If $f(x) + f'(x) le 0$ with $f(0) = 0$ and $f'(0) = 0$, then for what preimage we have $f(x) < 0$?
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
$exists x$ such that $f(x) < 0$ ?- for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)
- 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
asked Nov 28 at 12:36
Analysis Newbie
41627
add a comment |
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
$exists x$ such that $f(x) < 0$ ?- for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)
- 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
asked Nov 28 at 12:36
Analysis Newbie
41627
What about the function $f(x) = 0$?
– Arthur
Nov 28 at 12:52
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
– Analysis Newbie
Nov 28 at 12:58
add a comment |
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
$exists x$ such that $f(x) < 0$ ?- for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)
- 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
asked Nov 28 at 12:36
Analysis Newbie
41627
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
$exists x$ such that $f(x) < 0$ ?- for $x in B_{epsilon}(0)$ there holds $f(x) < 0$ ? ($B_epsilon$ denotes some neighborhood within $epsilon$)
- 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
asked Nov 28 at 12:36
Analysis Newbie
41627
asked Nov 28 at 12:36
Analysis Newbie
41627
edited Nov 28 at 12:57
asked Nov 28 at 12:36
Analysis Newbie
41627
asked Nov 28 at 12:36
Analysis Newbie
41627
asked Nov 28 at 12:36
Analysis Newbie
41627
41627
What about the function $f(x) = 0$?
– Arthur
Nov 28 at 12:52
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
– Analysis Newbie
Nov 28 at 12:58
add a comment |
What about the function $f(x) = 0$?
– Arthur
Nov 28 at 12:52
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
– Analysis Newbie
Nov 28 at 12:58
What about the function $f(x) = 0$?
– Arthur
Nov 28 at 12:52
What about the function $f(x) = 0$?
– Arthur
Nov 28 at 12:52
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
– Analysis Newbie
Nov 28 at 12:58
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
– Analysis Newbie
Nov 28 at 12:58
add a comment |
1 Answer
1
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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 $f$.
answered Nov 28 at 12:57
LutzL
55.4k42053
$f in C^1$ is enough for this conclusion?
– Analysis Newbie
Nov 28 at 13:11
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.
– LutzL
Nov 28 at 13:36
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
– Analysis Newbie
Nov 28 at 13:36
Thank you very much!
– Analysis Newbie
Nov 28 at 13:36
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 $f(x) = xf'(a) < 0$ ?
– Analysis Newbie
Nov 28 at 15:05
|
show 1 more comment
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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 $f$.
answered Nov 28 at 12:57
LutzL
55.4k42053
$f in C^1$ is enough for this conclusion?
– Analysis Newbie
Nov 28 at 13:11
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.
– LutzL
Nov 28 at 13:36
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
– Analysis Newbie
Nov 28 at 13:36
Thank you very much!
– Analysis Newbie
Nov 28 at 13:36
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 $f(x) = xf'(a) < 0$ ?
– Analysis Newbie
Nov 28 at 15:05
|
show 1 more comment
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 $f$.
answered Nov 28 at 12:57
LutzL
55.4k42053
$f in C^1$ is enough for this conclusion?
– Analysis Newbie
Nov 28 at 13:11
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.
– LutzL
Nov 28 at 13:36
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
– Analysis Newbie
Nov 28 at 13:36
Thank you very much!
– Analysis Newbie
Nov 28 at 13:36
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 $f(x) = xf'(a) < 0$ ?
– Analysis Newbie
Nov 28 at 15:05
|
show 1 more comment
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 $f$.
answered Nov 28 at 12:57
LutzL
55.4k42053
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 $f$.
answered Nov 28 at 12:57
LutzL
55.4k42053
edited Nov 28 at 13:02
answered Nov 28 at 12:57
LutzL
55.4k42053
answered Nov 28 at 12:57
LutzL
55.4k42053
answered Nov 28 at 12:57
LutzL
55.4k42053
55.4k42053
$f in C^1$ is enough for this conclusion?
– Analysis Newbie
Nov 28 at 13:11
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.
– LutzL
Nov 28 at 13:36
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
– Analysis Newbie
Nov 28 at 13:36
Thank you very much!
– Analysis Newbie
Nov 28 at 13:36
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 $f(x) = xf'(a) < 0$ ?
– Analysis Newbie
Nov 28 at 15:05
|
show 1 more comment
$f in C^1$ is enough for this conclusion?
– Analysis Newbie
Nov 28 at 13:11
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.
– LutzL
Nov 28 at 13:36
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
– Analysis Newbie
Nov 28 at 13:36
Thank you very much!
– Analysis Newbie
Nov 28 at 13:36
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 $f(x) = xf'(a) < 0$ ?
– Analysis Newbie
Nov 28 at 15:05
$f in C^1$ is enough for this conclusion?
– Analysis Newbie
Nov 28 at 13:11
$f in C^1$ is enough for this conclusion?
– Analysis Newbie
Nov 28 at 13:11
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.
– LutzL
Nov 28 at 13:36
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.
– LutzL
Nov 28 at 13:36
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
– Analysis Newbie
Nov 28 at 13:36
Is $f in C^1$ enough for this conclusion? I thought about some stronger condition like $f$ must be uniformly or Lipschitz continuous.
– Analysis Newbie
Nov 28 at 13:36
Thank you very much!
– Analysis Newbie
Nov 28 at 13:36
Thank you very much!
– Analysis Newbie
Nov 28 at 13:36
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 $f(x) = xf'(a) < 0$ ?
– Analysis Newbie
Nov 28 at 15:05
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 $f(x) = xf'(a) < 0$ ?
– Analysis Newbie
Nov 28 at 15:05
|
show 1 more comment
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What about the function $f(x) = 0$?
– Arthur
Nov 28 at 12:52
Sorry, I missed a condition that $f(x) + f'(x) = 0$ if and only if $x = 0$.
– Analysis Newbie
Nov 28 at 12:58