Limits of a multiple integral function
$begingroup$
Problem
Let $f(x)in L^{1}(mathbb{R}^N)cap L^{infty}(mathbb{R}^N)$ and $S_t=left{xinmathbb{R}^N: |x_1|le tright}$ with $t>0$. Let $phi(t)$ the integral function $$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx$$
Find $$lim_{tto 0^{+}}phi(t) mbox{and} lim_{tto +infty}phi(t)$$
My approach
Since $f(x)in L^{1}(mathbb{R}^{N})cap L^{infty}(mathbb{R}^N)$ then $f(x)in L^{2}(mathbb{R}^{N})$.
Infact, by definition of integral
$$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx=int_{mathbb{R}^{N}}chi_{mathbb{R}^Nsetminus S_{t}}(x)|f(x)|^2dxle\ \ le int_{mathbb{R}^{N}}|f(x)|^2dxle |f|_{infty}int_{mathbb{R}^{N}}|f(x)|dx=|f|_{infty}|f|_{1}<+infty$$
This means that $phi(t)$ is a well posed and bounded function:
$$0le phi(t)le |f|_{infty}|f|_{1} forall tin (0,+infty)$$
Note that for $0<t_1<t_2$, $S_{t_1}subset S_{t_2} impliesmathbb{R}^Nsetminus S_{t_1}supsetmathbb{R}^Nsetminus S_{t_2}$ so:
$$phi(t_1)=int_{mathbb{R}^{N}setminus S_{t_1}}|f(x)|^2dxge int_{mathbb{R}^{N}setminus S_{t_2}}|f(x)|^2dx=phi(t_2)$$
Hence $phi(t)$ is a bounded decreasing function, so $lim_{tto 0^{+}}phi(t)$ and $lim_{tto +infty}phi(t)$ exist and are finite.
Now, I would like to use the sequential criterion for a limit of a function:
$$lim_{tto 0^{+}}phi(t)=lim_{nto +infty}phileft(frac{1}{n}right)=lim_{nto +infty}int_{mathbb{R}^{N}setminus S_{frac{1}{n}}}|f(x)|^2dx=\ \ \ =lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx$$
Since $chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)le 1$ the sequence
$$g_{n}(x):=chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2le |f(x)|^2 forall xinmathbb{R}^{N}$$
I can use the dominated convergence theorem
$$lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}lim_{nto+infty}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}|f(x)|^2dx=|f|_{2}^2$$
For the limit $lim_{tto +infty}phi(t)$ I use the sequence $t_{n}=n$ so
$$lim_{tto +infty}phi(t)=lim_{nto +infty}int_{mathbb{R}^N}chi_{mathbb{R}^{N}setminus S_{n}}(x)|f(x)|^2dx=int_{mathbb{R}^N}overbrace{lim_{nto +infty}chi_{mathbb{R}^{N}setminus S_{n}}(x)}^{=0}|f(x)|^2dx=0$$
Is this approach ok? Thanks.
real-analysis integration functional-analysis measure-theory lp-spaces
asked Dec 12 '18 at 11:11
IxionIxion
768621
$endgroup$
add a comment |
$begingroup$
Problem
Let $f(x)in L^{1}(mathbb{R}^N)cap L^{infty}(mathbb{R}^N)$ and $S_t=left{xinmathbb{R}^N: |x_1|le tright}$ with $t>0$. Let $phi(t)$ the integral function $$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx$$
Find $$lim_{tto 0^{+}}phi(t) mbox{and} lim_{tto +infty}phi(t)$$
My approach
Since $f(x)in L^{1}(mathbb{R}^{N})cap L^{infty}(mathbb{R}^N)$ then $f(x)in L^{2}(mathbb{R}^{N})$.
