Group of shift operators not uniformly continuous
Let $X = { u in C(mathbb{R}) : lim_{t rightarrow pm infty} u(t) = 0 }$ with the supremum norm on X.
Given the operators $S(t): X rightarrow X, u mapsto S(t)u := u(bullet -t)$, so S(t) is the rightshift operator.
I want to show that $(S(t))_{t in mathbb{R}}$ is a group that is not uniformly continuous.
Uniform continuity on a group is defined as:
$ Vert S(t) - Id Vert_{L(X)}$ as $t rightarrow 0$, where L(X) is the norm defined as $sup_{u in X} frac{Vert S(t)u Vert_infty}{Vert u Vert_infty}$
Just by the definition I get:
$ Vert S(t) - Id Vert_{L(X)} = sup_{u in X} frac{ sup_{x in mathbb{R}} | u(x-t) - u(x) |}{sup_{x} |u(x)|}$
Now I guess it might be possible to choose some $u_0 in X$ for which one can show that this is larger than some $epsilon > 0$, but to me it seems as if this shouldn't be possible, since $u_0$ is supposed to be continuous. Can someone give some help please?
real-analysis functional-analysis operator-theory
asked Nov 26 at 14:44
eager2learn
1,23811430
add a comment |
Let $X = { u in C(mathbb{R}) : lim_{t rightarrow pm infty} u(t) = 0 }$ with the supremum norm on X.
Given the operators $S(t): X rightarrow X, u mapsto S(t)u := u(bullet -t)$, so S(t) is the rightshift operator.
I want to show that $(S(t))_{t in mathbb{R}}$ is a group that is not uniformly continuous.
Uniform continuity on a group is defined as:
$ Vert S(t) - Id Vert_{L(X)}$ as $t rightarrow 0$, where L(X) is the norm defined as $sup_{u in X} frac{Vert S(t)u Vert_infty}{Vert u Vert_infty}$
Just by the definition I get:
$ Vert S(t) - Id Vert_{L(X)} = sup_{u in X} frac{ sup_{x in mathbb{R}} | u(x-t) - u(x) |}{sup_{x} |u(x)|}$
Now I guess it might be possible to choose some $u_0 in X$ for which one can show that this is larger than some $epsilon > 0$, but to me it seems as if this shouldn't be possible, since $u_0$ is supposed to be continuous. Can someone give some help please?
real-analysis functional-analysis operator-theory
asked Nov 26 at 14:44
eager2learn
1,23811430
add a comment |
Let $X = { u in C(mathbb{R}) : lim_{t rightarrow pm infty} u(t) = 0 }$ with the supremum norm on X.
Given the operators $S(t): X rightarrow X, u mapsto S(t)u := u(bullet -t)$, so S(t) is the rightshift operator.
I want to show that $(S(t))_{t in mathbb{R}}$ is a group that is not uniformly continuous.
Uniform continuity on a group is defined as:
$ Vert S(t) - Id Vert_{L(X)}$ as $t rightarrow 0$, where L(X) is the norm defined as $sup_{u in X} frac{Vert S(t)u Vert_infty}{Vert u Vert_infty}$
Just by the definition I get:
$ Vert S(t) - Id Vert_{L(X)} = sup_{u in X} frac{ sup_{x in mathbb{R}} | u(x-t) - u(x) |}{sup_{x} |u(x)|}$
Now I guess it might be possible to choose some $u_0 in X$ for which one can show that this is larger than some $epsilon > 0$, but to me it seems as if this shouldn't be possible, since $u_0$ is supposed to be continuous. Can someone give some help please?
real-analysis functional-analysis operator-theory
asked Nov 26 at 14:44
eager2learn
1,23811430
Let $X = { u in C(mathbb{R}) : lim_{t rightarrow pm infty} u(t) = 0 }$ with the supremum norm on X.
Given the operators $S(t): X rightarrow X, u mapsto S(t)u := u(bullet -t)$, so S(t) is the rightshift operator.
I want to show that $(S(t))_{t in mathbb{R}}$ is a group that is not uniformly continuous.
