Applications of the lack of compactness of the closed unit ball in infinite-dimensional Banach spaces











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I am writing a paper on the compactness of closed balls in Banach spaces, with particular attention paid to the following theorem




Let $V$ be a Banach space over $mathbb R$ or $mathbb C$. The closed unit ball in $V$ is compact if and only if $V$ is finite-dimensional.




I am looking for some consequences or applications of this theorem (probably mostly related to the part which asserts that: if $V$ is infinite-dimensional, then the closed unit ball is not compact). I do prove the immediate corollary of this theorem, which is basically replacing "the closed unit ball" with "the closed ball of radius $r>0$ around $x_0in V$" in the statement of the theorem. I have also been looking at the notion of weak convergence, and how this can allow for compactness (in the weak sense) in infinite-dimensional spaces. Other than those two, I am looking for some other applications of this theorem. In particular, are there any specific interesting examples one can look at that follow from this theorem?



Any feedback is appreciated.










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    up vote
    3
    down vote

    favorite












    I am writing a paper on the compactness of closed balls in Banach spaces, with particular attention paid to the following theorem




    Let $V$ be a Banach space over $mathbb R$ or $mathbb C$. The closed unit ball in $V$ is compact if and only if $V$ is finite-dimensional.




    I am looking for some consequences or applications of this theorem (probably mostly related to the part which asserts that: if $V$ is infinite-dimensional, then the closed unit ball is not compact). I do prove the immediate corollary of this theorem, which is basically replacing "the closed unit ball" with "the closed ball of radius $r>0$ around $x_0in V$" in the statement of the theorem. I have also been looking at the notion of weak convergence, and how this can allow for compactness (in the weak sense) in infinite-dimensional spaces. Other than those two, I am looking for some other applications of this theorem. In particular, are there any specific interesting examples one can look at that follow from this theorem?



    Any feedback is appreciated.










    share|cite|improve this question


























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I am writing a paper on the compactness of closed balls in Banach spaces, with particular attention paid to the following theorem




      Let $V$ be a Banach space over $mathbb R$ or $mathbb C$. The closed unit ball in $V$ is compact if and only if $V$ is finite-dimensional.




      I am looking for some consequences or applications of this theorem (probably mostly related to the part which asserts that: if $V$ is infinite-dimensional, then the closed unit ball is not compact). I do prove the immediate corollary of this theorem, which is basically replacing "the closed unit ball" with "the closed ball of radius $r>0$ around $x_0in V$" in the statement of the theorem. I have also been looking at the notion of weak convergence, and how this can allow for compactness (in the weak sense) in infinite-dimensional spaces. Other than those two, I am looking for some other applications of this theorem. In particular, are there any specific interesting examples one can look at that follow from this theorem?



      Any feedback is appreciated.










      share|cite|improve this question















      I am writing a paper on the compactness of closed balls in Banach spaces, with particular attention paid to the following theorem




      Let $V$ be a Banach space over $mathbb R$ or $mathbb C$. The closed unit ball in $V$ is compact if and only if $V$ is finite-dimensional.




      I am looking for some consequences or applications of this theorem (probably mostly related to the part which asserts that: if $V$ is infinite-dimensional, then the closed unit ball is not compact). I do prove the immediate corollary of this theorem, which is basically replacing "the closed unit ball" with "the closed ball of radius $r>0$ around $x_0in V$" in the statement of the theorem. I have also been looking at the notion of weak convergence, and how this can allow for compactness (in the weak sense) in infinite-dimensional spaces. Other than those two, I am looking for some other applications of this theorem. In particular, are there any specific interesting examples one can look at that follow from this theorem?



      Any feedback is appreciated.







      functional-analysis analysis vector-spaces compactness examples-counterexamples






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      edited Nov 17 at 21:39

























      asked Nov 17 at 21:28









      Dave

      8,41811033




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          One of the simplest consequences is that continuous function doesn't attain their minimum on the ball, fact that is always true in finite dimensional spaces for Weierstrass theorem. Consider for example the integral functional $int_0^1|cdot|:(C([0,1]),VertcdotVert_{infty})tomathbb{R}$, whose infimum over the ball is zero but zero is never attained on the ball.



