Mixing
Lebesgue integral limit
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Suppose $f:mathbbR^+ to mathbbR$ is Lebesgue integrable on every finite interval and
$$I(x;alpha) = int_alpha^alpha+x f(t) dt$$
If $lim_alpha to inftyI(x; alpha) = F(x)$ exists for every $x$, then show $F(x) = Cx$ for some real number $C$.
My attempt is to first consider a continuous function with a finite limit at infinity. Then by mean value theorem there exists a point $xi_alpha > alpha$ where
$$lim_alpha to inftyint_alpha^alpha+x f(t) dt = lim_alpha to inftyf(xi_alpha)x = [lim_tto inftyf(t)]x $$
I do not see how to prove the result without these strong assumptions.
convergence lebesgue-integral
asked Jul 26 at 23:42
user28763
523
add a comment |Â
up vote
3
down vote
favorite
Suppose $f:mathbbR^+ to mathbbR$ is Lebesgue integrable on every finite interval and
$$I(x;alpha) = int_alpha^alpha+x f(t) dt$$
If $lim_alpha to inftyI(x; alpha) = F(x)$ exists for every $x$, then show $F(x) = Cx$ for some real number $C$.
My attempt is to first consider a continuous function with a finite limit at infinity. Then by mean value theorem there exists a point $xi_alpha > alpha$ where
$$lim_alpha to inftyint_alpha^alpha+x f(t) dt = lim_alpha to inftyf(xi_alpha)x = [lim_tto inftyf(t)]x $$
I do not see how to prove the result without these strong assumptions.
convergence lebesgue-integral
asked Jul 26 at 23:42
user28763
523
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose $f:mathbbR^+ to mathbbR$ is Lebesgue integrable on every finite interval and
$$I(x;alpha) = int_alpha^alpha+x f(t) dt$$
If $lim_alpha to inftyI(x; alpha) = F(x)$ exists for every $x$, then show $F(x) = Cx$ for some real number $C$.
My attempt is to first consider a continuous function with a finite limit at infinity. Then by mean value theorem there exists a point $xi_alpha > alpha$ where
$$lim_alpha to inftyint_alpha^alpha+x f(t) dt = lim_alpha to inftyf(xi_alpha)x = [lim_tto inftyf(t)]x $$
I do not see how to prove the result without these strong assumptions.
convergence lebesgue-integral
asked Jul 26 at 23:42
user28763
523
Suppose $f:mathbbR^+ to mathbbR$ is Lebesgue integrable on every finite interval and
$$I(x;alpha) = int_alpha^alpha+x f(t) dt$$
If $lim_alpha to inftyI(x; alpha) = F(x)$ exists for every $x$, then show $F(x) = Cx$ for some real number $C$.
My attempt is to first consider a continuous function with a finite limit at infinity. Then by mean value theorem there exists a point $xi_alpha > alpha$ where
$$lim_alpha to inftyint_alpha^alpha+x f(t) dt = lim_alpha to inftyf(xi_alpha)x = [lim_tto inftyf(t)]x $$
I do not see how to prove the result without these strong assumptions.
convergence lebesgue-integral
asked Jul 26 at 23:42
user28763
523
asked Jul 26 at 23:42
user28763
523
asked Jul 26 at 23:42
user28763
523
asked Jul 26 at 23:42
user28763
523
523
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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3
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accepted
Note that
$$int_alpha^alpha + x_1 +x_2 f(t) , dt = int_alpha^alpha + x_1 f(t) , dt + int_alpha+x_1^(alpha + x_1)+ x_2 f(t) , dt$$
Since $alpha + x_1 to infty$ iff $alpha to infty$ it follows that $F$ is additive:
$F(x_1+x_2) = F(x_1) + F(x_2)$.
With any sequence $alpha_n to infty$, we have $F(x) = lim_n to inftyI(x;alpha_n)$ and $F(x)$ is a pointwise limit of a sequence of continuous functions. Using Baire's theorem, it can be shown that $F$ is continuous on an everywhere dense set.
However, we only need continuity at one point to conclude that $F$, as an additive function, is everywhere continuous with $F(x) = F(1)cdot x$.
