recursal/Fanatic-Fandom huggingface dataset:
Ex3-9

**Chapter 3 Question 9**

Let

{

E
k

}

# k
1

∞

{\displaystyle \{ E\_k \}\_{k=1}^\infty}

 be a sequence of sets so that

∑

E
k

e

{\displaystyle \sum|E\_k|\_e}

 is finite. Then for any positive number

ϵ
>
0

{\displaystyle \epsilon > 0}

, there exists an integer

j
>
0

{\displaystyle j>0}

 such that

∑

# k
j

∞

E
k

e

<
ϵ

{\displaystyle \begin{align}

\sum\_{k=j}^{\infty}\left|E\_k\right|\_e<\epsilon

\end{align}
}

From Theorem 3.4, we have

⋃

# k
j

∞

E
k

e

≤

∑

# k
j

∞

E
k

e

<
ϵ

{\displaystyle \begin{align}

\end{align}}

From the definition of

lim
sup

{\displaystyle \lim \sup}

lim
sup

E
k

=

⋂

# i
1

∞

(

⋃

# k
i

∞

E
k

)

{\displaystyle \begin{align}

\lim \sup E\_k=\bigcap\_{i=1}^{\infty}\left(\bigcup\_{k=i}^{\infty}E\_k\right)

\end{align}}

gives us that for each

i

{\displaystyle i}

,

lim
sup

E
k

⊂

⋃

# k
i

∞

E
k

{\displaystyle \begin{align}

\lim \sup E\_k\subset \bigcup\_{k=i}^{\infty} E\_k

\end{align}}

Then from the above part, we know for each

ϵ
>
0

{\displaystyle \epsilon > 0}

, there is a

j

{\displaystyle j}

 such that

⋃

# k
j

∞

E
k

e

<
ϵ

{\displaystyle \begin{align}

\left|\bigcup\_{k=j}^{\infty} E\_k\right|\_e<\epsilon

\end{align}}

which implies

lim
sup

E
k

<
ϵ
.

{\displaystyle \begin{align}

\left|\lim \sup E\_k\right|<\epsilon.

\end{align}}

Since

ϵ

{\displaystyle \epsilon}

 can be arbitrary small, we have proven that

lim
sup

E
k

{\displaystyle \lim \sup E\_k}

 has measure zero.

Since

lim
inf

E
k

⊂
lim
sup

E
k

{\displaystyle \begin{align}

\lim \inf E\_k \subset \lim\sup E\_k

\end{align}}

we can also get that

lim
inf

E
k

{\displaystyle \lim \inf E\_k}

 has measure zero.

Another proof(proof in other words)

Let

# E
lim sup

E

k

{\displaystyle E=\limsup E\_{k}}

 and

F

j

=

⋃

# k
j

∞

E

k

{\displaystyle F\_{j}=\bigcup\_{k=j}^{\infty} E\_{k}}

. Then,

F
j

{\displaystyle F\_j}

 is a decreasing sequence of sets whose limit is

lim sup

E

k

{\displaystyle \limsup E\_{k}}

.

Then

E
⊂

⋂

# j
n

∞

F

j

⊆

F

n

{\displaystyle E \subset \bigcap\_{j=n}^{\infty} F\_{j} \subseteq F\_{n}}

 for all n, so

E

e

≤

F

n

e

 for all n by monotonicity of outer measure.

Moreover,

F

n

e

≤

∑

# k
n

∞

E

k

e

 for all n. Since

F

n

e

≤

∑

# k
n

∞

E

k

e

 converges, for every

ε
>
0

{\displaystyle \varepsilon > 0}

, there exists

N

{\displaystyle N}

 such that

∑

# k
n

∞

E

k

e

<
ε

{\displaystyle \sum\_{k=n}^{\infty}\left|E\_{k}\right|\_{e}<\varepsilon}

 for all

n
≥
N

{\displaystyle n \geq N}

.

Thus, for every

ε
>
0
,

{\displaystyle \varepsilon>0,}

 we have

E

e

≤
ε

{\displaystyle |E|\_{e} \leq \varepsilon}

, which implies that

E

e

=
0

{\displaystyle |E|\_{e}=0}

.

Hence,

# E
lim sup

E

k

{\displaystyle E=\limsup E\_{k}}

 is measurable and has measure 0. Since

lim inf

E

k

⊆
lim sup

E

k

{\displaystyle \liminf E\_{k} \subseteq \limsup E\_{k}}

,

lim inf

E

k

{\displaystyle \liminf E\_{k}}

 also has measure zero.
7-24

For easy typing, we let

# f
ϕ

{\displaystyle f = \phi}

 in solution. Note that since

f

{\displaystyle f}

has a strictly increasing derivative on

(
a
,
ξ
)
∪
(
ξ
,
b
)

{\displaystyle (a,\xi)\cup(\xi,b)}

it is enough to check the convexity inequality on

ξ

{\displaystyle \xi}

.

Since

ξ
∈
(
a
,
b
)

{\displaystyle \xi \in (a, b)}

, for

t
∈
[
0
,
1
]

{\displaystyle t \in [0,1]}

, we can write

# ξ
t
a
(
1
−
t
)
b

{\displaystyle \xi = ta + (1-t)b}

.

