Mixing
Negative Entropy $xlog x$ is convex
$begingroup$
can someone tell me (or show me), where can I find a proof for [f:mathbb{R}_{geq0}tomathbb{R},quad xmapsto xlog x] is convex without using the first or second derivative trick?
We call a map convex, then [f(lambda x+(1-lambda)tilde{x})leqlambda f(x)+(1-lambda)f(tilde{x})quadtext{for}~lambdain(0,1)]
real-analysis
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
$endgroup$
|
show 1 more comment
$begingroup$
can someone tell me (or show me), where can I find a proof for [f:mathbb{R}_{geq0}tomathbb{R},quad xmapsto xlog x] is convex without using the first or second derivative trick?
We call a map convex, then [f(lambda x+(1-lambda)tilde{x})leqlambda f(x)+(1-lambda)f(tilde{x})quadtext{for}~lambdain(0,1)]
real-analysis
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
$endgroup$
$begingroup$
I know, that the function is convex and I also know that the reason is that the seond derivative is positiv for all $x>0$. What I search is a proof without the second derivative argument.
$endgroup$
– FuncAna09
Jan 26 at 15:36
$begingroup$
Welcome @FuncAna09! Could you write the definition of convex function you know? It would be easier to help you. :)
$endgroup$
– Ixion
Jan 26 at 15:38
$begingroup$
Why are you searching for a proof that avoids the second derivative? In this case, it's by far the cleanest argument.
$endgroup$
– Misha Lavrov
Jan 26 at 15:51
$begingroup$
Simply because I want to know how the evidence would go without the derivation.
$endgroup$
– FuncAna09
Jan 26 at 15:58
$begingroup$
I think it might be helpful to consider a slight perturbation of the subtitle of Stanley Kubrick's movie "Doctor Strangelove": "How I learned to stop worrying and love the second derivative trick".
$endgroup$
– Lee Mosher
Jan 26 at 16:52
|
show 1 more comment
$begingroup$
can someone tell me (or show me), where can I find a proof for [f:mathbb{R}_{geq0}tomathbb{R},quad xmapsto xlog x] is convex without using the first or second derivative trick?
We call a map convex, then [f(lambda x+(1-lambda)tilde{x})leqlambda f(x)+(1-lambda)f(tilde{x})quadtext{for}~lambdain(0,1)]
real-analysis
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
$endgroup$
can someone tell me (or show me), where can I find a proof for [f:mathbb{R}_{geq0}tomathbb{R},quad xmapsto xlog x] is convex without using the first or second derivative trick?
We call a map convex, then [f(lambda x+(1-lambda)tilde{x})leqlambda f(x)+(1-lambda)f(tilde{x})quadtext{for}~lambdain(0,1)]
real-analysis
real-analysis
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
edited Jan 26 at 22:24
Ethan Bolker
45.2k553120
45.2k553120
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
asked Jan 26 at 15:29
FuncAna09FuncAna09
113
113
$begingroup$
I know, that the function is convex and I also know that the reason is that the seond derivative is positiv for all $x>0$. What I search is a proof without the second derivative argument.
$endgroup$
– FuncAna09
Jan 26 at 15:36
$begingroup$
Welcome @FuncAna09! Could you write the definition of convex function you know? It would be easier to help you. :)
$endgroup$
– Ixion
Jan 26 at 15:38
$begingroup$
Why are you searching for a proof that avoids the second derivative? In this case, it's by far the cleanest argument.
$endgroup$
– Misha Lavrov
Jan 26 at 15:51
$begingroup$
Simply because I want to know how the evidence would go without the derivation.
$endgroup$
– FuncAna09
Jan 26 at 15:58
$begingroup$
I think it might be helpful to consider a slight perturbation of the subtitle of Stanley Kubrick's movie "Doctor Strangelove": "How I learned to stop worrying and love the second derivative trick".