Addendum
To elaborate, it is easy to prove that an additive function has the property $F(rx) = rF(x)$ for any rational number $r$. Hence, for any real number $xi$, taking a sequence of rationals $r_n to xi$, we have if $F$ is continuous,
$$xi F(x) = lim_n to inftyr_nF(x) = lim_n to inftyF(r_n x) = F(xi x)$$
Also, if $F$ is additive and continuous at one point, say $c$, then it is continuous everywhere since
$$F(x + delta) - F(x) = F(delta) = F(c + delta) - F(c)$$
That $F$ must be continuous at one point is a somewhat deeper result requiring a lengthy argument. For example, see here.
answered Jul 26 at 23:54
RRL
43.5k42260
I see the first part. Can you explain more the last two statements.
– user28763
Jul 26 at 23:59
@user28763: I added some more explanation in the answer.
– RRL
Jul 27 at 0:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Note that
$$int_alpha^alpha + x_1 +x_2 f(t) , dt = int_alpha^alpha + x_1 f(t) , dt + int_alpha+x_1^(alpha + x_1)+ x_2 f(t) , dt$$
Since $alpha + x_1 to infty$ iff $alpha to infty$ it follows that $F$ is additive:
$F(x_1+x_2) = F(x_1) + F(x_2)$.
With any sequence $alpha_n to infty$, we have $F(x) = lim_n to inftyI(x;alpha_n)$ and $F(x)$ is a pointwise limit of a sequence of continuous functions. Using Baire's theorem, it can be shown that $F$ is continuous on an everywhere dense set.
However, we only need continuity at one point to conclude that $F$, as an additive function, is everywhere continuous with $F(x) = F(1)cdot x$.
Addendum
To elaborate, it is easy to prove that an additive function has the property $F(rx) = rF(x)$ for any rational number $r$. Hence, for any real number $xi$, taking a sequence of rationals $r_n to xi$, we have if $F$ is continuous,
$$xi F(x) = lim_n to inftyr_nF(x) = lim_n to inftyF(r_n x) = F(xi x)$$
Also, if $F$ is additive and continuous at one point, say $c$, then it is continuous everywhere since
$$F(x + delta) - F(x) = F(delta) = F(c + delta) - F(c)$$
That $F$ must be continuous at one point is a somewhat deeper result requiring a lengthy argument. For example, see here.
answered Jul 26 at 23:54
RRL
43.5k42260
I see the first part. Can you explain more the last two statements.
– user28763
Jul 26 at 23:59
@user28763: I added some more explanation in the answer.
– RRL
Jul 27 at 0:15
add a comment |Â
up vote
3
down vote
accepted
Note that
$$int_alpha^alpha + x_1 +x_2 f(t) , dt = int_alpha^alpha + x_1 f(t) , dt + int_alpha+x_1^(alpha + x_1)+ x_2 f(t) , dt$$
Since $alpha + x_1 to infty$ iff $alpha to infty$ it follows that $F$ is additive:
$F(x_1+x_2) = F(x_1) + F(x_2)$.
With any sequence $alpha_n to infty$, we have $F(x) = lim_n to inftyI(x;alpha_n)$ and $F(x)$ is a pointwise limit of a sequence of continuous functions. Using Baire's theorem, it can be shown that $F$ is continuous on an everywhere dense set.
However, we only need continuity at one point to conclude that $F$, as an additive function, is everywhere continuous with $F(x) = F(1)cdot x$.
Addendum
To elaborate, it is easy to prove that an additive function has the property $F(rx) = rF(x)$ for any rational number $r$. Hence, for any real number $xi$, taking a sequence of rationals $r_n to xi$, we have if $F$ is continuous,
$$xi F(x) = lim_n to inftyr_nF(x) = lim_n to inftyF(r_n x) = F(xi x)$$
Also, if $F$ is additive and continuous at one point, say $c$, then it is continuous everywhere since
$$F(x + delta) - F(x) = F(delta) = F(c + delta) - F(c)$$
That $F$ must be continuous at one point is a somewhat deeper result requiring a lengthy argument. For example, see here.
answered Jul 26 at 23:54
RRL
43.5k42260
I see the first part. Can you explain more the last two statements.
