Compactness of set of indicator functions












6












$begingroup$


Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
$$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?




Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.






fa.functional-analysis real-analysis measure-theory convex-analysis compactness







share|cite|improve this question









$endgroup$

















    6












    $begingroup$


    Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
    $$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
    Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?




    Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.






    fa.functional-analysis real-analysis measure-theory convex-analysis compactness







    share|cite|improve this question









    $endgroup$















      6












      6








      6


      2



      $begingroup$


      Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
      $$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
      Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?




      Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.






      fa.functional-analysis real-analysis measure-theory convex-analysis compactness







      share|cite|improve this question









      $endgroup$




      Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
      $$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
      Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?




      Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.






      fa.functional-analysis real-analysis measure-theory convex-analysis compactness




      fa.functional-analysis real-analysis measure-theory convex-analysis compactness






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked yesterday









      Sara WinsletSara Winslet

      1438




      1438






















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          $begingroup$

          This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.



          On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
            $endgroup$
            – Willie Wong
            yesterday






          • 1




            $begingroup$
            @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
            $endgroup$
            – Mateusz Wasilewski
            yesterday












          • $begingroup$
            After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
            $endgroup$
            – Willie Wong
            7 hours ago











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          $begingroup$

          This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.



          On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
            $endgroup$
            – Willie Wong
            yesterday






          • 1




            $begingroup$
            @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
            $endgroup$
            – Mateusz Wasilewski
            yesterday












          • $begingroup$
            After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
            $endgroup$
            – Willie Wong
            7 hours ago
















          10












          $begingroup$

          This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.



          On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
            $endgroup$
            – Willie Wong
            yesterday






          • 1




            $begingroup$
            @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
            $endgroup$
            – Mateusz Wasilewski
            yesterday












          • $begingroup$
            After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
            $endgroup$
            – Willie Wong
            7 hours ago














          10












          10








          10





          $begingroup$

          This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.



          On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.






          share|cite|improve this answer









          $endgroup$



          This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.



          On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Mateusz WasilewskiMateusz Wasilewski

          3,1551023




          3,1551023












          • $begingroup$
            In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
            $endgroup$
            – Willie Wong
            yesterday






          • 1




            $begingroup$
            @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
            $endgroup$
            – Mateusz Wasilewski
            yesterday












          • $begingroup$
            After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
            $endgroup$
            – Willie Wong
            7 hours ago


















          • $begingroup$
            In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
            $endgroup$
            – Willie Wong
            yesterday






          • 1




            $begingroup$
            @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
            $endgroup$
            – Mateusz Wasilewski
            yesterday












          • $begingroup$
            After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
            $endgroup$
            – Willie Wong
            7 hours ago
















          $begingroup$
          In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
          $endgroup$
          – Willie Wong
          yesterday




          $begingroup$
          In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
          $endgroup$
          – Willie Wong
          yesterday




          1




          1




          $begingroup$
          @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
          $endgroup$
          – Mateusz Wasilewski
          yesterday






          $begingroup$
          @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
          $endgroup$
          – Mateusz Wasilewski
          yesterday














          $begingroup$
          After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
          $endgroup$
          – Willie Wong
          7 hours ago




          $begingroup$
          After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
          $endgroup$
          – Willie Wong
          7 hours ago


















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