Is $(f_n)$ pointwise convergent?
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Let $f_n(x)$, for all n>=1, be a sequence of non-negative continuous functions on [0,1] such that
$$lim_{n→infty}int^1_0 f_n (x)dx=0$$
Which of the following is always correct ?
A. $f_n→0$ uniformly on $[0,1]$
B. $f_n$ may not converge uniformly but converges to $0$ pointwise
C. $f_n$ will converge point-wise and limit may be non zero
D. $f_n$ is not guaranteed to have pointwise limit.
I can find example where all four statements are true but can't find any counterexample.
Option A is certainly false. Consider a function (geometrically) a isosceles triangle whose one vertex is on $0$, other on $2/n$. And height $1$ unit. Its area is $2/n →0$ . But $f$ is not uniformly continuous
For others, i am completely stuck. I need ideas to solve this question rather than solution.please help!.
real-analysis sequences-and-series sequence-of-function
add a comment |
up vote
0
down vote
favorite
Let $f_n(x)$, for all n>=1, be a sequence of non-negative continuous functions on [0,1] such that
$$lim_{n→infty}int^1_0 f_n (x)dx=0$$
Which of the following is always correct ?
A. $f_n→0$ uniformly on $[0,1]$
B. $f_n$ may not converge uniformly but converges to $0$ pointwise
C. $f_n$ will converge point-wise and limit may be non zero
D. $f_n$ is not guaranteed to have pointwise limit.
I can find example where all four statements are true but can't find any counterexample.
Option A is certainly false. Consider a function (geometrically) a isosceles triangle whose one vertex is on $0$, other on $2/n$. And height $1$ unit. Its area is $2/n →0$ . But $f$ is not uniformly continuous
For others, i am completely stuck. I need ideas to solve this question rather than solution.please help!.
real-analysis sequences-and-series sequence-of-function
It's better to to tell how to tackle this kind of question rather than some abstract counter example, thanks
– Cloud JR
Nov 22 at 18:19
1
Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
– Daniel
Nov 22 at 18:19
@Daniel, thatnks , it help a lot... Consider adding your comment as answer coz it is really helpful for others too
– Cloud JR
Nov 22 at 18:30
If you can find examples to all 4, then the answer must be...
– Will M.
Nov 22 at 20:12
@will M option 4.
– Cloud JR
Nov 23 at 4:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f_n(x)$, for all n>=1, be a sequence of non-negative continuous functions on [0,1] such that
$$lim_{n→infty}int^1_0 f_n (x)dx=0$$
Which of the following is always correct ?
A. $f_n→0$ uniformly on $[0,1]$
B. $f_n$ may not converge uniformly but converges to $0$ pointwise
C. $f_n$ will converge point-wise and limit may be non zero
D. $f_n$ is not guaranteed to have pointwise limit.
I can find example where all four statements are true but can't find any counterexample.
Option A is certainly false. Consider a function (geometrically) a isosceles triangle whose one vertex is on $0$, other on $2/n$. And height $1$ unit. Its area is $2/n →0$ . But $f$ is not uniformly continuous
For others, i am completely stuck. I need ideas to solve this question rather than solution.please help!.
real-analysis sequences-and-series sequence-of-function
Let $f_n(x)$, for all n>=1, be a sequence of non-negative continuous functions on [0,1] such that
$$lim_{n→infty}int^1_0 f_n (x)dx=0$$
Which of the following is always correct ?
A. $f_n→0$ uniformly on $[0,1]$
B. $f_n$ may not converge uniformly but converges to $0$ pointwise
C. $f_n$ will converge point-wise and limit may be non zero
D. $f_n$ is not guaranteed to have pointwise limit.
I can find example where all four statements are true but can't find any counterexample.
Option A is certainly false. Consider a function (geometrically) a isosceles triangle whose one vertex is on $0$, other on $2/n$. And height $1$ unit. Its area is $2/n →0$ . But $f$ is not uniformly continuous
For others, i am completely stuck. I need ideas to solve this question rather than solution.please help!.
real-analysis sequences-and-series sequence-of-function
real-analysis sequences-and-series sequence-of-function
edited Nov 22 at 20:03
Bernard
117k637109
117k637109
asked Nov 22 at 18:10
Cloud JR
806417
806417
It's better to to tell how to tackle this kind of question rather than some abstract counter example, thanks
– Cloud JR
Nov 22 at 18:19
1
Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
– Daniel
Nov 22 at 18:19
@Daniel, thatnks , it help a lot... Consider adding your comment as answer coz it is really helpful for others too
– Cloud JR
Nov 22 at 18:30
If you can find examples to all 4, then the answer must be...
