Smoothness means that the gradient is Lipschitz continuous

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Let $f:mathbbR^drightarrow mathbbR$ be a convex and differentiable function. We are saying that $f$ is smooth (with parameter $L$) if $forall mathbfx, mathbfy in mathbbR^d: f(mathbfy) leq f(mathbfx) + nabla f(mathbfx)^T (mathbfy - mathbfx) + fracL2|mathbfx-mathbfy|^2$



I am trying to prove the following Lemma:



Let $f:mathbbR^d rightarrow mathbbR$ be convex and differentiable. The following two statements are equivalent:



(i) f is smooth with parameter $L$



(ii) $|nabla f(mathbfx) - nabla f(mathbfy)| leq L|mathbfx-mathbfy| forall mathbfx, mathbfy in mathbbR^d$



I've managed to do (ii) $rightarrow$ (i) by using the first order characterization of convexity, subtracting and adding the term $nabla f(mathbfx)^T(mathbfy-mathbfx)$ and then by using the Cauchy-Schwarz inequality.



However I am stuck in (i) $rightarrow$ (ii). I am taking the smoothness condition one time for $mathbfx, mathbfy$ and one time for $mathbfy, mathbfx$ and I am adding the inequalities produced leading to



$(nabla f(mathbfx) - nabla f(mathbfy))^T (mathbfx - mathbfy) leq L |mathbfx - mathbfy |^2 $



but this doesn't seem to lead somewhere. Any ideas on how to prove this?




optimization convex-analysis lipschitz-functions




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  • This is explained in the first lecture entitled "Gradient method" in Vandenberghe's UCLA 236c notes.
    – littleO
    Jul 23 at 7:20










  • Thanks! Didn't have the notion of co-coercivity in our lecture notes!
    – dimkou
    Jul 23 at 7:23














up vote
1
down vote

favorite
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Let $f:mathbbR^drightarrow mathbbR$ be a convex and differentiable function. We are saying that $f$ is smooth (with parameter $L$) if $forall mathbfx, mathbfy in mathbbR^d: f(mathbfy) leq f(mathbfx) + nabla f(mathbfx)^T (mathbfy - mathbfx) + fracL2|mathbfx-mathbfy|^2$



I am trying to prove the following Lemma:



Let $f:mathbbR^d rightarrow mathbbR$ be convex and differentiable. The following two statements are equivalent:



(i) f is smooth with parameter $L$



(ii) $|nabla f(mathbfx) - nabla f(mathbfy)| leq L|mathbfx-mathbfy| forall mathbfx, mathbfy in mathbbR^d$



I've managed to do (ii) $rightarrow$ (i) by using the first order characterization of convexity, subtracting and adding the term $nabla f(mathbfx)^T(mathbfy-mathbfx)$ and then by using the Cauchy-Schwarz inequality.



However I am stuck in (i) $rightarrow$ (ii). I am taking the smoothness condition one time for $mathbfx, mathbfy$ and one time for $mathbfy, mathbfx$ and I am adding the inequalities produced leading to



$(nabla f(mathbfx) - nabla f(mathbfy))^T (mathbfx - mathbfy) leq L |mathbfx - mathbfy |^2 $



but this doesn't seem to lead somewhere. Any ideas on how to prove this?




optimization convex-analysis lipschitz-functions




share|cite|improve this question



















  • This is explained in the first lecture entitled "Gradient method" in Vandenberghe's UCLA 236c notes.
    – littleO
    Jul 23 at 7:20










  • Thanks! Didn't have the notion of co-coercivity in our lecture notes!
    – dimkou
    Jul 23 at 7:23












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Let $f:mathbbR^drightarrow mathbbR$ be a convex and differentiable function. We are saying that $f$ is smooth (with parameter $L$) if $forall mathbfx, mathbfy in mathbbR^d: f(mathbfy) leq f(mathbfx) + nabla f(mathbfx)^T (mathbfy - mathbfx) + fracL2|mathbfx-mathbfy|^2$



I am trying to prove the following Lemma:



Let $f:mathbbR^d rightarrow mathbbR$ be convex and differentiable. The following two statements are equivalent:



(i) f is smooth with parameter $L$



(ii) $|nabla f(mathbfx) - nabla f(mathbfy)| leq L|mathbfx-mathbfy| forall mathbfx, mathbfy in mathbbR^d$



I've managed to do (ii) $rightarrow$ (i) by using the first order characterization of convexity, subtracting and adding the term $nabla f(mathbfx)^T(mathbfy-mathbfx)$ and then by using the Cauchy-Schwarz inequality.



However I am stuck in (i) $rightarrow$ (ii). I am taking the smoothness condition one time for $mathbfx, mathbfy$ and one time for $mathbfy, mathbfx$ and I am adding the inequalities produced leading to



$(nabla f(mathbfx) - nabla f(mathbfy))^T (mathbfx - mathbfy) leq L |mathbfx - mathbfy |^2 $



but this doesn't seem to lead somewhere. Any ideas on how to prove this?




optimization convex-analysis lipschitz-functions




share|cite|improve this question











Let $f:mathbbR^drightarrow mathbbR$ be a convex and differentiable function. We are saying that $f$ is smooth (with parameter $L$) if $forall mathbfx, mathbfy in mathbbR^d: f(mathbfy) leq f(mathbfx) + nabla f(mathbfx)^T (mathbfy - mathbfx) + fracL2|mathbfx-mathbfy|^2$



I am trying to prove the following Lemma:



Let $f:mathbbR^d rightarrow mathbbR$ be convex and differentiable. The following two statements are equivalent:



(i) f is smooth with parameter $L$



(ii) $|nabla f(mathbfx) - nabla f(mathbfy)| leq L|mathbfx-mathbfy| forall mathbfx, mathbfy in mathbbR^d$



I've managed to do (ii) $rightarrow$ (i) by using the first order characterization of convexity, subtracting and adding the term $nabla f(mathbfx)^T(mathbfy-mathbfx)$ and then by using the Cauchy-Schwarz inequality.



However I am stuck in (i) $rightarrow$ (ii). I am taking the smoothness condition one time for $mathbfx, mathbfy$ and one time for $mathbfy, mathbfx$ and I am adding the inequalities produced leading to



$(nabla f(mathbfx) - nabla f(mathbfy))^T (mathbfx - mathbfy) leq L |mathbfx - mathbfy |^2 $



but this doesn't seem to lead somewhere. Any ideas on how to prove this?





optimization convex-analysis lipschitz-functions





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 23 at 7:01









dimkou

535




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  • This is explained in the first lecture entitled "Gradient method" in Vandenberghe's UCLA 236c notes.
    – littleO
    Jul 23 at 7:20










  • Thanks! Didn't have the notion of co-coercivity in our lecture notes!
    – dimkou
    Jul 23 at 7:23
















  • This is explained in the first lecture entitled "Gradient method" in Vandenberghe's UCLA 236c notes.
    – littleO
    Jul 23 at 7:20










  • Thanks! Didn't have the notion of co-coercivity in our lecture notes!
    – dimkou
    Jul 23 at 7:23















This is explained in the first lecture entitled "Gradient method" in Vandenberghe's UCLA 236c notes.
– littleO
Jul 23 at 7:20




This is explained in the first lecture entitled "Gradient method" in Vandenberghe's UCLA 236c notes.
– littleO
Jul 23 at 7:20












Thanks! Didn't have the notion of co-coercivity in our lecture notes!
– dimkou
Jul 23 at 7:23




Thanks! Didn't have the notion of co-coercivity in our lecture notes!
– dimkou
Jul 23 at 7:23















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