Uniform convergence for operator of translation












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$begingroup$


For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.






functional-analysis convergence







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    1












    $begingroup$


    For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.






    functional-analysis convergence







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.






      functional-analysis convergence







      share|cite|improve this question









      $endgroup$




      For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.






      functional-analysis convergence




      functional-analysis convergence






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 12 at 10:53









      HanaHana

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          Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
            $endgroup$
            – Hana
            Jan 15 at 7:25










          • $begingroup$
            If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 7:28












          • $begingroup$
            Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
            $endgroup$
            – Hana
            Jan 15 at 16:03










          • $begingroup$
            @Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 23:21










          • $begingroup$
            Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
            $endgroup$
            – Hana
            Jan 17 at 19:25











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          $begingroup$

          Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
            $endgroup$
            – Hana
            Jan 15 at 7:25










          • $begingroup$
            If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 7:28












          • $begingroup$
            Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
            $endgroup$
            – Hana
            Jan 15 at 16:03










          • $begingroup$
            @Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 23:21










          • $begingroup$
            Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
            $endgroup$
            – Hana
            Jan 17 at 19:25
















          0












          $begingroup$

          Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
            $endgroup$
            – Hana
            Jan 15 at 7:25










          • $begingroup$
            If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 7:28












          • $begingroup$
            Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
            $endgroup$
            – Hana
            Jan 15 at 16:03










          • $begingroup$
            @Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 23:21










          • $begingroup$
            Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
            $endgroup$
            – Hana
            Jan 17 at 19:25














          0












          0








          0





          $begingroup$

          Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.






          share|cite|improve this answer











          $endgroup$



          Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 15 at 23:20

























          answered Jan 12 at 13:10









          Kavi Rama MurthyKavi Rama Murthy

          56k42158




          56k42158












          • $begingroup$
            Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
            $endgroup$
            – Hana
            Jan 15 at 7:25










          • $begingroup$
            If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 7:28












          • $begingroup$
            Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
            $endgroup$
            – Hana
            Jan 15 at 16:03










          • $begingroup$
            @Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 23:21










          • $begingroup$
            Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
            $endgroup$
            – Hana
            Jan 17 at 19:25


















          • $begingroup$
            Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
            $endgroup$
            – Hana
            Jan 15 at 7:25










          • $begingroup$
            If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 7:28












          • $begingroup$
            Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
            $endgroup$
            – Hana
            Jan 15 at 16:03










          • $begingroup$
            @Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 23:21










          • $begingroup$
            Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
            $endgroup$
            – Hana
            Jan 17 at 19:25
















          $begingroup$
          Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
          $endgroup$
          – Hana
          Jan 15 at 7:25




          $begingroup$
          Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
          $endgroup$
          – Hana
          Jan 15 at 7:25












          $begingroup$
          If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
          $endgroup$
          – Kavi Rama Murthy
          Jan 15 at 7:28






          $begingroup$
          If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
          $endgroup$
          – Kavi Rama Murthy
          Jan 15 at 7:28














          $begingroup$
          Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
          $endgroup$
          – Hana
          Jan 15 at 16:03




          $begingroup$
          Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
          $endgroup$
          – Hana
          Jan 15 at 16:03












          $begingroup$
          @Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
          $endgroup$
          – Kavi Rama Murthy
          Jan 15 at 23:21




          $begingroup$
          @Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
          $endgroup$
          – Kavi Rama Murthy
          Jan 15 at 23:21












          $begingroup$
          Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
          $endgroup$
          – Hana
          Jan 17 at 19:25




          $begingroup$
          Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
          $endgroup$
          – Hana
          Jan 17 at 19:25


















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