Conditions to obtain that the limit of a sequence satisfies the equation of the elememts of the sequence












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If we define by $f_{n}(t)=int_{0}^{t}g(y,f_{n}(y))dy$ and prove that the limit of $f_{n}$ exists. What do we need to assume about $g$ in order for the limit to satisfy the above integral equation aswell? In other words,



$f(t)=int_{0}^{t}g(y,f(y))dy$










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$endgroup$












  • $begingroup$
    If $g$ is continuous and bounded, you can apply the dominated convergence theorem to get $f(t) = int_0^t g(y,f(y)) , mathrm{d} y$.
    $endgroup$
    – p4sch
    Jan 12 at 19:54












  • $begingroup$
    @p4sch thanks, do you know if pointwise convergence of $f_{n}$? Or does it need to be uniform.
    $endgroup$
    – user7534
    Jan 12 at 20:04










  • $begingroup$
    Pointwise convergence is sufficient.
    $endgroup$
    – p4sch
    Jan 12 at 20:45










  • $begingroup$
    @p4sch right, we just think about it as $g_{n}$ essentially to use dominated. Then continuity to get the limit inside $g$?
    $endgroup$
    – user7534
    Jan 13 at 5:50








  • 1




    $begingroup$
    Yes. (In fact, you can weaken the assumptation by $g(x,y)$ is measurable in $x$ for fixed $y$ and continuous in $y$ for fixed $x$.)
    $endgroup$
    – p4sch
    Jan 13 at 11:51
















0












$begingroup$


If we define by $f_{n}(t)=int_{0}^{t}g(y,f_{n}(y))dy$ and prove that the limit of $f_{n}$ exists. What do we need to assume about $g$ in order for the limit to satisfy the above integral equation aswell? In other words,



$f(t)=int_{0}^{t}g(y,f(y))dy$










share|cite|improve this question











$endgroup$












  • $begingroup$
    If $g$ is continuous and bounded, you can apply the dominated convergence theorem to get $f(t) = int_0^t g(y,f(y)) , mathrm{d} y$.
    $endgroup$
    – p4sch
    Jan 12 at 19:54












  • $begingroup$
    @p4sch thanks, do you know if pointwise convergence of $f_{n}$? Or does it need to be uniform.
    $endgroup$
    – user7534
    Jan 12 at 20:04










  • $begingroup$
    Pointwise convergence is sufficient.
    $endgroup$
    – p4sch
    Jan 12 at 20:45










  • $begingroup$
    @p4sch right, we just think about it as $g_{n}$ essentially to use dominated. Then continuity to get the limit inside $g$?
    $endgroup$
    – user7534
    Jan 13 at 5:50








  • 1




    $begingroup$
    Yes. (In fact, you can weaken the assumptation by $g(x,y)$ is measurable in $x$ for fixed $y$ and continuous in $y$ for fixed $x$.)
    $endgroup$
    – p4sch
    Jan 13 at 11:51














0












0








0





$begingroup$


If we define by $f_{n}(t)=int_{0}^{t}g(y,f_{n}(y))dy$ and prove that the limit of $f_{n}$ exists. What do we need to assume about $g$ in order for the limit to satisfy the above integral equation aswell? In other words,



$f(t)=int_{0}^{t}g(y,f(y))dy$










share|cite|improve this question











$endgroup$




If we define by $f_{n}(t)=int_{0}^{t}g(y,f_{n}(y))dy$ and prove that the limit of $f_{n}$ exists. What do we need to assume about $g$ in order for the limit to satisfy the above integral equation aswell? In other words,



$f(t)=int_{0}^{t}g(y,f(y))dy$







real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 19:43







user7534

















asked Jan 12 at 19:37









user7534user7534

665




665












  • $begingroup$
    If $g$ is continuous and bounded, you can apply the dominated convergence theorem to get $f(t) = int_0^t g(y,f(y)) , mathrm{d} y$.
    $endgroup$
    – p4sch
    Jan 12 at 19:54












