Convergence of a series given properties on uniformly bounded functions
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I have the following problem from "Bollobas, Linear analysis. An introductory course"
Let $phi_n:[0,1]tomathbb{R}^+$ $(n=1,2,dots)$ be uniformly bounded continuous function such that $$int_{0}^1phi_n(x)dxgeq c$$ for some $c>0$. Suppose $c_ngeq 0$, $(n=1,2,dots)$ and $$sum_{ngeq 1}c_nphi_n(x)<infty$$ for every $xin[0,1]$. Prove that $$sum_{ngeq 1}c_n<infty$$
I don't know how to proceed. Any hints will be useful !
Thanks in advance
functional-analysis
asked Dec 6 at 2:37
jnaf
512211
add a comment |
up vote
4
down vote
favorite
I have the following problem from "Bollobas, Linear analysis. An introductory course"
Let $phi_n:[0,1]tomathbb{R}^+$ $(n=1,2,dots)$ be uniformly bounded continuous function such that $$int_{0}^1phi_n(x)dxgeq c$$ for some $c>0$. Suppose $c_ngeq 0$, $(n=1,2,dots)$ and $$sum_{ngeq 1}c_nphi_n(x)<infty$$ for every $xin[0,1]$. Prove that $$sum_{ngeq 1}c_n<infty$$
I don't know how to proceed. Any hints will be useful !
Thanks in advance
functional-analysis
asked Dec 6 at 2:37
jnaf
512211
Really the only traction you have on $phi_n$ is its integral on $[0,1]$. Have you tried integrating your $c_n phi_n$ sum over $[0,1]$ to see what happens?
– Eric Towers
Dec 6 at 2:42
@EricTowers It is not given that $sum c_n phi_n$ is integrable.
– Kavi Rama Murthy
Dec 6 at 5:38
@KaviRamaMurthy : Lebesgue tells me jnaf can (Riemann) integrate each $c_nphi_n$. I would wonder if jnaf can justify swapping the order of integration and summation, but thinking about this aspect reminds me that $phi_n(x) in ell^infty$ for each $x$ and the $phi_n$ are uniformly bounded away from zero.
– Eric Towers
Dec 6 at 13:53
... transposed -- not exactly "uniformly bounded away from zero".
– Eric Towers
Dec 6 at 14:19
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have the following problem from "Bollobas, Linear analysis. An introductory course"
Let $phi_n:[0,1]tomathbb{R}^+$ $(n=1,2,dots)$ be uniformly bounded continuous function such that $$int_{0}^1phi_n(x)dxgeq c$$ for some $c>0$. Suppose $c_ngeq 0$, $(n=1,2,dots)$ and $$sum_{ngeq 1}c_nphi_n(x)<infty$$ for every $xin[0,1]$. Prove that $$sum_{ngeq 1}c_n<infty$$
I don't know how to proceed. Any hints will be useful !
Thanks in advance
functional-analysis
asked Dec 6 at 2:37
jnaf
512211
I have the following problem from "Bollobas, Linear analysis. An introductory course"
Let $phi_n:[0,1]tomathbb{R}^+$ $(n=1,2,dots)$ be uniformly bounded continuous function such that $$int_{0}^1phi_n(x)dxgeq c$$ for some $c>0$. Suppose $c_ngeq 0$, $(n=1,2,dots)$ and $$sum_{ngeq 1}c_nphi_n(x)<infty$$ for every $xin[0,1]$. Prove that $$sum_{ngeq 1}c_n<infty$$
I don't know how to proceed. Any hints will be useful !
Thanks in advance
functional-analysis
functional-analysis
asked Dec 6 at 2:37
jnaf
512211
asked Dec 6 at 2:37
jnaf
512211
edited Dec 6 at 2:53
asked Dec 6 at 2:37
jnaf
512211
asked Dec 6 at 2:37
jnaf
512211
asked Dec 6 at 2:37
jnaf
512211
512211
Really the only traction you have on $phi_n$ is its integral on $[0,1]$. Have you tried integrating your $c_n phi_n$ sum over $[0,1]$ to see what happens?
– Eric Towers
Dec 6 at 2:42
@EricTowers It is not given that $sum c_n phi_n$ is integrable.
– Kavi Rama Murthy
Dec 6 at 5:38
@KaviRamaMurthy : Lebesgue tells me jnaf can (Riemann) integrate each $c_nphi_n$. I would wonder if jnaf can justify swapping the order of integration and summation, but thinking about this aspect reminds me that $phi_n(x) in ell^infty$ for each $x$ and the $phi_n$ are uniformly bounded away from zero.