– Will M.
Nov 22 at 20:12
@will M option 4.
– Cloud JR
Nov 23 at 4:13
add a comment |
It's better to to tell how to tackle this kind of question rather than some abstract counter example, thanks
– Cloud JR
Nov 22 at 18:19
1
Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
– Daniel
Nov 22 at 18:19
@Daniel, thatnks , it help a lot... Consider adding your comment as answer coz it is really helpful for others too
– Cloud JR
Nov 22 at 18:30
If you can find examples to all 4, then the answer must be...
– Will M.
Nov 22 at 20:12
@will M option 4.
– Cloud JR
Nov 23 at 4:13
It's better to to tell how to tackle this kind of question rather than some abstract counter example, thanks
– Cloud JR
Nov 22 at 18:19
It's better to to tell how to tackle this kind of question rather than some abstract counter example, thanks
– Cloud JR
Nov 22 at 18:19
1
1
Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
– Daniel
Nov 22 at 18:19
Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
– Daniel
Nov 22 at 18:19
@Daniel, thatnks , it help a lot... Consider adding your comment as answer coz it is really helpful for others too
– Cloud JR
Nov 22 at 18:30
@Daniel, thatnks , it help a lot... Consider adding your comment as answer coz it is really helpful for others too
– Cloud JR
Nov 22 at 18:30
If you can find examples to all 4, then the answer must be...
– Will M.
Nov 22 at 20:12
If you can find examples to all 4, then the answer must be...
– Will M.
Nov 22 at 20:12
@will M option 4.
– Cloud JR
Nov 23 at 4:13
@will M option 4.
– Cloud JR
Nov 23 at 4:13
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Expanding on the comment I made.
Hint: Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
Solution: A more elaborate construction of isosceles triangles suffice to build a sequence of $f_n$ that does not converge. Each $f_n$ can be for instance: non-zero only on an interval $I_n = (a_n - delta_n, a_n + delta_n)$, piecewise linear, with $f_n(x)$ being $0$ on the extremes of the interval and $f_n(a_n) = 1$. Notice that we have freedom to choose $a_n$ and $delta_n$. Since $int f_n = delta_n$, we just need to ask $delta_n to 0$ for the limit above to hold.
The main point to force that $f_n$ do not converge pointwise is the choice of $a_n$. If for instance $a_{2n} = frac13$ and $a_{2n+1} = frac23$, then $f_n(frac13)$ and $f_n(frac23)$ will oscilate between $0$ and $1$ infinitely many times, while for all other $x$ we have $f_n(x) = 0$.
To Think: If you want to build a sequence of functions with infinitely many points that do not converge, you can try a different choice of $a_n$. For instance, let $q_i$ be an enumeration of $mathbb{Q} cap (0, 1)$ and $p_i$ be the $i$-th prime number. Then, we can try defining $a_n = q_i$ iff $p_i$ is the smallest prime that divides $n$.
add a comment |
up vote
1
down vote
A. You could also consider $f_n(x) = x^n.$
B. Let $f_n$ be the line-segment graph connecting the points $(0,1), (1/n,0), (1,0).$ This sequence does not converge to $0$ at $x=0.$
C. Let $f_n$ be as in B. Then consider the sequence $f_1,0,f_2,0,f_3,0,dots.$ At $x=0,$ this sequence is $1,0,1,0,dots.$
D. See C.
Sorry , but what does $chi$ mean
– Cloud JR
Nov 22 at 18:23
$chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else.
– zhw.
Nov 22 at 18:26
Consider $chi [0,1/2] $ . Is this continous?
– Cloud JR
Nov 22 at 18:28
It is discontinous at 1/2
– Cloud JR
Nov 22 at 18:29
Oops I completely missed continuous.
– zhw.
Nov 22 at 18:42
|
show 3 more comments
up vote
1
down vote
I can find example where all four statements are true ...
All four?
It seems that the statements are worded in a way that (independent of the rest of the question) exactly one of the four statements must be true.
... but can't find any counterexample.
I even found an example for $f_n$ that does not converge in any point $x$:
Let $f^*_{c,w}(x)$ with $xin [-10,10]$ be defined as the function with the graph connecting the points:
$(-10,0)$, $((frac c{2w}-frac 1w),0)$, $((frac c{2w}),1)$, $((frac c{2w}+frac 1w),0)$, $(10,0)$.
