Uniform convergence of $f_n(x)=frac{sqrt{1+{(nx)}^2}}{n}$ and ${f_n}^prime$












0












$begingroup$


Discuss the uniform convergence of $f_n(x)=frac{sqrt{1+{(nx)}^2}}{n}$ and its first derivative on real line.



I think both $f_n$ and ${f_n}^prime$ do not converge uniformly. If we put $x=0,1$ in $f_n$ then we get point wise limit 0 and 1 respectively. So $f_n$ doesn't converge uniformly.



Now,
${f_n}^prime(x)=frac{nx}{sqrt{1+(nx)^2}}.$ If we put $x=1/n$ then it becomes $1/sqrt{2}$ . So sup norm never becomes zero. But point wise limit is zero. So it is not uniformly convergent. Correct? Thanks.






real-analysis uniform-convergence







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$endgroup$








  • 1




    $begingroup$
    Your answer for $f_n$ is incorrect. You have proved that there are two points $x,y$ such that $f_n(x)$ and $f_n(y)$ converge to different values, but this doesn't mean the convergence is non-uniform. It turns out that $f_n$ does converge uniformly. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:13


















0












$begingroup$


Discuss the uniform convergence of $f_n(x)=frac{sqrt{1+{(nx)}^2}}{n}$ and its first derivative on real line.



I think both $f_n$ and ${f_n}^prime$ do not converge uniformly. If we put $x=0,1$ in $f_n$ then we get point wise limit 0 and 1 respectively. So $f_n$ doesn't converge uniformly.



Now,
${f_n}^prime(x)=frac{nx}{sqrt{1+(nx)^2}}.$ If we put $x=1/n$ then it becomes $1/sqrt{2}$ . So sup norm never becomes zero. But point wise limit is zero. So it is not uniformly convergent. Correct? Thanks.






real-analysis uniform-convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your answer for $f_n$ is incorrect. You have proved that there are two points $x,y$ such that $f_n(x)$ and $f_n(y)$ converge to different values, but this doesn't mean the convergence is non-uniform. It turns out that $f_n$ does converge uniformly. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:13
















0












0








0





$begingroup$


Discuss the uniform convergence of $f_n(x)=frac{sqrt{1+{(nx)}^2}}{n}$ and its first derivative on real line.



I think both $f_n$ and ${f_n}^prime$ do not converge uniformly. If we put $x=0,1$ in $f_n$ then we get point wise limit 0 and 1 respectively. So $f_n$ doesn't converge uniformly.



Now,
${f_n}^prime(x)=frac{nx}{sqrt{1+(nx)^2}}.$ If we put $x=1/n$ then it becomes $1/sqrt{2}$ . So sup norm never becomes zero. But point wise limit is zero. So it is not uniformly convergent. Correct? Thanks.






real-analysis uniform-convergence







share|cite|improve this question









$endgroup$




Discuss the uniform convergence of $f_n(x)=frac{sqrt{1+{(nx)}^2}}{n}$ and its first derivative on real line.



I think both $f_n$ and ${f_n}^prime$ do not converge uniformly. If we put $x=0,1$ in $f_n$ then we get point wise limit 0 and 1 respectively. So $f_n$ doesn't converge uniformly.



Now,
${f_n}^prime(x)=frac{nx}{sqrt{1+(nx)^2}}.$ If we put $x=1/n$ then it becomes $1/sqrt{2}$ . So sup norm never becomes zero. But point wise limit is zero. So it is not uniformly convergent. Correct? Thanks.






real-analysis uniform-convergence




real-analysis uniform-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 6:24









ramanujanramanujan

719713




719713








  • 1




    $begingroup$
    Your answer for $f_n$ is incorrect. You have proved that there are two points $x,y$ such that $f_n(x)$ and $f_n(y)$ converge to different values, but this doesn't mean the convergence is non-uniform. It turns out that $f_n$ does converge uniformly. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:13
















  • 1




    $begingroup$
    Your answer for $f_n$ is incorrect. You have proved that there are two points $x,y$ such that $f_n(x)$ and $f_n(y)$ converge to different values, but this doesn't mean the convergence is non-uniform. It turns out that $f_n$ does converge uniformly. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:13