Since

f
′

{\displaystyle f'}

 exists and finite, by Mean Value Theorem, there exists

m
,
n
∈
(
a
,
b
)

{\displaystyle m,n \in (a,b)
}

 such that

f
′

(
m
# )

f
(
ξ
)
−
f
(
a
)

ξ
−
a

and

f
′

(
n
# )

f
(
b
)
−
f
(
ξ
)

b
−
ξ

{\displaystyle f'(m) = \frac{f(\xi) - f(a)}{\xi - a} ~\text{and} ~ f'(n) = \frac{f(b) - f(\xi)}{b - \xi}}

.(here we assume that

m
<
n

{\displaystyle m < n}

)

Since

f
′

{\displaystyle f'}

 increasing, we have

f
′

(
m
)
≤

f
′

(
n
)

{\displaystyle f'(m) \le f'(n)}

. Then

f
(
ξ
# )
f
(
a
)

f
′

(
m
)
(
ξ
−
a
)
≤
f
(
a
)

f
′

(
n
)
(
ξ
−
a
# )
f
(
a
)

f
(
b
)
−
f
(
ξ
)

b
−
ξ

(
ξ
−
a
)

{\displaystyle f(\xi) = f(a) + f'(m)(\xi - a) \le f(a) + f'(n)(\xi - a) = f(a) + \frac{f(b) - f(\xi)}{b - \xi}(\xi - a)}

.

Note that

b
−
# ξ
t
(
b
−
a
)

and

ξ
−
# a
−
(
b
−
a
)
(
t
−
1
)

{\displaystyle b-\xi = t(b-a) ~\text{and} ~ \xi - a = -(b-a)(t-1)}

, the computation part is easy and left it to reader.

Multiply

b
−
# ξ
t
(
b
−
a
)

{\displaystyle b-\xi = t(b-a)}

to both sides, then we get

f
(
ξ
)
t
(
b
−
a
)
≤
f
(
a
)
t
(
b
−
a
)
[
f
(
b
)
−
f
(
ξ
)
]
(
−
(
b
−
a
)
(
t
−
1
)
)

{\displaystyle f(\xi)t(b-a) \le f(a)t(b-a) + [f(b) - f(\xi)](-(b-a)(t-1))}

.

After simplify, we get

f
(
ξ
)
(
b
−
a
)
≤
t
(
f
(
a
)
−
f
(
b
)
(
b
−
a
)
−
(
t
−
1
)
f
(
b
)
(
b
−
a
)

⟹

f
(
ξ
)
≤
t
f
(
a
)
(
1
−
2
t
)
f
(
b
)

{\displaystyle f(\xi)(b-a) \le t(f(a) - f(b)(b-a) - (t-1)f(b)(b-a) \implies f(\xi) \le tf(a) + (1-2t)f(b)}

.

Since

t
∈
[
0
,
1
]
,
1
−
2
t
≤
1
−
t

{\displaystyle t \in [0,1], 1-2t \le 1-t}

. So, we have

f
(
ξ
# )
f
(
t
a
(
1
−
t
)
b
)
≤
t
f
(
a
)
(
1
−
t
)
f
(
b
)

{\displaystyle f(\xi) = f(ta + (1-t)b) \le tf(a) + (1-t)f(b)}

.

So,

# f
ϕ

{\displaystyle f = \phi}

 is convex on

(
a
,
b
)

{\displaystyle (a,b)}

.
Runo Misaki

* BB
* NV Casual
* GI
* MSA1 & MSA2
* MSA2 Casual

15 (New Vestroia)

16 (Gundalian Invaders)

17 (Mechtanium Surge)

18 (BioTechnic Metal)

19 (Master Quest)

27 (Hyperion Leonidas)

900 (Sentanine)

1000 (Malzord)

Novax or Aerogan (1100)

### Voice Actor

Julie Lemieux (English) (Nelvana Network)Kara Edwards (English) (Capcom Co., Ltd.)

Eri Sendai (Japanese)

### Guardian Bakugan

Blade Tigrerra (BB & NV)Slicer Wolfunoid (NV)

Sentanine (GI)

Malzord (MSA1)

Novax (MS2A)

Aerogan (MSA2) (Temporary)

### Home Palace

Wardington City, Earth (BB - GI)Bayview City, Earth (MSA1)

Bakugan City, Earth (MSA2)

### Faction

Battle BrawlersBakugan Battle Brawlers Resistance

Order of the Castle Knights (Formerly)

Team Grey (Formerly)

### Other Appearances

Mega Man Grey 2: Reploids AdventMega Man Grey 3: New Vestroia

Mega Man Grey 4: Gundalian Invaders

Mega Man Grey 5: Mechtanium Surge Arc 1

Mega Man Grey 6: Mechtanium Surge Arc 2

Mega Man Grey 7: BioTechnic Metal

Mega Man Grey 8: Master Quest

Bakugan: Battle Brawlers the Video Game

Bakugan: Battle Trainer

### Relatives

Hikaru (Younger Brother)Dan (Former Mate)

Ren (Husband)
5-12

Question: Give an example of a bounded continuous

f
(
x
)

{\textstyle f(x)}

on

(
0
,
∞
)

{\textstyle (0,\infty )}

 such that

lim

x
→
∞

f
(
x
# )
0

{\textstyle \lim \_{x\to \infty }f(x)=0}

 but

f
(
x
)
∉

L

p

(
0
,
∞
)

{\textstyle f(x)\notin L\_{p}(0,\infty )}

 for any

p
>
0

{\textstyle p>0}

.

