Hyperbolic tangents as a dense subset of smooth functions satisfying certain conditions
$begingroup$
Edit: The original question contained some errors that lead to some comments and answers that do not apply anymore. I thank those that took time to read the question and pointed out the errors. I tried to be more precise this time.
Let $V$ be the set of smooth (infinitely differentiable) functions $f:[0,+infty)rightarrowmathbb{R}$ that satisfy the following conditions:
- $f(0)=0$
- $f'(x)geq 0, forall x$
$f''(x)leq 0$, $forall x$
$lim_{xrightarrow+infty}f(x)$ is a real number
I'd like to find a dense subset $Asubseteq V$ (w.r.t. to the $L^2$ norm, say), where $A$ is a family of functions depending only on some real-valued parameters.
My first guess was to take the family ${atanh(sx): a,sgeq0}$, but after some numerical computations I saw it doesn't work: e.g. $arctan(x)in V$ cannot be arbitrarily approximated by functions of the form $atanh(sx)$.
My questions are:
- Can we find a dense $A$ as described above (depending only on some real parameters).
- In case that the answer to the previous question is "yes", can the hyperbolic tangents family I described be easily modified to become a dense subset?
- What if we change the definition domain from $[0,+infty)$ to $[0,B]$ for some real number $B>0$? Does this compactness condition help in being able to find a dense subset?
functional-analysis banach-spaces
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
$endgroup$
add a comment |
$begingroup$
Edit: The original question contained some errors that lead to some comments and answers that do not apply anymore. I thank those that took time to read the question and pointed out the errors. I tried to be more precise this time.
Let $V$ be the set of smooth (infinitely differentiable) functions $f:[0,+infty)rightarrowmathbb{R}$ that satisfy the following conditions:
- $f(0)=0$
- $f'(x)geq 0, forall x$
$f''(x)leq 0$, $forall x$
$lim_{xrightarrow+infty}f(x)$ is a real number
I'd like to find a dense subset $Asubseteq V$ (w.r.t. to the $L^2$ norm, say), where $A$ is a family of functions depending only on some real-valued parameters.
My first guess was to take the family ${atanh(sx): a,sgeq0}$, but after some numerical computations I saw it doesn't work: e.g. $arctan(x)in V$ cannot be arbitrarily approximated by functions of the form $atanh(sx)$.
My questions are:
- Can we find a dense $A$ as described above (depending only on some real parameters).
- In case that the answer to the previous question is "yes", can the hyperbolic tangents family I described be easily modified to become a dense subset?
- What if we change the definition domain from $[0,+infty)$ to $[0,B]$ for some real number $B>0$? Does this compactness condition help in being able to find a dense subset?
functional-analysis banach-spaces
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
$endgroup$
$begingroup$
But functions with $f''(x) > 0$ do not form a vector space, multiplying a function $f in V$ by a negative scalar reverses the inequality. Do you perhaps mean $V$ to be some sort of a cone?
$endgroup$
– Bartosz Malman
Nov 29 '18 at 16:08
$begingroup$
@BartoszMalman: You're right. I edited the question.
$endgroup$
– A. Bellmunt
Nov 29 '18 at 18:00
add a comment |
$begingroup$
Edit: The original question contained some errors that lead to some comments and answers that do not apply anymore. I thank those that took time to read the question and pointed out the errors. I tried to be more precise this time.
Let $V$ be the set of smooth (infinitely differentiable) functions $f:[0,+infty)rightarrowmathbb{R}$ that satisfy the following conditions:
- $f(0)=0$
- $f'(x)geq 0, forall x$
$f''(x)leq 0$, $forall x$
$lim_{xrightarrow+infty}f(x)$ is a real number
I'd like to find a dense subset $Asubseteq V$ (w.r.t. to the $L^2$ norm, say), where $A$ is a family of functions depending only on some real-valued parameters.
My first guess was to take the family ${atanh(sx): a,sgeq0}$, but after some numerical computations I saw it doesn't work: e.g. $arctan(x)in V$ cannot be arbitrarily approximated by functions of the form $atanh(sx)$.
My questions are:
- Can we find a dense $A$ as described above (depending only on some real parameters).
- In case that the answer to the previous question is "yes", can the hyperbolic tangents family I described be easily modified to become a dense subset?
- What if we change the definition domain from $[0,+infty)$ to $[0,B]$ for some real number $B>0$? Does this compactness condition help in being able to find a dense subset?
functional-analysis banach-spaces
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
$endgroup$
Edit: The original question contained some errors that lead to some comments and answers that do not apply anymore. I thank those that took time to read the question and pointed out the errors. I tried to be more precise this time.