Uniform continuity on a group is defined as:
$ Vert S(t) - Id Vert_{L(X)}$ as $t rightarrow 0$, where L(X) is the norm defined as $sup_{u in X} frac{Vert S(t)u Vert_infty}{Vert u Vert_infty}$
Just by the definition I get:
$ Vert S(t) - Id Vert_{L(X)} = sup_{u in X} frac{ sup_{x in mathbb{R}} | u(x-t) - u(x) |}{sup_{x} |u(x)|}$
Now I guess it might be possible to choose some $u_0 in X$ for which one can show that this is larger than some $epsilon > 0$, but to me it seems as if this shouldn't be possible, since $u_0$ is supposed to be continuous. Can someone give some help please?
real-analysis functional-analysis operator-theory
real-analysis functional-analysis operator-theory
asked Nov 26 at 14:44
eager2learn
1,23811430
asked Nov 26 at 14:44
eager2learn
1,23811430
asked Nov 26 at 14:44
eager2learn
1,23811430
asked Nov 26 at 14:44
eager2learn
1,23811430
asked Nov 26 at 14:44
eager2learn
1,23811430
1,23811430
add a comment |
add a comment |
1 Answer
1
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oldest
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Let $u$ be an element of $X$ whose supremum norm is $1$ . Define $u_n(x):= u(nx)$. Then for all $t$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant 1}sup_{xinmathbb R}
lvert u_n(x-t)-u_n(x)rvert=sup_{ngeqslant 1}
sup_{xinmathbb R}lvert u(nx-nt)-u(nx)rvert=sup_{ngeqslant 1}
sup_{x'inmathbb R}lvert u(x'-nt)-u(x')rvert
$$
hence for each integer $N$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant N}
sup_{x inmathbb R}lvert u(x -nt)-u( x)rvert
geqslant sup_{ngeqslant N}
sup_{x lt 0}lvert u(x -nt)-u( x)rvertgeqslant sup_{ngeqslant N}left(
sup_{x lt 0}lvert u( x)rvert -sup_{x lt 0}lvert u( x-nt)rvertright).
$$
which gives, in view of $sup_{ngeqslant N}sup_{x lt 0}lvert u( x-nt)rvertleqslant sup_{sleqslant -Nt}lvert u( s)rvert$ that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert -sup_{sleqslant -Nt}lvert u( s)rvert
$$
for $tgt 0$ and $Ngeqslant 1$. Letting $N$ going to infinity shows that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
$$
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
Thanks for your answer. At first I thought I understood the last step, but actually I don't see how this is true. Can you please explain this step?
– eager2learn
Dec 5 at 16:31
What do you call the last step? The justification of the last inequality?
– Davide Giraudo
Dec 6 at 10:09
Yes, I don't see why that is true.
– eager2learn
Dec 7 at 11:20
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
Let $u$ be an element of $X$ whose supremum norm is $1$ . Define $u_n(x):= u(nx)$. Then for all $t$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant 1}sup_{xinmathbb R}
lvert u_n(x-t)-u_n(x)rvert=sup_{ngeqslant 1}
sup_{xinmathbb R}lvert u(nx-nt)-u(nx)rvert=sup_{ngeqslant 1}
sup_{x'inmathbb R}lvert u(x'-nt)-u(x')rvert
$$
hence for each integer $N$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant N}
sup_{x inmathbb R}lvert u(x -nt)-u( x)rvert
geqslant sup_{ngeqslant N}
sup_{x lt 0}lvert u(x -nt)-u( x)rvertgeqslant sup_{ngeqslant N}left(
sup_{x lt 0}lvert u( x)rvert -sup_{x lt 0}lvert u( x-nt)rvertright).
$$
which gives, in view of $sup_{ngeqslant N}sup_{x lt 0}lvert u( x-nt)rvertleqslant sup_{sleqslant -Nt}lvert u( s)rvert$ that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert -sup_{sleqslant -Nt}lvert u( s)rvert
$$
for $tgt 0$ and $Ngeqslant 1$. Letting $N$ going to infinity shows that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
$$
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
Thanks for your answer. At first I thought I understood the last step, but actually I don't see how this is true. Can you please explain this step?
– eager2learn
Dec 5 at 16:31
What do you call the last step? The justification of the last inequality?
– Davide Giraudo
Dec 6 at 10:09
Yes, I don't see why that is true.
– eager2learn
Dec 7 at 11:20
add a comment |
Let $u$ be an element of $X$ whose supremum norm is $1$ . Define $u_n(x):= u(nx)$. Then for all $t$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant 1}sup_{xinmathbb R}
lvert u_n(x-t)-u_n(x)rvert=sup_{ngeqslant 1}
sup_{xinmathbb R}lvert u(nx-nt)-u(nx)rvert=sup_{ngeqslant 1}
sup_{x'inmathbb R}lvert u(x'-nt)-u(x')rvert
$$
hence for each integer $N$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant N}
sup_{x inmathbb R}lvert u(x -nt)-u( x)rvert
geqslant sup_{ngeqslant N}
sup_{x lt 0}lvert u(x -nt)-u( x)rvertgeqslant sup_{ngeqslant N}left(
sup_{x lt 0}lvert u( x)rvert -sup_{x lt 0}lvert u( x-nt)rvertright).
$$
which gives, in view of $sup_{ngeqslant N}sup_{x lt 0}lvert u( x-nt)rvertleqslant sup_{sleqslant -Nt}lvert u( s)rvert$ that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert -sup_{sleqslant -Nt}lvert u( s)rvert
$$
for $tgt 0$ and $Ngeqslant 1$. Letting $N$ going to infinity shows that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
$$
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
Thanks for your answer. At first I thought I understood the last step, but actually I don't see how this is true. Can you please explain this step?