Infact, by definition of integral
$$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx=int_{mathbb{R}^{N}}chi_{mathbb{R}^Nsetminus S_{t}}(x)|f(x)|^2dxle\ \ le int_{mathbb{R}^{N}}|f(x)|^2dxle |f|_{infty}int_{mathbb{R}^{N}}|f(x)|dx=|f|_{infty}|f|_{1}<+infty$$
This means that $phi(t)$ is a well posed and bounded function:
$$0le phi(t)le |f|_{infty}|f|_{1} forall tin (0,+infty)$$
Note that for $0<t_1<t_2$, $S_{t_1}subset S_{t_2} impliesmathbb{R}^Nsetminus S_{t_1}supsetmathbb{R}^Nsetminus S_{t_2}$ so:
$$phi(t_1)=int_{mathbb{R}^{N}setminus S_{t_1}}|f(x)|^2dxge int_{mathbb{R}^{N}setminus S_{t_2}}|f(x)|^2dx=phi(t_2)$$
Hence $phi(t)$ is a bounded decreasing function, so $lim_{tto 0^{+}}phi(t)$ and $lim_{tto +infty}phi(t)$ exist and are finite.
Now, I would like to use the sequential criterion for a limit of a function:
$$lim_{tto 0^{+}}phi(t)=lim_{nto +infty}phileft(frac{1}{n}right)=lim_{nto +infty}int_{mathbb{R}^{N}setminus S_{frac{1}{n}}}|f(x)|^2dx=\ \ \ =lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx$$
Since $chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)le 1$ the sequence
$$g_{n}(x):=chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2le |f(x)|^2 forall xinmathbb{R}^{N}$$
I can use the dominated convergence theorem
$$lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}lim_{nto+infty}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}|f(x)|^2dx=|f|_{2}^2$$
For the limit $lim_{tto +infty}phi(t)$ I use the sequence $t_{n}=n$ so
$$lim_{tto +infty}phi(t)=lim_{nto +infty}int_{mathbb{R}^N}chi_{mathbb{R}^{N}setminus S_{n}}(x)|f(x)|^2dx=int_{mathbb{R}^N}overbrace{lim_{nto +infty}chi_{mathbb{R}^{N}setminus S_{n}}(x)}^{=0}|f(x)|^2dx=0$$
Is this approach ok? Thanks.
real-analysis integration functional-analysis measure-theory lp-spaces
asked Dec 12 '18 at 11:11
IxionIxion
768621
$endgroup$
1
$begingroup$
Your argument is correct: the proof of the boundedness of $phi(t)$ is compact and elegant, while the calculation of the limits is a bit more lengthy but formally correct.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 19:39
$begingroup$
@DanieleTampieri: thank you! :) I would like to find the "fastest" way to evaluate the limits. Maybe there exists a theorem that could be useful...? Could you give me an hint?
$endgroup$
– Ixion
Dec 12 '18 at 21:09
$begingroup$
Mmm, let me see what can I do. I'll look at it.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 21:16
add a comment |
$begingroup$
Problem
Let $f(x)in L^{1}(mathbb{R}^N)cap L^{infty}(mathbb{R}^N)$ and $S_t=left{xinmathbb{R}^N: |x_1|le tright}$ with $t>0$. Let $phi(t)$ the integral function $$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx$$
Find $$lim_{tto 0^{+}}phi(t) mbox{and} lim_{tto +infty}phi(t)$$
My approach
Since $f(x)in L^{1}(mathbb{R}^{N})cap L^{infty}(mathbb{R}^N)$ then $f(x)in L^{2}(mathbb{R}^{N})$.
Infact, by definition of integral
$$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx=int_{mathbb{R}^{N}}chi_{mathbb{R}^Nsetminus S_{t}}(x)|f(x)|^2dxle\ \ le int_{mathbb{R}^{N}}|f(x)|^2dxle |f|_{infty}int_{mathbb{R}^{N}}|f(x)|dx=|f|_{infty}|f|_{1}<+infty$$
This means that $phi(t)$ is a well posed and bounded function:
$$0le phi(t)le |f|_{infty}|f|_{1} forall tin (0,+infty)$$
Note that for $0<t_1<t_2$, $S_{t_1}subset S_{t_2} impliesmathbb{R}^Nsetminus S_{t_1}supsetmathbb{R}^Nsetminus S_{t_2}$ so:
$$phi(t_1)=int_{mathbb{R}^{N}setminus S_{t_1}}|f(x)|^2dxge int_{mathbb{R}^{N}setminus S_{t_2}}|f(x)|^2dx=phi(t_2)$$
Hence $phi(t)$ is a bounded decreasing function, so $lim_{tto 0^{+}}phi(t)$ and $lim_{tto +infty}phi(t)$ exist and are finite.