Answer:

Define

f
(
x
# )

{

1
,

if

x
≤
e

1

/

ln
⁡
x

,

if

x
≥
e

{\displaystyle f(x)={\begin{cases}1,&{\text{if }}x\leq e\\1/{\ln x},&{\text{if }}x\geq e\end{cases}}}

It is obvious that function is bounded by 1 and it is continuous since both 1 and

1

/

ln
⁡
x

{\displaystyle 1/\ln x}

 are continuous and the the function is continuous at

# x
e

{\displaystyle x=e}

. Moreover, we have

lim

x
→
∞

f
(
x
# )

lim

x
→
∞

1

ln
⁡
x

=

{\displaystyle \lim \_{x\to \infty }f(x)=\lim \_{x\to \infty }{\frac {1}{\ln x}}=0.}

Now for any

p
>
0

{\displaystyle p > 0}

, since

l
n
(
x
# )
O
(

x

1
p

)

{\displaystyle ln(x)=O(x^{\frac {1}{p}})}

, we can find

K

p

{\displaystyle K\_{p}}

 such that for

x
≥

K

p

{\displaystyle x\geq K\_{p}}

,

ln
⁡
x
≤

x

1
p

{\displaystyle \ln x\leq x^{\frac {1}{p}}}

, then

∫

0

∞

f
(
x
)

p

≥

∫

K

p

∞

(

1

ln
⁡
x

)

≥

∫

K

p

∞

1
x

d
# x
∞

{\displaystyle \int \_{0}^{\infty }|f(x)|^{p}\geq \int \_{K\_{p}}^{\infty }\left({\frac {1}{\ln x}}\right)\geq \int \_{K\_{p}}^{\infty }{\frac {1}{x}}dx=+\infty }

this shows that

f
(
x
)
∉

L

p

(
0
,
∞
)
.

{\displaystyle f(x)\notin L\_{p}(0,\infty ).}
Zenet Surrow

* Gundalian Invaders
* Mechtanium Surge
* True Form

18 (BioTechnic Metal)

19 (Master Quest)

27 (Hyperion Leonidas)

### Voice Actor

Barbara Mamabolo (English) (Nelvana Network)Lara Woodhull (English) (Capcom Co., Ltd.)

Yuka Saito (Japanese)

### Other Appearances

Mega Man Grey 5: Mechtanium Surge Arc 1Mega Man Grey 6: Mechtanium Surge Arc 2

Mega Man Grey 7: BioTechnic Metal

Mega Man Grey 8: Master Quest

### Side

Rebel

**Zenet Surrow**, (Japanese version: **Zenet Sirone** (ゼネット・シローネ, *Zeneto Shirone***?**) is a Haos brawler from Gundalia and part of the Gundalian Agents. She is partnered with Contestir.

### Personality

Zenet is a Gundalian shapeshifter; while other Gundalians have demonstrated the ability to shift between a Human and Gundalian form, she has taken other people's appearances as well. She is afraid of losing and slowing her teammates down, believing that she will screw things up. She was once loyal to Barodius; after Gill disposed of her, she was placed under Kazarina's hypnosis. Once she was freed, she felt no loyalty towards him and called him a "weasel". Like Ren and Nurzak, she hated Barodius for using them and the other Gundalians for his own selfish gain.

She seems to be rather cynical, telling Contestir that Gundalians have no loyalty to each other. It seems that she jumps to conclusions easily, as her initial reaction upon seeing Nurzak and Mason Brown alive and well was to mistake them as ghosts or zombies.

Zenet seems to have a huge crush on Ren and wants him to return his feelings to her though he has yet to do so. Out of all the Gundalian Agents, she is the most loyal to Ren and considers him and the rest of their team her true friends.

Zenet hates betrayals and mistreatment by others as she points out to Contestir that Barodius and Gill care nothing about her loyalty or their people. Zenet prefers to be treated fairly and harbors a great deal of hatred towards Barodius for treating her and the rest of the Gundalians like expendable pawns.
Ex4-2

**Solution to Chapter 4, Exercise 2**

*Question: Let

f

{\displaystyle f}

be a simple function, taking its distinct values on disjoint sets

E
1

,
⋯
,

E
N

{\displaystyle E\_1, \cdots, E\_N}

. Show that

f

{\displaystyle f}

is measurable if and only if

E
1

,
⋯
,

E
N

{\displaystyle E\_1, \cdots, E\_N}

are measurable.*

Solution: Let the distinct values be

a
1

,
⋯
,

a
N

{\displaystyle a\_1,\cdots,a\_N}

so that

f
(
x

# k
1

N

a
k

χ

E
k

(
x
)

{\displaystyle f(x)=\sum\_{k=1}^N a\_k\chi\_{E\_k}(x)}

on

# k
1

N

E
k

{\displaystyle E=\bigcup\_{k=1}^N E\_k}

.

Suppose

f

{\displaystyle f}

is measurable. Then for any

a

{\displaystyle a}

,

{
f
≥
a
}

{\displaystyle \{f \geq a\}}

and

{
f
>
a
}

{\displaystyle \{f>a\}}

are measurable. Then

{
# }
{
f
≥
a
}
−
{
f
>
a
}

{\displaystyle \{f=a\}=\{f \geq a\}-\{f>a\}}

is measurable for any

a

{\displaystyle a}

. Since

E
k

=
{
# f

a
k

}

{\displaystyle E\_k=\{f=a\_k\}}

, each

E
k

{\displaystyle E\_k}

is measurable.

Conversely, suppose

E
1

,
⋯
,

E
N

{\displaystyle E\_1, \cdots, E\_N}

are measurable. Then for any

a

{\displaystyle a}

, if

a
≥
max
{

a
1

,
⋯
,

a
N

}

{\displaystyle a \geq \max\{a\_1,\cdots,a\_N\}}

, then

{
f
>
a
# }
∅

{\displaystyle \{f>a\}=\emptyset}

is measurable so

f

{\displaystyle f}

is measurable. If

a
<
max
{

a
1

,
⋯
,

a
N

}

{\displaystyle a<\max\{a\_1,\cdots,a\_N\}}

, then

{
f
>
a
# }

⋃

k

:

a
k

>
a

E
k

{\displaystyle \{f>a\}=\bigcup\_{k~:~a\_k>a}E\_k}

is a finite union of measurable sets so it is measurable. Thus,

f

{\displaystyle f}

is measurable.
Julie Makimoto

15 (New Vestroia)

16 (Gundalian Invaders)

17 (Mechtanium Surge)

18 (BioTechnic Metal)

19 (Master Quest)

27 (Hyperion Leonidas)

900 (Taurus)

1000 (Roaden)

1100 (Adgaran)

### Voice Actor

Katie Griffin (English) (Nelvana Network)Lara Woodhull (English) (Capcom Co., Ltd.)