$endgroup$
– Lee Mosher
Jan 26 at 16:52
|
show 1 more comment
$begingroup$
I know, that the function is convex and I also know that the reason is that the seond derivative is positiv for all $x>0$. What I search is a proof without the second derivative argument.
$endgroup$
– FuncAna09
Jan 26 at 15:36
$begingroup$
Welcome @FuncAna09! Could you write the definition of convex function you know? It would be easier to help you. :)
$endgroup$
– Ixion
Jan 26 at 15:38
$begingroup$
Why are you searching for a proof that avoids the second derivative? In this case, it's by far the cleanest argument.
$endgroup$
– Misha Lavrov
Jan 26 at 15:51
$begingroup$
Simply because I want to know how the evidence would go without the derivation.
$endgroup$
– FuncAna09
Jan 26 at 15:58
$begingroup$
I think it might be helpful to consider a slight perturbation of the subtitle of Stanley Kubrick's movie "Doctor Strangelove": "How I learned to stop worrying and love the second derivative trick".
$endgroup$
– Lee Mosher
Jan 26 at 16:52
$begingroup$
I know, that the function is convex and I also know that the reason is that the seond derivative is positiv for all $x>0$. What I search is a proof without the second derivative argument.
$endgroup$
– FuncAna09
Jan 26 at 15:36
$begingroup$
I know, that the function is convex and I also know that the reason is that the seond derivative is positiv for all $x>0$. What I search is a proof without the second derivative argument.
$endgroup$
– FuncAna09
Jan 26 at 15:36
$begingroup$
Welcome @FuncAna09! Could you write the definition of convex function you know? It would be easier to help you. :)
$endgroup$
– Ixion
Jan 26 at 15:38
$begingroup$
Welcome @FuncAna09! Could you write the definition of convex function you know? It would be easier to help you. :)
$endgroup$
– Ixion
Jan 26 at 15:38
$begingroup$
Why are you searching for a proof that avoids the second derivative? In this case, it's by far the cleanest argument.
$endgroup$
– Misha Lavrov
Jan 26 at 15:51
$begingroup$
Why are you searching for a proof that avoids the second derivative? In this case, it's by far the cleanest argument.
$endgroup$
– Misha Lavrov
Jan 26 at 15:51
$begingroup$
Simply because I want to know how the evidence would go without the derivation.
$endgroup$
– FuncAna09
Jan 26 at 15:58
$begingroup$
Simply because I want to know how the evidence would go without the derivation.
$endgroup$
– FuncAna09
Jan 26 at 15:58
$begingroup$
I think it might be helpful to consider a slight perturbation of the subtitle of Stanley Kubrick's movie "Doctor Strangelove": "How I learned to stop worrying and love the second derivative trick".
$endgroup$
– Lee Mosher
Jan 26 at 16:52
$begingroup$
I think it might be helpful to consider a slight perturbation of the subtitle of Stanley Kubrick's movie "Doctor Strangelove": "How I learned to stop worrying and love the second derivative trick".
$endgroup$
– Lee Mosher
Jan 26 at 16:52
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Answer to previous version of this question:
This is hair-splitting, but: the first derivative of your function is an increasing function, so your function is convex.
Hair splitting, because the first derivative trick is not the second derivative trick, and "increasing first derivative" is not exactly "non-negative second derivative", but yet implies convexity, too.
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
$endgroup$
$begingroup$
Okay that was my mistake. The question is as follows: Is there any evidence that only goes with the definition?
$endgroup$
– FuncAna09
Jan 26 at 16:38
add a comment |
$begingroup$
So I tried the proof, to prove that a function or set is convex we take two points $x$ and $y$ and we try proving that , $$f(lambda x + (1-lambda) y) leq lambda f(x) + (1-lambda) f(y)$$
The concept is simple if we joing two points $x$ and $y$ by a line then all the points lying on the line should also lie inside the set, the line being $lambda x + (1-lambda) y$ , if we vary lambda from $(0,1)$ we find at $lambda$ = $0$ we get the point $x$ at $lambda = 1$ we get the point $y$ and for the value of $lambda = 0.5$ we get the midpoint of $x$ and $y$.
