Why taking integral of both sides of matrix inequality is allowed?
$begingroup$
How to show if $nabla^2 f(x) succeq alpha I$, then the function is $alpha$-strongly convex?
In my optimization notes I have
$$nabla^2 f(x) succeq alpha I rightarrow alphatext{-strongly convex} ,,,,,,forall x$$
where $x in mathbb{R}^n$ and $Asucceq B$ means $A-B$ is positive semi-definite.
For the proof we use mean value theorem
$$
nabla f(y)-nabla f(y) = int_0^1 nabla^2 f(x+t(y-x))(y-x)dt
$$
where $x, y in mathbb{R}^n$ and $0 leq t leq 1$.
$$
langle nabla f(y)-nabla f(y) ,y-x rangle= langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle
$$
Since
$$nabla^2 f(x+t(y-x)) succeq alpha I tag{1}$$
and the fact that $langle Ad,d rangle geq c|d|^2 leftrightarrow A succeq cI tag{2}$
$$
langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle geq alpha |y-x|^2 tag{3}
$$
I do not understand why how we can use $(1)$ to get $(2)$? Because I do not know why we can use $(1)$ and write
$$
A=int_0^1 nabla^2 f(x+t(y-x))dt succeq int_0^1 alpha I dt= alpha I
$$
and then use $(2)$.
Is it possible to take integral from any matrix inequality?
integration matrices positive-semidefinite
asked Jan 17 at 19:24
SaeedSaeed
1,149310
$endgroup$
add a comment |
$begingroup$
How to show if $nabla^2 f(x) succeq alpha I$, then the function is $alpha$-strongly convex?
In my optimization notes I have
$$nabla^2 f(x) succeq alpha I rightarrow alphatext{-strongly convex} ,,,,,,forall x$$
where $x in mathbb{R}^n$ and $Asucceq B$ means $A-B$ is positive semi-definite.
For the proof we use mean value theorem
$$
nabla f(y)-nabla f(y) = int_0^1 nabla^2 f(x+t(y-x))(y-x)dt
$$
where $x, y in mathbb{R}^n$ and $0 leq t leq 1$.
$$
langle nabla f(y)-nabla f(y) ,y-x rangle= langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle
$$
Since
$$nabla^2 f(x+t(y-x)) succeq alpha I tag{1}$$
and the fact that $langle Ad,d rangle geq c|d|^2 leftrightarrow A succeq cI tag{2}$
$$
langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle geq alpha |y-x|^2 tag{3}
$$
I do not understand why how we can use $(1)$ to get $(2)$? Because I do not know why we can use $(1)$ and write
$$
A=int_0^1 nabla^2 f(x+t(y-x))dt succeq int_0^1 alpha I dt= alpha I
$$
and then use $(2)$.
Is it possible to take integral from any matrix inequality?
integration matrices positive-semidefinite
asked Jan 17 at 19:24
SaeedSaeed
1,149310
$endgroup$
add a comment |
$begingroup$
How to show if $nabla^2 f(x) succeq alpha I$, then the function is $alpha$-strongly convex?
In my optimization notes I have
$$nabla^2 f(x) succeq alpha I rightarrow alphatext{-strongly convex} ,,,,,,forall x$$
where $x in mathbb{R}^n$ and $Asucceq B$ means $A-B$ is positive semi-definite.
For the proof we use mean value theorem
$$
nabla f(y)-nabla f(y) = int_0^1 nabla^2 f(x+t(y-x))(y-x)dt
$$
where $x, y in mathbb{R}^n$ and $0 leq t leq 1$.
$$
langle nabla f(y)-nabla f(y) ,y-x rangle= langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle
$$
Since
$$nabla^2 f(x+t(y-x)) succeq alpha I tag{1}$$
and the fact that $langle Ad,d rangle geq c|d|^2 leftrightarrow A succeq cI tag{2}$
$$
langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle geq alpha |y-x|^2 tag{3}
$$
I do not understand why how we can use $(1)$ to get $(2)$? Because I do not know why we can use $(1)$ and write
$$
A=int_0^1 nabla^2 f(x+t(y-x))dt succeq int_0^1 alpha I dt= alpha I
$$
and then use $(2)$.
Is it possible to take integral from any matrix inequality?
integration matrices positive-semidefinite
asked Jan 17 at 19:24
SaeedSaeed
1,149310
$endgroup$
How to show if $nabla^2 f(x) succeq alpha I$, then the function is $alpha$-strongly convex?
In my optimization notes I have
$$nabla^2 f(x) succeq alpha I rightarrow alphatext{-strongly convex} ,,,,,,forall x$$
where $x in mathbb{R}^n$ and $Asucceq B$ means $A-B$ is positive semi-definite.
