If $f(x) + f'(x) le 0$ with $f(0) = 0$ and $f'(0) = 0$, then for what preimage we have $f(x) < 0$?












0












$begingroup$


Let $f(x) in C^1(mathbb{R})$ be a real-valued function such that $f(0) = 0$ and $f'(0) = 0$. If, in addition, we have $f(x) + f'(x) le 0$ (equality holds if and only if $x = 0$), can we conclude that





  1. $exists x$ such that $f(x) < 0$ ?

  2. for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)

  3. for $forall x in mathbb{R}$ there holds $f(x) < 0$ ?


If not, which extra conditions are required for each proposition to be true?



(The extra conditions may be, for example, $f in C^infty$, or $f$ is Lipschitz continuous.)






calculus real-analysis differential-equations analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What about the function $f(x) = 0$?
    $endgroup$
    – Arthur
    Nov 28 '18 at 12:52










  • $begingroup$
    Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 12:58
















0












$begingroup$


Let $f(x) in C^1(mathbb{R})$ be a real-valued function such that $f(0) = 0$ and $f'(0) = 0$. If, in addition, we have $f(x) + f'(x) le 0$ (equality holds if and only if $x = 0$), can we conclude that





  1. $exists x$ such that $f(x) < 0$ ?

  2. for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)

  3. for $forall x in mathbb{R}$ there holds $f(x) < 0$ ?


If not, which extra conditions are required for each proposition to be true?



(The extra conditions may be, for example, $f in C^infty$, or $f$ is Lipschitz continuous.)






calculus real-analysis differential-equations analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What about the function $f(x) = 0$?
    $endgroup$
    – Arthur
    Nov 28 '18 at 12:52










  • $begingroup$
    Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 12:58














0












0








0





$begingroup$


Let $f(x) in C^1(mathbb{R})$ be a real-valued function such that $f(0) = 0$ and $f'(0) = 0$. If, in addition, we have $f(x) + f'(x) le 0$ (equality holds if and only if $x = 0$), can we conclude that





  1. $exists x$ such that $f(x) < 0$ ?

  2. for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)

  3. for $forall x in mathbb{R}$ there holds $f(x) < 0$ ?


If not, which extra conditions are required for each proposition to be true?



(The extra conditions may be, for example, $f in C^infty$, or $f$ is Lipschitz continuous.)






calculus real-analysis differential-equations analysis







share|cite|improve this question











$endgroup$




Let $f(x) in C^1(mathbb{R})$ be a real-valued function such that $f(0) = 0$ and $f'(0) = 0$. If, in addition, we have $f(x) + f'(x) le 0$ (equality holds if and only if $x = 0$), can we conclude that





  1. $exists x$ such that $f(x) < 0$ ?

  2. for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)

  3. for $forall x in mathbb{R}$ there holds $f(x) < 0$ ?


If not, which extra conditions are required for each proposition to be true?



(The extra conditions may be, for example, $f in C^infty$, or $f$ is Lipschitz continuous.)






calculus real-analysis differential-equations analysis




calculus real-analysis differential-equations analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 12:57







Analysis Newbie

















asked Nov 28 '18 at 12:36









Analysis NewbieAnalysis Newbie

41627




41627












  • $begingroup$
    What about the function $f(x) = 0$?
    $endgroup$
    – Arthur
    Nov 28 '18 at 12:52










  • $begingroup$
    Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 12:58


















  • $begingroup$
    What about the function $f(x) = 0$?
    $endgroup$
    – Arthur
    Nov 28 '18 at 12:52










  • $begingroup$
    Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 12:58
















$begingroup$
What about the function $f(x) = 0$?
$endgroup$
– Arthur
Nov 28 '18 at 12:52




$begingroup$
What about the function $f(x) = 0$?
$endgroup$
– Arthur
Nov 28 '18 at 12:52












$begingroup$
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
$endgroup$
– Analysis Newbie
Nov 28 '18 at 12:58




$begingroup$
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
$endgroup$
– Analysis Newbie
Nov 28 '18 at 12:58










1 Answer
1






active

oldest

votes


















2












$begingroup$

Let $g(x)=e^xf(x)$. Then the conditions read $g(0)=0$, $g'(0)=0$ and $g'(x)< 0$ for $xne0$. The condition $f(x)<0$ is then equivalent to $g(x)<0$.