Now, I would like to use the sequential criterion for a limit of a function:
$$lim_{tto 0^{+}}phi(t)=lim_{nto +infty}phileft(frac{1}{n}right)=lim_{nto +infty}int_{mathbb{R}^{N}setminus S_{frac{1}{n}}}|f(x)|^2dx=\ \ \ =lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx$$
Since $chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)le 1$ the sequence
$$g_{n}(x):=chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2le |f(x)|^2 forall xinmathbb{R}^{N}$$
I can use the dominated convergence theorem
$$lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}lim_{nto+infty}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}|f(x)|^2dx=|f|_{2}^2$$
For the limit $lim_{tto +infty}phi(t)$ I use the sequence $t_{n}=n$ so
$$lim_{tto +infty}phi(t)=lim_{nto +infty}int_{mathbb{R}^N}chi_{mathbb{R}^{N}setminus S_{n}}(x)|f(x)|^2dx=int_{mathbb{R}^N}overbrace{lim_{nto +infty}chi_{mathbb{R}^{N}setminus S_{n}}(x)}^{=0}|f(x)|^2dx=0$$
Is this approach ok? Thanks.
real-analysis integration functional-analysis measure-theory lp-spaces
asked Dec 12 '18 at 11:11
IxionIxion
768621
$endgroup$
Problem
Let $f(x)in L^{1}(mathbb{R}^N)cap L^{infty}(mathbb{R}^N)$ and $S_t=left{xinmathbb{R}^N: |x_1|le tright}$ with $t>0$. Let $phi(t)$ the integral function $$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx$$
Find $$lim_{tto 0^{+}}phi(t) mbox{and} lim_{tto +infty}phi(t)$$
My approach
Since $f(x)in L^{1}(mathbb{R}^{N})cap L^{infty}(mathbb{R}^N)$ then $f(x)in L^{2}(mathbb{R}^{N})$.
Infact, by definition of integral
$$phi(t)=int_{mathbb{R}^Nsetminus S_t}|f(x)|^2dx=int_{mathbb{R}^{N}}chi_{mathbb{R}^Nsetminus S_{t}}(x)|f(x)|^2dxle\ \ le int_{mathbb{R}^{N}}|f(x)|^2dxle |f|_{infty}int_{mathbb{R}^{N}}|f(x)|dx=|f|_{infty}|f|_{1}<+infty$$
This means that $phi(t)$ is a well posed and bounded function:
$$0le phi(t)le |f|_{infty}|f|_{1} forall tin (0,+infty)$$
Note that for $0<t_1<t_2$, $S_{t_1}subset S_{t_2} impliesmathbb{R}^Nsetminus S_{t_1}supsetmathbb{R}^Nsetminus S_{t_2}$ so:
$$phi(t_1)=int_{mathbb{R}^{N}setminus S_{t_1}}|f(x)|^2dxge int_{mathbb{R}^{N}setminus S_{t_2}}|f(x)|^2dx=phi(t_2)$$
Hence $phi(t)$ is a bounded decreasing function, so $lim_{tto 0^{+}}phi(t)$ and $lim_{tto +infty}phi(t)$ exist and are finite.