– Eric Towers
Dec 6 at 13:53
... transposed -- not exactly "uniformly bounded away from zero".
– Eric Towers
Dec 6 at 14:19
add a comment |
Really the only traction you have on $phi_n$ is its integral on $[0,1]$. Have you tried integrating your $c_n phi_n$ sum over $[0,1]$ to see what happens?
– Eric Towers
Dec 6 at 2:42
@EricTowers It is not given that $sum c_n phi_n$ is integrable.
– Kavi Rama Murthy
Dec 6 at 5:38
@KaviRamaMurthy : Lebesgue tells me jnaf can (Riemann) integrate each $c_nphi_n$. I would wonder if jnaf can justify swapping the order of integration and summation, but thinking about this aspect reminds me that $phi_n(x) in ell^infty$ for each $x$ and the $phi_n$ are uniformly bounded away from zero.
– Eric Towers
Dec 6 at 13:53
... transposed -- not exactly "uniformly bounded away from zero".
– Eric Towers
Dec 6 at 14:19
Really the only traction you have on $phi_n$ is its integral on $[0,1]$. Have you tried integrating your $c_n phi_n$ sum over $[0,1]$ to see what happens?
– Eric Towers
Dec 6 at 2:42
Really the only traction you have on $phi_n$ is its integral on $[0,1]$. Have you tried integrating your $c_n phi_n$ sum over $[0,1]$ to see what happens?
– Eric Towers
Dec 6 at 2:42
@EricTowers It is not given that $sum c_n phi_n$ is integrable.
– Kavi Rama Murthy
Dec 6 at 5:38
@EricTowers It is not given that $sum c_n phi_n$ is integrable.
– Kavi Rama Murthy
Dec 6 at 5:38
@KaviRamaMurthy : Lebesgue tells me jnaf can (Riemann) integrate each $c_nphi_n$. I would wonder if jnaf can justify swapping the order of integration and summation, but thinking about this aspect reminds me that $phi_n(x) in ell^infty$ for each $x$ and the $phi_n$ are uniformly bounded away from zero.
– Eric Towers
Dec 6 at 13:53
@KaviRamaMurthy : Lebesgue tells me jnaf can (Riemann) integrate each $c_nphi_n$. I would wonder if jnaf can justify swapping the order of integration and summation, but thinking about this aspect reminds me that $phi_n(x) in ell^infty$ for each $x$ and the $phi_n$ are uniformly bounded away from zero.
– Eric Towers
Dec 6 at 13:53
... transposed -- not exactly "uniformly bounded away from zero".
– Eric Towers
Dec 6 at 14:19
... transposed -- not exactly "uniformly bounded away from zero".
– Eric Towers
Dec 6 at 14:19
add a comment |
1 Answer
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We know that the sequence $(varphi_n)_{n in mathbb{N}}$ is uniformly bounded, i.e. we have $|varphi_n|_infty le M$ for all $n in mathbb{N}$ for some $M>0$ independent of the index $n$. Moreover, we write $B_n:= {varphi_n ne 0 }$ and let $$X := bigcup_{n=1}^infty B_n.$$ Note that $X$ is open and not empty. In fact, we have $B_n neq emptyset$ for all $n in mathbb{N}$, because of $int_0^1 varphi_n(t) , dt ge c$. In particular, $0< lambda(X) le 1$. (Moreover, $X$ is a Baire-space as an open subset of a complete metric space. But we will not need this.)
Define
$$h(x) := sum_{k=1}^infty c_k varphi_k(x).$$
This series is pointwise convergent, as assumed, and therefore also measurable. Let $$A_n := {x in X : n < h(x) le n+1}$$ and note that $X = bigcup_{n=1}^infty A_n$.
By $sigma$-additivity of the measure, we have
$$ lambda(X) = sum_{k=1}^infty lambda(A_k).$$
Hence the series is convegent and we find for some $N in mathbb{N}$ that
$$sum_{k=N}^infty lambda(A_k) < frac{c}{2M}.$$
Set $Y:= bigcup_{k=1}^N A_k$. Note that for any $x in Y$ we have $|h(x)| le N+1$ and $lambda(X setminus Y) < c/(2M)$. Therefore we get
$$int_Y varphi_n(x) ,dx ge c- int_{X setminus Y} varphi_n(x) ,dx ge frac{c}{2}.$$
The monotone convergence theorem implies now
$$frac{c}{2} sum_{n=1}^infty c_n le sum_{k=1}^infty c_k int_Y varphi_k(x) , dx = int_Y sum_{k=1}^infty c_k varphi_k(x) , dx = int_Y h(x) , dx le N+1.$$
Thus the series in the last equation is convergent.
