Uniform convergence as $epsilonto 0^+$
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Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.
Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.
What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?
I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.
Thanks a lot in advance.
real-analysis uniform-convergence sequence-of-function
asked Jan 12 at 16:46
eleguitareleguitar
140114
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add a comment |
$begingroup$
Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.
Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.
What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?
I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.
Thanks a lot in advance.
real-analysis uniform-convergence sequence-of-function
asked Jan 12 at 16:46
eleguitareleguitar
140114
$endgroup$
add a comment |
$begingroup$
Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.
Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.
What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?
I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.
Thanks a lot in advance.
real-analysis uniform-convergence sequence-of-function
asked Jan 12 at 16:46
eleguitareleguitar
140114
$endgroup$
Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.
Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.
What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?
I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.
Thanks a lot in advance.
real-analysis uniform-convergence sequence-of-function
real-analysis uniform-convergence sequence-of-function
asked Jan 12 at 16:46
eleguitareleguitar
140114
asked Jan 12 at 16:46
eleguitareleguitar
140114
asked Jan 12 at 16:46
eleguitareleguitar
140114
asked Jan 12 at 16:46
eleguitareleguitar
140114
asked Jan 12 at 16:46
eleguitareleguitar
140114
140114
add a comment |
add a comment |
1 Answer
1
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Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
$$
sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
$$
for all $hin Scap(0,delta)$.
Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
$$
hat u_{1/n} := u_n.
$$
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
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$begingroup$
That's completely clear! Thank you so much!
$endgroup$
– eleguitar
Jan 12 at 18:45
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
$$
sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
$$
for all $hin Scap(0,delta)$.
Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
$$
hat u_{1/n} := u_n.
$$
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
$endgroup$
$begingroup$
That's completely clear! Thank you so much!
$endgroup$
– eleguitar
Jan 12 at 18:45
add a comment |
$begingroup$
Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
$$
sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
$$
for all $hin Scap(0,delta)$.
Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
$$
hat u_{1/n} := u_n.
$$
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
$endgroup$
$begingroup$
That's completely clear! Thank you so much!
$endgroup$
– eleguitar
Jan 12 at 18:45
add a comment |
$begingroup$
Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
$$
sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
$$
for all $hin Scap(0,delta)$.
Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
$$
hat u_{1/n} := u_n.
$$
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
$endgroup$
Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
$$
sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
$$
for all $hin Scap(0,delta)$.
Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
$$
hat u_{1/n} := u_n.
$$
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
edited Jan 12 at 17:06
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
answered Jan 12 at 17:00
BigbearZzzBigbearZzz
9,07421753
9,07421753
$begingroup$
That's completely clear! Thank you so much!
$endgroup$
– eleguitar
Jan 12 at 18:45
add a comment |
$begingroup$
That's completely clear! Thank you so much!
$endgroup$
– eleguitar
Jan 12 at 18:45
$begingroup$
That's completely clear! Thank you so much!
$endgroup$
– eleguitar
Jan 12 at 18:45
$begingroup$
That's completely clear! Thank you so much!
$endgroup$
– eleguitar
Jan 12 at 18:45
add a comment |
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