Uniform convergence of $f_n(x)=frac{sqrt{1+{(nx)}^2}}{n}$ and ${f_n}^prime$
$begingroup$
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
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
$endgroup$
add a comment |
$begingroup$
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
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
$endgroup$
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
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
asked Dec 19 '18 at 6:24
ramanujanramanujan
729713
729713
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $mathbb R$.
answered Dec 19 '18 at 7:08
User8128User8128
10.8k1622
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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votes
$begingroup$
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 $mathbb R$.
answered Dec 19 '18 at 7:08
User8128User8128
10.8k1622
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mathbb R$.
answered Dec 19 '18 at 7:08
User8128User8128
10.8k1622
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mathbb R$.
answered Dec 19 '18 at 7:08
User8128User8128
10.8k1622
$endgroup$
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 $mathbb R$.
answered Dec 19 '18 at 7:08
User8128User8128
10.8k1622
edited Dec 19 '18 at 7:21
answered Dec 19 '18 at 7:08
User8128User8128
10.8k1622
answered Dec 19 '18 at 7:08
User8128User8128
10.8k1622
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
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.
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Dec 19 '18 at 6:34
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
71.5k53170
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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