Strong convergence of rotation by 1/n operators

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let $ A_n: L^2([-pi,pi]) rightarrow L^2([-pi,pi])$ be $ A_nf(t)=f(t-frac1n) $.



Does $(A_n)$ converge in the strong operator topology?




functional-analysis operator-theory




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  • This is translation, not rotation ;)
    – daw
    Jul 21 at 19:45










  • I am thinking of it as a rotation because I actually meant that for example, $A_nf(-pi + frac12n) = f(pi - frac12n)$
    – MCL
    Jul 21 at 19:52














up vote
1
down vote

favorite












let $ A_n: L^2([-pi,pi]) rightarrow L^2([-pi,pi])$ be $ A_nf(t)=f(t-frac1n) $.



Does $(A_n)$ converge in the strong operator topology?




functional-analysis operator-theory




share|cite|improve this question



















  • This is translation, not rotation ;)
    – daw
    Jul 21 at 19:45










  • I am thinking of it as a rotation because I actually meant that for example, $A_nf(-pi + frac12n) = f(pi - frac12n)$
    – MCL
    Jul 21 at 19:52












up vote
1
down vote

favorite









up vote
1
down vote

favorite











let $ A_n: L^2([-pi,pi]) rightarrow L^2([-pi,pi])$ be $ A_nf(t)=f(t-frac1n) $.



Does $(A_n)$ converge in the strong operator topology?




functional-analysis operator-theory




share|cite|improve this question











let $ A_n: L^2([-pi,pi]) rightarrow L^2([-pi,pi])$ be $ A_nf(t)=f(t-frac1n) $.



Does $(A_n)$ converge in the strong operator topology?





functional-analysis operator-theory





share|cite|improve this question










share|cite|improve this question




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asked Jul 21 at 19:38









MCL

997




997











  • This is translation, not rotation ;)
    – daw
    Jul 21 at 19:45










  • I am thinking of it as a rotation because I actually meant that for example, $A_nf(-pi + frac12n) = f(pi - frac12n)$
    – MCL
    Jul 21 at 19:52
















  • This is translation, not rotation ;)
    – daw
    Jul 21 at 19:45










  • I am thinking of it as a rotation because I actually meant that for example, $A_nf(-pi + frac12n) = f(pi - frac12n)$
    – MCL
    Jul 21 at 19:52















This is translation, not rotation ;)
– daw
Jul 21 at 19:45




This is translation, not rotation ;)
– daw
Jul 21 at 19:45












I am thinking of it as a rotation because I actually meant that for example, $A_nf(-pi + frac12n) = f(pi - frac12n)$
– MCL
Jul 21 at 19:52




I am thinking of it as a rotation because I actually meant that for example, $A_nf(-pi + frac12n) = f(pi - frac12n)$
– MCL
Jul 21 at 19:52










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It does converge in the strong operator topology. You can approximate $f$ in $L^2$ by a $2pi$-periodic, continuous function $g$, and the strong convergence is clear for such a $g$.






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    It does converge in the strong operator topology. You can approximate $f$ in $L^2$ by a $2pi$-periodic, continuous function $g$, and the strong convergence is clear for such a $g$.






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      up vote
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      It does converge in the strong operator topology. You can approximate $f$ in $L^2$ by a $2pi$-periodic, continuous function $g$, and the strong convergence is clear for such a $g$.






      share|cite|improve this answer























        up vote
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        down vote










        up vote
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        down vote









        It does converge in the strong operator topology. You can approximate $f$ in $L^2$ by a $2pi$-periodic, continuous function $g$, and the strong convergence is clear for such a $g$.






        share|cite|improve this answer













        It does converge in the strong operator topology. You can approximate $f$ in $L^2$ by a $2pi$-periodic, continuous function $g$, and the strong convergence is clear for such a $g$.







        share|cite|improve this answer













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        answered Jul 23 at 2:47









        DisintegratingByParts

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