Mixing
Sup inequalites
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This is a follow-up to the question here with, hopefully, the correct inequalities.
I am trying to check if these following inequalities are true
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| leq
sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
and
$$
sup_0 < s < 1 left| int_0^s f(u) d mu(u) right| leq
sup_0 < s < 1/2left| int_0^s f(u) d mu(u) right|
+ sup_1/2 < s < 1 left| int_1/2^s f(u) d mu(u) right|.
$$
where $mu$ is the Lebesgue measure. Unlike my previous question here, the absolute values are inside the sup.
I think both inequalities are true in general. I argue about them as follows: assume that the maximum is at $s^*$, then
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| =
left| f(s^*) + g(s^*) right| leq
left| f(s^*) right| + left| g(s^*) right|
leq sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
I can also argue for the second inequality using a similar idea by considering the two cases: when $s^*$ is in $[0,1/2]$ and when it's in $[1/2,1]$. However, I don't think this proof is correct in general, since such an $s^*$ might not exist.
How would I go about proving these inequalities in general?
Also, is the inequality with $mu$ true for other measures as well?
inequality supremum-and-infimum
asked Jul 31 at 9:12
Tohiko
1245
add a comment |Â
up vote
1
down vote
favorite
This is a follow-up to the question here with, hopefully, the correct inequalities.
I am trying to check if these following inequalities are true
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| leq
sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
and
$$
sup_0 < s < 1 left| int_0^s f(u) d mu(u) right| leq
sup_0 < s < 1/2left| int_0^s f(u) d mu(u) right|
+ sup_1/2 < s < 1 left| int_1/2^s f(u) d mu(u) right|.
$$
where $mu$ is the Lebesgue measure. Unlike my previous question here, the absolute values are inside the sup.
I think both inequalities are true in general. I argue about them as follows: assume that the maximum is at $s^*$, then
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| =
left| f(s^*) + g(s^*) right| leq
left| f(s^*) right| + left| g(s^*) right|
leq sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
I can also argue for the second inequality using a similar idea by considering the two cases: when $s^*$ is in $[0,1/2]$ and when it's in $[1/2,1]$. However, I don't think this proof is correct in general, since such an $s^*$ might not exist.
How would I go about proving these inequalities in general?
Also, is the inequality with $mu$ true for other measures as well?
inequality supremum-and-infimum
asked Jul 31 at 9:12
Tohiko
1245
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is a follow-up to the question here with, hopefully, the correct inequalities.
I am trying to check if these following inequalities are true
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| leq
sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
and
$$
sup_0 < s < 1 left| int_0^s f(u) d mu(u) right| leq
sup_0 < s < 1/2left| int_0^s f(u) d mu(u) right|
+ sup_1/2 < s < 1 left| int_1/2^s f(u) d mu(u) right|.
$$
where $mu$ is the Lebesgue measure. Unlike my previous question here, the absolute values are inside the sup.
I think both inequalities are true in general. I argue about them as follows: assume that the maximum is at $s^*$, then
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| =
left| f(s^*) + g(s^*) right| leq
left| f(s^*) right| + left| g(s^*) right|
leq sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
I can also argue for the second inequality using a similar idea by considering the two cases: when $s^*$ is in $[0,1/2]$ and when it's in $[1/2,1]$. However, I don't think this proof is correct in general, since such an $s^*$ might not exist.
How would I go about proving these inequalities in general?
Also, is the inequality with $mu$ true for other measures as well?
inequality supremum-and-infimum
asked Jul 31 at 9:12
Tohiko
1245
This is a follow-up to the question here with, hopefully, the correct inequalities.
I am trying to check if these following inequalities are true
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| leq
sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
and
$$
sup_0 < s < 1 left| int_0^s f(u) d mu(u) right| leq
sup_0 < s < 1/2left| int_0^s f(u) d mu(u) right|
+ sup_1/2 < s < 1 left| int_1/2^s f(u) d mu(u) right|.
$$
where $mu$ is the Lebesgue measure. Unlike my previous question here, the absolute values are inside the sup.
I think both inequalities are true in general. I argue about them as follows: assume that the maximum is at $s^*$, then
$$
sup_0 < s < 1 left| (f(s) + g(s)) right| =
left| f(s^*) + g(s^*) right| leq
left| f(s^*) right| + left| g(s^*) right|
leq sup_0 < s < 1 left| f(s) right| + sup_0 < s < 1 left| g(s) right|
$$
I can also argue for the second inequality using a similar idea by considering the two cases: when $s^*$ is in $[0,1/2]$ and when it's in $[1/2,1]$. However, I don't think this proof is correct in general, since such an $s^*$ might not exist.
How would I go about proving these inequalities in general?
Also, is the inequality with $mu$ true for other measures as well?
inequality supremum-and-infimum
asked Jul 31 at 9:12
Tohiko
1245
asked Jul 31 at 9:12
Tohiko
1245
asked Jul 31 at 9:12
Tohiko
1245
asked Jul 31 at 9:12
Tohiko
1245
1245
add a comment |Â
add a comment |Â
1 Answer
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Your argument is wrong because the two supremums may be attained at different points. A correct proof is as follows: $|f(t)+g(t)| leq |f(t)|+|g(t)| leq sup : 0<s<1+sup g(s) $. This is true for all $t$. Take sup over $t$ to get the first inequality. The second one is similar since the integral over $(0,s)$ splits into integral over $(0,1/2)$ and $(1/2,s)$ is $s >1/2$ (and you can bound by just the first term on the right side when $s<1/2$.
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your argument is wrong because the two supremums may be attained at different points. A correct proof is as follows: $|f(t)+g(t)| leq |f(t)|+|g(t)| leq sup : 0<s<1+sup g(s) $. This is true for all $t$. Take sup over $t$ to get the first inequality. The second one is similar since the integral over $(0,s)$ splits into integral over $(0,1/2)$ and $(1/2,s)$ is $s >1/2$ (and you can bound by just the first term on the right side when $s<1/2$.
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
2
down vote
accepted
Your argument is wrong because the two supremums may be attained at different points. A correct proof is as follows: $|f(t)+g(t)| leq |f(t)|+|g(t)| leq sup : 0<s<1+sup g(s) $. This is true for all $t$. Take sup over $t$ to get the first inequality. The second one is similar since the integral over $(0,s)$ splits into integral over $(0,1/2)$ and $(1/2,s)$ is $s >1/2$ (and you can bound by just the first term on the right side when $s<1/2$.
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your argument is wrong because the two supremums may be attained at different points. A correct proof is as follows: $|f(t)+g(t)| leq |f(t)|+|g(t)| leq sup : 0<s<1+sup g(s) $. This is true for all $t$. Take sup over $t$ to get the first inequality. The second one is similar since the integral over $(0,s)$ splits into integral over $(0,1/2)$ and $(1/2,s)$ is $s >1/2$ (and you can bound by just the first term on the right side when $s<1/2$.
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
Your argument is wrong because the two supremums may be attained at different points. A correct proof is as follows: $|f(t)+g(t)| leq |f(t)|+|g(t)| leq sup : 0<s<1+sup g(s) $. This is true for all $t$. Take sup over $t$ to get the first inequality. The second one is similar since the integral over $(0,s)$ splits into integral over $(0,1/2)$ and $(1/2,s)$ is $s >1/2$ (and you can bound by just the first term on the right side when $s<1/2$.
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 9:22
Kavi Rama Murthy
19.5k2829
19.5k2829
add a comment |Â
add a comment |Â
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