Now for the proof:
$$f(x) = x log x$$
so $$f(lambda x + (1-lambda) y) = ({lambda x + (1-lambda) y})space log ({lambda x +(1-lambda)y})$$
The RHS becomes $$lambda x log (lambda x + (1-lambda) y)+ (1-lambda) y log (lambda x + (1-lambda) y$$
Take the first part and you will notice $lambda x log(lambda x + (1-lambda)y) geq lambda x log(lambda x) $ since $log(A+B) geq log A$
Again $$lambda x log lambda x = lambda x (log lambda + log x) geq lambda x log x$$
So this leads to $$lambda x log(lambda x + (1-lambda)y) geq lambda x log x$$
Similarly, $$(1-lambda) y log(lambda x + (1-lambda)y) geq (1-lambda) y log y$$
Adding these two equations you get,
**$$lambda x log(lambda x + (1-lambda)y) + (1-lambda) y log(lambda x + (1-lambda)y) geq lambda x log x + (1-lambda) y log y $$
This proves $$f(lambda x + (1-lambda) y) geq lambda f(x) + (1-lambda) f(y)$$ meaning that $f(x) = xlogx$ is not convex meaning that $-f(x)=-xlog x $ is convex.
Hope this helps …….
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
1
$begingroup$
This can not be, because the second derivative is $f''(x)=frac{1}{x}>0~forall x>0$.
$endgroup$
– FuncAna09
Jan 26 at 18:14
1
$begingroup$
$log(lambda x)<log(x)$ because $0<lambda<1$. This was your mistake.
$endgroup$
– FuncAna09
Jan 26 at 18:28
add a comment |
$begingroup$
That's what I figured out: Let $lambdain (0,1)$ and $x,yinmathbb{R}>0$. Then we get
begin{align*}
f(lambda x+(1-lambda)y)&=[lambda x+(1-lambda)y]log(lambda x+(1-lambda)y)\
\
&=lambda xlog(lambda x+(1-lambda)y)+(1-lambda)ylog(lambda x+(1-lambda)y).
end{align*}
For the first term of RHS we get
[lambda xlog(lambda x+(1-lambda)y)leqlambda xlog(x)+lambda(1-lambda)y]
and for the second term we get
[(1-lambda)ylog(lambda x+(1-lambda)y)leq (1-lambda)ylog(y)+lambda(1-lambda)x]
In summary, this results in
[f(lambda x+(1-lambda)y)leq lambda f(x)+(1-lambda)f(y)+lambda(1-lambda)(x+y).]
The last term is the problem now.
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Answer to previous version of this question:
This is hair-splitting, but: the first derivative of your function is an increasing function, so your function is convex.
Hair splitting, because the first derivative trick is not the second derivative trick, and "increasing first derivative" is not exactly "non-negative second derivative", but yet implies convexity, too.
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
$endgroup$
$begingroup$
Okay that was my mistake. The question is as follows: Is there any evidence that only goes with the definition?
$endgroup$
– FuncAna09
Jan 26 at 16:38
add a comment |
$begingroup$
Answer to previous version of this question:
This is hair-splitting, but: the first derivative of your function is an increasing function, so your function is convex.
Hair splitting, because the first derivative trick is not the second derivative trick, and "increasing first derivative" is not exactly "non-negative second derivative", but yet implies convexity, too.
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
$endgroup$
$begingroup$
Okay that was my mistake. The question is as follows: Is there any evidence that only goes with the definition?
$endgroup$
– FuncAna09
Jan 26 at 16:38
add a comment |
$begingroup$
Answer to previous version of this question:
This is hair-splitting, but: the first derivative of your function is an increasing function, so your function is convex.
Hair splitting, because the first derivative trick is not the second derivative trick, and "increasing first derivative" is not exactly "non-negative second derivative", but yet implies convexity, too.