– eager2learn
Dec 5 at 16:31
What do you call the last step? The justification of the last inequality?
– Davide Giraudo
Dec 6 at 10:09
Yes, I don't see why that is true.
– eager2learn
Dec 7 at 11:20
add a comment |
Let $u$ be an element of $X$ whose supremum norm is $1$ . Define $u_n(x):= u(nx)$. Then for all $t$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant 1}sup_{xinmathbb R}
lvert u_n(x-t)-u_n(x)rvert=sup_{ngeqslant 1}
sup_{xinmathbb R}lvert u(nx-nt)-u(nx)rvert=sup_{ngeqslant 1}
sup_{x'inmathbb R}lvert u(x'-nt)-u(x')rvert
$$
hence for each integer $N$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant N}
sup_{x inmathbb R}lvert u(x -nt)-u( x)rvert
geqslant sup_{ngeqslant N}
sup_{x lt 0}lvert u(x -nt)-u( x)rvertgeqslant sup_{ngeqslant N}left(
sup_{x lt 0}lvert u( x)rvert -sup_{x lt 0}lvert u( x-nt)rvertright).
$$
which gives, in view of $sup_{ngeqslant N}sup_{x lt 0}lvert u( x-nt)rvertleqslant sup_{sleqslant -Nt}lvert u( s)rvert$ that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert -sup_{sleqslant -Nt}lvert u( s)rvert
$$
for $tgt 0$ and $Ngeqslant 1$. Letting $N$ going to infinity shows that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
$$
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
Let $u$ be an element of $X$ whose supremum norm is $1$ . Define $u_n(x):= u(nx)$. Then for all $t$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant 1}sup_{xinmathbb R}
lvert u_n(x-t)-u_n(x)rvert=sup_{ngeqslant 1}
sup_{xinmathbb R}lvert u(nx-nt)-u(nx)rvert=sup_{ngeqslant 1}
sup_{x'inmathbb R}lvert u(x'-nt)-u(x')rvert
$$
hence for each integer $N$,
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslant sup_{ngeqslant N}
sup_{x inmathbb R}lvert u(x -nt)-u( x)rvert
geqslant sup_{ngeqslant N}
sup_{x lt 0}lvert u(x -nt)-u( x)rvertgeqslant sup_{ngeqslant N}left(
sup_{x lt 0}lvert u( x)rvert -sup_{x lt 0}lvert u( x-nt)rvertright).
$$
which gives, in view of $sup_{ngeqslant N}sup_{x lt 0}lvert u( x-nt)rvertleqslant sup_{sleqslant -Nt}lvert u( s)rvert$ that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert -sup_{sleqslant -Nt}lvert u( s)rvert
$$
for $tgt 0$ and $Ngeqslant 1$. Letting $N$ going to infinity shows that
$$
leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
$$
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
edited Dec 6 at 10:15
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
answered Nov 26 at 15:06
Davide Giraudo
125k16150259
125k16150259
Thanks for your answer. At first I thought I understood the last step, but actually I don't see how this is true. Can you please explain this step?
– eager2learn
Dec 5 at 16:31
What do you call the last step? The justification of the last inequality?
– Davide Giraudo
Dec 6 at 10:09
Yes, I don't see why that is true.
– eager2learn
Dec 7 at 11:20
add a comment |
Thanks for your answer. At first I thought I understood the last step, but actually I don't see how this is true. Can you please explain this step?
– eager2learn
Dec 5 at 16:31
What do you call the last step? The justification of the last inequality?
– Davide Giraudo
Dec 6 at 10:09
Yes, I don't see why that is true.
– eager2learn
Dec 7 at 11:20
Thanks for your answer. At first I thought I understood the last step, but actually I don't see how this is true. Can you please explain this step?
– eager2learn
Dec 5 at 16:31
Thanks for your answer. At first I thought I understood the last step, but actually I don't see how this is true. Can you please explain this step?
– eager2learn
Dec 5 at 16:31
What do you call the last step? The justification of the last inequality?
– Davide Giraudo
Dec 6 at 10:09
What do you call the last step? The justification of the last inequality?
– Davide Giraudo
Dec 6 at 10:09
Yes, I don't see why that is true.
– eager2learn
Dec 7 at 11:20
Yes, I don't see why that is true.
– eager2learn
Dec 7 at 11:20
add a comment |
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