1




1




$begingroup$
Your answer for $f_n$ is incorrect. You have proved that there are two points $x,y$ such that $f_n(x)$ and $f_n(y)$ converge to different values, but this doesn't mean the convergence is non-uniform. It turns out that $f_n$ does converge uniformly. See my answer below.
$endgroup$
– User8128
Dec 19 '18 at 7:13






$begingroup$
Your answer for $f_n$ is incorrect. You have proved that there are two points $x,y$ such that $f_n(x)$ and $f_n(y)$ converge to different values, but this doesn't mean the convergence is non-uniform. It turns out that $f_n$ does converge uniformly. See my answer below.
$endgroup$
– User8128
Dec 19 '18 at 7:13












2 Answers
2






active

oldest

votes


















1












$begingroup$

You're answer for $f_n$ is NOT correct actually. The functions $f_n$ do converge uniformly to the function $f(x) = lvert x rvert $. Indeed, we see that $$f_n(x) = sqrt{frac{1}{n^2} + x^2}$$ and thus using the inequality $$lvert b rvert le sqrt{a^2 + b^2} le lvert a rvert + lvert b rvert,$$ which holds for all $a,bin mathbb R$, we have $$lvert x rvert le f_n(x) le frac 1 n + lvert x rvert, ,,,,,,, forall x inmathbb R.$$ Sending $n to infty$ clearly shows that $f_n$ converges uniformly to $f(x) = lvert x rvert$ on $mathbb R$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent.
    $endgroup$
    – ramanujan
    Dec 19 '18 at 7:21










  • $begingroup$
    This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable).
    $endgroup$
    – ramanujan
    Dec 20 '18 at 13:36



















2












$begingroup$

Pointwise limit of $f_n'(x)$ is $0$ for $x=0$, $1$ for $x >0$ and $-1$ for $x<0$. The limit function is not continuous and hence the convergence is not uniform.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Am I correct about$f_n$?
    $endgroup$
    – ramanujan
    Dec 19 '18 at 6:41






  • 1




    $begingroup$
    @KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:09













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You're answer for $f_n$ is NOT correct actually. The functions $f_n$ do converge uniformly to the function $f(x) = lvert x rvert $. Indeed, we see that $$f_n(x) = sqrt{frac{1}{n^2} + x^2}$$ and thus using the inequality $$lvert b rvert le sqrt{a^2 + b^2} le lvert a rvert + lvert b rvert,$$ which holds for all $a,bin mathbb R$, we have $$lvert x rvert le f_n(x) le frac 1 n + lvert x rvert, ,,,,,,, forall x inmathbb R.$$ Sending $n to infty$ clearly shows that $f_n$ converges uniformly to $f(x) = lvert x rvert$ on $mathbb R$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent.
    $endgroup$
    – ramanujan
    Dec 19 '18 at 7:21










  • $begingroup$
    This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable).
    $endgroup$
    – ramanujan
    Dec 20 '18 at 13:36
















1












$begingroup$

You're answer for $f_n$ is NOT correct actually. The functions $f_n$ do converge uniformly to the function $f(x) = lvert x rvert $. Indeed, we see that $$f_n(x) = sqrt{frac{1}{n^2} + x^2}$$ and thus using the inequality $$lvert b rvert le sqrt{a^2 + b^2} le lvert a rvert + lvert b rvert,$$ which holds for all $a,bin mathbb R$, we have $$lvert x rvert le f_n(x) le frac 1 n + lvert x rvert, ,,,,,,, forall x inmathbb R.$$ Sending $n to infty$ clearly shows that $f_n$ converges uniformly to $f(x) = lvert x rvert$ on $mathbb R$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent.
    $endgroup$
    – ramanujan
    Dec 19 '18 at 7:21










  • $begingroup$
    This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable).
    $endgroup$
    – ramanujan
    Dec 20 '18 at 13:36














1












1








1





$begingroup$

You're answer for $f_n$ is NOT correct actually. The functions $f_n$ do converge uniformly to the function $f(x) = lvert x rvert $. Indeed, we see that $$f_n(x) = sqrt{frac{1}{n^2} + x^2}$$ and thus using the inequality $$lvert b rvert le sqrt{a^2 + b^2} le lvert a rvert + lvert b rvert,$$ which holds for all $a,bin mathbb R$, we have $$lvert x rvert le f_n(x) le frac 1 n + lvert x rvert, ,,,,,,, forall x inmathbb R.$$ Sending $n to infty$ clearly shows that $f_n$ converges uniformly to $f(x) = lvert x rvert$ on $mathbb R$.