Now, I would like to use the sequential criterion for a limit of a function:
$$lim_{tto 0^{+}}phi(t)=lim_{nto +infty}phileft(frac{1}{n}right)=lim_{nto +infty}int_{mathbb{R}^{N}setminus S_{frac{1}{n}}}|f(x)|^2dx=\ \ \ =lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx$$
Since $chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)le 1$ the sequence
$$g_{n}(x):=chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2le |f(x)|^2 forall xinmathbb{R}^{N}$$
I can use the dominated convergence theorem
$$lim_{nto +infty}int_{mathbb{R}^{N}}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}lim_{nto+infty}chi_{mathbb{R}^{N}setminus S_{frac{1}{n}}}(x)|f(x)|^2dx=int_{mathbb{R}^{N}}|f(x)|^2dx=|f|_{2}^2$$
For the limit $lim_{tto +infty}phi(t)$ I use the sequence $t_{n}=n$ so
$$lim_{tto +infty}phi(t)=lim_{nto +infty}int_{mathbb{R}^N}chi_{mathbb{R}^{N}setminus S_{n}}(x)|f(x)|^2dx=int_{mathbb{R}^N}overbrace{lim_{nto +infty}chi_{mathbb{R}^{N}setminus S_{n}}(x)}^{=0}|f(x)|^2dx=0$$
Is this approach ok? Thanks.
real-analysis integration functional-analysis measure-theory lp-spaces
real-analysis integration functional-analysis measure-theory lp-spaces
asked Dec 12 '18 at 11:11
IxionIxion
768621
asked Dec 12 '18 at 11:11
IxionIxion
768621
asked Dec 12 '18 at 11:11
IxionIxion
768621
asked Dec 12 '18 at 11:11
IxionIxion
768621
asked Dec 12 '18 at 11:11
IxionIxion
768621
768621
1
$begingroup$
Your argument is correct: the proof of the boundedness of $phi(t)$ is compact and elegant, while the calculation of the limits is a bit more lengthy but formally correct.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 19:39
$begingroup$
@DanieleTampieri: thank you! :) I would like to find the "fastest" way to evaluate the limits. Maybe there exists a theorem that could be useful...? Could you give me an hint?
$endgroup$
– Ixion
Dec 12 '18 at 21:09
$begingroup$
Mmm, let me see what can I do. I'll look at it.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 21:16
add a comment |
1
$begingroup$
Your argument is correct: the proof of the boundedness of $phi(t)$ is compact and elegant, while the calculation of the limits is a bit more lengthy but formally correct.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 19:39
$begingroup$
@DanieleTampieri: thank you! :) I would like to find the "fastest" way to evaluate the limits. Maybe there exists a theorem that could be useful...? Could you give me an hint?
$endgroup$
– Ixion
Dec 12 '18 at 21:09
$begingroup$
Mmm, let me see what can I do. I'll look at it.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 21:16
1
1
$begingroup$
Your argument is correct: the proof of the boundedness of $phi(t)$ is compact and elegant, while the calculation of the limits is a bit more lengthy but formally correct.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 19:39
$begingroup$
Your argument is correct: the proof of the boundedness of $phi(t)$ is compact and elegant, while the calculation of the limits is a bit more lengthy but formally correct.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 19:39
$begingroup$
@DanieleTampieri: thank you! :) I would like to find the "fastest" way to evaluate the limits. Maybe there exists a theorem that could be useful...? Could you give me an hint?
$endgroup$
– Ixion
Dec 12 '18 at 21:09
$begingroup$
@DanieleTampieri: thank you! :) I would like to find the "fastest" way to evaluate the limits. Maybe there exists a theorem that could be useful...? Could you give me an hint?
$endgroup$
– Ixion
Dec 12 '18 at 21:09
$begingroup$
Mmm, let me see what can I do. I'll look at it.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 21:16
$begingroup$
Mmm, let me see what can I do. I'll look at it.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 21:16
add a comment |
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$begingroup$
Your argument is correct: the proof of the boundedness of $phi(t)$ is compact and elegant, while the calculation of the limits is a bit more lengthy but formally correct.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 19:39
$begingroup$
@DanieleTampieri: thank you! :) I would like to find the "fastest" way to evaluate the limits. Maybe there exists a theorem that could be useful...? Could you give me an hint?
$endgroup$
– Ixion
Dec 12 '18 at 21:09
$begingroup$
Mmm, let me see what can I do. I'll look at it.
$endgroup$
– Daniele Tampieri
Dec 12 '18 at 21:16