Use $f_{c,w}(x)=f^*_{c,w}(x)$ for $xin[0,1]$.
Now use the following sequence:
$f_{0,1}$, $f_{1,1}$, $f_{2,1}$,
$f_{0,2}$, $f_{1,2}$, $f_{2,2}$, $f_{3,2}$, $f_{4,2}$,
$f_{0,4}$, $f_{1,4}$, ..., $f_{8,4}$,
$f_{0,8}$, $f_{2,8}$, ..., $f_{16,8}$,
...
$f_{0,(2^k)}$, $f_{1,(2^k)}$, ..., $f_{(2^{k+1}-1),(2^k)}$, $f_{(2^{k+1}),(2^k)}$,
...
If I'm correct this example can be modified by using $f_{c,w}(x)=sqrt wf^*_{c,w}(x)$. The resulting sequence should meet the conditions and for every value of $xin[0,1]$ the sequence $f_1(x)$, $f_2(x)$, $f_3(x)$ ... should even be unbounded!
(However maybe I made a mistake here.)
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Expanding on the comment I made.
Hint: Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
Solution: A more elaborate construction of isosceles triangles suffice to build a sequence of $f_n$ that does not converge. Each $f_n$ can be for instance: non-zero only on an interval $I_n = (a_n - delta_n, a_n + delta_n)$, piecewise linear, with $f_n(x)$ being $0$ on the extremes of the interval and $f_n(a_n) = 1$. Notice that we have freedom to choose $a_n$ and $delta_n$. Since $int f_n = delta_n$, we just need to ask $delta_n to 0$ for the limit above to hold.
The main point to force that $f_n$ do not converge pointwise is the choice of $a_n$. If for instance $a_{2n} = frac13$ and $a_{2n+1} = frac23$, then $f_n(frac13)$ and $f_n(frac23)$ will oscilate between $0$ and $1$ infinitely many times, while for all other $x$ we have $f_n(x) = 0$.
To Think: If you want to build a sequence of functions with infinitely many points that do not converge, you can try a different choice of $a_n$. For instance, let $q_i$ be an enumeration of $mathbb{Q} cap (0, 1)$ and $p_i$ be the $i$-th prime number. Then, we can try defining $a_n = q_i$ iff $p_i$ is the smallest prime that divides $n$.
add a comment |
up vote
1
down vote
accepted
Expanding on the comment I made.
Hint: Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
Solution: A more elaborate construction of isosceles triangles suffice to build a sequence of $f_n$ that does not converge. Each $f_n$ can be for instance: non-zero only on an interval $I_n = (a_n - delta_n, a_n + delta_n)$, piecewise linear, with $f_n(x)$ being $0$ on the extremes of the interval and $f_n(a_n) = 1$. Notice that we have freedom to choose $a_n$ and $delta_n$. Since $int f_n = delta_n$, we just need to ask $delta_n to 0$ for the limit above to hold.
The main point to force that $f_n$ do not converge pointwise is the choice of $a_n$. If for instance $a_{2n} = frac13$ and $a_{2n+1} = frac23$, then $f_n(frac13)$ and $f_n(frac23)$ will oscilate between $0$ and $1$ infinitely many times, while for all other $x$ we have $f_n(x) = 0$.
To Think: If you want to build a sequence of functions with infinitely many points that do not converge, you can try a different choice of $a_n$. For instance, let $q_i$ be an enumeration of $mathbb{Q} cap (0, 1)$ and $p_i$ be the $i$-th prime number. Then, we can try defining $a_n = q_i$ iff $p_i$ is the smallest prime that divides $n$.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Expanding on the comment I made.
Hint: Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
Solution: A more elaborate construction of isosceles triangles suffice to build a sequence of $f_n$ that does not converge. Each $f_n$ can be for instance: non-zero only on an interval $I_n = (a_n - delta_n, a_n + delta_n)$, piecewise linear, with $f_n(x)$ being $0$ on the extremes of the interval and $f_n(a_n) = 1$. Notice that we have freedom to choose $a_n$ and $delta_n$. Since $int f_n = delta_n$, we just need to ask $delta_n to 0$ for the limit above to hold.
The main point to force that $f_n$ do not converge pointwise is the choice of $a_n$. If for instance $a_{2n} = frac13$ and $a_{2n+1} = frac23$, then $f_n(frac13)$ and $f_n(frac23)$ will oscilate between $0$ and $1$ infinitely many times, while for all other $x$ we have $f_n(x) = 0$.