Let $V$ be the set of smooth (infinitely differentiable) functions $f:[0,+infty)rightarrowmathbb{R}$ that satisfy the following conditions:
- $f(0)=0$
- $f'(x)geq 0, forall x$
$f''(x)leq 0$, $forall x$
$lim_{xrightarrow+infty}f(x)$ is a real number
I'd like to find a dense subset $Asubseteq V$ (w.r.t. to the $L^2$ norm, say), where $A$ is a family of functions depending only on some real-valued parameters.
My first guess was to take the family ${atanh(sx): a,sgeq0}$, but after some numerical computations I saw it doesn't work: e.g. $arctan(x)in V$ cannot be arbitrarily approximated by functions of the form $atanh(sx)$.
My questions are:
- Can we find a dense $A$ as described above (depending only on some real parameters).
- In case that the answer to the previous question is "yes", can the hyperbolic tangents family I described be easily modified to become a dense subset?
- What if we change the definition domain from $[0,+infty)$ to $[0,B]$ for some real number $B>0$? Does this compactness condition help in being able to find a dense subset?
functional-analysis banach-spaces
functional-analysis banach-spaces
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
edited Nov 30 '18 at 15:18
A. Bellmunt
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
asked Nov 29 '18 at 15:32
A. BellmuntA. Bellmunt
895515
895515
$begingroup$
But functions with $f''(x) > 0$ do not form a vector space, multiplying a function $f in V$ by a negative scalar reverses the inequality. Do you perhaps mean $V$ to be some sort of a cone?
$endgroup$
– Bartosz Malman
Nov 29 '18 at 16:08
$begingroup$
@BartoszMalman: You're right. I edited the question.
$endgroup$
– A. Bellmunt
Nov 29 '18 at 18:00
add a comment |
$begingroup$
But functions with $f''(x) > 0$ do not form a vector space, multiplying a function $f in V$ by a negative scalar reverses the inequality. Do you perhaps mean $V$ to be some sort of a cone?
$endgroup$
– Bartosz Malman
Nov 29 '18 at 16:08
$begingroup$
@BartoszMalman: You're right. I edited the question.
$endgroup$
– A. Bellmunt
Nov 29 '18 at 18:00
$begingroup$
But functions with $f''(x) > 0$ do not form a vector space, multiplying a function $f in V$ by a negative scalar reverses the inequality. Do you perhaps mean $V$ to be some sort of a cone?
$endgroup$
– Bartosz Malman
Nov 29 '18 at 16:08
$begingroup$
But functions with $f''(x) > 0$ do not form a vector space, multiplying a function $f in V$ by a negative scalar reverses the inequality. Do you perhaps mean $V$ to be some sort of a cone?
$endgroup$
– Bartosz Malman
Nov 29 '18 at 16:08
$begingroup$
@BartoszMalman: You're right. I edited the question.
$endgroup$
– A. Bellmunt
Nov 29 '18 at 18:00
$begingroup$
@BartoszMalman: You're right. I edited the question.
$endgroup$
– A. Bellmunt
Nov 29 '18 at 18:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Edit: The following argument has been rendered moot by an edit to the question.
The claim is vacuously true; $V$ is empty.
From the limit condition at $infty$, there is some $u>0$ with $f(u)>0$. Then, by the Mean Value Theorem, there is some $vin (0,u)$ with $f'(v) = frac{f(u)}{u}>0$. Now, for all $x>v$, $f(x)ge f(v) + (x-v)f'(v)$, which goes to $infty$ as $xtoinfty$. Our function, which was supposed to have a finite limit at $infty$, instead goes to $infty$. There's nothing there - including that $f_{a,s}$ is never in $V$. In fact, for positive $a$ and $s$, the second derivative of $f_{a,s}$ is negative on $(0,infty)$.
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
$endgroup$
$begingroup$
Oh, god... you're absolutely right. The inequality of the second derivative should be the other way around... I already edited the question.
$endgroup$
– A. Bellmunt
Nov 30 '18 at 9:03
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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$begingroup$
Edit: The following argument has been rendered moot by an edit to the question.
The claim is vacuously true; $V$ is empty.