Risa Mizuno (Japanese)

### Guardian Bakugan

Hammer Gorem (BB & NV)Stone Dyrex (NV)

Taurus (GI)

Roaden (MSA1)

Adgaran (MSA2)

### Home Palace

Wardington City, Earth (BB - NV)Bayview City, Earth (GI - MSA1)

Bakugan City, Earth (MSA2)

### Faction

Battle BrawlersBakugan Battle Brawlers Resistance

Order of the Castle Knights (Formerly)
6-1

Question:

(a)Let

E

{\displaystyle E}

 be a measurable subset of

R
2

{\displaystyle \mathbf{R^2}}

 such that for almost every

x
∈

R
1

{\displaystyle x \in \mathbf{R^1}}

 , the set

{

y

(
x
,
y
)
∈
E

}

{\displaystyle \left \{ y|(x,y) \in E \right \} }

 has

R
1

{\displaystyle \mathbf{R^1}}

-measure zero. Show that

E

{\displaystyle E}

 has measure zero and that for almost every

y
∈

R
1

{\displaystyle y \in \mathbf{R^1}}

, the set

{

x

(
x
,
y
)
∈
E

}

{\displaystyle \left \{ x|(x,y) \in E \right \} }

 has measure zero.

(b) Let

f
(
x
,
y
)

{\displaystyle f(x, y)}

be nonnegative and measurable in

R
2

{\displaystyle \mathbf{R^2}}

. Suppose that for almost every

x
∈

R
1

{\displaystyle x \in \mathbf{R^1}}

,

f
(
x
,
y
)

{\displaystyle f(x, y)}

 is finite for almost every y. Show that for almost every

y
∈

R
1

{\displaystyle y \in \mathbf{R^1}}

,

f
(
x
,
y
)

{\displaystyle f(x, y)}

 is finite for almost every x.

Solution:

(a) Consider the indicator function

χ

E

{\displaystyle \chi\_{E}}

 on

R
2

{\displaystyle \mathbf{R^2}}

. This is a nonnegative measurable function, so by Fubini’s theorem we have

∬

R

2

χ

E

=

∫

R

(

∫

R

χ

E

(
x
,
y
)
d
y

)

d
# x

∫

R

(

∫

R

χ

E

(
x
,
y
)
d
x

)

d
y
.

{\displaystyle \iint\_{\mathbb{R}^{2}} \chi\_{E}=\int\_{\mathbb{R}}\left(\int\_{\mathbb{R}} \chi\_{E}(x, y) d y\right) d x=\int\_{\mathbb{R}}\left(\int\_{\mathbb{R}} \chi\_{E}(x, y) d x\right) d y .}

For almost every

x
∈
R

{\displaystyle x \in R }

, we have

∫

R

χ

E

(
x
,
y
)
d
# y

{
y
:
(
x
,
y
)
∈
E
}

=
0

{\displaystyle \int\_{\mathbb{R}} \chi\_{E}(x, y) d y=|\{y:(x, y) \in E\}|=0}

.

so all the integrals in the first line above are 0. Since

∬

R

2

χ

E

=

E

{\displaystyle \iint\_{\mathbb{R}^{2}} \chi\_{E} = | E|}

, we have

E

=
0

{\displaystyle | E|=0}

.

Moreover, since

∫

R

(

∫

R

χ

E

(
x
,
y
)
d
x

)

d
# y

∫

R

{
x
:
(
x
,
y
)
∈
E
}

d
y

{\displaystyle \int\_{\mathbb{R}}\left(\int\_{\mathbb{R}} \chi\_{E}(x, y) d x\right) d y=\int\_{\mathbb{R}}|\{x:(x, y) \in E\}| d y}

and since

{
x
:
(
x
,
y
)
∈
E
}

{\displaystyle |\{x:(x, y) \in E\}|}

 is a non-negative function of y whose integral over y is zero,it follows that

{
x
:
(
x
,
y
)
∈
E
}

=
0

{\displaystyle |\{x:(x, y) \in E\}| = 0}

 for almost every y.

(b) Let

E

{\displaystyle E}

 be the set of points on which f is not finite. This is a measurable subset of

R
2

{\displaystyle \mathbf{R^2}}

, and

E

x

=
{
y
:
(
x
,
y
)
∈
E
}

{\displaystyle E\_{x} = \{y:(x, y) \in E\} }

 has measure zero for almost every

x
∈
R

{\displaystyle x \in R }

. so by part (a), we could obtain

E

y

=
{
x
:
(
x
,
y
)
∈
E
}

{\displaystyle E\_{y} = \{x:(x, y) \in E\} }

 has measure 0 for almost every

y
∈
R

{\displaystyle y \in R }

, which is equivalent to what we want to prove.
7-1

**Solution 7.1.**

Let

f

{\displaystyle f}

 be measurable in

R

n

{\displaystyle \R^n}

 and different from zero in some set of positive measure. Show that there is a positive constant *c* such that

f

∗

(
x
)
≥
c

x

−
n

{\displaystyle f^{\*}(x) \geq c|x|^{-n}}

  for

x

≥
1

{\displaystyle |x| \geq 1}

 .

