Inequality for a function 4
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Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that
$$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.
Is it correct that
$$u(t)leqfrac1sqrt N+frac1N^frac14$$
for every $tgeq 0$?
How could I prove that?
real-analysis analysis inequality
edited Aug 26 at 21:12
zahbaz
7,65521636
asked Aug 26 at 21:01
user495333
816
add a comment |Â
up vote
1
down vote
favorite
Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that
$$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.
Is it correct that
$$u(t)leqfrac1sqrt N+frac1N^frac14$$
for every $tgeq 0$?
How could I prove that?
real-analysis analysis inequality
edited Aug 26 at 21:12
zahbaz
7,65521636
asked Aug 26 at 21:01
user495333
816
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that
$$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.
Is it correct that
$$u(t)leqfrac1sqrt N+frac1N^frac14$$
for every $tgeq 0$?
How could I prove that?
real-analysis analysis inequality
edited Aug 26 at 21:12
zahbaz
7,65521636
asked Aug 26 at 21:01
user495333
816
Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that
$$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.
Is it correct that
$$u(t)leqfrac1sqrt N+frac1N^frac14$$
for every $tgeq 0$?
How could I prove that?
real-analysis analysis inequality
edited Aug 26 at 21:12
zahbaz
7,65521636
asked Aug 26 at 21:01
user495333
816
edited Aug 26 at 21:12
zahbaz
7,65521636
edited Aug 26 at 21:12
zahbaz
7,65521636
edited Aug 26 at 21:12
zahbaz
7,65521636
7,65521636
asked Aug 26 at 21:01
user495333
816
asked Aug 26 at 21:01
user495333
816
asked Aug 26 at 21:01
user495333
816
816
add a comment |Â
add a comment |Â
1 Answer
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The inequality does not hold. As a counterexample, consider
$$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
Then one can show that $u$ satisfies the first inequality for $N = 1$:
$$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
and
$$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$
answered Aug 26 at 21:42
Trevor Norton
1,192410
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The inequality does not hold. As a counterexample, consider
$$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
Then one can show that $u$ satisfies the first inequality for $N = 1$:
$$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
and
$$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$
answered Aug 26 at 21:42
Trevor Norton
1,192410
add a comment |Â
up vote
0
down vote
The inequality does not hold. As a counterexample, consider
$$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
Then one can show that $u$ satisfies the first inequality for $N = 1$:
$$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
and
$$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$
answered Aug 26 at 21:42
Trevor Norton
1,192410
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The inequality does not hold. As a counterexample, consider
$$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
Then one can show that $u$ satisfies the first inequality for $N = 1$:
$$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
and
$$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$
answered Aug 26 at 21:42
Trevor Norton
1,192410
The inequality does not hold. As a counterexample, consider
$$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
Then one can show that $u$ satisfies the first inequality for $N = 1$:
$$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
and
$$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$
answered Aug 26 at 21:42
Trevor Norton
1,192410
answered Aug 26 at 21:42
Trevor Norton
1,192410
answered Aug 26 at 21:42
Trevor Norton
1,192410
answered Aug 26 at 21:42
Trevor Norton
1,192410
1,192410
add a comment |Â
add a comment |Â
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