Hyperbolic tangents as a dense subset of smooth functions satisfying certain conditions












0












$begingroup$


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:




  1. Can we find a dense $A$ as described above (depending only on some real parameters).

  2. 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?

  3. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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:




  1. Can we find a dense $A$ as described above (depending only on some real parameters).

  2. 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?

  3. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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:




  1. Can we find a dense $A$ as described above (depending only on some real parameters).

  2. 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?

  3. 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







share|cite|improve this question











$endgroup$




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:




  1. Can we find a dense $A$ as described above (depending only on some real parameters).

  2. 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?

  3. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 15:18







A. Bellmunt

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $(0,infty)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $(0,infty)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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 $(0,infty)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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 $(0,infty)$.






share|cite|improve this answer











$endgroup$



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 $(0,infty)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 11:17

























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


















  • $begingroup$
    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


















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