Mixing
Series and lim sup
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
In the course of some calculations, I've come upon the following question: When is it true that
$$ limsup_t to 0^+ sum_m |f_m(t)| leq sum_m sup_0 < t < 2^-m |f_m(t)|; ? $$
It's clear that this does not hold in general; one could take for instance $f_m(t) = t/m$, which makes the LHS infinite, while the RHS converges.
I've considered using the reverse Fatou's lemma/dominated convergence theorem, but I don't have a dominating function available in my context.
I'd like to conclude that the inequality holds when the LHS is finite, but would welcome counterexamples/other conditions which make it correct.
sequences-and-series limsup-and-liminf
asked Jul 25 at 16:09
ec92
1,112817
add a comment |Â
up vote
3
down vote
favorite
In the course of some calculations, I've come upon the following question: When is it true that
$$ limsup_t to 0^+ sum_m |f_m(t)| leq sum_m sup_0 < t < 2^-m |f_m(t)|; ? $$
It's clear that this does not hold in general; one could take for instance $f_m(t) = t/m$, which makes the LHS infinite, while the RHS converges.
I've considered using the reverse Fatou's lemma/dominated convergence theorem, but I don't have a dominating function available in my context.
I'd like to conclude that the inequality holds when the LHS is finite, but would welcome counterexamples/other conditions which make it correct.
sequences-and-series limsup-and-liminf
asked Jul 25 at 16:09
ec92
1,112817
Yes, thanks. Corrected.
– ec92
Jul 25 at 16:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In the course of some calculations, I've come upon the following question: When is it true that
$$ limsup_t to 0^+ sum_m |f_m(t)| leq sum_m sup_0 < t < 2^-m |f_m(t)|; ? $$
It's clear that this does not hold in general; one could take for instance $f_m(t) = t/m$, which makes the LHS infinite, while the RHS converges.
I've considered using the reverse Fatou's lemma/dominated convergence theorem, but I don't have a dominating function available in my context.
I'd like to conclude that the inequality holds when the LHS is finite, but would welcome counterexamples/other conditions which make it correct.
sequences-and-series limsup-and-liminf
asked Jul 25 at 16:09
ec92
1,112817
In the course of some calculations, I've come upon the following question: When is it true that
$$ limsup_t to 0^+ sum_m |f_m(t)| leq sum_m sup_0 < t < 2^-m |f_m(t)|; ? $$
It's clear that this does not hold in general; one could take for instance $f_m(t) = t/m$, which makes the LHS infinite, while the RHS converges.
I've considered using the reverse Fatou's lemma/dominated convergence theorem, but I don't have a dominating function available in my context.
I'd like to conclude that the inequality holds when the LHS is finite, but would welcome counterexamples/other conditions which make it correct.
sequences-and-series limsup-and-liminf
asked Jul 25 at 16:09
ec92
1,112817
edited Jul 25 at 16:36
asked Jul 25 at 16:09
ec92
1,112817
asked Jul 25 at 16:09
ec92
1,112817
asked Jul 25 at 16:09
ec92
1,112817
1,112817
Yes, thanks. Corrected.
– ec92
Jul 25 at 16:36
add a comment |Â
Yes, thanks. Corrected.
– ec92
Jul 25 at 16:36
Yes, thanks. Corrected.
– ec92
Jul 25 at 16:36
Yes, thanks. Corrected.
– ec92
Jul 25 at 16:36
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Of course it's true if, say, $|f_m(t)|$ is a decreasing function of $t$. But that's sort of stupid. I doubt that there is any "interesting" condition that makes this true, since the LHS involves $f_m(t)$ for pairs $(m,t)$ that simply don't come up on the RHS.
No, assuming the LHS is finite is not enough. Say $$f_m(t)=begincases
1,&(t=2^-m),
\0,&(tne 2^-m).endcases$$
Then LSH $=1$, RHS $=0$. You could convert this to a continuous counterexample: Say $f_m$ is a continuous function supported on $[2^-m,3cdot 2^-m]$ with $0le f_mle 1$ and $f_m(2cdot 2^-m)=1$.
answered Jul 25 at 16:51
David C. Ullrich
54k33481
add a comment |Â
up vote
1
down vote
This does not hold in general even if both sides are finite.
To build a counterexample, it suffices to put a significant weight on the part that the RHS does not see. I will build an example with non continuous functions, but it is not difficult to smooth them.
I assume $mgeq 1$.
First, define $f_m(t)$ on the interval $[0,2^-m)$ by $$f_m(t)=2^-mqquad textif $0leq t<2^-m$$$
Clearly, your RHS is equal to $1$. Now choose the remaining parts of the $f_m$ large enough so as to make the LHS bigger than $1$, but small enough to make it finite. For instance:
$$f_m(t)=cases1&if $2^-mleq t<2^-m+1$\0&if $2^-m+1leq t$$$
will result in the LHS being equal to $2$.
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Of course it's true if, say, $|f_m(t)|$ is a decreasing function of $t$. But that's sort of stupid. I doubt that there is any "interesting" condition that makes this true, since the LHS involves $f_m(t)$ for pairs $(m,t)$ that simply don't come up on the RHS.
No, assuming the LHS is finite is not enough. Say $$f_m(t)=begincases
1,&(t=2^-m),
\0,&(tne 2^-m).endcases$$
Then LSH $=1$, RHS $=0$. You could convert this to a continuous counterexample: Say $f_m$ is a continuous function supported on $[2^-m,3cdot 2^-m]$ with $0le f_mle 1$ and $f_m(2cdot 2^-m)=1$.
answered Jul 25 at 16:51
David C. Ullrich
54k33481
add a comment |Â
up vote
2
down vote
Of course it's true if, say, $|f_m(t)|$ is a decreasing function of $t$. But that's sort of stupid. I doubt that there is any "interesting" condition that makes this true, since the LHS involves $f_m(t)$ for pairs $(m,t)$ that simply don't come up on the RHS.
