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Why this highly oscillatory function is not Lebesgue integrable?
Showing that the integral of $x^nf(x)=0$ where $f$ is Lebesgue Integrable.Find a non-negative function on [0,1] such that $tcdot m(x:f(x) geq t) to 0$ that is not Lebesgue IntegrableLebesgue-integrable and existence of integralWhy are some convergent Lebesgue integrals 'undefined'?Show a function is Lebesgue integrableAre $1/sqrtx$ or $1/x$ Lebesgue integrable on $(0,1)$? If so, why?Showing $f$ is Lebesgue integrableIs the function $f(x,y)$ integrable (Fubini Tonelli).Show that $int_0^1 fracxlog x(1+x)^2 dx$ is Lebesgue integrableLimes of lebesgue-integrable functions
$begingroup$
I am reading the following note:
https://arxiv.org/pdf/1309.3112v1.pdf
Please see Ch. 1.3 (p.6~p.7).
On p.7, it says the infimum is not attained with a control law $u(t)$ belonging to the space of Lebesgue integrable functions.
But for $tin [0,1]$, $$int_0^1 |u_k(t)| dt < infty,$$ for $krightarrow infty $. I am confused why $u(t)$ may not be Lebesgue integrable.
real-analysis measure-theory lebesgue-integral
edited 22 hours ago
J.G.
32.2k23250
asked 22 hours ago
sleeve chensleeve chen
3,15742155
$endgroup$
|
show 2 more comments
$begingroup$
I am reading the following note:
https://arxiv.org/pdf/1309.3112v1.pdf
Please see Ch. 1.3 (p.6~p.7).
On p.7, it says the infimum is not attained with a control law $u(t)$ belonging to the space of Lebesgue integrable functions.
But for $tin [0,1]$, $$int_0^1 |u_k(t)| dt < infty,$$ for $krightarrow infty $. I am confused why $u(t)$ may not be Lebesgue integrable.
real-analysis measure-theory lebesgue-integral
edited 22 hours ago
J.G.
32.2k23250
asked 22 hours ago
sleeve chensleeve chen
3,15742155
$endgroup$
$begingroup$
p.6-p.7 not available in the link.
$endgroup$
– Kavi Rama Murthy
22 hours ago
$begingroup$
@KaviRamaMurthy Just fix it. Thanks!
$endgroup$
– sleeve chen
22 hours ago
1
$begingroup$
I've fixed the link, but you should make your question self-contained. I'd do it for you if I wasn't about to go offline.
$endgroup$
– J.G.
22 hours ago
$begingroup$
The limit function $u$ is integrable, however it is not a minimizer.
$endgroup$
– daw
22 hours ago
$begingroup$
What do you mean when you refer to $u(t)$ in the last sentence? You have not actually defined any function $u(t)$, nor does the text define one.
$endgroup$
– Eric Wofsey
22 hours ago
|
show 2 more comments
$begingroup$
I am reading the following note:
https://arxiv.org/pdf/1309.3112v1.pdf
Please see Ch. 1.3 (p.6~p.7).
On p.7, it says the infimum is not attained with a control law $u(t)$ belonging to the space of Lebesgue integrable functions.
But for $tin [0,1]$, $$int_0^1 |u_k(t)| dt < infty,$$ for $krightarrow infty $. I am confused why $u(t)$ may not be Lebesgue integrable.
real-analysis measure-theory lebesgue-integral
edited 22 hours ago
J.G.
32.2k23250
asked 22 hours ago
sleeve chensleeve chen
3,15742155
$endgroup$
I am reading the following note:
https://arxiv.org/pdf/1309.3112v1.pdf
Please see Ch. 1.3 (p.6~p.7).
On p.7, it says the infimum is not attained with a control law $u(t)$ belonging to the space of Lebesgue integrable functions.
But for $tin [0,1]$, $$int_0^1 |u_k(t)| dt < infty,$$ for $krightarrow infty $. I am confused why $u(t)$ may not be Lebesgue integrable.
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
edited 22 hours ago
J.G.
32.2k23250
asked 22 hours ago
sleeve chensleeve chen
3,15742155
edited 22 hours ago
J.G.
32.2k23250
asked 22 hours ago
sleeve chensleeve chen
3,15742155
edited 22 hours ago
J.G.
32.2k23250
edited 22 hours ago
J.G.
32.2k23250
edited 22 hours ago
J.G.
32.2k23250
32.2k23250
asked 22 hours ago
sleeve chensleeve chen
3,15742155
asked 22 hours ago
sleeve chensleeve chen
3,15742155
asked 22 hours ago
sleeve chensleeve chen
3,15742155
3,15742155
$begingroup$
p.6-p.7 not available in the link.
$endgroup$
– Kavi Rama Murthy
22 hours ago
$begingroup$
@KaviRamaMurthy Just fix it. Thanks!
$endgroup$
– sleeve chen
22 hours ago
1
$begingroup$
I've fixed the link, but you should make your question self-contained. I'd do it for you if I wasn't about to go offline.
$endgroup$
– J.G.
22 hours ago
$begingroup$
The limit function $u$ is integrable, however it is not a minimizer.
$endgroup$
– daw
22 hours ago
$begingroup$
What do you mean when you refer to $u(t)$ in the last sentence? You have not actually defined any function $u(t)$, nor does the text define one.
$endgroup$
– Eric Wofsey
22 hours ago
|
show 2 more comments
$begingroup$
p.6-p.7 not available in the link.
$endgroup$
– Kavi Rama Murthy
22 hours ago
$begingroup$
@KaviRamaMurthy Just fix it. Thanks!
