Banach spaces, $C^1([-1,1])$












1














Denote by $C^1([a,b])$ the vector space of continuously differentiable functions with derivatives at $a,b$ defined as one-sided limits. I was wondering whether the linear subspace $C^1([-1,1])$ of $C^0([-1,1])$ closed with respect to $||cdot||_{C^0}$?



I thought it was not, and I wanted to prove that $|x|$ is a limit point of $C^1([-1,1])$ implying $C^1([-1,1])$ is not closed. However, I have troubles finding a sequence of differentiable functions which converges to $|x|$. Any suggestions would be appreciated!






sequences-and-series functional-analysis banach-spaces







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    1














    Denote by $C^1([a,b])$ the vector space of continuously differentiable functions with derivatives at $a,b$ defined as one-sided limits. I was wondering whether the linear subspace $C^1([-1,1])$ of $C^0([-1,1])$ closed with respect to $||cdot||_{C^0}$?



    I thought it was not, and I wanted to prove that $|x|$ is a limit point of $C^1([-1,1])$ implying $C^1([-1,1])$ is not closed. However, I have troubles finding a sequence of differentiable functions which converges to $|x|$. Any suggestions would be appreciated!






    sequences-and-series functional-analysis banach-spaces







    share|cite|improve this question

























      1












      1








      1


      1





      Denote by $C^1([a,b])$ the vector space of continuously differentiable functions with derivatives at $a,b$ defined as one-sided limits. I was wondering whether the linear subspace $C^1([-1,1])$ of $C^0([-1,1])$ closed with respect to $||cdot||_{C^0}$?



      I thought it was not, and I wanted to prove that $|x|$ is a limit point of $C^1([-1,1])$ implying $C^1([-1,1])$ is not closed. However, I have troubles finding a sequence of differentiable functions which converges to $|x|$. Any suggestions would be appreciated!






      sequences-and-series functional-analysis banach-spaces







      share|cite|improve this question













      Denote by $C^1([a,b])$ the vector space of continuously differentiable functions with derivatives at $a,b$ defined as one-sided limits. I was wondering whether the linear subspace $C^1([-1,1])$ of $C^0([-1,1])$ closed with respect to $||cdot||_{C^0}$?



      I thought it was not, and I wanted to prove that $|x|$ is a limit point of $C^1([-1,1])$ implying $C^1([-1,1])$ is not closed. However, I have troubles finding a sequence of differentiable functions which converges to $|x|$. Any suggestions would be appreciated!






      sequences-and-series functional-analysis banach-spaces




      sequences-and-series functional-analysis banach-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Nov 26 '18 at 19:48









      James

      745218




      745218






















          2 Answers
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          Hint: If $xin[-1,1]$, what is $displaystylelim_{ntoinfty}sqrt{x^2+frac1{n^2}}$?






          share|cite|improve this answer





















          • Then one would have to show that $||sqrt{x^2+frac{1}{n^2}}-sqrt{x^2}||_{C^0}to 0$ as $ntoinfty$. Could one use uniform convergence for that?
            – James
            Nov 26 '18 at 20:19










          • One can and one should.
            – José Carlos Santos
            Nov 26 '18 at 20:22



















          1














          It's not closed, which is a result of the Weierstrass Approxmation Theorem: Given any $fin C^0[-1,1]$ and any $epsilon >0$, there is a polynomial $p$ such that $|f-p|_{C^0} < epsilon$. In fact this shows that $C^infty$ is not a closed subspace of $C^0$ since polynomials are smooth.