No, assuming the LHS is finite is not enough. Say $$f_m(t)=begincases
1,&(t=2^-m),
\0,&(tne 2^-m).endcases$$
Then LSH $=1$, RHS $=0$. You could convert this to a continuous counterexample: Say $f_m$ is a continuous function supported on $[2^-m,3cdot 2^-m]$ with $0le f_mle 1$ and $f_m(2cdot 2^-m)=1$.
answered Jul 25 at 16:51
David C. Ullrich
54k33481
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Of course it's true if, say, $|f_m(t)|$ is a decreasing function of $t$. But that's sort of stupid. I doubt that there is any "interesting" condition that makes this true, since the LHS involves $f_m(t)$ for pairs $(m,t)$ that simply don't come up on the RHS.
No, assuming the LHS is finite is not enough. Say $$f_m(t)=begincases
1,&(t=2^-m),
\0,&(tne 2^-m).endcases$$
Then LSH $=1$, RHS $=0$. You could convert this to a continuous counterexample: Say $f_m$ is a continuous function supported on $[2^-m,3cdot 2^-m]$ with $0le f_mle 1$ and $f_m(2cdot 2^-m)=1$.
answered Jul 25 at 16:51
David C. Ullrich
54k33481
Of course it's true if, say, $|f_m(t)|$ is a decreasing function of $t$. But that's sort of stupid. I doubt that there is any "interesting" condition that makes this true, since the LHS involves $f_m(t)$ for pairs $(m,t)$ that simply don't come up on the RHS.
No, assuming the LHS is finite is not enough. Say $$f_m(t)=begincases
1,&(t=2^-m),
\0,&(tne 2^-m).endcases$$
Then LSH $=1$, RHS $=0$. You could convert this to a continuous counterexample: Say $f_m$ is a continuous function supported on $[2^-m,3cdot 2^-m]$ with $0le f_mle 1$ and $f_m(2cdot 2^-m)=1$.
answered Jul 25 at 16:51
David C. Ullrich
54k33481
answered Jul 25 at 16:51
David C. Ullrich
54k33481
answered Jul 25 at 16:51
David C. Ullrich
54k33481
answered Jul 25 at 16:51
David C. Ullrich
54k33481
54k33481
add a comment |Â
add a comment |Â
up vote
1
down vote
This does not hold in general even if both sides are finite.
To build a counterexample, it suffices to put a significant weight on the part that the RHS does not see. I will build an example with non continuous functions, but it is not difficult to smooth them.
I assume $mgeq 1$.
First, define $f_m(t)$ on the interval $[0,2^-m)$ by $$f_m(t)=2^-mqquad textif $0leq t<2^-m$$$
Clearly, your RHS is equal to $1$. Now choose the remaining parts of the $f_m$ large enough so as to make the LHS bigger than $1$, but small enough to make it finite. For instance:
$$f_m(t)=cases1&if $2^-mleq t<2^-m+1$\0&if $2^-m+1leq t$$$
will result in the LHS being equal to $2$.
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
add a comment |Â
up vote
1
down vote
This does not hold in general even if both sides are finite.
To build a counterexample, it suffices to put a significant weight on the part that the RHS does not see. I will build an example with non continuous functions, but it is not difficult to smooth them.
I assume $mgeq 1$.
First, define $f_m(t)$ on the interval $[0,2^-m)$ by $$f_m(t)=2^-mqquad textif $0leq t<2^-m$$$
Clearly, your RHS is equal to $1$. Now choose the remaining parts of the $f_m$ large enough so as to make the LHS bigger than $1$, but small enough to make it finite. For instance:
$$f_m(t)=cases1&if $2^-mleq t<2^-m+1$\0&if $2^-m+1leq t$$$
will result in the LHS being equal to $2$.
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This does not hold in general even if both sides are finite.
To build a counterexample, it suffices to put a significant weight on the part that the RHS does not see. I will build an example with non continuous functions, but it is not difficult to smooth them.
I assume $mgeq 1$.
First, define $f_m(t)$ on the interval $[0,2^-m)$ by $$f_m(t)=2^-mqquad textif $0leq t<2^-m$$$
Clearly, your RHS is equal to $1$. Now choose the remaining parts of the $f_m$ large enough so as to make the LHS bigger than $1$, but small enough to make it finite. For instance:
$$f_m(t)=cases1&if $2^-mleq t<2^-m+1$\0&if $2^-m+1leq t$$$
will result in the LHS being equal to $2$.
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
This does not hold in general even if both sides are finite.
To build a counterexample, it suffices to put a significant weight on the part that the RHS does not see. I will build an example with non continuous functions, but it is not difficult to smooth them.
I assume $mgeq 1$.
First, define $f_m(t)$ on the interval $[0,2^-m)$ by $$f_m(t)=2^-mqquad textif $0leq t<2^-m$$$
Clearly, your RHS is equal to $1$. Now choose the remaining parts of the $f_m$ large enough so as to make the LHS bigger than $1$, but small enough to make it finite. For instance:
$$f_m(t)=cases1&if $2^-mleq t<2^-m+1$\0&if $2^-m+1leq t$$$
will result in the LHS being equal to $2$.
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
answered Jul 25 at 16:52
Arnaud Mortier
18.8k22159
18.8k22159
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862567%2fseries-and-lim-sup%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Yes, thanks. Corrected.
– ec92
Jul 25 at 16:36