$endgroup$
– sleeve chen
22 hours ago
1
$begingroup$
I've fixed the link, but you should make your question self-contained. I'd do it for you if I wasn't about to go offline.
$endgroup$
– J.G.
22 hours ago
$begingroup$
The limit function $u$ is integrable, however it is not a minimizer.
$endgroup$
– daw
22 hours ago
$begingroup$
What do you mean when you refer to $u(t)$ in the last sentence? You have not actually defined any function $u(t)$, nor does the text define one.
$endgroup$
– Eric Wofsey
22 hours ago
$begingroup$
p.6-p.7 not available in the link.
$endgroup$
– Kavi Rama Murthy
22 hours ago
$begingroup$
p.6-p.7 not available in the link.
$endgroup$
– Kavi Rama Murthy
22 hours ago
$begingroup$
@KaviRamaMurthy Just fix it. Thanks!
$endgroup$
– sleeve chen
22 hours ago
$begingroup$
@KaviRamaMurthy Just fix it. Thanks!
$endgroup$
– sleeve chen
22 hours ago
1
1
$begingroup$
I've fixed the link, but you should make your question self-contained. I'd do it for you if I wasn't about to go offline.
$endgroup$
– J.G.
22 hours ago
$begingroup$
I've fixed the link, but you should make your question self-contained. I'd do it for you if I wasn't about to go offline.
$endgroup$
– J.G.
22 hours ago
$begingroup$
The limit function $u$ is integrable, however it is not a minimizer.
$endgroup$
– daw
22 hours ago
$begingroup$
The limit function $u$ is integrable, however it is not a minimizer.
$endgroup$
– daw
22 hours ago
$begingroup$
What do you mean when you refer to $u(t)$ in the last sentence? You have not actually defined any function $u(t)$, nor does the text define one.
$endgroup$
– Eric Wofsey
22 hours ago
$begingroup$
What do you mean when you refer to $u(t)$ in the last sentence? You have not actually defined any function $u(t)$, nor does the text define one.
$endgroup$
– Eric Wofsey
22 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The minimizing sequence $u_k$ is bounded in $L^1$, oscillatory, which means that it converges weakly to the zero function. However, the functional is not weakly lower semi-continuous with respect to $u$, so in the limit the function $u$ is not a minimizer.
This shows that the infimum of the functional is zero. If there is $(x,u)$ such that the functional is zero, then necessarily $xequiv 0$, $|u|equiv 1$, a contradiction to $dot x = u$.
answered 22 hours ago
dawdaw
24.8k1745
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The minimizing sequence $u_k$ is bounded in $L^1$, oscillatory, which means that it converges weakly to the zero function. However, the functional is not weakly lower semi-continuous with respect to $u$, so in the limit the function $u$ is not a minimizer.
This shows that the infimum of the functional is zero. If there is $(x,u)$ such that the functional is zero, then necessarily $xequiv 0$, $|u|equiv 1$, a contradiction to $dot x = u$.
answered 22 hours ago
dawdaw
24.8k1745
$endgroup$
add a comment |
$begingroup$
The minimizing sequence $u_k$ is bounded in $L^1$, oscillatory, which means that it converges weakly to the zero function. However, the functional is not weakly lower semi-continuous with respect to $u$, so in the limit the function $u$ is not a minimizer.
This shows that the infimum of the functional is zero. If there is $(x,u)$ such that the functional is zero, then necessarily $xequiv 0$, $|u|equiv 1$, a contradiction to $dot x = u$.
answered 22 hours ago
dawdaw
24.8k1745
$endgroup$
add a comment |
$begingroup$
The minimizing sequence $u_k$ is bounded in $L^1$, oscillatory, which means that it converges weakly to the zero function. However, the functional is not weakly lower semi-continuous with respect to $u$, so in the limit the function $u$ is not a minimizer.
This shows that the infimum of the functional is zero. If there is $(x,u)$ such that the functional is zero, then necessarily $xequiv 0$, $|u|equiv 1$, a contradiction to $dot x = u$.
answered 22 hours ago
dawdaw
24.8k1745
$endgroup$
The minimizing sequence $u_k$ is bounded in $L^1$, oscillatory, which means that it converges weakly to the zero function. However, the functional is not weakly lower semi-continuous with respect to $u$, so in the limit the function $u$ is not a minimizer.
This shows that the infimum of the functional is zero. If there is $(x,u)$ such that the functional is zero, then necessarily $xequiv 0$, $|u|equiv 1$, a contradiction to $dot x = u$.
answered 22 hours ago
dawdaw
24.8k1745
answered 22 hours ago
dawdaw
24.8k1745
answered 22 hours ago
dawdaw
24.8k1745
answered 22 hours ago
dawdaw
24.8k1745
24.8k1745
add a comment |
add a comment |
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$begingroup$
p.6-p.7 not available in the link.
$endgroup$
– Kavi Rama Murthy
22 hours ago
$begingroup$
@KaviRamaMurthy Just fix it. Thanks!
$endgroup$
– sleeve chen
22 hours ago
1
$begingroup$
I've fixed the link, but you should make your question self-contained. I'd do it for you if I wasn't about to go offline.
$endgroup$
– J.G.
22 hours ago
$begingroup$
The limit function $u$ is integrable, however it is not a minimizer.
$endgroup$
– daw
22 hours ago
$begingroup$
What do you mean when you refer to $u(t)$ in the last sentence? You have not actually defined any function $u(t)$, nor does the text define one.
$endgroup$
– Eric Wofsey
22 hours ago