To Think: If you want to build a sequence of functions with infinitely many points that do not converge, you can try a different choice of $a_n$. For instance, let $q_i$ be an enumeration of $mathbb{Q} cap (0, 1)$ and $p_i$ be the $i$-th prime number. Then, we can try defining $a_n = q_i$ iff $p_i$ is the smallest prime that divides $n$.
Expanding on the comment I made.
Hint: Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
Solution: A more elaborate construction of isosceles triangles suffice to build a sequence of $f_n$ that does not converge. Each $f_n$ can be for instance: non-zero only on an interval $I_n = (a_n - delta_n, a_n + delta_n)$, piecewise linear, with $f_n(x)$ being $0$ on the extremes of the interval and $f_n(a_n) = 1$. Notice that we have freedom to choose $a_n$ and $delta_n$. Since $int f_n = delta_n$, we just need to ask $delta_n to 0$ for the limit above to hold.
The main point to force that $f_n$ do not converge pointwise is the choice of $a_n$. If for instance $a_{2n} = frac13$ and $a_{2n+1} = frac23$, then $f_n(frac13)$ and $f_n(frac23)$ will oscilate between $0$ and $1$ infinitely many times, while for all other $x$ we have $f_n(x) = 0$.
To Think: If you want to build a sequence of functions with infinitely many points that do not converge, you can try a different choice of $a_n$. For instance, let $q_i$ be an enumeration of $mathbb{Q} cap (0, 1)$ and $p_i$ be the $i$-th prime number. Then, we can try defining $a_n = q_i$ iff $p_i$ is the smallest prime that divides $n$.
answered Nov 22 at 18:58
Daniel
1,516210
1,516210
add a comment |
add a comment |
up vote
1
down vote
A. You could also consider $f_n(x) = x^n.$
B. Let $f_n$ be the line-segment graph connecting the points $(0,1), (1/n,0), (1,0).$ This sequence does not converge to $0$ at $x=0.$
C. Let $f_n$ be as in B. Then consider the sequence $f_1,0,f_2,0,f_3,0,dots.$ At $x=0,$ this sequence is $1,0,1,0,dots.$
D. See C.
Sorry , but what does $chi$ mean
– Cloud JR
Nov 22 at 18:23
$chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else.
– zhw.
Nov 22 at 18:26
Consider $chi [0,1/2] $ . Is this continous?
– Cloud JR
Nov 22 at 18:28
It is discontinous at 1/2
– Cloud JR
Nov 22 at 18:29
Oops I completely missed continuous.
– zhw.
Nov 22 at 18:42
|
show 3 more comments
up vote
1
down vote
A. You could also consider $f_n(x) = x^n.$
B. Let $f_n$ be the line-segment graph connecting the points $(0,1), (1/n,0), (1,0).$ This sequence does not converge to $0$ at $x=0.$
C. Let $f_n$ be as in B. Then consider the sequence $f_1,0,f_2,0,f_3,0,dots.$ At $x=0,$ this sequence is $1,0,1,0,dots.$
D. See C.
Sorry , but what does $chi$ mean
– Cloud JR
Nov 22 at 18:23
$chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else.
– zhw.
Nov 22 at 18:26
Consider $chi [0,1/2] $ . Is this continous?
– Cloud JR
Nov 22 at 18:28
It is discontinous at 1/2
– Cloud JR
Nov 22 at 18:29
Oops I completely missed continuous.
– zhw.
Nov 22 at 18:42
|
show 3 more comments
up vote
1
down vote
up vote
1
down vote
A. You could also consider $f_n(x) = x^n.$
B. Let $f_n$ be the line-segment graph connecting the points $(0,1), (1/n,0), (1,0).$ This sequence does not converge to $0$ at $x=0.$
C. Let $f_n$ be as in B. Then consider the sequence $f_1,0,f_2,0,f_3,0,dots.$ At $x=0,$ this sequence is $1,0,1,0,dots.$
D. See C.
A. You could also consider $f_n(x) = x^n.$
B. Let $f_n$ be the line-segment graph connecting the points $(0,1), (1/n,0), (1,0).$ This sequence does not converge to $0$ at $x=0.$
C. Let $f_n$ be as in B. Then consider the sequence $f_1,0,f_2,0,f_3,0,dots.$ At $x=0,$ this sequence is $1,0,1,0,dots.$
D. See C.
edited Nov 22 at 18:52
answered Nov 22 at 18:20
zhw.
