decreases so it cannot have limit $infty $ as $x to 0$.
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
The question was modified after I posted this answer. There is no such function: $f(x) >1$ and $frac f(x) f'(x) >1$ for $0<x<r$ for some $r$. In particular $f'(x) >0$ for $0<x<r$ so $f$ is increasing in $(0,r)$. As $x$ decreases $f(x) $ decreases so it cannot have limit $infty $ as $x to 0$.
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
edited Aug 23 at 9:40
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
answered Aug 23 at 8:51
Kavi Rama Murthy
24.5k31133
24.5k31133
Sorry for missing the absolute value.
– S. Bryant
Aug 23 at 9:33
add a comment |Â
Sorry for missing the absolute value.
– S. Bryant
Aug 23 at 9:33
Sorry for missing the absolute value.
– S. Bryant
Aug 23 at 9:33
Sorry for missing the absolute value.
– S. Bryant
Aug 23 at 9:33
add a comment |Â
What? Putting this question on hold, after all the stimulating discussion it generated, and the good answers, is ridiculous.
– Giuseppe Negro
Aug 24 at 10:27