Mixing
Why is $frac{x^n}{x^3+x^4}chi_{[0,1[} in mathcal{L}^1$
$begingroup$
In one of our exams it was written without explanation that $f_{n}(x):=frac{x^n}{x^3+x^4}chi_{[0,1[} in mathcal{L}^1$
But I fail to see how this does not warrant an explanation.
My ideas
$frac{x^n}{x^3+x^4}$ is continuous on $]0,1[$, $forall n in mathbb N$.
I basically want to find a function $g in mathcal{L}^1$ such that $f_{n} leq g, forall n in mathbb N$
on $]0,1[$ this is not difficult as $frac{x^n}{x^3+x^4}chi_{]0,1[}leqfrac{1}{x^3+x^4}$
But what about at $x = 0$, I mean $f_{n}(0)$ is not defined so how can I create a plausible inequality? I do realize that ${0}$ has measure zero but how do I successfuly incorporate that into proving that $f_{n} in mathcal{L}^1$
real-analysis integration measure-theory continuity
asked Jan 4 at 10:58
SABOYSABOY
567311
$endgroup$
add a comment |
$begingroup$
In one of our exams it was written without explanation that $f_{n}(x):=frac{x^n}{x^3+x^4}chi_{[0,1[} in mathcal{L}^1$
But I fail to see how this does not warrant an explanation.
My ideas
$frac{x^n}{x^3+x^4}$ is continuous on $]0,1[$, $forall n in mathbb N$.
I basically want to find a function $g in mathcal{L}^1$ such that $f_{n} leq g, forall n in mathbb N$
on $]0,1[$ this is not difficult as $frac{x^n}{x^3+x^4}chi_{]0,1[}leqfrac{1}{x^3+x^4}$
But what about at $x = 0$, I mean $f_{n}(0)$ is not defined so how can I create a plausible inequality? I do realize that ${0}$ has measure zero but how do I successfuly incorporate that into proving that $f_{n} in mathcal{L}^1$
real-analysis integration measure-theory continuity
asked Jan 4 at 10:58
SABOYSABOY
567311
$endgroup$
1
$begingroup$
Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
$endgroup$
– Botond
Jan 4 at 11:16
add a comment |
$begingroup$
In one of our exams it was written without explanation that $f_{n}(x):=frac{x^n}{x^3+x^4}chi_{[0,1[} in mathcal{L}^1$
But I fail to see how this does not warrant an explanation.
My ideas
$frac{x^n}{x^3+x^4}$ is continuous on $]0,1[$, $forall n in mathbb N$.
I basically want to find a function $g in mathcal{L}^1$ such that $f_{n} leq g, forall n in mathbb N$
on $]0,1[$ this is not difficult as $frac{x^n}{x^3+x^4}chi_{]0,1[}leqfrac{1}{x^3+x^4}$
But what about at $x = 0$, I mean $f_{n}(0)$ is not defined so how can I create a plausible inequality? I do realize that ${0}$ has measure zero but how do I successfuly incorporate that into proving that $f_{n} in mathcal{L}^1$
real-analysis integration measure-theory continuity
asked Jan 4 at 10:58
SABOYSABOY
567311
$endgroup$
In one of our exams it was written without explanation that $f_{n}(x):=frac{x^n}{x^3+x^4}chi_{[0,1[} in mathcal{L}^1$
But I fail to see how this does not warrant an explanation.
My ideas
$frac{x^n}{x^3+x^4}$ is continuous on $]0,1[$, $forall n in mathbb N$.
I basically want to find a function $g in mathcal{L}^1$ such that $f_{n} leq g, forall n in mathbb N$
on $]0,1[$ this is not difficult as $frac{x^n}{x^3+x^4}chi_{]0,1[}leqfrac{1}{x^3+x^4}$
But what about at $x = 0$, I mean $f_{n}(0)$ is not defined so how can I create a plausible inequality? I do realize that ${0}$ has measure zero but how do I successfuly incorporate that into proving that $f_{n} in mathcal{L}^1$
real-analysis integration measure-theory continuity
real-analysis integration measure-theory continuity
asked Jan 4 at 10:58
SABOYSABOY
567311
asked Jan 4 at 10:58
SABOYSABOY
567311
asked Jan 4 at 10:58
SABOYSABOY
567311
asked Jan 4 at 10:58
SABOYSABOY
567311
asked Jan 4 at 10:58
SABOYSABOY
567311
567311
1
$begingroup$
Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
$endgroup$
– Botond
Jan 4 at 11:16
add a comment |
1
$begingroup$
Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
$endgroup$
– Botond
Jan 4 at 11:16
1
1
$begingroup$
Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
$endgroup$
– Botond
Jan 4 at 11:16
$begingroup$
Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
$endgroup$
– Botond
Jan 4 at 11:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Most likely, the goal is to study the convergence in $mathcal L^1$ of the sequence $left(f_nright)_{ngeqslant 1}$ hence the fact that $f_n$ is not integrable for $nin{0;1;2}$ is not a problem.