71.3k43075
71.3k43075
Sorry , but what does $chi$ mean
– Cloud JR
Nov 22 at 18:23
$chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else.
– zhw.
Nov 22 at 18:26
Consider $chi [0,1/2] $ . Is this continous?
– Cloud JR
Nov 22 at 18:28
It is discontinous at 1/2
– Cloud JR
Nov 22 at 18:29
Oops I completely missed continuous.
– zhw.
Nov 22 at 18:42
|
show 3 more comments
Sorry , but what does $chi$ mean
– Cloud JR
Nov 22 at 18:23
$chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else.
– zhw.
Nov 22 at 18:26
Consider $chi [0,1/2] $ . Is this continous?
– Cloud JR
Nov 22 at 18:28
It is discontinous at 1/2
– Cloud JR
Nov 22 at 18:29
Oops I completely missed continuous.
– zhw.
Nov 22 at 18:42
Sorry , but what does $chi$ mean
– Cloud JR
Nov 22 at 18:23
Sorry , but what does $chi$ mean
– Cloud JR
Nov 22 at 18:23
$chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else.
– zhw.
Nov 22 at 18:26
$chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else.
– zhw.
Nov 22 at 18:26
Consider $chi [0,1/2] $ . Is this continous?
– Cloud JR
Nov 22 at 18:28
Consider $chi [0,1/2] $ . Is this continous?
– Cloud JR
Nov 22 at 18:28
It is discontinous at 1/2
– Cloud JR
Nov 22 at 18:29
It is discontinous at 1/2
– Cloud JR
Nov 22 at 18:29
Oops I completely missed continuous.
– zhw.
Nov 22 at 18:42
Oops I completely missed continuous.
– zhw.
Nov 22 at 18:42
|
show 3 more comments
up vote
1
down vote
I can find example where all four statements are true ...
All four?
It seems that the statements are worded in a way that (independent of the rest of the question) exactly one of the four statements must be true.
... but can't find any counterexample.
I even found an example for $f_n$ that does not converge in any point $x$:
Let $f^*_{c,w}(x)$ with $xin [-10,10]$ be defined as the function with the graph connecting the points:
$(-10,0)$, $((frac c{2w}-frac 1w),0)$, $((frac c{2w}),1)$, $((frac c{2w}+frac 1w),0)$, $(10,0)$.
Use $f_{c,w}(x)=f^*_{c,w}(x)$ for $xin[0,1]$.
Now use the following sequence:
$f_{0,1}$, $f_{1,1}$, $f_{2,1}$,
$f_{0,2}$, $f_{1,2}$, $f_{2,2}$, $f_{3,2}$, $f_{4,2}$,
$f_{0,4}$, $f_{1,4}$, ..., $f_{8,4}$,
$f_{0,8}$, $f_{2,8}$, ..., $f_{16,8}$,
...
$f_{0,(2^k)}$, $f_{1,(2^k)}$, ..., $f_{(2^{k+1}-1),(2^k)}$, $f_{(2^{k+1}),(2^k)}$,
...
If I'm correct this example can be modified by using $f_{c,w}(x)=sqrt wf^*_{c,w}(x)$. The resulting sequence should meet the conditions and for every value of $xin[0,1]$ the sequence $f_1(x)$, $f_2(x)$, $f_3(x)$ ... should even be unbounded!
(However maybe I made a mistake here.)
add a comment |
up vote
1
down vote
I can find example where all four statements are true ...
All four?
It seems that the statements are worded in a way that (independent of the rest of the question) exactly one of the four statements must be true.
... but can't find any counterexample.
I even found an example for $f_n$ that does not converge in any point $x$:
Let $f^*_{c,w}(x)$ with $xin [-10,10]$ be defined as the function with the graph connecting the points:
$(-10,0)$, $((frac c{2w}-frac 1w),0)$, $((frac c{2w}),1)$, $((frac c{2w}+frac 1w),0)$, $(10,0)$.
Use $f_{c,w}(x)=f^*_{c,w}(x)$ for $xin[0,1]$.
Now use the following sequence:
$f_{0,1}$, $f_{1,1}$, $f_{2,1}$,
$f_{0,2}$, $f_{1,2}$, $f_{2,2}$, $f_{3,2}$, $f_{4,2}$,
$f_{0,4}$, $f_{1,4}$, ..., $f_{8,4}$,
$f_{0,8}$, $f_{2,8}$, ..., $f_{16,8}$,
...