From the limit condition at $infty$, there is some $u>0$ with $f(u)>0$. Then, by the Mean Value Theorem, there is some $vin (0,u)$ with $f'(v) = frac{f(u)}{u}>0$. Now, for all $x>v$, $f(x)ge f(v) + (x-v)f'(v)$, which goes to $infty$ as $xtoinfty$. Our function, which was supposed to have a finite limit at $infty$, instead goes to $infty$. There's nothing there - including that $f_{a,s}$ is never in $V$. In fact, for positive $a$ and $s$, the second derivative of $f_{a,s}$ is negative on $(0,infty)$.
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
$endgroup$
$begingroup$
Oh, god... you're absolutely right. The inequality of the second derivative should be the other way around... I already edited the question.
$endgroup$
– A. Bellmunt
Nov 30 '18 at 9:03
add a comment |
$begingroup$
Edit: The following argument has been rendered moot by an edit to the question.
The claim is vacuously true; $V$ is empty.
From the limit condition at $infty$, there is some $u>0$ with $f(u)>0$. Then, by the Mean Value Theorem, there is some $vin (0,u)$ with $f'(v) = frac{f(u)}{u}>0$. Now, for all $x>v$, $f(x)ge f(v) + (x-v)f'(v)$, which goes to $infty$ as $xtoinfty$. Our function, which was supposed to have a finite limit at $infty$, instead goes to $infty$. There's nothing there - including that $f_{a,s}$ is never in $V$. In fact, for positive $a$ and $s$, the second derivative of $f_{a,s}$ is negative on $(0,infty)$.
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
$endgroup$
$begingroup$
Oh, god... you're absolutely right. The inequality of the second derivative should be the other way around... I already edited the question.
$endgroup$
– A. Bellmunt
Nov 30 '18 at 9:03
add a comment |
$begingroup$
Edit: The following argument has been rendered moot by an edit to the question.
The claim is vacuously true; $V$ is empty.
From the limit condition at $infty$, there is some $u>0$ with $f(u)>0$. Then, by the Mean Value Theorem, there is some $vin (0,u)$ with $f'(v) = frac{f(u)}{u}>0$. Now, for all $x>v$, $f(x)ge f(v) + (x-v)f'(v)$, which goes to $infty$ as $xtoinfty$. Our function, which was supposed to have a finite limit at $infty$, instead goes to $infty$. There's nothing there - including that $f_{a,s}$ is never in $V$. In fact, for positive $a$ and $s$, the second derivative of $f_{a,s}$ is negative on $(0,infty)$.
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
$endgroup$
Edit: The following argument has been rendered moot by an edit to the question.
The claim is vacuously true; $V$ is empty.
From the limit condition at $infty$, there is some $u>0$ with $f(u)>0$. Then, by the Mean Value Theorem, there is some $vin (0,u)$ with $f'(v) = frac{f(u)}{u}>0$. Now, for all $x>v$, $f(x)ge f(v) + (x-v)f'(v)$, which goes to $infty$ as $xtoinfty$. Our function, which was supposed to have a finite limit at $infty$, instead goes to $infty$. There's nothing there - including that $f_{a,s}$ is never in $V$. In fact, for positive $a$ and $s$, the second derivative of $f_{a,s}$ is negative on $(0,infty)$.
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
edited Nov 30 '18 at 11:17
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
answered Nov 29 '18 at 20:07
jmerryjmerry
3,737514
3,737514
$begingroup$
Oh, god... you're absolutely right. The inequality of the second derivative should be the other way around... I already edited the question.
$endgroup$
– A. Bellmunt
Nov 30 '18 at 9:03
add a comment |
$begingroup$
Oh, god... you're absolutely right. The inequality of the second derivative should be the other way around... I already edited the question.
$endgroup$
– A. Bellmunt
Nov 30 '18 at 9:03
$begingroup$
Oh, god... you're absolutely right. The inequality of the second derivative should be the other way around... I already edited the question.
$endgroup$
– A. Bellmunt
Nov 30 '18 at 9:03
$begingroup$
Oh, god... you're absolutely right. The inequality of the second derivative should be the other way around... I already edited the question.
$endgroup$
– A. Bellmunt
Nov 30 '18 at 9:03
add a comment |
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$begingroup$
But functions with $f''(x) > 0$ do not form a vector space, multiplying a function $f in V$ by a negative scalar reverses the inequality. Do you perhaps mean $V$ to be some sort of a cone?
$endgroup$
– Bartosz Malman
Nov 29 '18 at 16:08
$begingroup$
@BartoszMalman: You're right. I edited the question.
$endgroup$
– A. Bellmunt
Nov 29 '18 at 18:00