From Hessian to Lipschitz continuity
$begingroup$
Let $f$ be convex and twice differentiable. I can't prove that
$$
nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
$$
I only established the converse.
edit:
There exists a $zintext{conv}{x,y}$ such that
$$
nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
$$
By multiplying on the left by $(y-x)^T$ we get
$$
begin{align}
(y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
&leq (y-x)^T nabla f(x) + L||y-x||_2^2
end{align}
$$
Now how do we prove that
$$
(nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
$$
I think I'm missing something very easy.
convex-analysis
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
$endgroup$
add a comment |
$begingroup$
Let $f$ be convex and twice differentiable. I can't prove that
$$
nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
$$
I only established the converse.
edit:
There exists a $zintext{conv}{x,y}$ such that
$$
nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
$$
By multiplying on the left by $(y-x)^T$ we get
$$
begin{align}
(y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
&leq (y-x)^T nabla f(x) + L||y-x||_2^2
end{align}
$$
Now how do we prove that
$$
(nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
$$
I think I'm missing something very easy.
convex-analysis
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
$endgroup$
add a comment |
$begingroup$
Let $f$ be convex and twice differentiable. I can't prove that
$$
nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
$$
I only established the converse.
edit:
There exists a $zintext{conv}{x,y}$ such that
$$
nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
$$
By multiplying on the left by $(y-x)^T$ we get
$$
begin{align}
(y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
&leq (y-x)^T nabla f(x) + L||y-x||_2^2
end{align}
$$
Now how do we prove that
$$
(nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
$$
I think I'm missing something very easy.
convex-analysis
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
$endgroup$
Let $f$ be convex and twice differentiable. I can't prove that
$$
nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
$$
I only established the converse.
edit:
There exists a $zintext{conv}{x,y}$ such that
$$
nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
$$
By multiplying on the left by $(y-x)^T$ we get
$$
begin{align}
(y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
&leq (y-x)^T nabla f(x) + L||y-x||_2^2
end{align}
$$
Now how do we prove that
$$
(nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
$$
I think I'm missing something very easy.
convex-analysis
convex-analysis
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
edited May 23 '17 at 12:26
Kiuhnm
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
asked May 23 '17 at 1:24
KiuhnmKiuhnm
361210
361210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
begin{align}
|nabla f(y) - nabla f(x)|
&=
|g(1) - g(0)|
\
&= left| int_0^1 g'(t) mathop{dt}right|
\
&le int_0^1 |g'(t)| mathop{dt}
\
&= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
\
&le int_0^1 L |x-y| mathop{dt}
\
&= L|x-y|.
end{align}
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
$endgroup$
$begingroup$
What property do you use for the last $leq$? Thanks.
$endgroup$
– Kiuhnm
May 23 '17 at 23:06
1
$begingroup$
@Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
$endgroup$
– angryavian
May 23 '17 at 23:21
$begingroup$
Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
$endgroup$
– Kiuhnm
May 24 '17 at 0:12
1
$begingroup$
Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
$endgroup$
– angryavian
May 24 '17 at 1:33
add a comment |
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1 Answer
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active
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1 Answer
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active
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oldest
votes
$begingroup$
Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
begin{align}
|nabla f(y) - nabla f(x)|
&=
|g(1) - g(0)|
\
&= left| int_0^1 g'(t) mathop{dt}right|
\
&le int_0^1 |g'(t)| mathop{dt}
\
&= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
\
&le int_0^1 L |x-y| mathop{dt}
\
&= L|x-y|.
end{align}
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
$endgroup$
$begingroup$
What property do you use for the last $leq$? Thanks.
$endgroup$
– Kiuhnm
May 23 '17 at 23:06
1
$begingroup$
@Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
$endgroup$
– angryavian
May 23 '17 at 23:21
$begingroup$
Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
$endgroup$
– Kiuhnm
May 24 '17 at 0:12
1
$begingroup$
Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
$endgroup$
– angryavian
May 24 '17 at 1:33
add a comment |
$begingroup$
Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
begin{align}
|nabla f(y) - nabla f(x)|
&=
|g(1) - g(0)|
\
&= left| int_0^1 g'(t) mathop{dt}right|
\
&le int_0^1 |g'(t)| mathop{dt}
\
&= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
\
&le int_0^1 L |x-y| mathop{dt}
\
&= L|x-y|.
end{align}
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
$endgroup$
$begingroup$
What property do you use for the last $leq$? Thanks.
$endgroup$
– Kiuhnm
May 23 '17 at 23:06
1
$begingroup$
@Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
$endgroup$
– angryavian
May 23 '17 at 23:21
$begingroup$
Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
$endgroup$
– Kiuhnm
May 24 '17 at 0:12
1
$begingroup$
Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
$endgroup$
– angryavian
May 24 '17 at 1:33
add a comment |
$begingroup$
Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
begin{align}
|nabla f(y) - nabla f(x)|
&=
|g(1) - g(0)|
\
&= left| int_0^1 g'(t) mathop{dt}right|
\
&le int_0^1 |g'(t)| mathop{dt}
\
&= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
\
&le int_0^1 L |x-y| mathop{dt}
\
&= L|x-y|.
end{align}
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
$endgroup$
Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
begin{align}
|nabla f(y) - nabla f(x)|
&=
|g(1) - g(0)|
\
&= left| int_0^1 g'(t) mathop{dt}right|
\
&le int_0^1 |g'(t)| mathop{dt}
\
&= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
\
&le int_0^1 L |x-y| mathop{dt}
\
&= L|x-y|.
end{align}
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
answered May 23 '17 at 21:36
angryavianangryavian
41.4k23380
41.4k23380
$begingroup$
What property do you use for the last $leq$? Thanks.
$endgroup$
– Kiuhnm
May 23 '17 at 23:06
1
$begingroup$
@Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
$endgroup$
– angryavian
May 23 '17 at 23:21
$begingroup$
Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
$endgroup$
– Kiuhnm
May 24 '17 at 0:12
1
$begingroup$
Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
$endgroup$
– angryavian
May 24 '17 at 1:33
add a comment |
$begingroup$
What property do you use for the last $leq$? Thanks.
$endgroup$
– Kiuhnm
May 23 '17 at 23:06
1
$begingroup$
@Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
$endgroup$
– angryavian
May 23 '17 at 23:21
$begingroup$
Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
$endgroup$
– Kiuhnm
May 24 '17 at 0:12
1
$begingroup$
Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
$endgroup$
– angryavian
May 24 '17 at 1:33
$begingroup$
What property do you use for the last $leq$? Thanks.
$endgroup$
– Kiuhnm
May 23 '17 at 23:06
$begingroup$
What property do you use for the last $leq$? Thanks.
$endgroup$
– Kiuhnm
May 23 '17 at 23:06
1
1
$begingroup$
@Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
$endgroup$
– angryavian
May 23 '17 at 23:21
$begingroup$
@Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
$endgroup$
– angryavian
May 23 '17 at 23:21
$begingroup$
Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
$endgroup$
– Kiuhnm
May 24 '17 at 0:12
$begingroup$
Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
$endgroup$
– Kiuhnm
May 24 '17 at 0:12
1
1
$begingroup$
Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
$endgroup$
– angryavian
May 24 '17 at 1:33
$begingroup$
Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
$endgroup$
– angryavian
May 24 '17 at 1:33
add a comment |
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