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
$endgroup$
Answer to previous version of this question:
This is hair-splitting, but: the first derivative of your function is an increasing function, so your function is convex.
Hair splitting, because the first derivative trick is not the second derivative trick, and "increasing first derivative" is not exactly "non-negative second derivative", but yet implies convexity, too.
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
edited Jan 26 at 16:44
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
answered Jan 26 at 16:29
kimchi loverkimchi lover
11.5k31229
11.5k31229
$begingroup$
Okay that was my mistake. The question is as follows: Is there any evidence that only goes with the definition?
$endgroup$
– FuncAna09
Jan 26 at 16:38
add a comment |
$begingroup$
Okay that was my mistake. The question is as follows: Is there any evidence that only goes with the definition?
$endgroup$
– FuncAna09
Jan 26 at 16:38
$begingroup$
Okay that was my mistake. The question is as follows: Is there any evidence that only goes with the definition?
$endgroup$
– FuncAna09
Jan 26 at 16:38
$begingroup$
Okay that was my mistake. The question is as follows: Is there any evidence that only goes with the definition?
$endgroup$
– FuncAna09
Jan 26 at 16:38
add a comment |
$begingroup$
So I tried the proof, to prove that a function or set is convex we take two points $x$ and $y$ and we try proving that , $$f(lambda x + (1-lambda) y) leq lambda f(x) + (1-lambda) f(y)$$
The concept is simple if we joing two points $x$ and $y$ by a line then all the points lying on the line should also lie inside the set, the line being $lambda x + (1-lambda) y$ , if we vary lambda from $(0,1)$ we find at $lambda$ = $0$ we get the point $x$ at $lambda = 1$ we get the point $y$ and for the value of $lambda = 0.5$ we get the midpoint of $x$ and $y$.
Now for the proof:
$$f(x) = x log x$$
so $$f(lambda x + (1-lambda) y) = ({lambda x + (1-lambda) y})space log ({lambda x +(1-lambda)y})$$
The RHS becomes $$lambda x log (lambda x + (1-lambda) y)+ (1-lambda) y log (lambda x + (1-lambda) y$$
Take the first part and you will notice $lambda x log(lambda x + (1-lambda)y) geq lambda x log(lambda x) $ since $log(A+B) geq log A$
Again $$lambda x log lambda x = lambda x (log lambda + log x) geq lambda x log x$$
So this leads to $$lambda x log(lambda x + (1-lambda)y) geq lambda x log x$$
Similarly, $$(1-lambda) y log(lambda x + (1-lambda)y) geq (1-lambda) y log y$$
Adding these two equations you get,
**$$lambda x log(lambda x + (1-lambda)y) + (1-lambda) y log(lambda x + (1-lambda)y) geq lambda x log x + (1-lambda) y log y $$
This proves $$f(lambda x + (1-lambda) y) geq lambda f(x) + (1-lambda) f(y)$$ meaning that $f(x) = xlogx$ is not convex meaning that $-f(x)=-xlog x $ is convex.
Hope this helps …….
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
1
$begingroup$
This can not be, because the second derivative is $f''(x)=frac{1}{x}>0~forall x>0$.
$endgroup$
– FuncAna09
Jan 26 at 18:14
1
$begingroup$
$log(lambda x)<log(x)$ because $0<lambda<1$. This was your mistake.
$endgroup$
– FuncAna09
Jan 26 at 18:28
add a comment |
$begingroup$
So I tried the proof, to prove that a function or set is convex we take two points $x$ and $y$ and we try proving that , $$f(lambda x + (1-lambda) y) leq lambda f(x) + (1-lambda) f(y)$$
The concept is simple if we joing two points $x$ and $y$ by a line then all the points lying on the line should also lie inside the set, the line being $lambda x + (1-lambda) y$ , if we vary lambda from $(0,1)$ we find at $lambda$ = $0$ we get the point $x$ at $lambda = 1$ we get the point $y$ and for the value of $lambda = 0.5$ we get the midpoint of $x$ and $y$.