As one can see, $g$ has to be falling, and in fact strongly so, by the mean value theorem. Thus $g(x)<0$ for all $x>0$. By the same argument, $g(x)>0$ for $x<0$, and the same holds for $f$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f in C^1$ is enough for this conclusion?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:11










  • $begingroup$
    Yes, as then also $gin C^1$ and $frac{g(x_2)-g(x_1)}{x_2-x_1}<0$ for all $x_1<x_2le0$ and $0le x_1<x_2$ by mean value theorem, proving the strictly falling property.
    $endgroup$
    – LutzL
    Nov 28 '18 at 13:36










  • $begingroup$
    Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36










  • $begingroup$
    One more question, if we have $f'(x)+f(x)<0$ for $x > 0$, by mean value theorem $f(x)=xf'(a)$ for some $a in (0,x)$, then we get $f'(x)+xf'(a) < 0$ for all $0<a<x$, can we directly obtain that $f'(x)<0$ for all $x > 0$, then by mean value theorem $f(x) = xf'(a) < 0$ ?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 15:05













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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

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2












$begingroup$

Let $g(x)=e^xf(x)$. Then the conditions read $g(0)=0$, $g'(0)=0$ and $g'(x)< 0$ for $xne0$. The condition $f(x)<0$ is then equivalent to $g(x)<0$.



As one can see, $g$ has to be falling, and in fact strongly so, by the mean value theorem. Thus $g(x)<0$ for all $x>0$. By the same argument, $g(x)>0$ for $x<0$, and the same holds for $f$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f in C^1$ is enough for this conclusion?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:11










  • $begingroup$
    Yes, as then also $gin C^1$ and $frac{g(x_2)-g(x_1)}{x_2-x_1}<0$ for all $x_1<x_2le0$ and $0le x_1<x_2$ by mean value theorem, proving the strictly falling property.
    $endgroup$
    – LutzL
    Nov 28 '18 at 13:36










  • $begingroup$
    Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36










  • $begingroup$
    One more question, if we have $f'(x)+f(x)<0$ for $x > 0$, by mean value theorem $f(x)=xf'(a)$ for some $a in (0,x)$, then we get $f'(x)+xf'(a) < 0$ for all $0<a<x$, can we directly obtain that $f'(x)<0$ for all $x > 0$, then by mean value theorem $f(x) = xf'(a) < 0$ ?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 15:05


















2












$begingroup$

Let $g(x)=e^xf(x)$. Then the conditions read $g(0)=0$, $g'(0)=0$ and $g'(x)< 0$ for $xne0$. The condition $f(x)<0$ is then equivalent to $g(x)<0$.



As one can see, $g$ has to be falling, and in fact strongly so, by the mean value theorem. Thus $g(x)<0$ for all $x>0$. By the same argument, $g(x)>0$ for $x<0$, and the same holds for $f$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f in C^1$ is enough for this conclusion?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:11










  • $begingroup$
    Yes, as then also $gin C^1$ and $frac{g(x_2)-g(x_1)}{x_2-x_1}<0$ for all $x_1<x_2le0$ and $0le x_1<x_2$ by mean value theorem, proving the strictly falling property.
    $endgroup$
    – LutzL
    Nov 28 '18 at 13:36










  • $begingroup$
    Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36










  • $begingroup$
    One more question, if we have $f'(x)+f(x)<0$ for $x > 0$, by mean value theorem $f(x)=xf'(a)$ for some $a in (0,x)$, then we get $f'(x)+xf'(a) < 0$ for all $0<a<x$, can we directly obtain that $f'(x)<0$ for all $x > 0$, then by mean value theorem $f(x) = xf'(a) < 0$ ?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 15:05
















2












2








2





$begingroup$

Let $g(x)=e^xf(x)$. Then the conditions read $g(0)=0$, $g'(0)=0$ and $g'(x)< 0$ for $xne0$. The condition $f(x)<0$ is then equivalent to $g(x)<0$.



As one can see, $g$ has to be falling, and in fact strongly so, by the mean value theorem. Thus $g(x)<0$ for all $x>0$. By the same argument, $g(x)>0$ for $x<0$, and the same holds for $f$.






share|cite|improve this answer











$endgroup$



Let $g(x)=e^xf(x)$. Then the conditions read $g(0)=0$, $g'(0)=0$ and $g'(x)< 0$ for $xne0$. The condition $f(x)<0$ is then equivalent to $g(x)<0$.



