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 $[0,C]^2$?
real-analysis optimization non-convex-optimization
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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 $[0,C]^2$?
real-analysis optimization non-convex-optimization
1
$f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
â Kavi Rama Murthy
Aug 30 at 7:13
$f(0,t)$ is not defined ! ?
â Fred
Aug 30 at 7:13
add a comment |Â
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 $[0,C]^2$?
real-analysis optimization non-convex-optimization
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
real-analysis optimization non-convex-optimization
asked Aug 30 at 7:07
Dave
19919
19919
1
$f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
â Kavi Rama Murthy
Aug 30 at 7:13
$f(0,t)$ is not defined ! ?
â Fred
Aug 30 at 7:13
add a comment |Â
1
$f$ does not even have a gradient at $(0,t)$ for $t>0$. (even if you define $f(0,t)$ to be $0$).
â Kavi Rama Murthy
Aug 30 at 7:13
$f(0,t)$ is not defined ! ?
â 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 $0$).
â 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 $0$).
â Kavi Rama Murthy
Aug 30 at 7:13
$f(0,t)$ is not defined ! ?
â Fred
Aug 30 at 7:13
$f(0,t)$ is not defined ! ?
â Fred
Aug 30 at 7:13
add a comment |Â
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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 $0$).
â Kavi Rama Murthy
Aug 30 at 7:13
$f(0,t)$ is not defined ! ?
â Fred
Aug 30 at 7:13