Now for the proof:
$$f(x) = x log x$$
so $$f(lambda x + (1-lambda) y) = ({lambda x + (1-lambda) y})space log ({lambda x +(1-lambda)y})$$
The RHS becomes $$lambda x log (lambda x + (1-lambda) y)+ (1-lambda) y log (lambda x + (1-lambda) y$$
Take the first part and you will notice $lambda x log(lambda x + (1-lambda)y) geq lambda x log(lambda x) $ since $log(A+B) geq log A$
Again $$lambda x log lambda x = lambda x (log lambda + log x) geq lambda x log x$$
So this leads to $$lambda x log(lambda x + (1-lambda)y) geq lambda x log x$$
Similarly, $$(1-lambda) y log(lambda x + (1-lambda)y) geq (1-lambda) y log y$$
Adding these two equations you get,
**$$lambda x log(lambda x + (1-lambda)y) + (1-lambda) y log(lambda x + (1-lambda)y) geq lambda x log x + (1-lambda) y log y $$
This proves $$f(lambda x + (1-lambda) y) geq lambda f(x) + (1-lambda) f(y)$$ meaning that $f(x) = xlogx$ is not convex meaning that $-f(x)=-xlog x $ is convex.
Hope this helps …….
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
1
$begingroup$
This can not be, because the second derivative is $f''(x)=frac{1}{x}>0~forall x>0$.
$endgroup$
– FuncAna09
Jan 26 at 18:14
1
$begingroup$
$log(lambda x)<log(x)$ because $0<lambda<1$. This was your mistake.
$endgroup$
– FuncAna09
Jan 26 at 18:28
add a comment |
$begingroup$
So I tried the proof, to prove that a function or set is convex we take two points $x$ and $y$ and we try proving that , $$f(lambda x + (1-lambda) y) leq lambda f(x) + (1-lambda) f(y)$$
The concept is simple if we joing two points $x$ and $y$ by a line then all the points lying on the line should also lie inside the set, the line being $lambda x + (1-lambda) y$ , if we vary lambda from $(0,1)$ we find at $lambda$ = $0$ we get the point $x$ at $lambda = 1$ we get the point $y$ and for the value of $lambda = 0.5$ we get the midpoint of $x$ and $y$.
Now for the proof:
$$f(x) = x log x$$
so $$f(lambda x + (1-lambda) y) = ({lambda x + (1-lambda) y})space log ({lambda x +(1-lambda)y})$$
The RHS becomes $$lambda x log (lambda x + (1-lambda) y)+ (1-lambda) y log (lambda x + (1-lambda) y$$
Take the first part and you will notice $lambda x log(lambda x + (1-lambda)y) geq lambda x log(lambda x) $ since $log(A+B) geq log A$
Again $$lambda x log lambda x = lambda x (log lambda + log x) geq lambda x log x$$
So this leads to $$lambda x log(lambda x + (1-lambda)y) geq lambda x log x$$
Similarly, $$(1-lambda) y log(lambda x + (1-lambda)y) geq (1-lambda) y log y$$
Adding these two equations you get,
**$$lambda x log(lambda x + (1-lambda)y) + (1-lambda) y log(lambda x + (1-lambda)y) geq lambda x log x + (1-lambda) y log y $$
This proves $$f(lambda x + (1-lambda) y) geq lambda f(x) + (1-lambda) f(y)$$ meaning that $f(x) = xlogx$ is not convex meaning that $-f(x)=-xlog x $ is convex.
Hope this helps …….
