Converse of Weierstrass M-Test: counterexample to the statement
The statement is If $sum_{n=1}^{infty} f_n $ converges uniformly on a domain say $A$, then there exist constants $M_n$ such that $|f_n(x)| leq M_n$ for all $n$ and $x in A$ and $sum M_n$ converges. My intuition is that the statement is false because if it is true then the converse statement of Weierstrass M-Test is true. I know that $sum 1/n$ diverges so my attempt is to produce $sum f_n$ converges where $|f_n(x)| < 1/n$. How can I produce such a series for that?
real-analysis
asked Nov 29 '18 at 0:37
Dong Le
517
add a comment |
The statement is If $sum_{n=1}^{infty} f_n $ converges uniformly on a domain say $A$, then there exist constants $M_n$ such that $|f_n(x)| leq M_n$ for all $n$ and $x in A$ and $sum M_n$ converges. My intuition is that the statement is false because if it is true then the converse statement of Weierstrass M-Test is true. I know that $sum 1/n$ diverges so my attempt is to produce $sum f_n$ converges where $|f_n(x)| < 1/n$. How can I produce such a series for that?
real-analysis
asked Nov 29 '18 at 0:37
Dong Le
517
Just pick the constant functions $1,-1,frac{1}{2},-frac{1}{2},ldots$ for the purpose.
– AdditIdent
Nov 29 '18 at 0:48
Ah!! Thanks, I see what you mean.
– Dong Le
Nov 29 '18 at 0:50
1
The more interesting (harder) question is if the series converges uniformly and absolutely, does such $M_n$ exist -- because the Weierstrass test always gives absolute convergence. Otherwise we have trivial examples.
– RRL
Nov 29 '18 at 0:52
add a comment |
The statement is If $sum_{n=1}^{infty} f_n $ converges uniformly on a domain say $A$, then there exist constants $M_n$ such that $|f_n(x)| leq M_n$ for all $n$ and $x in A$ and $sum M_n$ converges. My intuition is that the statement is false because if it is true then the converse statement of Weierstrass M-Test is true. I know that $sum 1/n$ diverges so my attempt is to produce $sum f_n$ converges where $|f_n(x)| < 1/n$. How can I produce such a series for that?
real-analysis
asked Nov 29 '18 at 0:37
Dong Le
517
The statement is If $sum_{n=1}^{infty} f_n $ converges uniformly on a domain say $A$, then there exist constants $M_n$ such that $|f_n(x)| leq M_n$ for all $n$ and $x in A$ and $sum M_n$ converges. My intuition is that the statement is false because if it is true then the converse statement of Weierstrass M-Test is true. I know that $sum 1/n$ diverges so my attempt is to produce $sum f_n$ converges where $|f_n(x)| < 1/n$. How can I produce such a series for that?
real-analysis
real-analysis
asked Nov 29 '18 at 0:37
Dong Le
517
asked Nov 29 '18 at 0:37
Dong Le
517
asked Nov 29 '18 at 0:37
Dong Le
517
asked Nov 29 '18 at 0:37
Dong Le
517
asked Nov 29 '18 at 0:37
Dong Le
517
517
Just pick the constant functions $1,-1,frac{1}{2},-frac{1}{2},ldots$ for the purpose.
– AdditIdent
Nov 29 '18 at 0:48
Ah!! Thanks, I see what you mean.
– Dong Le
Nov 29 '18 at 0:50
1
The more interesting (harder) question is if the series converges uniformly and absolutely, does such $M_n$ exist -- because the Weierstrass test always gives absolute convergence. Otherwise we have trivial examples.
– RRL
Nov 29 '18 at 0:52
add a comment |
Just pick the constant functions $1,-1,frac{1}{2},-frac{1}{2},ldots$ for the purpose.
– AdditIdent
Nov 29 '18 at 0:48
Ah!! Thanks, I see what you mean.
– Dong Le
Nov 29 '18 at 0:50
1
The more interesting (harder) question is if the series converges uniformly and absolutely, does such $M_n$ exist -- because the Weierstrass test always gives absolute convergence. Otherwise we have trivial examples.
– RRL
Nov 29 '18 at 0:52
Just pick the constant functions $1,-1,frac{1}{2},-frac{1}{2},ldots$ for the purpose.
– AdditIdent
Nov 29 '18 at 0:48
Just pick the constant functions $1,-1,frac{1}{2},-frac{1}{2},ldots$ for the purpose.
– AdditIdent
Nov 29 '18 at 0:48
Ah!! Thanks, I see what you mean.
– Dong Le
Nov 29 '18 at 0:50
Ah!! Thanks, I see what you mean.
