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Lipschitz continuity of the gradient of a Quadratic upper bound convex function?
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According to Prof. Vandenberg's lecture notes [http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf], page 11, a function is called Lipschitz continuous gradient when
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
Note that the definition does not assume that $f$ is even a convex function.
However, if $f$ is a convex function we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
which mean the convex function is bounded by a quadradic function.
Can we prove the reverse, i.e., if we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
then
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
convex-analysis convex-optimization machine-learning
asked Jul 18 at 1:06
Saeed
887
add a comment |Â
up vote
-1
down vote
favorite
According to Prof. Vandenberg's lecture notes [http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf], page 11, a function is called Lipschitz continuous gradient when
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
Note that the definition does not assume that $f$ is even a convex function.
However, if $f$ is a convex function we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
which mean the convex function is bounded by a quadradic function.
Can we prove the reverse, i.e., if we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
then
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
convex-analysis convex-optimization machine-learning
asked Jul 18 at 1:06
Saeed
887
Please show what you've tried and where you're stuck to get positive response.
– Alex Francisco
Jul 18 at 1:54
The definition of $alpha$ strongly convex function is at page 135 of this monograph [cs.huji.ac.il/~shais/papers/OLsurvey.pdf]. My question is how to find an upper bound for deviation of gradient of an $alpha$-strongly convex function?
– Saeed
Jul 18 at 16:13
Sorry, I was confused. The exact statement of the problem has been posted. forget about strong convexity of $f$.
– Saeed
Jul 18 at 17:49
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
According to Prof. Vandenberg's lecture notes [http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf], page 11, a function is called Lipschitz continuous gradient when
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
Note that the definition does not assume that $f$ is even a convex function.
However, if $f$ is a convex function we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
which mean the convex function is bounded by a quadradic function.
Can we prove the reverse, i.e., if we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
then
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
convex-analysis convex-optimization machine-learning
asked Jul 18 at 1:06
Saeed
887
According to Prof. Vandenberg's lecture notes [http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf], page 11, a function is called Lipschitz continuous gradient when
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
Note that the definition does not assume that $f$ is even a convex function.
However, if $f$ is a convex function we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
which mean the convex function is bounded by a quadradic function.
Can we prove the reverse, i.e., if we have
$$
f(y) leq f(x) + langle nabla f(x),y-x rangle + fracalpha2|y-x|_2^2
$$
then
$$
|nabla f(y)-nabla f(x)|_2 leq alpha|y-x|_2
$$
convex-analysis convex-optimization machine-learning
asked Jul 18 at 1:06
Saeed
887
edited Jul 18 at 17:48
asked Jul 18 at 1:06
Saeed
887
asked Jul 18 at 1:06
Saeed
887
asked Jul 18 at 1:06
Saeed
887
887
Please show what you've tried and where you're stuck to get positive response.
– Alex Francisco
Jul 18 at 1:54
The definition of $alpha$ strongly convex function is at page 135 of this monograph [cs.huji.ac.il/~shais/papers/OLsurvey.pdf]. My question is how to find an upper bound for deviation of gradient of an $alpha$-strongly convex function?
– Saeed
Jul 18 at 16:13
Sorry, I was confused. The exact statement of the problem has been posted. forget about strong convexity of $f$.
– Saeed
Jul 18 at 17:49
add a comment |Â
Please show what you've tried and where you're stuck to get positive response.
– Alex Francisco
Jul 18 at 1:54
The definition of $alpha$ strongly convex function is at page 135 of this monograph [cs.huji.ac.il/~shais/papers/OLsurvey.pdf]. My question is how to find an upper bound for deviation of gradient of an $alpha$-strongly convex function?
– Saeed
Jul 18 at 16:13
Sorry, I was confused. The exact statement of the problem has been posted. forget about strong convexity of $f$.
– Saeed
Jul 18 at 17:49
Please show what you've tried and where you're stuck to get positive response.
– Alex Francisco
Jul 18 at 1:54
Please show what you've tried and where you're stuck to get positive response.
– Alex Francisco
Jul 18 at 1:54
The definition of $alpha$ strongly convex function is at page 135 of this monograph [cs.huji.ac.il/~shais/papers/OLsurvey.pdf]. My question is how to find an upper bound for deviation of gradient of an $alpha$-strongly convex function?
– Saeed
Jul 18 at 16:13
The definition of $alpha$ strongly convex function is at page 135 of this monograph [cs.huji.ac.il/~shais/papers/OLsurvey.pdf]. My question is how to find an upper bound for deviation of gradient of an $alpha$-strongly convex function?
– Saeed
Jul 18 at 16:13
Sorry, I was confused. The exact statement of the problem has been posted. forget about strong convexity of $f$.
– Saeed
Jul 18 at 17:49
Sorry, I was confused. The exact statement of the problem has been posted. forget about strong convexity of $f$.
– Saeed
Jul 18 at 17:49
add a comment |Â
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Please show what you've tried and where you're stuck to get positive response.
– Alex Francisco
Jul 18 at 1:54
The definition of $alpha$ strongly convex function is at page 135 of this monograph [cs.huji.ac.il/~shais/papers/OLsurvey.pdf]. My question is how to find an upper bound for deviation of gradient of an $alpha$-strongly convex function?
– Saeed
Jul 18 at 16:13
Sorry, I was confused. The exact statement of the problem has been posted. forget about strong convexity of $f$.
– Saeed
Jul 18 at 17:49