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
So I tried the proof, to prove that a function or set is convex we take two points $x$ and $y$ and we try proving that , $$f(lambda x + (1-lambda) y) leq lambda f(x) + (1-lambda) f(y)$$
The concept is simple if we joing two points $x$ and $y$ by a line then all the points lying on the line should also lie inside the set, the line being $lambda x + (1-lambda) y$ , if we vary lambda from $(0,1)$ we find at $lambda$ = $0$ we get the point $x$ at $lambda = 1$ we get the point $y$ and for the value of $lambda = 0.5$ we get the midpoint of $x$ and $y$.
Now for the proof:
$$f(x) = x log x$$
so $$f(lambda x + (1-lambda) y) = ({lambda x + (1-lambda) y})space log ({lambda x +(1-lambda)y})$$
The RHS becomes $$lambda x log (lambda x + (1-lambda) y)+ (1-lambda) y log (lambda x + (1-lambda) y$$
Take the first part and you will notice $lambda x log(lambda x + (1-lambda)y) geq lambda x log(lambda x) $ since $log(A+B) geq log A$
Again $$lambda x log lambda x = lambda x (log lambda + log x) geq lambda x log x$$
So this leads to $$lambda x log(lambda x + (1-lambda)y) geq lambda x log x$$
Similarly, $$(1-lambda) y log(lambda x + (1-lambda)y) geq (1-lambda) y log y$$
Adding these two equations you get,
**$$lambda x log(lambda x + (1-lambda)y) + (1-lambda) y log(lambda x + (1-lambda)y) geq lambda x log x + (1-lambda) y log y $$
This proves $$f(lambda x + (1-lambda) y) geq lambda f(x) + (1-lambda) f(y)$$ meaning that $f(x) = xlogx$ is not convex meaning that $-f(x)=-xlog x $ is convex.
Hope this helps …….
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 26 at 17:35
SNEHIL SANYALSNEHIL SANYAL
654110
654110
1
$begingroup$
This can not be, because the second derivative is $f''(x)=frac{1}{x}>0~forall x>0$.
$endgroup$
– FuncAna09
Jan 26 at 18:14
1
$begingroup$
$log(lambda x)<log(x)$ because $0<lambda<1$. This was your mistake.
$endgroup$
– FuncAna09
Jan 26 at 18:28
add a comment |
1
$begingroup$
This can not be, because the second derivative is $f''(x)=frac{1}{x}>0~forall x>0$.
$endgroup$
– FuncAna09
Jan 26 at 18:14
1
$begingroup$
$log(lambda x)<log(x)$ because $0<lambda<1$. This was your mistake.
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– FuncAna09
Jan 26 at 18:28
1
1
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This can not be, because the second derivative is $f''(x)=frac{1}{x}>0~forall x>0$.
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– FuncAna09
Jan 26 at 18:14
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This can not be, because the second derivative is $f''(x)=frac{1}{x}>0~forall x>0$.
$endgroup$
– FuncAna09
Jan 26 at 18:14
1
1
$begingroup$
$log(lambda x)<log(x)$ because $0<lambda<1$. This was your mistake.
$endgroup$
– FuncAna09
Jan 26 at 18:28
$begingroup$
$log(lambda x)<log(x)$ because $0<lambda<1$. This was your mistake.
$endgroup$
– FuncAna09
Jan 26 at 18:28
add a comment |
$begingroup$
That's what I figured out: Let $lambdain (0,1)$ and $x,yinmathbb{R}>0$. Then we get
begin{align*}
f(lambda x+(1-lambda)y)&=[lambda x+(1-lambda)y]log(lambda x+(1-lambda)y)\
\
&=lambda xlog(lambda x+(1-lambda)y)+(1-lambda)ylog(lambda x+(1-lambda)y).
end{align*}
For the first term of RHS we get
[lambda xlog(lambda x+(1-lambda)y)leqlambda xlog(x)+lambda(1-lambda)y]
and for the second term we get
[(1-lambda)ylog(lambda x+(1-lambda)y)leq (1-lambda)ylog(y)+lambda(1-lambda)x]
In summary, this results in
[f(lambda x+(1-lambda)y)leq lambda f(x)+(1-lambda)f(y)+lambda(1-lambda)(x+y).]