As one can see, $g$ has to be falling, and in fact strongly so, by the mean value theorem. Thus $g(x)<0$ for all $x>0$. By the same argument, $g(x)>0$ for $x<0$, and the same holds for $f$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 13:02

























answered Nov 28 '18 at 12:57









LutzLLutzL

56.7k42054




56.7k42054












  • $begingroup$
    $f in C^1$ is enough for this conclusion?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:11










  • $begingroup$
    Yes, as then also $gin C^1$ and $frac{g(x_2)-g(x_1)}{x_2-x_1}<0$ for all $x_1<x_2le0$ and $0le x_1<x_2$ by mean value theorem, proving the strictly falling property.
    $endgroup$
    – LutzL
    Nov 28 '18 at 13:36










  • $begingroup$
    Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36










  • $begingroup$
    One more question, if we have $f'(x)+f(x)<0$ for $x > 0$, by mean value theorem $f(x)=xf'(a)$ for some $a in (0,x)$, then we get $f'(x)+xf'(a) < 0$ for all $0<a<x$, can we directly obtain that $f'(x)<0$ for all $x > 0$, then by mean value theorem $f(x) = xf'(a) < 0$ ?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 15:05




















  • $begingroup$
    $f in C^1$ is enough for this conclusion?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:11










  • $begingroup$
    Yes, as then also $gin C^1$ and $frac{g(x_2)-g(x_1)}{x_2-x_1}<0$ for all $x_1<x_2le0$ and $0le x_1<x_2$ by mean value theorem, proving the strictly falling property.
    $endgroup$
    – LutzL
    Nov 28 '18 at 13:36










  • $begingroup$
    Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 13:36










  • $begingroup$
    One more question, if we have $f'(x)+f(x)<0$ for $x > 0$, by mean value theorem $f(x)=xf'(a)$ for some $a in (0,x)$, then we get $f'(x)+xf'(a) < 0$ for all $0<a<x$, can we directly obtain that $f'(x)<0$ for all $x > 0$, then by mean value theorem $f(x) = xf'(a) < 0$ ?
    $endgroup$
    – Analysis Newbie
    Nov 28 '18 at 15:05


















$begingroup$
$f in C^1$ is enough for this conclusion?
$endgroup$
– Analysis Newbie
Nov 28 '18 at 13:11




$begingroup$
$f in C^1$ is enough for this conclusion?
$endgroup$
– Analysis Newbie
Nov 28 '18 at 13:11












$begingroup$
Yes, as then also $gin C^1$ and $frac{g(x_2)-g(x_1)}{x_2-x_1}<0$ for all $x_1<x_2le0$ and $0le x_1<x_2$ by mean value theorem, proving the strictly falling property.
$endgroup$
– LutzL
Nov 28 '18 at 13:36




$begingroup$
Yes, as then also $gin C^1$ and $frac{g(x_2)-g(x_1)}{x_2-x_1}<0$ for all $x_1<x_2le0$ and $0le x_1<x_2$ by mean value theorem, proving the strictly falling property.
$endgroup$
– LutzL
Nov 28 '18 at 13:36












$begingroup$
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
$endgroup$
– Analysis Newbie
Nov 28 '18 at 13:36






$begingroup$
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
$endgroup$
– Analysis Newbie
Nov 28 '18 at 13:36














$begingroup$
Thank you very much!
$endgroup$
– Analysis Newbie
Nov 28 '18 at 13:36




$begingroup$
Thank you very much!
$endgroup$
– Analysis Newbie
Nov 28 '18 at 13:36












$begingroup$
One more question, if we have $f'(x)+f(x)<0$ for $x > 0$, by mean value theorem $f(x)=xf'(a)$ for some $a in (0,x)$, then we get $f'(x)+xf'(a) < 0$ for all $0<a<x$, can we directly obtain that $f'(x)<0$ for all $x > 0$, then by mean value theorem $f(x) = xf'(a) < 0$ ?
$endgroup$
– Analysis Newbie
Nov 28 '18 at 15:05






$begingroup$
One more question, if we have $f'(x)+f(x)<0$ for $x > 0$, by mean value theorem $f(x)=xf'(a)$ for some $a in (0,x)$, then we get $f'(x)+xf'(a) < 0$ for all $0<a<x$, can we directly obtain that $f'(x)<0$ for all $x > 0$, then by mean value theorem $f(x) = xf'(a) < 0$ ?
$endgroup$
– Analysis Newbie
Nov 28 '18 at 15:05




















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