If $f$ is a concave function then $f$ is increasing The Next CEO...
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If $f$ is a concave function then $f$ is increasing
The Next CEO of Stack OverflowDual of concave function is convexIf $limlimits_{x to infty} f(x)$ is a finite real number and $f''(x)$ is bounded, then $limlimits_{x to infty} f'(x) = 0$Prove the concavity of the transformation from a concave function to anotherIs the function $ln(ax + b)$ increasing/decreasing, concave/convex?prove that if $lim limits_{n to infty}F( a_n)=ell$, then $lim limits_{x to infty}F( x)=ell$differentiable functions, concave or convex?concave and strictly increasing function without differentiableIs $f(e^x)$ log-concave if $f$ is positive decreasing convex function on $[0,a]$?Is maximum of increasing concave functions quasi-concave?Extension of concave function definition with 3 non-negative constants: a+b+c=1
$begingroup$
My question is how to prove the following assertion:
If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.
I know that if $f$ is differentiable then the mentioned hypothesis ($f'$ is decreasing) and using M.V.T imply that firstly $limlimits_{xtoinfty}f'(x)=0$ also $f'geq0$ and $f$ is increasing. but here we do not have the differentiability.
Anyone can help me. Thanks.
calculus convex-analysis
edited Mar 16 at 7:52
Saad
20.2k92352
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
$endgroup$
add a comment |
$begingroup$
My question is how to prove the following assertion:
If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.
I know that if $f$ is differentiable then the mentioned hypothesis ($f'$ is decreasing) and using M.V.T imply that firstly $limlimits_{xtoinfty}f'(x)=0$ also $f'geq0$ and $f$ is increasing. but here we do not have the differentiability.
Anyone can help me. Thanks.
calculus convex-analysis
edited Mar 16 at 7:52
Saad
20.2k92352
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
$endgroup$
add a comment |
$begingroup$
My question is how to prove the following assertion:
If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.
I know that if $f$ is differentiable then the mentioned hypothesis ($f'$ is decreasing) and using M.V.T imply that firstly $limlimits_{xtoinfty}f'(x)=0$ also $f'geq0$ and $f$ is increasing. but here we do not have the differentiability.
Anyone can help me. Thanks.
calculus convex-analysis
edited Mar 16 at 7:52
Saad
20.2k92352
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
$endgroup$
My question is how to prove the following assertion:
If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.
I know that if $f$ is differentiable then the mentioned hypothesis ($f'$ is decreasing) and using M.V.T imply that firstly $limlimits_{xtoinfty}f'(x)=0$ also $f'geq0$ and $f$ is increasing. but here we do not have the differentiability.
Anyone can help me. Thanks.
calculus convex-analysis
calculus convex-analysis
edited Mar 16 at 7:52
Saad
20.2k92352
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
edited Mar 16 at 7:52
Saad
20.2k92352
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
edited Mar 16 at 7:52
Saad
20.2k92352
edited Mar 16 at 7:52
Saad
20.2k92352
edited Mar 16 at 7:52
Saad
20.2k92352
20.2k92352
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
asked Mar 16 at 7:33
soodehMehboodisoodehMehboodi
64238
64238
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
$$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.
Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
$$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
where $k$ is any positive integer such that $x_0+k>x_1$.
Can you take it from here?
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
@ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
$endgroup$
– soodehMehboodi
Mar 16 at 8:26
$begingroup$
@soodehMehboodi See my edit.
$endgroup$
– Robert Z
Mar 16 at 9:16
$begingroup$
@ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
$endgroup$
– soodehMehboodi
Mar 16 at 10:00
$begingroup$
One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
$endgroup$
– Robert Z
Mar 16 at 10:48
$begingroup$
@ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
$endgroup$
– soodehMehboodi
Mar 16 at 14:47
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
$$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.
Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
$$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
where $k$ is any positive integer such that $x_0+k>x_1$.
Can you take it from here?
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
@ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
$endgroup$
– soodehMehboodi
Mar 16 at 8:26
$begingroup$
@soodehMehboodi See my edit.
$endgroup$
– Robert Z
Mar 16 at 9:16
$begingroup$
@ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
$endgroup$
– soodehMehboodi
Mar 16 at 10:00
$begingroup$
One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
$endgroup$
– Robert Z
Mar 16 at 10:48
$begingroup$
@ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
$endgroup$
– soodehMehboodi
Mar 16 at 14:47
|
show 2 more comments
$begingroup$
Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
$$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.
Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
$$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
where $k$ is any positive integer such that $x_0+k>x_1$.
Can you take it from here?
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
@ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
$endgroup$
– soodehMehboodi
Mar 16 at 8:26
$begingroup$
@soodehMehboodi See my edit.
$endgroup$
– Robert Z
Mar 16 at 9:16
$begingroup$
@ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
$endgroup$
– soodehMehboodi
Mar 16 at 10:00
$begingroup$
One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
$endgroup$
– Robert Z
Mar 16 at 10:48
$begingroup$
@ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
$endgroup$
– soodehMehboodi
Mar 16 at 14:47
|
show 2 more comments
$begingroup$
Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
$$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.
Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
$$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
where $k$ is any positive integer such that $x_0+k>x_1$.
Can you take it from here?
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
$endgroup$
Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
$$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.
Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
$$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
where $k$ is any positive integer such that $x_0+k>x_1$.
Can you take it from here?
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
edited Mar 16 at 9:15
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
answered Mar 16 at 7:46
Robert ZRobert Z
101k1070143
101k1070143
$begingroup$
@ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
$endgroup$
– soodehMehboodi
Mar 16 at 8:26
$begingroup$
@soodehMehboodi See my edit.
$endgroup$
– Robert Z
Mar 16 at 9:16
$begingroup$
@ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
$endgroup$
– soodehMehboodi
Mar 16 at 10:00
$begingroup$
One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
$endgroup$
– Robert Z
Mar 16 at 10:48
$begingroup$
@ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
$endgroup$
– soodehMehboodi
Mar 16 at 14:47
|
show 2 more comments
$begingroup$
@ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
$endgroup$
– soodehMehboodi
Mar 16 at 8:26
$begingroup$
@soodehMehboodi See my edit.
$endgroup$
– Robert Z
Mar 16 at 9:16
$begingroup$
@ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
$endgroup$
– soodehMehboodi
Mar 16 at 10:00
$begingroup$
One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
$endgroup$
– Robert Z
Mar 16 at 10:48
$begingroup$
@ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
$endgroup$
– soodehMehboodi
Mar 16 at 14:47
$begingroup$
@ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
$endgroup$
– soodehMehboodi
Mar 16 at 8:26
$begingroup$
@ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
$endgroup$
– soodehMehboodi
Mar 16 at 8:26
$begingroup$
@soodehMehboodi See my edit.
$endgroup$
– Robert Z
Mar 16 at 9:16
$begingroup$
@soodehMehboodi See my edit.
$endgroup$
– Robert Z
Mar 16 at 9:16
$begingroup$
@ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
$endgroup$
– soodehMehboodi
Mar 16 at 10:00
$begingroup$
@ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
$endgroup$
– soodehMehboodi
Mar 16 at 10:00
$begingroup$
One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
$endgroup$
– Robert Z
Mar 16 at 10:48
$begingroup$
One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
$endgroup$
– Robert Z
Mar 16 at 10:48
$begingroup$
@ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
$endgroup$
– soodehMehboodi
Mar 16 at 14:47
$begingroup$
@ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
$endgroup$
– soodehMehboodi
Mar 16 at 14:47
|
show 2 more comments
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