For $ngeqslant 3$ and $xin(0,1)$,
$$
0leqslant frac{x^3}{x^3+x^4}leqslant frac{x^3+x^4}{x^3+x^4}=1
$$
hence
$$
0leqslant f_n(x)leqslant x^{n-3}leqslant 1
$$
and a bounded function on $(0,1)$ is integrable.
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
The fact that $n geq 3 $ is relevant is because $lim_{x to 0^{+}}f_{n}(x) = 0 $ when $n geq 3$ correct ? So indeed then $f_{n}$ is defined on $[0,1)$ and we can use $f_{n}leq 1$ on $[0,1)$ while $n geq 3$ correct??
$endgroup$
– SABOY
Jan 4 at 12:03
$begingroup$
Actually we do not need to define $f_n(0)$; the point is that $int_0^1 1/t dt$ and $int_0^1 1/t^2 dt$ are infinite.
$endgroup$
– Davide Giraudo
Jan 4 at 12:14
$begingroup$
Ok let me ask it plainly, does $f_{n} chi_{]0,1[} in mathcal{L}^1$ imply that $f_{n} chi_{[0,1[}in mathcal{L}^1$ because ${0}$ is of measure zero?
$endgroup$
– SABOY
Jan 4 at 12:24
$begingroup$
Yes. ${}{}{}{}$
$endgroup$
– Davide Giraudo
Jan 4 at 12:51
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
Most likely, the goal is to study the convergence in $mathcal L^1$ of the sequence $left(f_nright)_{ngeqslant 1}$ hence the fact that $f_n$ is not integrable for $nin{0;1;2}$ is not a problem.
For $ngeqslant 3$ and $xin(0,1)$,
$$
0leqslant frac{x^3}{x^3+x^4}leqslant frac{x^3+x^4}{x^3+x^4}=1
$$
hence
$$
0leqslant f_n(x)leqslant x^{n-3}leqslant 1
$$
and a bounded function on $(0,1)$ is integrable.
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
The fact that $n geq 3 $ is relevant is because $lim_{x to 0^{+}}f_{n}(x) = 0 $ when $n geq 3$ correct ? So indeed then $f_{n}$ is defined on $[0,1)$ and we can use $f_{n}leq 1$ on $[0,1)$ while $n geq 3$ correct??
$endgroup$
– SABOY
Jan 4 at 12:03
$begingroup$
Actually we do not need to define $f_n(0)$; the point is that $int_0^1 1/t dt$ and $int_0^1 1/t^2 dt$ are infinite.
$endgroup$
– Davide Giraudo
Jan 4 at 12:14
$begingroup$
Ok let me ask it plainly, does $f_{n} chi_{]0,1[} in mathcal{L}^1$ imply that $f_{n} chi_{[0,1[}in mathcal{L}^1$ because ${0}$ is of measure zero?
$endgroup$
– SABOY
Jan 4 at 12:24
$begingroup$
Yes. ${}{}{}{}$
$endgroup$
– Davide Giraudo
Jan 4 at 12:51
add a comment |
$begingroup$
Most likely, the goal is to study the convergence in $mathcal L^1$ of the sequence $left(f_nright)_{ngeqslant 1}$ hence the fact that $f_n$ is not integrable for $nin{0;1;2}$ is not a problem.
For $ngeqslant 3$ and $xin(0,1)$,
$$
0leqslant frac{x^3}{x^3+x^4}leqslant frac{x^3+x^4}{x^3+x^4}=1
$$
hence
$$
0leqslant f_n(x)leqslant x^{n-3}leqslant 1
$$
and a bounded function on $(0,1)$ is integrable.
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
The fact that $n geq 3 $ is relevant is because $lim_{x to 0^{+}}f_{n}(x) = 0 $ when $n geq 3$ correct ? So indeed then $f_{n}$ is defined on $[0,1)$ and we can use $f_{n}leq 1$ on $[0,1)$ while $n geq 3$ correct??
$endgroup$
– SABOY
Jan 4 at 12:03
$begingroup$
Actually we do not need to define $f_n(0)$; the point is that $int_0^1 1/t dt$ and $int_0^1 1/t^2 dt$ are infinite.
$endgroup$
– Davide Giraudo
Jan 4 at 12:14
$begingroup$
Ok let me ask it plainly, does $f_{n} chi_{]0,1[} in mathcal{L}^1$ imply that $f_{n} chi_{[0,1[}in mathcal{L}^1$ because ${0}$ is of measure zero?
$endgroup$
– SABOY
Jan 4 at 12:24
$begingroup$
Yes. ${}{}{}{}$
$endgroup$
– Davide Giraudo
Jan 4 at 12:51
add a comment |
$begingroup$
Most likely, the goal is to study the convergence in $mathcal L^1$ of the sequence $left(f_nright)_{ngeqslant 1}$ hence the fact that $f_n$ is not integrable for $nin{0;1;2}$ is not a problem.
For $ngeqslant 3$ and $xin(0,1)$,
$$
0leqslant frac{x^3}{x^3+x^4}leqslant frac{x^3+x^4}{x^3+x^4}=1
$$
hence
$$
0leqslant f_n(x)leqslant x^{n-3}leqslant 1
$$
and a bounded function on $(0,1)$ is integrable.