– Dong Le
Nov 29 '18 at 0:50
1
1
The more interesting (harder) question is if the series converges uniformly and absolutely, does such $M_n$ exist -- because the Weierstrass test always gives absolute convergence. Otherwise we have trivial examples.
– RRL
Nov 29 '18 at 0:52
The more interesting (harder) question is if the series converges uniformly and absolutely, does such $M_n$ exist -- because the Weierstrass test always gives absolute convergence. Otherwise we have trivial examples.
– RRL
Nov 29 '18 at 0:52
add a comment |
1 Answer
1
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oldest
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For an absolutely convergent example, take $f_n(x) = frac{1}{n}mathbf{1}_{(n-1,n)}$. The sums converge to the function:
$$
f(x) = frac{1}{lceil x rceil}
$$
because summation in this case is joining piecewise constant functions, each of height $frac{1}{n}$ on the interval $(n-1,n)$. Clearly, convergence is absolute and also uniform, but $sup f_n(x) = frac{1}{n}$ for any $f_n$, so the right hand side of the sum doesn't converge.
Moral of the story: you can create annoying counterexamples by partially defining a function on some domain and then setting it equal to $0$ everywhere else.
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
1
Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1)
– RRL
Nov 29 '18 at 1:05
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
For an absolutely convergent example, take $f_n(x) = frac{1}{n}mathbf{1}_{(n-1,n)}$. The sums converge to the function:
$$
f(x) = frac{1}{lceil x rceil}
$$
because summation in this case is joining piecewise constant functions, each of height $frac{1}{n}$ on the interval $(n-1,n)$. Clearly, convergence is absolute and also uniform, but $sup f_n(x) = frac{1}{n}$ for any $f_n$, so the right hand side of the sum doesn't converge.
Moral of the story: you can create annoying counterexamples by partially defining a function on some domain and then setting it equal to $0$ everywhere else.
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
1
Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1)
– RRL
Nov 29 '18 at 1:05
add a comment |
For an absolutely convergent example, take $f_n(x) = frac{1}{n}mathbf{1}_{(n-1,n)}$. The sums converge to the function:
$$
f(x) = frac{1}{lceil x rceil}
$$
because summation in this case is joining piecewise constant functions, each of height $frac{1}{n}$ on the interval $(n-1,n)$. Clearly, convergence is absolute and also uniform, but $sup f_n(x) = frac{1}{n}$ for any $f_n$, so the right hand side of the sum doesn't converge.
Moral of the story: you can create annoying counterexamples by partially defining a function on some domain and then setting it equal to $0$ everywhere else.
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
1
Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1)
– RRL
Nov 29 '18 at 1:05
add a comment |
For an absolutely convergent example, take $f_n(x) = frac{1}{n}mathbf{1}_{(n-1,n)}$. The sums converge to the function:
$$
f(x) = frac{1}{lceil x rceil}
$$
because summation in this case is joining piecewise constant functions, each of height $frac{1}{n}$ on the interval $(n-1,n)$. Clearly, convergence is absolute and also uniform, but $sup f_n(x) = frac{1}{n}$ for any $f_n$, so the right hand side of the sum doesn't converge.
Moral of the story: you can create annoying counterexamples by partially defining a function on some domain and then setting it equal to $0$ everywhere else.
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
For an absolutely convergent example, take $f_n(x) = frac{1}{n}mathbf{1}_{(n-1,n)}$. The sums converge to the function:
$$
f(x) = frac{1}{lceil x rceil}
$$
because summation in this case is joining piecewise constant functions, each of height $frac{1}{n}$ on the interval $(n-1,n)$. Clearly, convergence is absolute and also uniform, but $sup f_n(x) = frac{1}{n}$ for any $f_n$, so the right hand side of the sum doesn't converge.
Moral of the story: you can create annoying counterexamples by partially defining a function on some domain and then setting it equal to $0$ everywhere else.
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
answered Nov 29 '18 at 0:59
rubikscube09
1,174717
1,174717
1
Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1)
– RRL
Nov 29 '18 at 1:05
add a comment |
1
Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1)
– RRL
Nov 29 '18 at 1:05
1
1
Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1)
– RRL
Nov 29 '18 at 1:05
Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1)
– RRL
Nov 29 '18 at 1:05
add a comment |
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Just pick the constant functions $1,-1,frac{1}{2},-frac{1}{2},ldots$ for the purpose.
– AdditIdent
Nov 29 '18 at 0:48
Ah!! Thanks, I see what you mean.
– Dong Le
Nov 29 '18 at 0:50
1
The more interesting (harder) question is if the series converges uniformly and absolutely, does such $M_n$ exist -- because the Weierstrass test always gives absolute convergence. Otherwise we have trivial examples.
– RRL
Nov 29 '18 at 0:52