          Another interesting thing is that is easy to find an infinity of elements that lie on the ball which are $epsilon$-separated. For an example consider the characteristics functions ${f_n=chi_{[n,n+1]}(x)}_{ninmathbb{N}}subset L^infty(mathbb{R})$. One can easily check that they belong to the unit ball of $L^infty$ and that are 2-separated, i.e. $Vert f_n-f_mVert_infty=2$ for each $nne m$.






          share|cite|improve this answer























          • Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $Vert e_n-e_mVert=2$ for any orthonormal sequence $(e_n)_{ngeq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no?
            – Dave
            Nov 17 at 23:27










          • You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now
            – Marco
            Nov 17 at 23:32






          • 1




            but wouldn't the minimum still be attained with the zero function in $C[0,1]$?
            – Dave
            Nov 17 at 23:37










          • Yes, on the entire space, but the zero function doesn’t belong to the unit ball.
            – Marco
            Nov 18 at 8:27










          • The standard is to define the unit ball as ${xin X: |x|le 1}$ which surely contains $0$. Perhaps @Marco means the unit sphere ${xin X: |x|=1}$.
            – Jochen
            Nov 18 at 13:15











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote













          One of the simplest consequences is that continuous function doesn't attain their minimum on the ball, fact that is always true in finite dimensional spaces for Weierstrass theorem. Consider for example the integral functional $int_0^1|cdot|:(C([0,1]),VertcdotVert_{infty})tomathbb{R}$, whose infimum over the ball is zero but zero is never attained on the ball.



          Another interesting thing is that is easy to find an infinity of elements that lie on the ball which are $epsilon$-separated. For an example consider the characteristics functions ${f_n=chi_{[n,n+1]}(x)}_{ninmathbb{N}}subset L^infty(mathbb{R})$. One can easily check that they belong to the unit ball of $L^infty$ and that are 2-separated, i.e. $Vert f_n-f_mVert_infty=2$ for each $nne m$.






          share|cite|improve this answer























          • Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $Vert e_n-e_mVert=2$ for any orthonormal sequence $(e_n)_{ngeq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no?
            – Dave
            Nov 17 at 23:27










          • You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now
            – Marco
            Nov 17 at 23:32






          • 1




            but wouldn't the minimum still be attained with the zero function in $C[0,1]$?
            – Dave
            Nov 17 at 23:37










          • Yes, on the entire space, but the zero function doesn’t belong to the unit ball.
            – Marco
            Nov 18 at 8:27










          • The standard is to define the unit ball as ${xin X: |x|le 1}$ which surely contains $0$. Perhaps @Marco means the unit sphere ${xin X: |x|=1}$.
            – Jochen
            Nov 18 at 13:15















          up vote
          1
          down vote













          One of the simplest consequences is that continuous function doesn't attain their minimum on the ball, fact that is always true in finite dimensional spaces for Weierstrass theorem. Consider for example the integral functional $int_0^1|cdot|:(C([0,1]),VertcdotVert_{infty})tomathbb{R}$, whose infimum over the ball is zero but zero is never attained on the ball.



          Another interesting thing is that is easy to find an infinity of elements that lie on the ball which are $epsilon$-separated. For an example consider the characteristics functions ${f_n=chi_{[n,n+1]}(x)}_{ninmathbb{N}}subset L^infty(mathbb{R})$. One can easily check that they belong to the unit ball of $L^infty$ and that are 2-separated, i.e. $Vert f_n-f_mVert_infty=2$ for each $nne m$.






          share|cite|improve this answer























          • Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $Vert e_n-e_mVert=2$ for any orthonormal sequence $(e_n)_{ngeq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no?
            – Dave
            Nov 17 at 23:27










          • You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now
            – Marco
            Nov 17 at 23:32






          • 1




            but wouldn't the minimum still be attained with the zero function in $C[0,1]$?
            – Dave
            Nov 17 at 23:37










          • Yes, on the entire space, but the zero function doesn’t belong to the unit ball.
            – Marco
            Nov 18 at 8:27










          • The standard is to define the unit ball as ${xin X: |x|le 1}$ which surely contains $0$. Perhaps @Marco means the unit sphere ${xin X: |x|=1}$.
            – Jochen
            Nov 18 at 13:15













          up vote
          1
          down vote










          up vote
          1
          down vote









          One of the simplest consequences is that continuous function doesn't attain their minimum on the ball, fact that is always true in finite dimensional spaces for Weierstrass theorem. Consider for example the integral functional $int_0^1|cdot|:(C([0,1]),VertcdotVert_{infty})tomathbb{R}$, whose infimum over the ball is zero but zero is never attained on the ball.