**Solution:**

Let

E
⊂

R

n

{\displaystyle E \subset \R^n}

 be a bounded subset of positive measure on which

f

>
0

{\displaystyle |f|>0}

, let

x
∈

R

n

{\displaystyle x \in \mathbb{R}^n}

 and

x

≥
1

{\displaystyle |x| \geq 1}

, and let

Q

′

{\displaystyle Q^{\prime}}

 be the smallest cube centered at

x

{\displaystyle x}

 containing

E

{\displaystyle E}

. If the side length of

Q

′

{\displaystyle Q^{\prime}}

is

k

x

{\displaystyle k|x|}

, take

# c

∫

E

f

k

n

{\displaystyle c=\frac{\int\_{E}|f|}{k^{n}} }

. Then we have

f

∗

(
x
# )

sup

Q

1

Q

∫

Q

f

≥

sup

Q

1

Q

∫

Q
∩
E

f

≥

1

Q

′

∫

Q

′

∩
E

f

=

∫

E

f

k

n

x

n

=
c

x

−
n

,

where the sup is taken over all cubes centered at

x

{\displaystyle x}

.
7-8

Question: Prove the following converse of Theorem 7.31: If

f

{\displaystyle f}

 is of bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

, and if the function

V
(
x
# )
V
[
a
,
x
]

{\displaystyle V(x)=V[a,x]}

 is absolutely continuous on

[
a
,
b
]

{\displaystyle [a, b]}

, then

f

{\displaystyle f}

 is absolutely continuous on

[
a
,
b
]

{\displaystyle [a, b]}

.

Solution:

Since

f

{\displaystyle f}

 is of bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

, then we have

V
(
x
# )
V
[
a
,
x
]
≤
V
[
a
,
b
]
<
∞

{\displaystyle V(x)=V[a,x]\leq V[a,b]<\infty }

for all

a
≤
x
≤
b

{\displaystyle a \le x \le b}

.

Thus,

V
(

b

i

)
−
V
(

a

i

)

{\displaystyle V(b\_{i})-V(a\_{i})}

 is well-defined and so is

∑

V
(

b

i

)
−
V
(

a

i

)

{\displaystyle \sum |V(b\_{i})-V(a\_{i})|}

.

Since the function

V
(
x
# )
V
[
a
,
x
]

{\displaystyle V(x)=V[a,x]}

 is absolutely continuous on

[
a
,
b
]

{\displaystyle [a, b]}

, by the definition of absolutely continuous, for any

ϵ
>
0

{\displaystyle \epsilon > 0}

, there exists

δ
>
0

{\displaystyle \delta > 0}

 such that for any collection

{
[

a

i

,

b

i

]
}

{\displaystyle \{[a\_{i},b\_{i}]\}}

 of nonoverlapping subintervals of

[
a
,
b
]

{\displaystyle [a, b]}

,

∑

V
(

b

i

)
−
V
(

a

i

)

<
ϵ

if

∑
(

b

i

−

a

i

)
<
δ
.

{\displaystyle \sum |V(b\_{i})-V(a\_{i})|<\epsilon \quad {\text{if}}\quad \sum (b\_{i}-a\_{i})<\delta .}

From Chapter 2, we know that

f
(

b

i

)
−
f
(

a

i

)

≤
V
[

a

i

,

b

i

]

{\displaystyle |f(b\_{i})-f(a\_{i})|\leq V[a\_{i},b\_{i}]}

. So, if

∑
(

b

i

−

a

i

)
<
δ

{\displaystyle \sum (b\_{i}-a\_{i})<\delta }

,

∑

f
(

b

i

)
−
f
(

a

i

)

≤
∑
V
[

a

i

,

b

i

]
.

{\displaystyle \sum |f(b\_{i})-f(a\_{i})|\leq \sum V[a\_{i},b\_{i}].}

Since

V
[

a

i

,

b

i

# ]
V
[
a
,

b

i

]
−
V
[
a
,

a

i

# ]
V
(

b

i

)
−
V
(

a

i

)

{\displaystyle V[a\_{i},b\_{i}]=V[a,b\_{i}]-V[a,a\_{i}]=V(b\_{i})-V(a\_{i})}

, we then have

∑

f
(

b

i

)
−
f
(

a

i

)

≤
∑

V
(

b

i

)
−
V
(

a

i

)

<
ϵ

when

∑
(

b

i

−

a

i

)
<
δ

{\displaystyle \sum (b\_{i}-a\_{i})<\delta }

.

Thus, we show that

f

{\displaystyle f}

 is absolutely continuous on

[
a
,
b
]

{\displaystyle [a, b]}

.
She is not a lady

**She is not a lady** - 4 odcinek z serii Family Francis

## Fabuła

Andre przychodzi do swej matki i opowiada o swojej "połówce,,. Ta się wścieka i bierze w ręce kieliszek i... Nagle Francis osłania Andre i przyjmuje na Ciebie cios.

## Bohaterowie

* Andre
* Cecylia
* Beatrice
* Francis

## Wypływ na serie Family Francis

1. Francis od tego czasu ma bliznę
2. Wpłynął ten odcinek na relacje Cecylii z Francisem i Andre.
8-2

**Question 8-2**: Prove the converse of Hölder's inequality for

# p
1

{\displaystyle p = 1}

 and

∞

{\displaystyle \infty}

. Show also that for

1
≤
p
≤
∞

{\displaystyle 1\leq p\leq \infty}

, a real-valued measurable

f

{\displaystyle f}

 belongs to

L
p

(
E
)

{\displaystyle L^p(E)}

 if

f
g
∈

L
1

(
E
)

{\displaystyle fg\in L^1(E)}

 for every

g
∈

L

p
′

(
E
)

{\displaystyle g\in L^{p'}(E)}

,

1

/

p
1

/

p
′

=
1

{\displaystyle 1/p+1/p'=1}

. The negation is also of interest: if

f
∉

L
p

(
E
)

{\displaystyle f\notin L^p(E)}

, then there exists

g
∈

L

p
′

(
E
)

{\displaystyle g\in L^{p'}(E)}

 such that

f
g
∉

L
1

(
E
)

{\displaystyle fg\notin L^1(E)}

. (To verify the negation, construct g of the form

∑

a
k

g
k

{\displaystyle \sum a\_kg\_k}

 for appropriate

a
k

{\displaystyle a\_k}

and

g
k

{\displaystyle g\_k }

, with

g
k

{\displaystyle g\_k }

 satisfying

∫
E

f

g
k

→
∞

{\displaystyle \int\_E fg\_k\rightarrow +\infty}

.)

