Conditions to obtain that the limit of a sequence satisfies the equation of the elememts of the sequence
Multi tool use
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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,
$f(t)=int_{0}^{t}g(y,f(y))dy$
real-analysis
asked Jan 12 at 19:37
user7534user7534
665
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add a comment |
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,
$f(t)=int_{0}^{t}g(y,f(y))dy$
real-analysis
asked Jan 12 at 19:37
user7534user7534
665
$endgroup$
$begingroup$
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.
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– 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
add a comment |
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,
$f(t)=int_{0}^{t}g(y,f(y))dy$
real-analysis
asked Jan 12 at 19:37
user7534user7534
665
$endgroup$
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
real-analysis
asked Jan 12 at 19:37
user7534user7534
665
asked Jan 12 at 19:37
user7534user7534
665
edited Jan 12 at 19:43
user7534
asked Jan 12 at 19:37
user7534user7534
665
asked Jan 12 at 19:37
user7534user7534
665
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
add a comment |
$begingroup$
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
add a comment |
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cseJFm1
$begingroup$
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