Is the gradient of $f(x,t)=xe^t/x$ w.r.t. ($x,t$) a Lipschitz continuous function?

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Consider a function $f(x,t)=xe^t/x$ where $C ge x,tge0$ with $C$ being a positive constant. Then, is the gradient $Delta f$ a Lipschitz continuous function over the domain $[0,C]^2$?










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    $f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
    – Kavi Rama Murthy
    Aug 30 at 7:13











  • $f(0,t)$ is not defined ! ?
    – Fred
    Aug 30 at 7:13














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0
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Consider a function $f(x,t)=xe^t/x$ where $C ge x,tge0$ with $C$ being a positive constant. Then, is the gradient $Delta f$ a Lipschitz continuous function over the domain $[0,C]^2$?










share|cite|improve this question

















  • 1




    $f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
    – Kavi Rama Murthy
    Aug 30 at 7:13











  • $f(0,t)$ is not defined ! ?
    – Fred
    Aug 30 at 7:13












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider a function $f(x,t)=xe^t/x$ where $C ge x,tge0$ with $C$ being a positive constant. Then, is the gradient $Delta f$ a Lipschitz continuous function over the domain $[0,C]^2$?










share|cite|improve this question













Consider a function $f(x,t)=xe^t/x$ where $C ge x,tge0$ with $C$ being a positive constant. Then, is the gradient $Delta f$ a Lipschitz continuous function over the domain $[0,C]^2$?







real-analysis optimization non-convex-optimization






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asked Aug 30 at 7:07









Dave

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  • 1




    $f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
    – Kavi Rama Murthy
    Aug 30 at 7:13











  • $f(0,t)$ is not defined ! ?
    – Fred
    Aug 30 at 7:13












  • 1




    $f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
    – Kavi Rama Murthy
    Aug 30 at 7:13











  • $f(0,t)$ is not defined ! ?
    – Fred
    Aug 30 at 7:13







1




1




$f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
– Kavi Rama Murthy
Aug 30 at 7:13





$f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
– Kavi Rama Murthy
Aug 30 at 7:13













$f(0,t)$ is not defined ! ?
– Fred
Aug 30 at 7:13




$f(0,t)$ is not defined ! ?
– Fred
Aug 30 at 7:13















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