Is a composition of a Lipschitz continuous function with a Continuously Differentiable function Lipschitz over an unbounded set? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If $f:mathbb R^ntomathbb R$ is twice continuously differentiable, then $nabla f$ is Lipschitz continuousIs the partial derivative continuous w.r.t. other variables that locally Lipschitz continuous to the function?Continuous and almost everywhere continuously differentiable with bounded gradient implies Lipschitz?Is there a function on a compact interval that is differentiable but not Lipschitz continuous?Example for unbounded Lipschitz function on a bounded domainContinuously differentiable implies Lipschitz in one variable?'Uniformly' locally Lipschitz implies continuously differentiableConditions for a function to be Lipschitz or just uniformly continuousConcatenation of continuous function with uniformly convergent series converges uniformly?How to prove the following function is not Lipschitz continuous?
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Is a composition of a Lipschitz continuous function with a Continuously Differentiable function Lipschitz over an unbounded set?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If $f:mathbb R^ntomathbb R$ is twice continuously differentiable, then $nabla f$ is Lipschitz continuousIs the partial derivative continuous w.r.t. other variables that locally Lipschitz continuous to the function?Continuous and almost everywhere continuously differentiable with bounded gradient implies Lipschitz?Is there a function on a compact interval that is differentiable but not Lipschitz continuous?Example for unbounded Lipschitz function on a bounded domainContinuously differentiable implies Lipschitz in one variable?'Uniformly' locally Lipschitz implies continuously differentiableConditions for a function to be Lipschitz or just uniformly continuousConcatenation of continuous function with uniformly convergent series converges uniformly?How to prove the following function is not Lipschitz continuous?
$begingroup$
Assume the following:
$U subseteq mathbbR^m$ is a closed and connected set (not necessarily bounded),
$Xi in Lip_alpha(U,X)$, where $X subseteq mathbbR^n$,
$g in C^1(X,Y)$, where $X subseteq mathbbY^m$.
Is the function $G = g circ Xi: U rightarrow Y$ Lipschitz?
Up to now, I cannot find neither a way to prove it true, nor a counterexample to prove it wrong. The problem is that: $Xi(U)$ is not necessarily bounded, so I cannot assume $g$ Lipschitz on $X$, even if $g in C^1(X)$. If $g$ would be Lipschitz, then the problem is trivial.
general-topology continuity lipschitz-functions
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
$endgroup$
add a comment |
$begingroup$
Assume the following:
$U subseteq mathbbR^m$ is a closed and connected set (not necessarily bounded),
$Xi in Lip_alpha(U,X)$, where $X subseteq mathbbR^n$,
$g in C^1(X,Y)$, where $X subseteq mathbbY^m$.
Is the function $G = g circ Xi: U rightarrow Y$ Lipschitz?
Up to now, I cannot find neither a way to prove it true, nor a counterexample to prove it wrong. The problem is that: $Xi(U)$ is not necessarily bounded, so I cannot assume $g$ Lipschitz on $X$, even if $g in C^1(X)$. If $g$ would be Lipschitz, then the problem is trivial.
general-topology continuity lipschitz-functions
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
$endgroup$
3
$begingroup$
What about if the Lipschitz map is the identity map, and $g(x) = ||x||^2$? Is the composition Lipschitz?
$endgroup$
– George Dewhirst
Apr 1 at 13:40
$begingroup$
Mmm I think this is a good counterexample, I did not think about the identity map as Lipschitz function!
$endgroup$
– giovanni_13
Apr 1 at 13:55
add a comment |
$begingroup$
Assume the following:
$U subseteq mathbbR^m$ is a closed and connected set (not necessarily bounded),
$Xi in Lip_alpha(U,X)$, where $X subseteq mathbbR^n$,
$g in C^1(X,Y)$, where $X subseteq mathbbY^m$.
Is the function $G = g circ Xi: U rightarrow Y$ Lipschitz?
Up to now, I cannot find neither a way to prove it true, nor a counterexample to prove it wrong. The problem is that: $Xi(U)$ is not necessarily bounded, so I cannot assume $g$ Lipschitz on $X$, even if $g in C^1(X)$. If $g$ would be Lipschitz, then the problem is trivial.
general-topology continuity lipschitz-functions
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
$endgroup$
Assume the following:
$U subseteq mathbbR^m$ is a closed and connected set (not necessarily bounded),
$Xi in Lip_alpha(U,X)$, where $X subseteq mathbbR^n$,
$g in C^1(X,Y)$, where $X subseteq mathbbY^m$.
Is the function $G = g circ Xi: U rightarrow Y$ Lipschitz?
Up to now, I cannot find neither a way to prove it true, nor a counterexample to prove it wrong. The problem is that: $Xi(U)$ is not necessarily bounded, so I cannot assume $g$ Lipschitz on $X$, even if $g in C^1(X)$. If $g$ would be Lipschitz, then the problem is trivial.
general-topology continuity lipschitz-functions
general-topology continuity lipschitz-functions
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
asked Apr 1 at 13:08
giovanni_13giovanni_13
757
757
3
$begingroup$
What about if the Lipschitz map is the identity map, and $g(x) = ||x||^2$? Is the composition Lipschitz?
$endgroup$
– George Dewhirst
Apr 1 at 13:40
$begingroup$
Mmm I think this is a good counterexample, I did not think about the identity map as Lipschitz function!
$endgroup$
– giovanni_13
Apr 1 at 13:55
add a comment |
3
$begingroup$
What about if the Lipschitz map is the identity map, and $g(x) = ||x||^2$? Is the composition Lipschitz?
$endgroup$
– George Dewhirst
Apr 1 at 13:40
$begingroup$
Mmm I think this is a good counterexample, I did not think about the identity map as Lipschitz function!
$endgroup$
– giovanni_13
Apr 1 at 13:55
3
3
$begingroup$
What about if the Lipschitz map is the identity map, and $g(x) = ||x||^2$? Is the composition Lipschitz?
$endgroup$
– George Dewhirst
Apr 1 at 13:40
$begingroup$
What about if the Lipschitz map is the identity map, and $g(x) = ||x||^2$? Is the composition Lipschitz?
$endgroup$
– George Dewhirst
Apr 1 at 13:40
$begingroup$
Mmm I think this is a good counterexample, I did not think about the identity map as Lipschitz function!
$endgroup$
– giovanni_13
Apr 1 at 13:55
$begingroup$
Mmm I think this is a good counterexample, I did not think about the identity map as Lipschitz function!
$endgroup$
– giovanni_13
Apr 1 at 13:55
add a comment |
0
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3
$begingroup$
What about if the Lipschitz map is the identity map, and $g(x) = ||x||^2$? Is the composition Lipschitz?
$endgroup$
– George Dewhirst
Apr 1 at 13:40
$begingroup$
Mmm I think this is a good counterexample, I did not think about the identity map as Lipschitz function!
$endgroup$
– giovanni_13
Apr 1 at 13:55