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
Most likely, the goal is to study the convergence in $mathcal L^1$ of the sequence $left(f_nright)_{ngeqslant 1}$ hence the fact that $f_n$ is not integrable for $nin{0;1;2}$ is not a problem.
For $ngeqslant 3$ and $xin(0,1)$,
$$
0leqslant frac{x^3}{x^3+x^4}leqslant frac{x^3+x^4}{x^3+x^4}=1
$$
hence
$$
0leqslant f_n(x)leqslant x^{n-3}leqslant 1
$$
and a bounded function on $(0,1)$ is integrable.
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 4 at 11:38
Davide GiraudoDavide Giraudo
125k16150261
125k16150261
$begingroup$
The fact that $n geq 3 $ is relevant is because $lim_{x to 0^{+}}f_{n}(x) = 0 $ when $n geq 3$ correct ? So indeed then $f_{n}$ is defined on $[0,1)$ and we can use $f_{n}leq 1$ on $[0,1)$ while $n geq 3$ correct??
$endgroup$
– SABOY
Jan 4 at 12:03
$begingroup$
Actually we do not need to define $f_n(0)$; the point is that $int_0^1 1/t dt$ and $int_0^1 1/t^2 dt$ are infinite.
$endgroup$
– Davide Giraudo
Jan 4 at 12:14
$begingroup$
Ok let me ask it plainly, does $f_{n} chi_{]0,1[} in mathcal{L}^1$ imply that $f_{n} chi_{[0,1[}in mathcal{L}^1$ because ${0}$ is of measure zero?
$endgroup$
– SABOY
Jan 4 at 12:24
$begingroup$
Yes. ${}{}{}{}$
$endgroup$
– Davide Giraudo
Jan 4 at 12:51
add a comment |
$begingroup$
The fact that $n geq 3 $ is relevant is because $lim_{x to 0^{+}}f_{n}(x) = 0 $ when $n geq 3$ correct ? So indeed then $f_{n}$ is defined on $[0,1)$ and we can use $f_{n}leq 1$ on $[0,1)$ while $n geq 3$ correct??
$endgroup$
– SABOY
Jan 4 at 12:03
$begingroup$
Actually we do not need to define $f_n(0)$; the point is that $int_0^1 1/t dt$ and $int_0^1 1/t^2 dt$ are infinite.
$endgroup$
– Davide Giraudo
Jan 4 at 12:14
$begingroup$
Ok let me ask it plainly, does $f_{n} chi_{]0,1[} in mathcal{L}^1$ imply that $f_{n} chi_{[0,1[}in mathcal{L}^1$ because ${0}$ is of measure zero?
$endgroup$
– SABOY
Jan 4 at 12:24
$begingroup$
Yes. ${}{}{}{}$
$endgroup$
– Davide Giraudo
Jan 4 at 12:51
$begingroup$
The fact that $n geq 3 $ is relevant is because $lim_{x to 0^{+}}f_{n}(x) = 0 $ when $n geq 3$ correct ? So indeed then $f_{n}$ is defined on $[0,1)$ and we can use $f_{n}leq 1$ on $[0,1)$ while $n geq 3$ correct??
$endgroup$
– SABOY
Jan 4 at 12:03
$begingroup$
The fact that $n geq 3 $ is relevant is because $lim_{x to 0^{+}}f_{n}(x) = 0 $ when $n geq 3$ correct ? So indeed then $f_{n}$ is defined on $[0,1)$ and we can use $f_{n}leq 1$ on $[0,1)$ while $n geq 3$ correct??
$endgroup$
– SABOY
Jan 4 at 12:03
$begingroup$
Actually we do not need to define $f_n(0)$; the point is that $int_0^1 1/t dt$ and $int_0^1 1/t^2 dt$ are infinite.
$endgroup$
– Davide Giraudo
Jan 4 at 12:14
$begingroup$
Actually we do not need to define $f_n(0)$; the point is that $int_0^1 1/t dt$ and $int_0^1 1/t^2 dt$ are infinite.
$endgroup$
– Davide Giraudo
Jan 4 at 12:14
$begingroup$
Ok let me ask it plainly, does $f_{n} chi_{]0,1[} in mathcal{L}^1$ imply that $f_{n} chi_{[0,1[}in mathcal{L}^1$ because ${0}$ is of measure zero?
$endgroup$
– SABOY
Jan 4 at 12:24
$begingroup$
Ok let me ask it plainly, does $f_{n} chi_{]0,1[} in mathcal{L}^1$ imply that $f_{n} chi_{[0,1[}in mathcal{L}^1$ because ${0}$ is of measure zero?
$endgroup$
– SABOY
Jan 4 at 12:24
$begingroup$
Yes. ${}{}{}{}$
$endgroup$
– Davide Giraudo
Jan 4 at 12:51
$begingroup$
Yes. ${}{}{}{}$
$endgroup$
– Davide Giraudo
Jan 4 at 12:51
add a comment |
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$begingroup$
Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
$endgroup$
– Botond
Jan 4 at 11:16