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Extend a locally Lipschitz function to a globally Lipschitz
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I want to show that for any continuous function $f:R^Nto R^V$ of at most linear growth, i.e.
$$exists C<inftyforall xin R^N ||f(x)||leq C(1+||x||), $$
there exists a sequence $f_n:R^Nto R^V,ngeq1,$ such that
- $f_n$'s are of at most linear growth, uniformly in $n$.
- $f_n$'s are Lipschitz continuous.
- $f_nto f$ locally uniformly on $R^N$.
As a hint suggested I defined convolution functions $g_n(x)=n^Nint_R^Nf(y)zeta(n(x-y))dy,$ for some $zetain C^infty(R^N,R)$ with $zetageq0,$ supp$zetasubset$ unit ball, and $int_R^Nzeta(x)dx=1$ and I showed that $g_n$'s satisfy points 1. and 3. but only locally Lipschitz continuity. At this point I should modify the $g_n$'s to get globally Lipschitz condition, but I don't know how to proceed. I think I have to choose nice compact $Ksubset R^N$ (large enough to not lose the convergence) and set $f_n(x)=g_n(x)1_K+somethingcdot1_K^c$, (similarly as this question Extending a $k$-lipschitz function), but I'm not sure. Thanks in advance.
real-analysis uniform-convergence lipschitz-functions
asked Jul 25 at 16:33
Demetrio Masciurett
285
add a comment |Â
up vote
1
down vote
favorite
I want to show that for any continuous function $f:R^Nto R^V$ of at most linear growth, i.e.
$$exists C<inftyforall xin R^N ||f(x)||leq C(1+||x||), $$
there exists a sequence $f_n:R^Nto R^V,ngeq1,$ such that
- $f_n$'s are of at most linear growth, uniformly in $n$.
- $f_n$'s are Lipschitz continuous.
- $f_nto f$ locally uniformly on $R^N$.
As a hint suggested I defined convolution functions $g_n(x)=n^Nint_R^Nf(y)zeta(n(x-y))dy,$ for some $zetain C^infty(R^N,R)$ with $zetageq0,$ supp$zetasubset$ unit ball, and $int_R^Nzeta(x)dx=1$ and I showed that $g_n$'s satisfy points 1. and 3. but only locally Lipschitz continuity. At this point I should modify the $g_n$'s to get globally Lipschitz condition, but I don't know how to proceed. I think I have to choose nice compact $Ksubset R^N$ (large enough to not lose the convergence) and set $f_n(x)=g_n(x)1_K+somethingcdot1_K^c$, (similarly as this question Extending a $k$-lipschitz function), but I'm not sure. Thanks in advance.
real-analysis uniform-convergence lipschitz-functions
asked Jul 25 at 16:33
Demetrio Masciurett
285
1
It may be easier to modify the definition of $g_n$ and use $f(y)cdot varphi_n(y)$ where $varphi_n$ is a continuous function identically $1$ in the ball of radius $n$ (with centre $0$) and vanishes outside the ball of radius $n+1$ instead of just $f(y)$ in the convolution.
– Daniel Fischer♦
Jul 25 at 16:39
It works! Thank you very much!
– Demetrio Masciurett
Jul 25 at 17:21
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show that for any continuous function $f:R^Nto R^V$ of at most linear growth, i.e.
$$exists C<inftyforall xin R^N ||f(x)||leq C(1+||x||), $$
there exists a sequence $f_n:R^Nto R^V,ngeq1,$ such that
- $f_n$'s are of at most linear growth, uniformly in $n$.
- $f_n$'s are Lipschitz continuous.
- $f_nto f$ locally uniformly on $R^N$.
