Finding the Spectrum of an element of $ell^infty$

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I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
enter image description here



How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.



Am I missing something? Kindly Help!!!



Thanks & Regards in advance






c-star-algebras algebras







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$endgroup$

















    1












    $begingroup$


    I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
    enter image description here



    How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.



    Am I missing something? Kindly Help!!!



    Thanks & Regards in advance






    c-star-algebras algebras







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
      enter image description here



      How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.



      Am I missing something? Kindly Help!!!



      Thanks & Regards in advance






      c-star-algebras algebras







      share|cite|improve this question









      $endgroup$




      I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
      enter image description here



      How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.



      Am I missing something? Kindly Help!!!



      Thanks & Regards in advance






      c-star-algebras algebras




      c-star-algebras algebras






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 17 at 17:46









      Devendra Singh RanaDevendra Singh Rana

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      7641416






















          1 Answer
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          $begingroup$

          The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.



          As for the proof of the claim, here is a guideline for which you can fill in the details:



          Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.



          For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            If you want the details on how to prove the claim in this example, let me know and I will edit.
            $endgroup$
            – Aweygan
            Jan 17 at 17:58










          • $begingroup$
            A little hint of the proof will be helpful @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:15






          • 1




            $begingroup$
            @DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
            $endgroup$
            – Aweygan
            Jan 17 at 18:22










          • $begingroup$
            This is actually enough, thanks @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:34








          • 1




            $begingroup$
            You're welcome, glad to help!
            $endgroup$
            – Aweygan
            Jan 17 at 18:34











          Your Answer





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          active

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          $begingroup$

          The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.



          As for the proof of the claim, here is a guideline for which you can fill in the details:



          Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.



          For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            If you want the details on how to prove the claim in this example, let me know and I will edit.
            $endgroup$
            – Aweygan
            Jan 17 at 17:58










          • $begingroup$
            A little hint of the proof will be helpful @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:15






          • 1




            $begingroup$
            @DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
            $endgroup$
            – Aweygan
            Jan 17 at 18:22










          • $begingroup$
            This is actually enough, thanks @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:34








          • 1




            $begingroup$
            You're welcome, glad to help!
            $endgroup$
            – Aweygan
            Jan 17 at 18:34
















          2












          $begingroup$

          The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.



          As for the proof of the claim, here is a guideline for which you can fill in the details:



          Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.



          For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            If you want the details on how to prove the claim in this example, let me know and I will edit.
            $endgroup$
            – Aweygan
            Jan 17 at 17:58










          • $begingroup$
            A little hint of the proof will be helpful @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:15






          • 1




            $begingroup$
            @DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
            $endgroup$
            – Aweygan
            Jan 17 at 18:22










          • $begingroup$
            This is actually enough, thanks @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:34








          • 1




            $begingroup$
            You're welcome, glad to help!
            $endgroup$
            – Aweygan
            Jan 17 at 18:34














          2












          2








          2





          $begingroup$

          The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.



          As for the proof of the claim, here is a guideline for which you can fill in the details:



          Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.



          For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.






          share|cite|improve this answer











          $endgroup$



          The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.



          As for the proof of the claim, here is a guideline for which you can fill in the details:



          Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.



          For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 17 at 18:22

























          answered Jan 17 at 17:57









          AweyganAweygan

          14.2k21441




          14.2k21441








          • 1




            $begingroup$
            If you want the details on how to prove the claim in this example, let me know and I will edit.
            $endgroup$
            – Aweygan
            Jan 17 at 17:58










          • $begingroup$
            A little hint of the proof will be helpful @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:15






          • 1




            $begingroup$
            @DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
            $endgroup$
            – Aweygan
            Jan 17 at 18:22










          • $begingroup$
            This is actually enough, thanks @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:34








          • 1




            $begingroup$
            You're welcome, glad to help!
            $endgroup$
            – Aweygan
            Jan 17 at 18:34














          • 1




            $begingroup$
            If you want the details on how to prove the claim in this example, let me know and I will edit.
            $endgroup$
            – Aweygan
            Jan 17 at 17:58










          • $begingroup$
            A little hint of the proof will be helpful @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:15






          • 1




            $begingroup$
            @DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
            $endgroup$
            – Aweygan
            Jan 17 at 18:22










          • $begingroup$
            This is actually enough, thanks @Aweygan
            $endgroup$
            – Devendra Singh Rana
            Jan 17 at 18:34








          • 1




            $begingroup$
            You're welcome, glad to help!
            $endgroup$
            – Aweygan
            Jan 17 at 18:34








          1




          1




          $begingroup$
          If you want the details on how to prove the claim in this example, let me know and I will edit.
          $endgroup$
          – Aweygan
          Jan 17 at 17:58




          $begingroup$
          If you want the details on how to prove the claim in this example, let me know and I will edit.
          $endgroup$
          – Aweygan
          Jan 17 at 17:58












          $begingroup$
          A little hint of the proof will be helpful @Aweygan
          $endgroup$
          – Devendra Singh Rana
          Jan 17 at 18:15




          $begingroup$
          A little hint of the proof will be helpful @Aweygan
          $endgroup$
          – Devendra Singh Rana
          Jan 17 at 18:15




          1




          1




          $begingroup$
          @DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
          $endgroup$
          – Aweygan
          Jan 17 at 18:22




          $begingroup$
          @DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
          $endgroup$
          – Aweygan
          Jan 17 at 18:22












          $begingroup$
          This is actually enough, thanks @Aweygan
          $endgroup$
          – Devendra Singh Rana
          Jan 17 at 18:34






          $begingroup$
          This is actually enough, thanks @Aweygan
          $endgroup$
          – Devendra Singh Rana
          Jan 17 at 18:34






          1




          1




          $begingroup$
          You're welcome, glad to help!
          $endgroup$
          – Aweygan
          Jan 17 at 18:34




          $begingroup$
          You're welcome, glad to help!
          $endgroup$
          – Aweygan
          Jan 17 at 18:34


















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