  • $begingroup$
    @p4sch thanks, do you know if pointwise convergence of $f_{n}$? Or does it need to be uniform.
    $endgroup$
    – user7534
    Jan 12 at 20:04










  • $begingroup$
    Pointwise convergence is sufficient.
    $endgroup$
    – p4sch
    Jan 12 at 20:45










  • $begingroup$
    @p4sch right, we just think about it as $g_{n}$ essentially to use dominated. Then continuity to get the limit inside $g$?
    $endgroup$
    – user7534
    Jan 13 at 5:50








  • 1




    $begingroup$
    Yes. (In fact, you can weaken the assumptation by $g(x,y)$ is measurable in $x$ for fixed $y$ and continuous in $y$ for fixed $x$.)
    $endgroup$
    – p4sch
    Jan 13 at 11:51


















  • $begingroup$
    If $g$ is continuous and bounded, you can apply the dominated convergence theorem to get $f(t) = int_0^t g(y,f(y)) , mathrm{d} y$.
    $endgroup$
    – p4sch
    Jan 12 at 19:54












  • $begingroup$
    @p4sch thanks, do you know if pointwise convergence of $f_{n}$? Or does it need to be uniform.
    $endgroup$
    – user7534
    Jan 12 at 20:04










  • $begingroup$
    Pointwise convergence is sufficient.
    $endgroup$
    – p4sch
    Jan 12 at 20:45










  • $begingroup$
    @p4sch right, we just think about it as $g_{n}$ essentially to use dominated. Then continuity to get the limit inside $g$?
    $endgroup$
    – user7534
    Jan 13 at 5:50








  • 1




    $begingroup$
    Yes. (In fact, you can weaken the assumptation by $g(x,y)$ is measurable in $x$ for fixed $y$ and continuous in $y$ for fixed $x$.)
    $endgroup$
    – p4sch
    Jan 13 at 11:51
















$begingroup$
If $g$ is continuous and bounded, you can apply the dominated convergence theorem to get $f(t) = int_0^t g(y,f(y)) , mathrm{d} y$.
$endgroup$
– p4sch
Jan 12 at 19:54






$begingroup$
If $g$ is continuous and bounded, you can apply the dominated convergence theorem to get $f(t) = int_0^t g(y,f(y)) , mathrm{d} y$.
$endgroup$
– p4sch
Jan 12 at 19:54














$begingroup$
@p4sch thanks, do you know if pointwise convergence of $f_{n}$? Or does it need to be uniform.
$endgroup$
– user7534
Jan 12 at 20:04




$begingroup$
@p4sch thanks, do you know if pointwise convergence of $f_{n}$? Or does it need to be uniform.
$endgroup$
– user7534
Jan 12 at 20:04












$begingroup$
Pointwise convergence is sufficient.
$endgroup$
– p4sch
Jan 12 at 20:45




$begingroup$
Pointwise convergence is sufficient.
$endgroup$
– p4sch
Jan 12 at 20:45












$begingroup$
@p4sch right, we just think about it as $g_{n}$ essentially to use dominated. Then continuity to get the limit inside $g$?
$endgroup$
– user7534
Jan 13 at 5:50






$begingroup$
@p4sch right, we just think about it as $g_{n}$ essentially to use dominated. Then continuity to get the limit inside $g$?
$endgroup$
– user7534
Jan 13 at 5:50






1




1




$begingroup$
Yes. (In fact, you can weaken the assumptation by $g(x,y)$ is measurable in $x$ for fixed $y$ and continuous in $y$ for fixed $x$.)
$endgroup$
– p4sch
Jan 13 at 11:51




$begingroup$
Yes. (In fact, you can weaken the assumptation by $g(x,y)$ is measurable in $x$ for fixed $y$ and continuous in $y$ for fixed $x$.)
$endgroup$
– p4sch
Jan 13 at 11:51










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