**Answer**: Part 1 has a solution on Math Stackexchange here: https://math.stackexchange.com/questions/4179726/a-certain-converse-of-h%C3%B6lders-inequality-for-l-infty-space

**Part 2**: **Show also that for

1
≤
p
≤
∞

{\displaystyle 1\leq p\leq \infty}

, a real- valued measurable

f

{\displaystyle f}

 belongs to

L
p

(
E
)

{\displaystyle L^p(E)}

 if

f
g
∈

L
1

(
E
)

{\displaystyle fg\in L^1(E)}

 for every

g
∈

L

p
′

(
E
)

{\displaystyle g\in L^{p'}(E)}

,

1

/

p
1

/

p
′

=
1

{\displaystyle 1/p+1/p'=1}**.

**Lemma 8.1.1.** There exists

M
>
0

{\displaystyle M > 0}

 such that

‖
f
g

‖
1

≤
M

{\displaystyle \lVert fg \rVert\_1\leq M}

, for all

g
∈

L

p
′

(
E
)

{\displaystyle g\in L^{p'}(E)}

,

‖
g

‖

p
′

≤
1

{\displaystyle \lVert g \rVert\_{p'}\leq 1}

.

*proof.* Suppose not. Then we have a sequence of

L

p
′

{\displaystyle L^{p'}}

 functions

{

g
k

}

{\displaystyle \{g\_k\}}

 with

‖

g
k

‖

p
′

≤
1

{\displaystyle \lVert g\_k \rVert\_{p'}\leq 1}

 where

∫
E

f

g
k

>

3
k

{\displaystyle \int\_E|fg\_k|>3^k}

.

Let

# k
1

∞

2

−
k

g
k

{\displaystyle g=\sum\_{k=1}^\infty 2^{-k}|g\_k|}

. Then

‖
g

‖

p
′

≤

∑

# k
1

∞

2

−
k

=
1

{\displaystyle \lVert g \rVert\_{p'}\leq\sum\_{k=1}^\infty 2^{-k}=1}

 but

∫
E

f
g

=

∑

# k
1

∞

2

−
k

∫
E

f

g
k

>

∑

# k
1

∞

(

3
2

)
k

=
∞

,

so

f
g
∉

L
1

(
E
)

{\displaystyle fg\notin L^1(E)}

. This is a contradiction.

So

‖
f

‖
p

=

sup

‖
g

‖

p
′

≤
1

∫
E

f
g
≤

sup

‖
g

‖

p
′

≤
1

‖
f
g

‖
1

≤
M
<
∞

{\displaystyle \lVert f \rVert\_p=\sup\_{\lVert g \rVert\_{p'}\leq1}\int\_E fg\leq\sup\_{\lVert g \rVert\_{p'}\leq1}\lVert fg \rVert\_1\leq M<\infty}

.

Thus

f
∈

L
p

(
E
)

{\displaystyle f \in L^p(E)}

.

**Part 3: Show if

f
∉

L
p

(
E
)

{\displaystyle f\notin L^p(E)}

, then there exists

g
∈

L

p
′

(
E
)

{\displaystyle g\in L^{p'}(E)}

 such that

f
g
∉

L
1

(
E
)

{\displaystyle fg\notin L^1(E)}

.**
Shun Kazami

16 (New Vestroia)

17 (Gundalian Invaders)

18 (Mechtanium Surge)

19 (BioTechnic Metal)

20 (Master Quest)

28 (Hyperion Leonidas)

900 (Hawktor)

1000 (Taylean)

1100 (Jaakor)

### Voice Actor

Zachary Bennett (English) (Nelvana Network)Roger Craig Smith (English) (Capcom Co., Ltd.)

Chihiro Suzuki (Japanese)

### Guardian Bakugan

Storm Skyress (BB & NV)Master Ingram (NV)

Hawktor (GI)

Taylean (MSA1)

Jaakor (MSA2)

Skytruss (MSA2) (Temporary)

Orbeum (MSA2) (Temporary)

### Faction

Battle BrawlersBakugan Battle Brawlers Resistance

Order of the Castle Knights (Formerly)

### Parents

Shiori (Mother)Grandpa Kazami (Grandfather)
6-11

∫

R

n

e

−
‖
x

‖
2

=

π

n

/

2

{\displaystyle \int\_{\mathbb{R}^n}e^{-\|\text{x}\|^2}=\pi^{n/2}}

(For

# n
1

{\displaystyle n=1}

, write

(

∫

−
∞

∞

e

−

x
2

d
x

)

2

=

∫

−
∞

∞

∫

−
∞

∞

e

−

x
2

−

y
2

d
x

d
y

{\textstyle \left(\int^\infty\_{-\infty}e^{-x^2} \ dx\right)^2=\int^\infty\_{-\infty}\int^\infty\_{-\infty}e^{-x^2-y^2} \ dx \ dy}

 and use polar coordinates. For

n
>
1

{\displaystyle n>1}

, use the formula

e

−
‖
x

‖
2

=

e

−

x
1
2

⋯

e

−

x
n
2

{\displaystyle e^{-\|\text{x}\|^2}=e^{-x\_1^2}\cdots e^{-x\_n^2}}

 and Fubini's theorem to reduce to the case

# n
1

{\displaystyle n=1}

.)

