Prove uniform convergence of $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}$ to $f(x) = |x|$












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For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$




I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!






real-analysis sequences-and-series uniform-convergence







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    For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$




    I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!






    real-analysis sequences-and-series uniform-convergence







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      1












      1








      1





      $begingroup$



      For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$




      I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!






      real-analysis sequences-and-series uniform-convergence







      share|cite|improve this question











      $endgroup$





      For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$




      I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!






      real-analysis sequences-and-series uniform-convergence




      real-analysis sequences-and-series uniform-convergence






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      edited Dec 5 '18 at 9:36









      Chinnapparaj R

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      asked Dec 5 '18 at 9:28









      user613048user613048

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          Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.






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            If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.






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              Hint: We have
              $$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$






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                Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$




                Moral:



                Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!






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                  4 Answers
                  4






                  active

                  oldest

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                  4 Answers
                  4






                  active

                  oldest

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                  active

                  oldest

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                  active

                  oldest

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                  1












                  $begingroup$

                  Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.






                  share|cite|improve this answer









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                    1












                    $begingroup$

                    Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.






                    share|cite|improve this answer









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                      1












                      1








                      1





                      $begingroup$

                      Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.






                      share|cite|improve this answer









                      $endgroup$



                      Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.







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                      answered Dec 5 '18 at 9:35









                      José Carlos SantosJosé Carlos Santos

                      159k22126229




                      159k22126229























                          1












                          $begingroup$

                          If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.






                              share|cite|improve this answer









                              $endgroup$



                              If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 5 '18 at 9:31









                              Kavi Rama MurthyKavi Rama Murthy

                              57.7k42160




                              57.7k42160























                                  1












                                  $begingroup$

                                  Hint: We have
                                  $$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    Hint: We have
                                    $$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      Hint: We have
                                      $$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Hint: We have
                                      $$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 5 '18 at 9:32









                                      p4schp4sch

                                      5,245217




                                      5,245217























                                          1












                                          $begingroup$

                                          Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$




                                          Moral:



                                          Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!






                                          share|cite|improve this answer











                                          $endgroup$


















                                            1












                                            $begingroup$

                                            Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$




                                            Moral:



                                            Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!






                                            share|cite|improve this answer











                                            $endgroup$
















                                              1












                                              1








                                              1





                                              $begingroup$

                                              Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$




                                              Moral:



                                              Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!






                                              share|cite|improve this answer











                                              $endgroup$



                                              Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$




                                              Moral:



                                              Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!







                                              share|cite|improve this answer














                                              share|cite|improve this answer



                                              share|cite|improve this answer








                                              edited Dec 5 '18 at 9:42

























                                              answered Dec 5 '18 at 9:33









                                              Chinnapparaj RChinnapparaj R

                                              5,4331928




                                              5,4331928






























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