          Another interesting thing is that is easy to find an infinity of elements that lie on the ball which are $epsilon$-separated. For an example consider the characteristics functions ${f_n=chi_{[n,n+1]}(x)}_{ninmathbb{N}}subset L^infty(mathbb{R})$. One can easily check that they belong to the unit ball of $L^infty$ and that are 2-separated, i.e. $Vert f_n-f_mVert_infty=2$ for each $nne m$.






          share|cite|improve this answer














          One of the simplest consequences is that continuous function doesn't attain their minimum on the ball, fact that is always true in finite dimensional spaces for Weierstrass theorem. Consider for example the integral functional $int_0^1|cdot|:(C([0,1]),VertcdotVert_{infty})tomathbb{R}$, whose infimum over the ball is zero but zero is never attained on the ball.



          Another interesting thing is that is easy to find an infinity of elements that lie on the ball which are $epsilon$-separated. For an example consider the characteristics functions ${f_n=chi_{[n,n+1]}(x)}_{ninmathbb{N}}subset L^infty(mathbb{R})$. One can easily check that they belong to the unit ball of $L^infty$ and that are 2-separated, i.e. $Vert f_n-f_mVert_infty=2$ for each $nne m$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 17 at 23:33

























          answered Nov 17 at 22:01









          Marco

          1909




          1909












          • Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $Vert e_n-e_mVert=2$ for any orthonormal sequence $(e_n)_{ngeq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no?
            – Dave
            Nov 17 at 23:27










          • You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now
            – Marco
            Nov 17 at 23:32






          • 1




            but wouldn't the minimum still be attained with the zero function in $C[0,1]$?
            – Dave
            Nov 17 at 23:37










          • Yes, on the entire space, but the zero function doesn’t belong to the unit ball.
            – Marco
            Nov 18 at 8:27










          • The standard is to define the unit ball as ${xin X: |x|le 1}$ which surely contains $0$. Perhaps @Marco means the unit sphere ${xin X: |x|=1}$.
            – Jochen
            Nov 18 at 13:15


















          • Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $Vert e_n-e_mVert=2$ for any orthonormal sequence $(e_n)_{ngeq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no?
            – Dave
            Nov 17 at 23:27










          • You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now
            – Marco
            Nov 17 at 23:32






          • 1




            but wouldn't the minimum still be attained with the zero function in $C[0,1]$?
            – Dave
            Nov 17 at 23:37










          • Yes, on the entire space, but the zero function doesn’t belong to the unit ball.
            – Marco
            Nov 18 at 8:27










          • The standard is to define the unit ball as ${xin X: |x|le 1}$ which surely contains $0$. Perhaps @Marco means the unit sphere ${xin X: |x|=1}$.
            – Jochen
            Nov 18 at 13:15
















          Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $Vert e_n-e_mVert=2$ for any orthonormal sequence $(e_n)_{ngeq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no?
          – Dave
          Nov 17 at 23:27




          Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $Vert e_n-e_mVert=2$ for any orthonormal sequence $(e_n)_{ngeq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no?
          – Dave
          Nov 17 at 23:27












          You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now
          – Marco
          Nov 17 at 23:32




          You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now
          – Marco
          Nov 17 at 23:32




          1




          1




          but wouldn't the minimum still be attained with the zero function in $C[0,1]$?
          – Dave
          Nov 17 at 23:37




          but wouldn't the minimum still be attained with the zero function in $C[0,1]$?
          – Dave
          Nov 17 at 23:37












          Yes, on the entire space, but the zero function doesn’t belong to the unit ball.
          – Marco
          Nov 18 at 8:27




          Yes, on the entire space, but the zero function doesn’t belong to the unit ball.
          – Marco
          Nov 18 at 8:27












          The standard is to define the unit ball as ${xin X: |x|le 1}$ which surely contains $0$. Perhaps @Marco means the unit sphere ${xin X: |x|=1}$.
          – Jochen
          Nov 18 at 13:15




          The standard is to define the unit ball as ${xin X: |x|le 1}$ which surely contains $0$. Perhaps @Marco means the unit sphere ${xin X: |x|=1}$.
          – Jochen
          Nov 18 at 13:15


















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