# n
1

{\displaystyle n=1}

, we consider the integral

I
1

=

∫

−
∞

∞

e

−

x
2

d
x

{\displaystyle I\_1=\int^\infty\_{-\infty} e^{-x^2} \ dx}

 by evaluating

∫

R

2

e

−

x
2

−

y
2

d
x

d
y

=

∫

0

∞

∫
0

2
π

r

e

−

r
2

d
r

d
θ

=

∫
0

2
π

d
θ
⋅

1
2

∫

0

∞

e

−

r
2

d

r
2

=
θ

0

2
π

⋅

1
2

(

−

e

−

r
2

)

0

∞

=
π

{\displaystyle \begin{align}
\int\_{\mathbb{R}^2}e^{-x^2-y^2} \ dx \ dy
&= \int^\infty\_{0}\int^{2\pi}\_0 re^{-r^2} \ dr \ d\theta\\
&=\int^{2\pi}\_0 \ d\theta\cdot\dfrac{1}{2}\int^\infty\_{0} e^{-r^2} \ dr^2\\
&=\theta\ \biggr|^{2\pi}\_0 \cdot \dfrac{1}{2}\left(-e^{-r^2}\right)\biggr|^{\infty}\_0\\
&=\pi
\end{align}
}

By Fubini's theorem,

# π

∫

R

2

e

−

x
2

−

y
2

d
x

d
# y

∫

R

e

−

x
2

d
x
⋅

∫

R

e

−

y
2

d
# y

(

∫

R

e

−

x
2

d
x

)

2

=

I
1
2

{\displaystyle \pi=\int\_{\mathbb{R}^2}e^{-x^2-y^2} \ dx \ dy =\int\_{\mathbb{R}}e^{-x^2} \ dx \cdot \int\_{\mathbb{R}}e^{-y^2} \ dy=\left(\int\_{\mathbb{R}}e^{-x^2} \ dx\right)^2=I\_1^2
}

 implies that

I
1

=

π

1

/

2

{\displaystyle I\_1=\pi^{1/2}
}

. For any

n
>
1

{\displaystyle n>1}

, we always have

e

−
‖
x

‖
2

=

∏

# i
1

n

e

−

x
i
2

{\displaystyle e^{-\|x\|^2}=\prod^n\_{i=1}e^{-x\_i^2}
}

, thus assume that the statement holds for

n
−
1

{\displaystyle n-1}

. For the case

n

{\displaystyle n}

, we have

∫

R

n

e

−
‖
x

‖
2

d
# x

∫

R

e

−

x
1

d

x
1

⋅

∫

R

n
−
1

∏

# i
1

n
−
1

e

−

x
i
2

d

x
2

∧
⋯
∧
d

x
n

=

π

1

/

2

⋅

π

n
−
1

2

=

π

n

/

2

{\displaystyle \int\_{\mathbb{R}^n}e^{-\|x\|^2} \ dx=\int\_{\mathbb{R}}e^{-x\_1} \ dx\_1\cdot \int\_{\mathbb{R}^{n-1}}\prod\_{i=1}^{n-1} e^{-x\_i^2} \ dx\_2\wedge \cdots \wedge dx\_n=\pi^{1/2}\cdot\pi^{\frac{n-1}{2}}=\pi^{n/2}
}

by Fubini's theorem. This shows the assertion by induction on

n

{\displaystyle n}

.

◼

{\displaystyle \blacksquare }
Mega Man Grey Wiki

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Ex2-19

Let

Γ

1

,

Γ

2

{\displaystyle {\Gamma\_{1},\Gamma\_{2}}}

 be partitions of

[

a

′

,

b

′

]

{\displaystyle {[a^{\prime},b^{\prime}]}}

 and choose two families of intermediate points for

Γ

1

,

Γ

2

{\displaystyle {\Gamma\_{1},\Gamma\_{2}}}

 respectively. Also, let

Γ
,

Γ

′

{\displaystyle {\Gamma, \Gamma^{\prime}}}

 be partition of

[
a
,

a

′

]

{\displaystyle {[a, a^{\prime}]}}

 and

[

b

′

,
b
]

{\displaystyle {[b^{\prime}, b]}}

. Then

# U

Γ

1

∪
Γ
∪
Γ

′

{\displaystyle {U=\Gamma\_{1}\cup\Gamma\cup\Gamma{\prime}}}

 and

# V

Γ

2

∪
Γ
∪

Γ

′

{\displaystyle {V=\Gamma\_{2}\cup\Gamma\cup\Gamma^{\prime}}}

 are two partitions of

[
a
,
b
]

{\displaystyle {[a, b]}}

. Choose two more families of intermediate points for partitions

Γ
,

Γ

′

{\displaystyle {\Gamma, \Gamma^{\prime}}}

 and let the union of intermediate points of

Γ

1

,

Γ

2

,
Γ
,

Γ

′

{\displaystyle {\Gamma\_{1},\Gamma\_{2},\Gamma,\Gamma^{\prime}}}

 be the family of intermediate points of

U

{\displaystyle {U}}

 and

V

{\displaystyle {V}}

. We have

R

U

[
a
,
b
# ]

R

Γ

[
a
,

a

′

]

R

Γ

1

[

a

′

,

b

′

]

R

Γ

′

[

b

′

,
b
]

{\displaystyle {R\_{U}[a, b]=R\_{\Gamma}[a, a^{\prime}]+R\_{\Gamma\_{1}}[a^{\prime},b^{\prime}]+R\_{\Gamma^{\prime}}[b^{\prime},b]}}

 and

R

V

[
a
,
b
# ]