The last term is the problem now.
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
$endgroup$
add a comment |
$begingroup$
That's what I figured out: Let $lambdain (0,1)$ and $x,yinmathbb{R}>0$. Then we get
begin{align*}
f(lambda x+(1-lambda)y)&=[lambda x+(1-lambda)y]log(lambda x+(1-lambda)y)\
\
&=lambda xlog(lambda x+(1-lambda)y)+(1-lambda)ylog(lambda x+(1-lambda)y).
end{align*}
For the first term of RHS we get
[lambda xlog(lambda x+(1-lambda)y)leqlambda xlog(x)+lambda(1-lambda)y]
and for the second term we get
[(1-lambda)ylog(lambda x+(1-lambda)y)leq (1-lambda)ylog(y)+lambda(1-lambda)x]
In summary, this results in
[f(lambda x+(1-lambda)y)leq lambda f(x)+(1-lambda)f(y)+lambda(1-lambda)(x+y).]
The last term is the problem now.
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
$endgroup$
add a comment |
$begingroup$
That's what I figured out: Let $lambdain (0,1)$ and $x,yinmathbb{R}>0$. Then we get
begin{align*}
f(lambda x+(1-lambda)y)&=[lambda x+(1-lambda)y]log(lambda x+(1-lambda)y)\
\
&=lambda xlog(lambda x+(1-lambda)y)+(1-lambda)ylog(lambda x+(1-lambda)y).
end{align*}
For the first term of RHS we get
[lambda xlog(lambda x+(1-lambda)y)leqlambda xlog(x)+lambda(1-lambda)y]
and for the second term we get
[(1-lambda)ylog(lambda x+(1-lambda)y)leq (1-lambda)ylog(y)+lambda(1-lambda)x]
In summary, this results in
[f(lambda x+(1-lambda)y)leq lambda f(x)+(1-lambda)f(y)+lambda(1-lambda)(x+y).]
The last term is the problem now.
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
$endgroup$
That's what I figured out: Let $lambdain (0,1)$ and $x,yinmathbb{R}>0$. Then we get
begin{align*}
f(lambda x+(1-lambda)y)&=[lambda x+(1-lambda)y]log(lambda x+(1-lambda)y)\
\
&=lambda xlog(lambda x+(1-lambda)y)+(1-lambda)ylog(lambda x+(1-lambda)y).
end{align*}
For the first term of RHS we get
[lambda xlog(lambda x+(1-lambda)y)leqlambda xlog(x)+lambda(1-lambda)y]
and for the second term we get
[(1-lambda)ylog(lambda x+(1-lambda)y)leq (1-lambda)ylog(y)+lambda(1-lambda)x]
In summary, this results in
[f(lambda x+(1-lambda)y)leq lambda f(x)+(1-lambda)f(y)+lambda(1-lambda)(x+y).]
The last term is the problem now.
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
edited Jan 26 at 22:21
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
answered Jan 26 at 19:00
FuncAna09FuncAna09
113
113
add a comment |
add a comment |
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$begingroup$
I know, that the function is convex and I also know that the reason is that the seond derivative is positiv for all $x>0$. What I search is a proof without the second derivative argument.
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– FuncAna09
Jan 26 at 15:36
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Welcome @FuncAna09! Could you write the definition of convex function you know? It would be easier to help you. :)
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– Ixion
Jan 26 at 15:38
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Why are you searching for a proof that avoids the second derivative? In this case, it's by far the cleanest argument.
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– Misha Lavrov
Jan 26 at 15:51
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Simply because I want to know how the evidence would go without the derivation.
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– FuncAna09
Jan 26 at 15:58
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I think it might be helpful to consider a slight perturbation of the subtitle of Stanley Kubrick's movie "Doctor Strangelove": "How I learned to stop worrying and love the second derivative trick".
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– Lee Mosher
Jan 26 at 16:52