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A function have minimum or maximum
Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?
I could not even construct such an example satisfying the assumptions.
calculus convex-analysis
asked Nov 21 '18 at 11:16
xldd
1,315510
add a comment |
Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?
I could not even construct such an example satisfying the assumptions.
calculus convex-analysis
asked Nov 21 '18 at 11:16
xldd
1,315510
add a comment |
Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?
I could not even construct such an example satisfying the assumptions.
calculus convex-analysis
asked Nov 21 '18 at 11:16
xldd
1,315510
Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?
I could not even construct such an example satisfying the assumptions.
calculus convex-analysis
calculus convex-analysis
asked Nov 21 '18 at 11:16
xldd
1,315510
asked Nov 21 '18 at 11:16
xldd
1,315510
edited Nov 21 '18 at 11:47
asked Nov 21 '18 at 11:16
xldd
1,315510
asked Nov 21 '18 at 11:16
xldd
1,315510
asked Nov 21 '18 at 11:16
xldd
1,315510
1,315510
add a comment |
add a comment |
3 Answers
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If $$f(x)=-ln(1+e^{-x})-2x$$
then
$$f'(x)=-1-frac{1}{1+e^{-x}}$$
and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because
$$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$
for all $xinmathbb{R}$.
The function $f$ has no global maximum or minimum (its derivative is always negative).
Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).
(You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)
answered Nov 21 '18 at 12:38
smcc
4,297517
add a comment |
Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.
answered Nov 21 '18 at 12:06
P De Donato
4147
add a comment |
Take a generic function $f(x)$ that meets the required criteria.
Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.
As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.
answered Nov 21 '18 at 12:06
MoKo19
1914
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $$f(x)=-ln(1+e^{-x})-2x$$
then
$$f'(x)=-1-frac{1}{1+e^{-x}}$$
and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because
$$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$
for all $xinmathbb{R}$.
The function $f$ has no global maximum or minimum (its derivative is always negative).
Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).
(You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)
answered Nov 21 '18 at 12:38
smcc
4,297517
add a comment |
If $$f(x)=-ln(1+e^{-x})-2x$$
then
$$f'(x)=-1-frac{1}{1+e^{-x}}$$
and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because
$$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$
for all $xinmathbb{R}$.
The function $f$ has no global maximum or minimum (its derivative is always negative).
Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).
(You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)
answered Nov 21 '18 at 12:38
smcc
4,297517
add a comment |
If $$f(x)=-ln(1+e^{-x})-2x$$
then
$$f'(x)=-1-frac{1}{1+e^{-x}}$$
and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because
$$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$
for all $xinmathbb{R}$.
The function $f$ has no global maximum or minimum (its derivative is always negative).
Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).
(You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)
answered Nov 21 '18 at 12:38
smcc
4,297517
If $$f(x)=-ln(1+e^{-x})-2x$$
then
$$f'(x)=-1-frac{1}{1+e^{-x}}$$
and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because
$$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$
for all $xinmathbb{R}$.
The function $f$ has no global maximum or minimum (its derivative is always negative).
Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).
(You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)
answered Nov 21 '18 at 12:38
smcc
4,297517
edited Nov 21 '18 at 12:44
answered Nov 21 '18 at 12:38
smcc
4,297517
answered Nov 21 '18 at 12:38
smcc
4,297517
answered Nov 21 '18 at 12:38
smcc
4,297517
4,297517
add a comment |
add a comment |
Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.
answered Nov 21 '18 at 12:06
P De Donato
4147
add a comment |
Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.
answered Nov 21 '18 at 12:06
P De Donato
4147
add a comment |
Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.
answered Nov 21 '18 at 12:06
P De Donato
4147
Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.
answered Nov 21 '18 at 12:06
P De Donato
4147
answered Nov 21 '18 at 12:06
P De Donato
4147
answered Nov 21 '18 at 12:06
P De Donato
4147
answered Nov 21 '18 at 12:06
P De Donato
4147
4147
add a comment |
add a comment |
Take a generic function $f(x)$ that meets the required criteria.
Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.
As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.
answered Nov 21 '18 at 12:06
MoKo19
1914
add a comment |
Take a generic function $f(x)$ that meets the required criteria.
Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.
As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.
answered Nov 21 '18 at 12:06
MoKo19
1914
add a comment |
Take a generic function $f(x)$ that meets the required criteria.
Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.
As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.
answered Nov 21 '18 at 12:06
MoKo19
1914
Take a generic function $f(x)$ that meets the required criteria.
Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.
As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.
answered Nov 21 '18 at 12:06
MoKo19
1914
answered Nov 21 '18 at 12:06
MoKo19
1914
answered Nov 21 '18 at 12:06
MoKo19
1914
answered Nov 21 '18 at 12:06
MoKo19
1914
1914
add a comment |
add a comment |
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