As a hint suggested I defined convolution functions $g_n(x)=n^Nint_R^Nf(y)zeta(n(x-y))dy,$ for some $zetain C^infty(R^N,R)$ with $zetageq0,$ supp$zetasubset$ unit ball, and $int_R^Nzeta(x)dx=1$ and I showed that $g_n$'s satisfy points 1. and 3. but only locally Lipschitz continuity. At this point I should modify the $g_n$'s to get globally Lipschitz condition, but I don't know how to proceed. I think I have to choose nice compact $Ksubset R^N$ (large enough to not lose the convergence) and set $f_n(x)=g_n(x)1_K+somethingcdot1_K^c$, (similarly as this question Extending a $k$-lipschitz function), but I'm not sure. Thanks in advance.
real-analysis uniform-convergence lipschitz-functions
asked Jul 25 at 16:33
Demetrio Masciurett
285
I want to show that for any continuous function $f:R^Nto R^V$ of at most linear growth, i.e.
$$exists C<inftyforall xin R^N ||f(x)||leq C(1+||x||), $$
there exists a sequence $f_n:R^Nto R^V,ngeq1,$ such that
- $f_n$'s are of at most linear growth, uniformly in $n$.
- $f_n$'s are Lipschitz continuous.
- $f_nto f$ locally uniformly on $R^N$.
As a hint suggested I defined convolution functions $g_n(x)=n^Nint_R^Nf(y)zeta(n(x-y))dy,$ for some $zetain C^infty(R^N,R)$ with $zetageq0,$ supp$zetasubset$ unit ball, and $int_R^Nzeta(x)dx=1$ and I showed that $g_n$'s satisfy points 1. and 3. but only locally Lipschitz continuity. At this point I should modify the $g_n$'s to get globally Lipschitz condition, but I don't know how to proceed. I think I have to choose nice compact $Ksubset R^N$ (large enough to not lose the convergence) and set $f_n(x)=g_n(x)1_K+somethingcdot1_K^c$, (similarly as this question Extending a $k$-lipschitz function), but I'm not sure. Thanks in advance.
real-analysis uniform-convergence lipschitz-functions
asked Jul 25 at 16:33
Demetrio Masciurett
285
edited Jul 25 at 16:50
asked Jul 25 at 16:33
Demetrio Masciurett
285
asked Jul 25 at 16:33
Demetrio Masciurett
285
asked Jul 25 at 16:33
Demetrio Masciurett
285
285
1
It may be easier to modify the definition of $g_n$ and use $f(y)cdot varphi_n(y)$ where $varphi_n$ is a continuous function identically $1$ in the ball of radius $n$ (with centre $0$) and vanishes outside the ball of radius $n+1$ instead of just $f(y)$ in the convolution.
– Daniel Fischer♦
Jul 25 at 16:39
It works! Thank you very much!
– Demetrio Masciurett
Jul 25 at 17:21
add a comment |Â
1
It may be easier to modify the definition of $g_n$ and use $f(y)cdot varphi_n(y)$ where $varphi_n$ is a continuous function identically $1$ in the ball of radius $n$ (with centre $0$) and vanishes outside the ball of radius $n+1$ instead of just $f(y)$ in the convolution.
– Daniel Fischer♦
Jul 25 at 16:39
It works! Thank you very much!
– Demetrio Masciurett
Jul 25 at 17:21
1
1
It may be easier to modify the definition of $g_n$ and use $f(y)cdot varphi_n(y)$ where $varphi_n$ is a continuous function identically $1$ in the ball of radius $n$ (with centre $0$) and vanishes outside the ball of radius $n+1$ instead of just $f(y)$ in the convolution.
– Daniel Fischer♦
Jul 25 at 16:39
It may be easier to modify the definition of $g_n$ and use $f(y)cdot varphi_n(y)$ where $varphi_n$ is a continuous function identically $1$ in the ball of radius $n$ (with centre $0$) and vanishes outside the ball of radius $n+1$ instead of just $f(y)$ in the convolution.
– Daniel Fischer♦
Jul 25 at 16:39
It works! Thank you very much!
– Demetrio Masciurett
Jul 25 at 17:21
It works! Thank you very much!
– Demetrio Masciurett
Jul 25 at 17:21
add a comment |Â
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1
It may be easier to modify the definition of $g_n$ and use $f(y)cdot varphi_n(y)$ where $varphi_n$ is a continuous function identically $1$ in the ball of radius $n$ (with centre $0$) and vanishes outside the ball of radius $n+1$ instead of just $f(y)$ in the convolution.
– Daniel Fischer♦
Jul 25 at 16:39
It works! Thank you very much!
– Demetrio Masciurett
Jul 25 at 17:21