Mixing
$left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p = sum_{k=1}^{infty} left| frac{(-1)^k}{k}...
$begingroup$
Here's the problem statement, which is the context for my question,
Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by
$$
u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
For which $p geq 1$ does $int |u|^p dm < infty$?
The solution manual says that: Consider
$$
u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
$u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that
$$
|u|^p=sum_{k=1}^{infty}
left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
...
so, I don't get the above equality. How do I get from
$$
|u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
quad text{to} quad
sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
It seems that I at least have to use the triangle inequality, but that would ruin the equality.
real-analysis lebesgue-measure
asked Jan 28 at 20:31
TomTom
285
$endgroup$
add a comment |
$begingroup$
Here's the problem statement, which is the context for my question,
Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by
$$
u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
For which $p geq 1$ does $int |u|^p dm < infty$?
The solution manual says that: Consider
$$
u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
$u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that
$$
|u|^p=sum_{k=1}^{infty}
left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
...
so, I don't get the above equality. How do I get from
$$
|u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
quad text{to} quad
sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
It seems that I at least have to use the triangle inequality, but that would ruin the equality.
real-analysis lebesgue-measure
asked Jan 28 at 20:31
TomTom
285
$endgroup$
add a comment |
$begingroup$
Here's the problem statement, which is the context for my question,
Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by
$$
u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
For which $p geq 1$ does $int |u|^p dm < infty$?
The solution manual says that: Consider
$$
u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
$u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that
$$
|u|^p=sum_{k=1}^{infty}
left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
...
so, I don't get the above equality. How do I get from
$$
|u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
quad text{to} quad
sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
It seems that I at least have to use the triangle inequality, but that would ruin the equality.
real-analysis lebesgue-measure
asked Jan 28 at 20:31
TomTom
285
$endgroup$
Here's the problem statement, which is the context for my question,
Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by
$$
u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
For which $p geq 1$ does $int |u|^p dm < infty$?
The solution manual says that: Consider
$$
u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
$$
$u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that
$$
|u|^p=sum_{k=1}^{infty}
left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
...
so, I don't get the above equality. How do I get from
$$
|u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
quad text{to} quad
sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
It seems that I at least have to use the triangle inequality, but that would ruin the equality.
real-analysis lebesgue-measure
real-analysis lebesgue-measure
asked Jan 28 at 20:31
TomTom
285
asked Jan 28 at 20:31
TomTom
285
asked Jan 28 at 20:31
TomTom
285
asked Jan 28 at 20:31
TomTom
285
asked Jan 28 at 20:31
TomTom
285
285
add a comment |
add a comment |
1 Answer
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$begingroup$
If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
$$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
where $p=lfloor x rfloor$.
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
$$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
where $p=lfloor x rfloor$.
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
$begingroup$
If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
$$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
where $p=lfloor x rfloor$.
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
$begingroup$
If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
$$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
where $p=lfloor x rfloor$.
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
$endgroup$
If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
$$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
where $p=lfloor x rfloor$.
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
answered Jan 28 at 20:59
Stefan LafonStefan Lafon
3,005212
3,005212
add a comment |
add a comment |
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