answered Dec 6 at 14:01
p4sch
4,800217
add a comment |
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1 Answer
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1 Answer
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oldest
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active
oldest
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accepted
We know that the sequence $(varphi_n)_{n in mathbb{N}}$ is uniformly bounded, i.e. we have $|varphi_n|_infty le M$ for all $n in mathbb{N}$ for some $M>0$ independent of the index $n$. Moreover, we write $B_n:= {varphi_n ne 0 }$ and let $$X := bigcup_{n=1}^infty B_n.$$ Note that $X$ is open and not empty. In fact, we have $B_n neq emptyset$ for all $n in mathbb{N}$, because of $int_0^1 varphi_n(t) , dt ge c$. In particular, $0< lambda(X) le 1$. (Moreover, $X$ is a Baire-space as an open subset of a complete metric space. But we will not need this.)
Define
$$h(x) := sum_{k=1}^infty c_k varphi_k(x).$$
This series is pointwise convergent, as assumed, and therefore also measurable. Let $$A_n := {x in X : n < h(x) le n+1}$$ and note that $X = bigcup_{n=1}^infty A_n$.
By $sigma$-additivity of the measure, we have
$$ lambda(X) = sum_{k=1}^infty lambda(A_k).$$
Hence the series is convegent and we find for some $N in mathbb{N}$ that
$$sum_{k=N}^infty lambda(A_k) < frac{c}{2M}.$$
Set $Y:= bigcup_{k=1}^N A_k$. Note that for any $x in Y$ we have $|h(x)| le N+1$ and $lambda(X setminus Y) < c/(2M)$. Therefore we get
$$int_Y varphi_n(x) ,dx ge c- int_{X setminus Y} varphi_n(x) ,dx ge frac{c}{2}.$$
The monotone convergence theorem implies now
$$frac{c}{2} sum_{n=1}^infty c_n le sum_{k=1}^infty c_k int_Y varphi_k(x) , dx = int_Y sum_{k=1}^infty c_k varphi_k(x) , dx = int_Y h(x) , dx le N+1.$$
Thus the series in the last equation is convergent.
answered Dec 6 at 14:01
p4sch
4,800217
add a comment |
up vote
2
down vote
accepted
We know that the sequence $(varphi_n)_{n in mathbb{N}}$ is uniformly bounded, i.e. we have $|varphi_n|_infty le M$ for all $n in mathbb{N}$ for some $M>0$ independent of the index $n$. Moreover, we write $B_n:= {varphi_n ne 0 }$ and let $$X := bigcup_{n=1}^infty B_n.$$ Note that $X$ is open and not empty. In fact, we have $B_n neq emptyset$ for all $n in mathbb{N}$, because of $int_0^1 varphi_n(t) , dt ge c$. In particular, $0< lambda(X) le 1$. (Moreover, $X$ is a Baire-space as an open subset of a complete metric space. But we will not need this.)
Define
$$h(x) := sum_{k=1}^infty c_k varphi_k(x).$$
This series is pointwise convergent, as assumed, and therefore also measurable. Let $$A_n := {x in X : n < h(x) le n+1}$$ and note that $X = bigcup_{n=1}^infty A_n$.
By $sigma$-additivity of the measure, we have
$$ lambda(X) = sum_{k=1}^infty lambda(A_k).$$
Hence the series is convegent and we find for some $N in mathbb{N}$ that
$$sum_{k=N}^infty lambda(A_k) < frac{c}{2M}.$$
Set $Y:= bigcup_{k=1}^N A_k$. Note that for any $x in Y$ we have $|h(x)| le N+1$ and $lambda(X setminus Y) < c/(2M)$. Therefore we get
$$int_Y varphi_n(x) ,dx ge c- int_{X setminus Y} varphi_n(x) ,dx ge frac{c}{2}.$$
The monotone convergence theorem implies now
$$frac{c}{2} sum_{n=1}^infty c_n le sum_{k=1}^infty c_k int_Y varphi_k(x) , dx = int_Y sum_{k=1}^infty c_k varphi_k(x) , dx = int_Y h(x) , dx le N+1.$$
Thus the series in the last equation is convergent.
answered Dec 6 at 14:01
p4sch
4,800217
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We know that the sequence $(varphi_n)_{n in mathbb{N}}$ is uniformly bounded, i.e. we have $|varphi_n|_infty le M$ for all $n in mathbb{N}$ for some $M>0$ independent of the index $n$. Moreover, we write $B_n:= {varphi_n ne 0 }$ and let $$X := bigcup_{n=1}^infty B_n.$$ Note that $X$ is open and not empty. In fact, we have $B_n neq emptyset$ for all $n in mathbb{N}$, because of $int_0^1 varphi_n(t) , dt ge c$. In particular, $0< lambda(X) le 1$. (Moreover, $X$ is a Baire-space as an open subset of a complete metric space. But we will not need this.)
