Why is $frac{x^n}{x^3+x^4}chi_{[0,1[} in mathcal{L}^1$












1














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







share|cite|improve this question


















  • 1




    Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
    – Botond
    Jan 4 at 11:16


















1














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







share|cite|improve this question


















  • 1




    Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
    – Botond
    Jan 4 at 11:16
















1












1








1







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







share|cite|improve this question













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






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asked Jan 4 at 10:58









SABOYSABOY

508311




508311








  • 1




    Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
    – Botond
    Jan 4 at 11:16
















  • 1




    Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
    – Botond
    Jan 4 at 11:16










1




1




Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
– Botond
Jan 4 at 11:16






Is $n in mathbb{N}$? Because it's not really integrable for $n=0,1,2$.
– Botond
Jan 4 at 11:16












1 Answer
1






active

oldest

votes


















1














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.






share|cite|improve this answer





















  • 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??
    – SABOY
    Jan 4 at 12:03












  • 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.
    – Davide Giraudo
    Jan 4 at 12:14










  • 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?
    – SABOY
    Jan 4 at 12:24












  • Yes. ${}{}{}{}$
    – Davide Giraudo
    Jan 4 at 12:51











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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.






share|cite|improve this answer





















  • 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??
    – SABOY
    Jan 4 at 12:03












  • 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.
    – Davide Giraudo
    Jan 4 at 12:14










  • 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?
    – SABOY
    Jan 4 at 12:24












  • Yes. ${}{}{}{}$
    – Davide Giraudo
    Jan 4 at 12:51
















1














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.






share|cite|improve this answer





















  • 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??
    – SABOY
    Jan 4 at 12:03












  • 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.
    – Davide Giraudo
    Jan 4 at 12:14










  • 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?
    – SABOY
    Jan 4 at 12:24












  • Yes. ${}{}{}{}$
    – Davide Giraudo
    Jan 4 at 12:51














1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 11:38









Davide GiraudoDavide Giraudo

125k16150261




125k16150261












  • 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??
    – SABOY
    Jan 4 at 12:03












  • 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.
    – Davide Giraudo
    Jan 4 at 12:14










  • 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?
    – SABOY
    Jan 4 at 12:24












  • Yes. ${}{}{}{}$
    – Davide Giraudo
    Jan 4 at 12:51


















  • 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??
    – SABOY
    Jan 4 at 12:03












  • 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.
    – Davide Giraudo
    Jan 4 at 12:14










  • 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?
    – SABOY
    Jan 4 at 12:24












  • Yes. ${}{}{}{}$
    – Davide Giraudo
    Jan 4 at 12:51
















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??
– SABOY
Jan 4 at 12:03






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??
– SABOY
Jan 4 at 12:03














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.
– Davide Giraudo
Jan 4 at 12:14




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.
– Davide Giraudo
Jan 4 at 12:14












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?
– SABOY
Jan 4 at 12:24






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?
– SABOY
Jan 4 at 12:24














Yes. ${}{}{}{}$
– Davide Giraudo
Jan 4 at 12:51




Yes. ${}{}{}{}$
– Davide Giraudo
Jan 4 at 12:51


















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