If a series of functions satisfies the Cauchy criterion uniformly, then the series converges uniformly on $S$
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Problem: If a series $sum_{k=0}^infty g_k$ of functions satisfies the Cauchy criterion uniformly on a set $S$, then the series converges uniformly on $S$.
If the series satisfies the Cauchy criterion uniformly on $S$, then
$$forall epsilon>0, exists N : n geq m geq N Rightarrow bigg|sum_{k=m}^n g_k(x) bigg| < epsilon , forall x in S$$
Therefore, we can define
$$f_l(x) = sum_{i=0}^l g_i(x)$$
Which transforms the criterion into
$$bigg| sum_{k=0}^ng_k(x) - sum_{k=0}^m g_k(x) bigg| = |f_n(x) - f_m(x)| < epsilon$$
Therefore, the definition of uniformly Cauchy is satisfied.
Is this correct? Is the converse true?
real-analysis sequences-and-series proof-verification convergence uniform-convergence
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
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add a comment |
$begingroup$
Problem: If a series $sum_{k=0}^infty g_k$ of functions satisfies the Cauchy criterion uniformly on a set $S$, then the series converges uniformly on $S$.
If the series satisfies the Cauchy criterion uniformly on $S$, then
$$forall epsilon>0, exists N : n geq m geq N Rightarrow bigg|sum_{k=m}^n g_k(x) bigg| < epsilon , forall x in S$$
Therefore, we can define
$$f_l(x) = sum_{i=0}^l g_i(x)$$
Which transforms the criterion into
$$bigg| sum_{k=0}^ng_k(x) - sum_{k=0}^m g_k(x) bigg| = |f_n(x) - f_m(x)| < epsilon$$
Therefore, the definition of uniformly Cauchy is satisfied.
Is this correct? Is the converse true?
real-analysis sequences-and-series proof-verification convergence uniform-convergence
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
$endgroup$
add a comment |
$begingroup$
Problem: If a series $sum_{k=0}^infty g_k$ of functions satisfies the Cauchy criterion uniformly on a set $S$, then the series converges uniformly on $S$.
If the series satisfies the Cauchy criterion uniformly on $S$, then
$$forall epsilon>0, exists N : n geq m geq N Rightarrow bigg|sum_{k=m}^n g_k(x) bigg| < epsilon , forall x in S$$
Therefore, we can define
$$f_l(x) = sum_{i=0}^l g_i(x)$$
Which transforms the criterion into
$$bigg| sum_{k=0}^ng_k(x) - sum_{k=0}^m g_k(x) bigg| = |f_n(x) - f_m(x)| < epsilon$$
Therefore, the definition of uniformly Cauchy is satisfied.
Is this correct? Is the converse true?
real-analysis sequences-and-series proof-verification convergence uniform-convergence
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
$endgroup$
Problem: If a series $sum_{k=0}^infty g_k$ of functions satisfies the Cauchy criterion uniformly on a set $S$, then the series converges uniformly on $S$.
If the series satisfies the Cauchy criterion uniformly on $S$, then
$$forall epsilon>0, exists N : n geq m geq N Rightarrow bigg|sum_{k=m}^n g_k(x) bigg| < epsilon , forall x in S$$
Therefore, we can define
$$f_l(x) = sum_{i=0}^l g_i(x)$$
Which transforms the criterion into
$$bigg| sum_{k=0}^ng_k(x) - sum_{k=0}^m g_k(x) bigg| = |f_n(x) - f_m(x)| < epsilon$$
Therefore, the definition of uniformly Cauchy is satisfied.
Is this correct? Is the converse true?
real-analysis sequences-and-series proof-verification convergence uniform-convergence
real-analysis sequences-and-series proof-verification convergence uniform-convergence
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
edited Dec 17 '18 at 12:32
The Bosco
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
asked Dec 17 '18 at 12:25
The BoscoThe Bosco
541212
541212
add a comment |
add a comment |
1 Answer
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If you write $f_l(x) = sum_{i=0}^l g_i(x)$ instead of $f_l(x) = sum_{i=0}^l f_i(x)$, then everything is correct.
And yes, the converse is also true. Try a proof !
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
$endgroup$
$begingroup$
Thank you! I fixed it. I was trying the converse. It seems to be one of those where you actually just go backwards from the original proof.
$endgroup$
– The Bosco
Dec 17 '18 at 12:33
add a comment |
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$begingroup$
If you write $f_l(x) = sum_{i=0}^l g_i(x)$ instead of $f_l(x) = sum_{i=0}^l f_i(x)$, then everything is correct.
And yes, the converse is also true. Try a proof !
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
$endgroup$
$begingroup$
Thank you! I fixed it. I was trying the converse. It seems to be one of those where you actually just go backwards from the original proof.
$endgroup$
– The Bosco
Dec 17 '18 at 12:33
add a comment |
$begingroup$
If you write $f_l(x) = sum_{i=0}^l g_i(x)$ instead of $f_l(x) = sum_{i=0}^l f_i(x)$, then everything is correct.
And yes, the converse is also true. Try a proof !
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
$endgroup$
$begingroup$
Thank you! I fixed it. I was trying the converse. It seems to be one of those where you actually just go backwards from the original proof.
$endgroup$
– The Bosco
Dec 17 '18 at 12:33
add a comment |
$begingroup$
If you write $f_l(x) = sum_{i=0}^l g_i(x)$ instead of $f_l(x) = sum_{i=0}^l f_i(x)$, then everything is correct.
And yes, the converse is also true. Try a proof !
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
$endgroup$
If you write $f_l(x) = sum_{i=0}^l g_i(x)$ instead of $f_l(x) = sum_{i=0}^l f_i(x)$, then everything is correct.
And yes, the converse is also true. Try a proof !
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
answered Dec 17 '18 at 12:28
FredFred
44.7k1846
44.7k1846
$begingroup$
Thank you! I fixed it. I was trying the converse. It seems to be one of those where you actually just go backwards from the original proof.
$endgroup$
– The Bosco
Dec 17 '18 at 12:33
add a comment |
$begingroup$
Thank you! I fixed it. I was trying the converse. It seems to be one of those where you actually just go backwards from the original proof.
$endgroup$
– The Bosco
Dec 17 '18 at 12:33
$begingroup$
Thank you! I fixed it. I was trying the converse. It seems to be one of those where you actually just go backwards from the original proof.
$endgroup$
– The Bosco
Dec 17 '18 at 12:33
$begingroup$
Thank you! I fixed it. I was trying the converse. It seems to be one of those where you actually just go backwards from the original proof.
$endgroup$
– The Bosco
Dec 17 '18 at 12:33
add a comment |
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