Define
$$h(x) := sum_{k=1}^infty c_k varphi_k(x).$$
This series is pointwise convergent, as assumed, and therefore also measurable. Let $$A_n := {x in X : n < h(x) le n+1}$$ and note that $X = bigcup_{n=1}^infty A_n$.
By $sigma$-additivity of the measure, we have
$$ lambda(X) = sum_{k=1}^infty lambda(A_k).$$
Hence the series is convegent and we find for some $N in mathbb{N}$ that
$$sum_{k=N}^infty lambda(A_k) < frac{c}{2M}.$$
Set $Y:= bigcup_{k=1}^N A_k$. Note that for any $x in Y$ we have $|h(x)| le N+1$ and $lambda(X setminus Y) < c/(2M)$. Therefore we get
$$int_Y varphi_n(x) ,dx ge c- int_{X setminus Y} varphi_n(x) ,dx ge frac{c}{2}.$$
The monotone convergence theorem implies now
$$frac{c}{2} sum_{n=1}^infty c_n le sum_{k=1}^infty c_k int_Y varphi_k(x) , dx = int_Y sum_{k=1}^infty c_k varphi_k(x) , dx = int_Y h(x) , dx le N+1.$$
Thus the series in the last equation is convergent.
answered Dec 6 at 14:01
p4sch
4,800217
We know that the sequence $(varphi_n)_{n in mathbb{N}}$ is uniformly bounded, i.e. we have $|varphi_n|_infty le M$ for all $n in mathbb{N}$ for some $M>0$ independent of the index $n$. Moreover, we write $B_n:= {varphi_n ne 0 }$ and let $$X := bigcup_{n=1}^infty B_n.$$ Note that $X$ is open and not empty. In fact, we have $B_n neq emptyset$ for all $n in mathbb{N}$, because of $int_0^1 varphi_n(t) , dt ge c$. In particular, $0< lambda(X) le 1$. (Moreover, $X$ is a Baire-space as an open subset of a complete metric space. But we will not need this.)
Define
$$h(x) := sum_{k=1}^infty c_k varphi_k(x).$$
This series is pointwise convergent, as assumed, and therefore also measurable. Let $$A_n := {x in X : n < h(x) le n+1}$$ and note that $X = bigcup_{n=1}^infty A_n$.
By $sigma$-additivity of the measure, we have
$$ lambda(X) = sum_{k=1}^infty lambda(A_k).$$
Hence the series is convegent and we find for some $N in mathbb{N}$ that
$$sum_{k=N}^infty lambda(A_k) < frac{c}{2M}.$$
Set $Y:= bigcup_{k=1}^N A_k$. Note that for any $x in Y$ we have $|h(x)| le N+1$ and $lambda(X setminus Y) < c/(2M)$. Therefore we get
$$int_Y varphi_n(x) ,dx ge c- int_{X setminus Y} varphi_n(x) ,dx ge frac{c}{2}.$$
The monotone convergence theorem implies now
$$frac{c}{2} sum_{n=1}^infty c_n le sum_{k=1}^infty c_k int_Y varphi_k(x) , dx = int_Y sum_{k=1}^infty c_k varphi_k(x) , dx = int_Y h(x) , dx le N+1.$$
Thus the series in the last equation is convergent.
answered Dec 6 at 14:01
p4sch
4,800217
answered Dec 6 at 14:01
p4sch
4,800217
answered Dec 6 at 14:01
p4sch
4,800217
answered Dec 6 at 14:01
p4sch
4,800217
4,800217
add a comment |
add a comment |
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Really the only traction you have on $phi_n$ is its integral on $[0,1]$. Have you tried integrating your $c_n phi_n$ sum over $[0,1]$ to see what happens?
– Eric Towers
Dec 6 at 2:42
@EricTowers It is not given that $sum c_n phi_n$ is integrable.
– Kavi Rama Murthy
Dec 6 at 5:38
@KaviRamaMurthy : Lebesgue tells me jnaf can (Riemann) integrate each $c_nphi_n$. I would wonder if jnaf can justify swapping the order of integration and summation, but thinking about this aspect reminds me that $phi_n(x) in ell^infty$ for each $x$ and the $phi_n$ are uniformly bounded away from zero.
– Eric Towers
Dec 6 at 13:53
... transposed -- not exactly "uniformly bounded away from zero".
– Eric Towers
Dec 6 at 14:19