Pointwise vs Uniform Convergence
Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that
- {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$
- {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.
I have proved that $f_n$ converges uniformly to $f(x) = 0$.
However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?
real-analysis
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
asked Dec 1 at 19:23
Ahmad Lamaa
1326
add a comment |
Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that
- {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$
- {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.
I have proved that $f_n$ converges uniformly to $f(x) = 0$.
However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?
real-analysis
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
asked Dec 1 at 19:23
Ahmad Lamaa
1326
add a comment |
Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that
- {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$
- {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.
I have proved that $f_n$ converges uniformly to $f(x) = 0$.
However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?
real-analysis
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
asked Dec 1 at 19:23
Ahmad Lamaa
1326
Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that
- {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$
- {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.
I have proved that $f_n$ converges uniformly to $f(x) = 0$.
However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?
real-analysis
real-analysis
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
asked Dec 1 at 19:23
Ahmad Lamaa
1326
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
asked Dec 1 at 19:23
Ahmad Lamaa
1326
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
edited Dec 1 at 19:31
Mostafa Ayaz
13.7k3836
13.7k3836
asked Dec 1 at 19:23
Ahmad Lamaa
1326
asked Dec 1 at 19:23
Ahmad Lamaa
1326
asked Dec 1 at 19:23
Ahmad Lamaa
1326
1326
add a comment |
add a comment |
1 Answer
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The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
$$g(x) = left{ array{0 &text{ if } xin (0,1] \
1 & text{ if } x=0 }right. $$
(Note that $f^prime_n(0) = 1$ for every $n$).
edited Dec 1 at 20:02
amWhy
191k28224439
answered Dec 1 at 19:46
Thomas
16.7k21631
That's correct. I have overlooked the case where x=0. Thanks!
– Ahmad Lamaa
Dec 1 at 19:50
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
$$g(x) = left{ array{0 &text{ if } xin (0,1] \
1 & text{ if } x=0 }right. $$
(Note that $f^prime_n(0) = 1$ for every $n$).
edited Dec 1 at 20:02
amWhy
191k28224439
answered Dec 1 at 19:46
Thomas
16.7k21631
That's correct. I have overlooked the case where x=0. Thanks!
– Ahmad Lamaa
Dec 1 at 19:50
add a comment |
The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
$$g(x) = left{ array{0 &text{ if } xin (0,1] \
1 & text{ if } x=0 }right. $$
(Note that $f^prime_n(0) = 1$ for every $n$).
edited Dec 1 at 20:02
amWhy
191k28224439
answered Dec 1 at 19:46
Thomas
16.7k21631
That's correct. I have overlooked the case where x=0. Thanks!
– Ahmad Lamaa
Dec 1 at 19:50
add a comment |
The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
$$g(x) = left{ array{0 &text{ if } xin (0,1] \
1 & text{ if } x=0 }right. $$
(Note that $f^prime_n(0) = 1$ for every $n$).
edited Dec 1 at 20:02
amWhy
191k28224439
answered Dec 1 at 19:46
Thomas
16.7k21631
The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
$$g(x) = left{ array{0 &text{ if } xin (0,1] \
1 & text{ if } x=0 }right. $$
(Note that $f^prime_n(0) = 1$ for every $n$).
edited Dec 1 at 20:02
amWhy
191k28224439
answered Dec 1 at 19:46
Thomas
16.7k21631
edited Dec 1 at 20:02
amWhy
191k28224439
edited Dec 1 at 20:02
amWhy
191k28224439
edited Dec 1 at 20:02
amWhy
191k28224439
191k28224439
answered Dec 1 at 19:46
Thomas
16.7k21631
answered Dec 1 at 19:46
Thomas
16.7k21631
answered Dec 1 at 19:46
Thomas
16.7k21631
16.7k21631
That's correct. I have overlooked the case where x=0. Thanks!
– Ahmad Lamaa
Dec 1 at 19:50
add a comment |
That's correct. I have overlooked the case where x=0. Thanks!
– Ahmad Lamaa
Dec 1 at 19:50
That's correct. I have overlooked the case where x=0. Thanks!
– Ahmad Lamaa
Dec 1 at 19:50
That's correct. I have overlooked the case where x=0. Thanks!
– Ahmad Lamaa
Dec 1 at 19:50
add a comment |
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