share|cite|improve this answer











$endgroup$



You're answer for $f_n$ is NOT correct actually. The functions $f_n$ do converge uniformly to the function $f(x) = lvert x rvert $. Indeed, we see that $$f_n(x) = sqrt{frac{1}{n^2} + x^2}$$ and thus using the inequality $$lvert b rvert le sqrt{a^2 + b^2} le lvert a rvert + lvert b rvert,$$ which holds for all $a,bin mathbb R$, we have $$lvert x rvert le f_n(x) le frac 1 n + lvert x rvert, ,,,,,,, forall x inmathbb R.$$ Sending $n to infty$ clearly shows that $f_n$ converges uniformly to $f(x) = lvert x rvert$ on $mathbb R$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 7:21

























answered Dec 19 '18 at 7:08









User8128User8128

10.8k1622




10.8k1622












  • $begingroup$
    Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent.
    $endgroup$
    – ramanujan
    Dec 19 '18 at 7:21










  • $begingroup$
    This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable).
    $endgroup$
    – ramanujan
    Dec 20 '18 at 13:36


















  • $begingroup$
    Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent.
    $endgroup$
    – ramanujan
    Dec 19 '18 at 7:21










  • $begingroup$
    This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable).
    $endgroup$
    – ramanujan
    Dec 20 '18 at 13:36
















$begingroup$
Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent.
$endgroup$
– ramanujan
Dec 19 '18 at 7:21




$begingroup$
Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent.
$endgroup$
– ramanujan
Dec 19 '18 at 7:21












$begingroup$
This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable).
$endgroup$
– ramanujan
Dec 20 '18 at 13:36




$begingroup$
This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable).
$endgroup$
– ramanujan
Dec 20 '18 at 13:36











2












$begingroup$

Pointwise limit of $f_n'(x)$ is $0$ for $x=0$, $1$ for $x >0$ and $-1$ for $x<0$. The limit function is not continuous and hence the convergence is not uniform.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Am I correct about$f_n$?
    $endgroup$
    – ramanujan
    Dec 19 '18 at 6:41






  • 1




    $begingroup$
    @KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:09


















2












$begingroup$

Pointwise limit of $f_n'(x)$ is $0$ for $x=0$, $1$ for $x >0$ and $-1$ for $x<0$. The limit function is not continuous and hence the convergence is not uniform.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Am I correct about$f_n$?
    $endgroup$
    – ramanujan
    Dec 19 '18 at 6:41






  • 1




    $begingroup$
    @KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:09
















2












2








2





$begingroup$

Pointwise limit of $f_n'(x)$ is $0$ for $x=0$, $1$ for $x >0$ and $-1$ for $x<0$. The limit function is not continuous and hence the convergence is not uniform.






share|cite|improve this answer









$endgroup$



Pointwise limit of $f_n'(x)$ is $0$ for $x=0$, $1$ for $x >0$ and $-1$ for $x<0$. The limit function is not continuous and hence the convergence is not uniform.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 6:34









Kavi Rama MurthyKavi Rama Murthy

68.3k53169




68.3k53169












  • $begingroup$
    Am I correct about$f_n$?
    $endgroup$
    – ramanujan
    Dec 19 '18 at 6:41






  • 1




    $begingroup$
    @KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:09




















  • $begingroup$
    Am I correct about$f_n$?
    $endgroup$
    – ramanujan
    Dec 19 '18 at 6:41






  • 1




    $begingroup$
    @KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below.
    $endgroup$
    – User8128
    Dec 19 '18 at 7:09


















$begingroup$
Am I correct about$f_n$?
$endgroup$
– ramanujan
Dec 19 '18 at 6:41




$begingroup$
Am I correct about$f_n$?
$endgroup$
– ramanujan
Dec 19 '18 at 6:41




1




1




$begingroup$
@KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below.
$endgroup$
– User8128
Dec 19 '18 at 7:09






$begingroup$
@KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below.
$endgroup$
– User8128
Dec 19 '18 at 7:09




















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