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
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            active

            oldest

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            2














            Hint: If $xin[-1,1]$, what is $displaystylelim_{ntoinfty}sqrt{x^2+frac1{n^2}}$?






            share|cite|improve this answer





















            • Then one would have to show that $||sqrt{x^2+frac{1}{n^2}}-sqrt{x^2}||_{C^0}to 0$ as $ntoinfty$. Could one use uniform convergence for that?
              – James
              Nov 26 '18 at 20:19










            • One can and one should.
              – José Carlos Santos
              Nov 26 '18 at 20:22
















            2














            Hint: If $xin[-1,1]$, what is $displaystylelim_{ntoinfty}sqrt{x^2+frac1{n^2}}$?






            share|cite|improve this answer





















            • Then one would have to show that $||sqrt{x^2+frac{1}{n^2}}-sqrt{x^2}||_{C^0}to 0$ as $ntoinfty$. Could one use uniform convergence for that?
              – James
              Nov 26 '18 at 20:19










            • One can and one should.
              – José Carlos Santos
              Nov 26 '18 at 20:22














            2












            2








            2






            Hint: If $xin[-1,1]$, what is $displaystylelim_{ntoinfty}sqrt{x^2+frac1{n^2}}$?






            share|cite|improve this answer












            Hint: If $xin[-1,1]$, what is $displaystylelim_{ntoinfty}sqrt{x^2+frac1{n^2}}$?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 '18 at 19:54









            José Carlos Santos

            150k22121221




            150k22121221












            • Then one would have to show that $||sqrt{x^2+frac{1}{n^2}}-sqrt{x^2}||_{C^0}to 0$ as $ntoinfty$. Could one use uniform convergence for that?
              – James
              Nov 26 '18 at 20:19










            • One can and one should.
              – José Carlos Santos
              Nov 26 '18 at 20:22


















            • Then one would have to show that $||sqrt{x^2+frac{1}{n^2}}-sqrt{x^2}||_{C^0}to 0$ as $ntoinfty$. Could one use uniform convergence for that?
              – James
              Nov 26 '18 at 20:19










            • One can and one should.
              – José Carlos Santos
              Nov 26 '18 at 20:22
















            Then one would have to show that $||sqrt{x^2+frac{1}{n^2}}-sqrt{x^2}||_{C^0}to 0$ as $ntoinfty$. Could one use uniform convergence for that?
            – James
            Nov 26 '18 at 20:19




            Then one would have to show that $||sqrt{x^2+frac{1}{n^2}}-sqrt{x^2}||_{C^0}to 0$ as $ntoinfty$. Could one use uniform convergence for that?
            – James
            Nov 26 '18 at 20:19












            One can and one should.
            – José Carlos Santos
            Nov 26 '18 at 20:22




            One can and one should.
            – José Carlos Santos
            Nov 26 '18 at 20:22











            1














            It's not closed, which is a result of the Weierstrass Approxmation Theorem: Given any $fin C^0[-1,1]$ and any $epsilon >0$, there is a polynomial $p$ such that $|f-p|_{C^0} < epsilon$. In fact this shows that $C^infty$ is not a closed subspace of $C^0$ since polynomials are smooth.






            share|cite|improve this answer


























              1














              It's not closed, which is a result of the Weierstrass Approxmation Theorem: Given any $fin C^0[-1,1]$ and any $epsilon >0$, there is a polynomial $p$ such that $|f-p|_{C^0} < epsilon$. In fact this shows that $C^infty$ is not a closed subspace of $C^0$ since polynomials are smooth.






              share|cite|improve this answer
























                1












                1








                1






                It's not closed, which is a result of the Weierstrass Approxmation Theorem: Given any $fin C^0[-1,1]$ and any $epsilon >0$, there is a polynomial $p$ such that $|f-p|_{C^0} < epsilon$. In fact this shows that $C^infty$ is not a closed subspace of $C^0$ since polynomials are smooth.






                share|cite|improve this answer












                It's not closed, which is a result of the Weierstrass Approxmation Theorem: Given any $fin C^0[-1,1]$ and any $epsilon >0$, there is a polynomial $p$ such that $|f-p|_{C^0} < epsilon$. In fact this shows that $C^infty$ is not a closed subspace of $C^0$ since polynomials are smooth.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 '18 at 19:54









                user25959

                1,573816




                1,573816






























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