$f_{0,(2^k)}$, $f_{1,(2^k)}$, ..., $f_{(2^{k+1}-1),(2^k)}$, $f_{(2^{k+1}),(2^k)}$,
...
If I'm correct this example can be modified by using $f_{c,w}(x)=sqrt wf^*_{c,w}(x)$. The resulting sequence should meet the conditions and for every value of $xin[0,1]$ the sequence $f_1(x)$, $f_2(x)$, $f_3(x)$ ... should even be unbounded!
(However maybe I made a mistake here.)
add a comment |
up vote
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down vote
up vote
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I can find example where all four statements are true ...
All four?
It seems that the statements are worded in a way that (independent of the rest of the question) exactly one of the four statements must be true.
... but can't find any counterexample.
I even found an example for $f_n$ that does not converge in any point $x$:
Let $f^*_{c,w}(x)$ with $xin [-10,10]$ be defined as the function with the graph connecting the points:
$(-10,0)$, $((frac c{2w}-frac 1w),0)$, $((frac c{2w}),1)$, $((frac c{2w}+frac 1w),0)$, $(10,0)$.
Use $f_{c,w}(x)=f^*_{c,w}(x)$ for $xin[0,1]$.
Now use the following sequence:
$f_{0,1}$, $f_{1,1}$, $f_{2,1}$,
$f_{0,2}$, $f_{1,2}$, $f_{2,2}$, $f_{3,2}$, $f_{4,2}$,
$f_{0,4}$, $f_{1,4}$, ..., $f_{8,4}$,
$f_{0,8}$, $f_{2,8}$, ..., $f_{16,8}$,
...
$f_{0,(2^k)}$, $f_{1,(2^k)}$, ..., $f_{(2^{k+1}-1),(2^k)}$, $f_{(2^{k+1}),(2^k)}$,
...
If I'm correct this example can be modified by using $f_{c,w}(x)=sqrt wf^*_{c,w}(x)$. The resulting sequence should meet the conditions and for every value of $xin[0,1]$ the sequence $f_1(x)$, $f_2(x)$, $f_3(x)$ ... should even be unbounded!
(However maybe I made a mistake here.)
I can find example where all four statements are true ...
All four?
It seems that the statements are worded in a way that (independent of the rest of the question) exactly one of the four statements must be true.
... but can't find any counterexample.
I even found an example for $f_n$ that does not converge in any point $x$:
Let $f^*_{c,w}(x)$ with $xin [-10,10]$ be defined as the function with the graph connecting the points:
$(-10,0)$, $((frac c{2w}-frac 1w),0)$, $((frac c{2w}),1)$, $((frac c{2w}+frac 1w),0)$, $(10,0)$.
Use $f_{c,w}(x)=f^*_{c,w}(x)$ for $xin[0,1]$.
Now use the following sequence:
$f_{0,1}$, $f_{1,1}$, $f_{2,1}$,
$f_{0,2}$, $f_{1,2}$, $f_{2,2}$, $f_{3,2}$, $f_{4,2}$,
$f_{0,4}$, $f_{1,4}$, ..., $f_{8,4}$,
$f_{0,8}$, $f_{2,8}$, ..., $f_{16,8}$,
...
$f_{0,(2^k)}$, $f_{1,(2^k)}$, ..., $f_{(2^{k+1}-1),(2^k)}$, $f_{(2^{k+1}),(2^k)}$,
...
If I'm correct this example can be modified by using $f_{c,w}(x)=sqrt wf^*_{c,w}(x)$. The resulting sequence should meet the conditions and for every value of $xin[0,1]$ the sequence $f_1(x)$, $f_2(x)$, $f_3(x)$ ... should even be unbounded!
(However maybe I made a mistake here.)
edited Nov 22 at 21:35
answered Nov 22 at 20:40
Martin Rosenau
1,134129
1,134129
add a comment |
add a comment |
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It's better to to tell how to tackle this kind of question rather than some abstract counter example, thanks
– Cloud JR
Nov 22 at 18:19
1
Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.
– Daniel
Nov 22 at 18:19
@Daniel, thatnks , it help a lot... Consider adding your comment as answer coz it is really helpful for others too
– Cloud JR
Nov 22 at 18:30
If you can find examples to all 4, then the answer must be...
– Will M.
Nov 22 at 20:12
@will M option 4.
– Cloud JR
Nov 23 at 4:13