Compact subset of space of Continuous functions












5














Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?



I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.






real-analysis general-topology functional-analysis







share|cite|improve this question






















  • Have you heard of Arzela-Ascoli theorem?
    – Thomas Shelby
    2 days ago












  • Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
    – Giuseppe Negro
    2 days ago










  • @ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
    – vidyarthi
    2 days ago








  • 1




    @ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
    – daw
    2 days ago






  • 2




    @ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
    – MaoWao
    2 days ago
















5














Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?



I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.






real-analysis general-topology functional-analysis







share|cite|improve this question






















  • Have you heard of Arzela-Ascoli theorem?
    – Thomas Shelby
    2 days ago












  • Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
    – Giuseppe Negro
    2 days ago










  • @ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
    – vidyarthi
    2 days ago








  • 1




    @ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
    – daw
    2 days ago






  • 2




    @ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
    – MaoWao
    2 days ago














5












5








5


2





Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?



I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.






real-analysis general-topology functional-analysis







share|cite|improve this question













Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?



I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.






real-analysis general-topology functional-analysis




real-analysis general-topology functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









vidyarthividyarthi

2,9341832




2,9341832












  • Have you heard of Arzela-Ascoli theorem?
    – Thomas Shelby
    2 days ago












  • Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
    – Giuseppe Negro
    2 days ago










  • @ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
    – vidyarthi
    2 days ago








  • 1




    @ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
    – daw
    2 days ago






  • 2




    @ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
    – MaoWao
    2 days ago


















  • Have you heard of Arzela-Ascoli theorem?
    – Thomas Shelby
    2 days ago












  • Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
    – Giuseppe Negro
    2 days ago










  • @ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
    – vidyarthi
    2 days ago








  • 1




    @ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
    – daw
    2 days ago






  • 2




    @ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
    – MaoWao
    2 days ago
















Have you heard of Arzela-Ascoli theorem?
– Thomas Shelby
2 days ago






Have you heard of Arzela-Ascoli theorem?
– Thomas Shelby
2 days ago














Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
– Giuseppe Negro
2 days ago




Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
– Giuseppe Negro
2 days ago












@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
– vidyarthi
2 days ago






@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
– vidyarthi
2 days ago






1




1




@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
– daw
2 days ago




@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
– daw
2 days ago




2




2




@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
– MaoWao
2 days ago




@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
– MaoWao
2 days ago










2 Answers
2






active

oldest

votes


















2














Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$



Hence $K$ isn't bounded so it cannot be compact.





An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.

If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
$$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
which is a contradiction.






share|cite|improve this answer





























    6














    Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?






    share|cite|improve this answer





















    • The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
      – vidyarthi
      2 days ago








    • 2




      @vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
      – Henno Brandsma
      2 days ago











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    2 Answers
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    2 Answers
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    active

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    2














    Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
    We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$



    Hence $K$ isn't bounded so it cannot be compact.





    An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.

    If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
    $$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
    which is a contradiction.






    share|cite|improve this answer


























      2














      Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
      We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$



      Hence $K$ isn't bounded so it cannot be compact.





      An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.

      If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
      $$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
      which is a contradiction.






      share|cite|improve this answer
























        2












        2








        2






        Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
        We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$



        Hence $K$ isn't bounded so it cannot be compact.





        An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.

        If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
        $$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
        which is a contradiction.






        share|cite|improve this answer












        Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
        We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$



        Hence $K$ isn't bounded so it cannot be compact.





        An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.

        If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
        $$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
        which is a contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        mechanodroidmechanodroid

        27k62446




        27k62446























            6














            Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?






            share|cite|improve this answer





















            • The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
              – vidyarthi
              2 days ago








            • 2




              @vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
              – Henno Brandsma
              2 days ago
















            6














            Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?






            share|cite|improve this answer





















            • The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
              – vidyarthi
              2 days ago








            • 2




              @vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
              – Henno Brandsma
              2 days ago














            6












            6








            6






            Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?






            share|cite|improve this answer












            Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            FredFred

            44.4k1845




            44.4k1845












            • The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
              – vidyarthi
              2 days ago








            • 2




              @vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
              – Henno Brandsma
              2 days ago


















            • The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
              – vidyarthi
              2 days ago








            • 2




              @vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
              – Henno Brandsma
              2 days ago
















            The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
            – vidyarthi
            2 days ago






            The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
            – vidyarthi
            2 days ago






            2




            2




            @vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
            – Henno Brandsma
            2 days ago




            @vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
            – Henno Brandsma
            2 days ago


















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