If $f>0$ and $(ln f)'=f'/f$ is Lipschitz continuous, are we able to conclude that $(ln f)'''$ is bounded?
Let $fin C^2(mathbb R)$ be positive and $g:=ln f$. Assume $$g'=frac{f'}f$$ is Lipschitz continuous and hence $g''$ is bounded.
Is $g'''$ bounded too?
calculus derivatives lipschitz-functions
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
add a comment |
Let $fin C^2(mathbb R)$ be positive and $g:=ln f$. Assume $$g'=frac{f'}f$$ is Lipschitz continuous and hence $g''$ is bounded.
Is $g'''$ bounded too?
calculus derivatives lipschitz-functions
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
add a comment |
Let $fin C^2(mathbb R)$ be positive and $g:=ln f$. Assume $$g'=frac{f'}f$$ is Lipschitz continuous and hence $g''$ is bounded.
Is $g'''$ bounded too?
calculus derivatives lipschitz-functions
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
Let $fin C^2(mathbb R)$ be positive and $g:=ln f$. Assume $$g'=frac{f'}f$$ is Lipschitz continuous and hence $g''$ is bounded.
Is $g'''$ bounded too?
calculus derivatives lipschitz-functions
calculus derivatives lipschitz-functions
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
asked Dec 11 '18 at 21:53
0xbadf00d
1,75741430
1,75741430
add a comment |
add a comment |
1 Answer
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No, of course. Let $$u(x)=begin{cases} e^{-x^2}x^2sin x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ and consider $$f(x)=expleft(int_0^x,dtint_0^t u(s),dsright)$$
Then, $(ln f)''=u$, which is bounded and continuous, but $$u'(x)=begin{cases} 2(1-x^2)xe^{-x^2}sin x^{-2}-frac2xe^{-x^2}cos x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ is unbounded.
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
What if we add the assumption that $int_mathbb Rf=1$, $int_mathbb Rfrac{left|f'right|^8}{f^7}<infty$ and $int_mathbb Rfrac{left|f''right|^4}{f^3}<infty$?
– 0xbadf00d
Dec 11 '18 at 22:36
I see no reason why you couldn't patch my example (or a slight variation) with a suitable fast-decaying function outside of $[-1,1]$, but I haven't put much thought into that.
– Saucy O'Path
Dec 11 '18 at 22:44
add a comment |
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1 Answer
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1 Answer
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No, of course. Let $$u(x)=begin{cases} e^{-x^2}x^2sin x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ and consider $$f(x)=expleft(int_0^x,dtint_0^t u(s),dsright)$$
Then, $(ln f)''=u$, which is bounded and continuous, but $$u'(x)=begin{cases} 2(1-x^2)xe^{-x^2}sin x^{-2}-frac2xe^{-x^2}cos x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ is unbounded.
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
What if we add the assumption that $int_mathbb Rf=1$, $int_mathbb Rfrac{left|f'right|^8}{f^7}<infty$ and $int_mathbb Rfrac{left|f''right|^4}{f^3}<infty$?
– 0xbadf00d
Dec 11 '18 at 22:36
I see no reason why you couldn't patch my example (or a slight variation) with a suitable fast-decaying function outside of $[-1,1]$, but I haven't put much thought into that.
– Saucy O'Path
Dec 11 '18 at 22:44
add a comment |
No, of course. Let $$u(x)=begin{cases} e^{-x^2}x^2sin x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ and consider $$f(x)=expleft(int_0^x,dtint_0^t u(s),dsright)$$
Then, $(ln f)''=u$, which is bounded and continuous, but $$u'(x)=begin{cases} 2(1-x^2)xe^{-x^2}sin x^{-2}-frac2xe^{-x^2}cos x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ is unbounded.
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
What if we add the assumption that $int_mathbb Rf=1$, $int_mathbb Rfrac{left|f'right|^8}{f^7}<infty$ and $int_mathbb Rfrac{left|f''right|^4}{f^3}<infty$?
– 0xbadf00d
Dec 11 '18 at 22:36
I see no reason why you couldn't patch my example (or a slight variation) with a suitable fast-decaying function outside of $[-1,1]$, but I haven't put much thought into that.
– Saucy O'Path
Dec 11 '18 at 22:44
add a comment |
No, of course. Let $$u(x)=begin{cases} e^{-x^2}x^2sin x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ and consider $$f(x)=expleft(int_0^x,dtint_0^t u(s),dsright)$$
Then, $(ln f)''=u$, which is bounded and continuous, but $$u'(x)=begin{cases} 2(1-x^2)xe^{-x^2}sin x^{-2}-frac2xe^{-x^2}cos x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ is unbounded.
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
No, of course. Let $$u(x)=begin{cases} e^{-x^2}x^2sin x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ and consider $$f(x)=expleft(int_0^x,dtint_0^t u(s),dsright)$$
Then, $(ln f)''=u$, which is bounded and continuous, but $$u'(x)=begin{cases} 2(1-x^2)xe^{-x^2}sin x^{-2}-frac2xe^{-x^2}cos x^{-2}&text{if }xne 0\ 0&text{if }x=0end{cases}$$ is unbounded.
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
edited Dec 11 '18 at 22:14
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
answered Dec 11 '18 at 22:06
Saucy O'Path
5,8341626
5,8341626
What if we add the assumption that $int_mathbb Rf=1$, $int_mathbb Rfrac{left|f'right|^8}{f^7}<infty$ and $int_mathbb Rfrac{left|f''right|^4}{f^3}<infty$?
– 0xbadf00d
Dec 11 '18 at 22:36
I see no reason why you couldn't patch my example (or a slight variation) with a suitable fast-decaying function outside of $[-1,1]$, but I haven't put much thought into that.
– Saucy O'Path
Dec 11 '18 at 22:44
add a comment |
What if we add the assumption that $int_mathbb Rf=1$, $int_mathbb Rfrac{left|f'right|^8}{f^7}<infty$ and $int_mathbb Rfrac{left|f''right|^4}{f^3}<infty$?
– 0xbadf00d
Dec 11 '18 at 22:36
I see no reason why you couldn't patch my example (or a slight variation) with a suitable fast-decaying function outside of $[-1,1]$, but I haven't put much thought into that.
– Saucy O'Path
Dec 11 '18 at 22:44
What if we add the assumption that $int_mathbb Rf=1$, $int_mathbb Rfrac{left|f'right|^8}{f^7}<infty$ and $int_mathbb Rfrac{left|f''right|^4}{f^3}<infty$?
– 0xbadf00d
Dec 11 '18 at 22:36
What if we add the assumption that $int_mathbb Rf=1$, $int_mathbb Rfrac{left|f'right|^8}{f^7}<infty$ and $int_mathbb Rfrac{left|f''right|^4}{f^3}<infty$?
– 0xbadf00d
Dec 11 '18 at 22:36
I see no reason why you couldn't patch my example (or a slight variation) with a suitable fast-decaying function outside of $[-1,1]$, but I haven't put much thought into that.
– Saucy O'Path
Dec 11 '18 at 22:44
I see no reason why you couldn't patch my example (or a slight variation) with a suitable fast-decaying function outside of $[-1,1]$, but I haven't put much thought into that.
– Saucy O'Path
Dec 11 '18 at 22:44
add a comment |
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