– user28763
Jul 26 at 23:59
@user28763: I added some more explanation in the answer.
– RRL
Jul 27 at 0:15
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Note that
$$int_alpha^alpha + x_1 +x_2 f(t) , dt = int_alpha^alpha + x_1 f(t) , dt + int_alpha+x_1^(alpha + x_1)+ x_2 f(t) , dt$$
Since $alpha + x_1 to infty$ iff $alpha to infty$ it follows that $F$ is additive:
$F(x_1+x_2) = F(x_1) + F(x_2)$.
With any sequence $alpha_n to infty$, we have $F(x) = lim_n to inftyI(x;alpha_n)$ and $F(x)$ is a pointwise limit of a sequence of continuous functions. Using Baire's theorem, it can be shown that $F$ is continuous on an everywhere dense set.
However, we only need continuity at one point to conclude that $F$, as an additive function, is everywhere continuous with $F(x) = F(1)cdot x$.
Addendum
To elaborate, it is easy to prove that an additive function has the property $F(rx) = rF(x)$ for any rational number $r$. Hence, for any real number $xi$, taking a sequence of rationals $r_n to xi$, we have if $F$ is continuous,
$$xi F(x) = lim_n to inftyr_nF(x) = lim_n to inftyF(r_n x) = F(xi x)$$
Also, if $F$ is additive and continuous at one point, say $c$, then it is continuous everywhere since
$$F(x + delta) - F(x) = F(delta) = F(c + delta) - F(c)$$
That $F$ must be continuous at one point is a somewhat deeper result requiring a lengthy argument. For example, see here.
answered Jul 26 at 23:54
RRL
43.5k42260
Note that
$$int_alpha^alpha + x_1 +x_2 f(t) , dt = int_alpha^alpha + x_1 f(t) , dt + int_alpha+x_1^(alpha + x_1)+ x_2 f(t) , dt$$
Since $alpha + x_1 to infty$ iff $alpha to infty$ it follows that $F$ is additive:
$F(x_1+x_2) = F(x_1) + F(x_2)$.
With any sequence $alpha_n to infty$, we have $F(x) = lim_n to inftyI(x;alpha_n)$ and $F(x)$ is a pointwise limit of a sequence of continuous functions. Using Baire's theorem, it can be shown that $F$ is continuous on an everywhere dense set.
However, we only need continuity at one point to conclude that $F$, as an additive function, is everywhere continuous with $F(x) = F(1)cdot x$.
Addendum
To elaborate, it is easy to prove that an additive function has the property $F(rx) = rF(x)$ for any rational number $r$. Hence, for any real number $xi$, taking a sequence of rationals $r_n to xi$, we have if $F$ is continuous,
$$xi F(x) = lim_n to inftyr_nF(x) = lim_n to inftyF(r_n x) = F(xi x)$$
Also, if $F$ is additive and continuous at one point, say $c$, then it is continuous everywhere since
$$F(x + delta) - F(x) = F(delta) = F(c + delta) - F(c)$$
That $F$ must be continuous at one point is a somewhat deeper result requiring a lengthy argument. For example, see here.
answered Jul 26 at 23:54
RRL
43.5k42260
edited Jul 27 at 0:26
answered Jul 26 at 23:54
RRL
43.5k42260
answered Jul 26 at 23:54
RRL
43.5k42260
answered Jul 26 at 23:54
RRL
43.5k42260
43.5k42260
I see the first part. Can you explain more the last two statements.
– user28763
Jul 26 at 23:59
@user28763: I added some more explanation in the answer.
– RRL
Jul 27 at 0:15
add a comment |Â
I see the first part. Can you explain more the last two statements.
– user28763
Jul 26 at 23:59
@user28763: I added some more explanation in the answer.
– RRL
Jul 27 at 0:15
I see the first part. Can you explain more the last two statements.
– user28763
Jul 26 at 23:59
I see the first part. Can you explain more the last two statements.
– user28763
Jul 26 at 23:59
@user28763: I added some more explanation in the answer.
– RRL
Jul 27 at 0:15
@user28763: I added some more explanation in the answer.
– RRL
Jul 27 at 0:15
add a comment |Â
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