Group of shift operators not uniformly continuous












2














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







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    2














    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







    share|cite|improve this question

























      2












      2








      2







      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







      share|cite|improve this question













      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






      share|cite|improve this question













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      asked Nov 26 at 14:44









      eager2learn

      1,23811430




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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
          $$
          leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
          $$






          share|cite|improve this answer























          • 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











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          1 Answer
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          1














          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
          $$
          leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
          $$






          share|cite|improve this answer























          • 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
















          1














          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
          $$
          leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
          $$






          share|cite|improve this answer























          • 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














          1












          1








          1






          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
          $$
          leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
          $$






          share|cite|improve this answer














          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
          $$
          leftlVert S(t)-operatorname{Id}rightrVert_Xgeqslantsup_{x lt 0}lvert u( x)rvert.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 at 10:15

























          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


















          • 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


















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