For the proof we use mean value theorem
$$
nabla f(y)-nabla f(y) = int_0^1 nabla^2 f(x+t(y-x))(y-x)dt
$$
where $x, y in mathbb{R}^n$ and $0 leq t leq 1$.
$$
langle nabla f(y)-nabla f(y) ,y-x rangle= langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle
$$
Since
$$nabla^2 f(x+t(y-x)) succeq alpha I tag{1}$$
and the fact that $langle Ad,d rangle geq c|d|^2 leftrightarrow A succeq cI tag{2}$
$$
langle int_0^1 nabla^2 f(x+t(y-x))(y-x)dt,y-x rangle geq alpha |y-x|^2 tag{3}
$$
I do not understand why how we can use $(1)$ to get $(2)$? Because I do not know why we can use $(1)$ and write
$$
A=int_0^1 nabla^2 f(x+t(y-x))dt succeq int_0^1 alpha I dt= alpha I
$$
and then use $(2)$.
Is it possible to take integral from any matrix inequality?
integration matrices positive-semidefinite
integration matrices positive-semidefinite
asked Jan 17 at 19:24
SaeedSaeed
1,149310
asked Jan 17 at 19:24
SaeedSaeed
1,149310
asked Jan 17 at 19:24
SaeedSaeed
1,149310
asked Jan 17 at 19:24
SaeedSaeed
1,149310
asked Jan 17 at 19:24
SaeedSaeed
1,149310
1,149310
add a comment |
add a comment |
1 Answer
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In your statement of the "mean value theorem" and in other parts of your post, you are taking the integral of a vector or a matrix, which does not make sense.
I think it would be better to state the integral equality as
$$langle nabla f(y) - nabla f(x), y-x rangle
= int_0^1 langle nabla^2 f(x + t(y-x)) (y-x), y-x rangle , dt.$$
(This is simply the fundamental theorem of calculus $g(1) - g(0) = int_0^1 g'(t) , dt$ applied to $g(t) := langle nabla f(x + t(y-x)), y-x rangle$.)
From here, you can immediately use
$nabla^2 f(x + t(y-x)) (y-x), y-x rangle ge alpha |y-x|^2$ from (2), as desired.
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
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add a comment |
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$begingroup$
In your statement of the "mean value theorem" and in other parts of your post, you are taking the integral of a vector or a matrix, which does not make sense.
I think it would be better to state the integral equality as
$$langle nabla f(y) - nabla f(x), y-x rangle
= int_0^1 langle nabla^2 f(x + t(y-x)) (y-x), y-x rangle , dt.$$
(This is simply the fundamental theorem of calculus $g(1) - g(0) = int_0^1 g'(t) , dt$ applied to $g(t) := langle nabla f(x + t(y-x)), y-x rangle$.)
From here, you can immediately use
$nabla^2 f(x + t(y-x)) (y-x), y-x rangle ge alpha |y-x|^2$ from (2), as desired.
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
$endgroup$
add a comment |
$begingroup$
In your statement of the "mean value theorem" and in other parts of your post, you are taking the integral of a vector or a matrix, which does not make sense.
I think it would be better to state the integral equality as
$$langle nabla f(y) - nabla f(x), y-x rangle
= int_0^1 langle nabla^2 f(x + t(y-x)) (y-x), y-x rangle , dt.$$
(This is simply the fundamental theorem of calculus $g(1) - g(0) = int_0^1 g'(t) , dt$ applied to $g(t) := langle nabla f(x + t(y-x)), y-x rangle$.)
From here, you can immediately use
$nabla^2 f(x + t(y-x)) (y-x), y-x rangle ge alpha |y-x|^2$ from (2), as desired.
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
$endgroup$
add a comment |
$begingroup$
In your statement of the "mean value theorem" and in other parts of your post, you are taking the integral of a vector or a matrix, which does not make sense.
I think it would be better to state the integral equality as
$$langle nabla f(y) - nabla f(x), y-x rangle
= int_0^1 langle nabla^2 f(x + t(y-x)) (y-x), y-x rangle , dt.$$
(This is simply the fundamental theorem of calculus $g(1) - g(0) = int_0^1 g'(t) , dt$ applied to $g(t) := langle nabla f(x + t(y-x)), y-x rangle$.)
From here, you can immediately use
$nabla^2 f(x + t(y-x)) (y-x), y-x rangle ge alpha |y-x|^2$ from (2), as desired.
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
$endgroup$
In your statement of the "mean value theorem" and in other parts of your post, you are taking the integral of a vector or a matrix, which does not make sense.
I think it would be better to state the integral equality as
$$langle nabla f(y) - nabla f(x), y-x rangle
= int_0^1 langle nabla^2 f(x + t(y-x)) (y-x), y-x rangle , dt.$$
(This is simply the fundamental theorem of calculus $g(1) - g(0) = int_0^1 g'(t) , dt$ applied to $g(t) := langle nabla f(x + t(y-x)), y-x rangle$.)
From here, you can immediately use
$nabla^2 f(x + t(y-x)) (y-x), y-x rangle ge alpha |y-x|^2$ from (2), as desired.
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
answered Jan 17 at 19:39
angryavianangryavian
42.6k23481
42.6k23481
add a comment |
add a comment |
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