R

Γ

[
a
,

a

′

]

R

Γ

2

[

a

′

,

b

′

]

R

Γ

′

[

b

′

,
b
]

{\displaystyle {R\_{V}[a, b]=R\_{\Gamma}[a, a^{\prime}]+R\_{\Gamma\_{2}}[a^{\prime},b^{\prime}]+R\_{\Gamma^{\prime}}[b^{\prime},b]}}

. Now, given any

ϵ
>
0

{\displaystyle {\epsilon>0}}

 such that

R

Γ

1

[

a

′

,

b

′

]
−

R

Γ

2

[

a

′

,

b

′

]

<
ϵ

{\displaystyle {|R\_{\Gamma\_{1}}[a^{\prime},b^{\prime}]-R\_{\Gamma\_{2}}[a^{\prime},b^{\prime}]|<\epsilon}}

, since

∫
a
b

f

d

ϕ

{\displaystyle {\int\_a^b \! f \, \mathrm{d}\phi}}

 exists, there is some

δ
>
0

{\displaystyle {\delta>0}}

 such that

U

,

V

<
δ

 implies

R

Γ

U

[
a
,
b
]
−

R

Γ

V

[
a
,
b
]

<
ϵ

{\displaystyle {|R\_{\Gamma\_{U}}[a,b]-R\_{\Gamma\_{V}}[a,b]|<\epsilon}}

. Finally, assuming

Γ

,

Γ

′

<
δ

 and suppose that

Γ

1

,

Γ

2

<
δ

, we have

U

=

Γ

1

∪
Γ
∪
Γ

′

,

V

=

Γ

2

∪
Γ
∪
Γ

′

<
δ

, this implies

R

Γ

U

[
a
,
b
]
−

R

Γ

V

[
a
,
b
]

=

R

Γ

1

[

a

′

,

b

′

]
−

R

Γ

2

[

a

′

,

b

′

]

<
ϵ

, which complete the proof
Assassin's Creed: Reborn

Przykro mi, ale opowieść jest jeszcze nieskończona...Artykuł ten opisuje wydarzenia lub jest związany z opowieścią pt. **Assassin's Creed: Reborn**, która nie została jeszcze w pełni ukończona. Kiedy opowieść ta zostanie skończona, ktoś na pewno uzupełni ten artykuł. Cierpliwości.
**Assassin's Creed: Reborn** (*pl.* Assassin’s Creed: Odrodzeni) – fanowska opowieść autorstwa użytkownika SqeriX, opowiadająca o losach młodego Bastiena Duponta – Francuza, który uciekając przed epidemią dżumy trafia do rządzonej przez Kazimierza Wielkiego Polski. Niedługo po przybyciu młodzieniec pakuje się w kłopoty, z których wyciąga go tajemniczy zakapturzony jegomość.

Aktualnie opowieść jest w procesie tworzenia od 8 kwietnia 2021 roku.

## Fabuła

### Współczesność

Jedna z niewielu komórek asasynów działających w Polsce spostrzega, iż Abstergo, które nie wyrażało zbyt dużego zainteresowania Polską, nagle zaczęło coraz częściej odwiedzać okolice Gdańska. Inna polska komórka, zdobywszy dane wywiadowcze, przekazuje im, że templariusze pozyskawszy próbki DNA polskiego asasyna Lecha Bogusławowicza, żyjącego w czasach Kazimierza Wielkiego, za pomocą Animusa śledzą jego wspomnienia genetyczne, które doprowadziły ich nad Bałtyk. Pozyskawszy informacje o życiorysie Bogusławowicza asasyni postanawiają odnaleźć szczątki jego bliskiego przyjaciela Bastiena Duponta, aby pozyskać z nich DNA i odtworzyć jego wspomnienia, aby sprawdzić, co odkrył XIV-wieczny asasyn i dlaczego Abstergo się tym interesuje.

### Przeszłość

Głównym bohaterem opowieści jest młody Bastien Dupont, mieszkający w małej miejscowości pod Paryżem. Pod koniec 1348 roku, gdy do jego domu dociera epidemia dżumy, a jego rodzice zostają zarażeni, ojciec chłopaka daje mu swoje ostatnie pieniądze i nakazuje mu uciekać jak najdalej od domu. Bastien wyrusza do Polski – ojczyzny swojego ojca, ponieważ ma nadzieję, że epidemia jeszcze tam nie dotarła.

Po długiej i wykańczającej podróży Bastien dociera do Krakowa, gdzie niedługo po przybyciu pakuje się w poważne kłopoty. Z pomocą przychodzi mu wtedy tajemniczy zakapturzony człowiek, który wyciąga go z kłopotów i pomaga mu uciec z miasta. Widząc w Bastienie potencjał, nieznajomy przygarnia go i wciąga w temat sekretnej wojny toczonej od zarania dziejów. Podczas swojej podróży duet walczy z zagrożeniami dla ludzkiej wolności i próbuje wyjaśnić, w jaki sposób pewien dobrze im znany zakon, który od kilkudziesięciu lat powinien już nie istnieć, istnieje dalej i kontynuuje swoje plany wobec świata.
Ex2-10

Question:

If

λ
1

<

λ
2

<
…
<

λ
m

{\displaystyle \lambda\_1 < \lambda\_2 < \ldots < \lambda\_m}

 is a finite sequence and

−
∞
<
s
<
∞

{\displaystyle -\infty<s<+\infty}

, write

∑
k

a
k

e

−
s

λ
k

{\displaystyle \sum\_k a\_k e^{-s\lambda\_k}}

 as a Riemann–Stieltjes integral.

Solutions:

Take the function

f
:
[
a
,
b
]
→

R

,
f
(
x
# )

e

−
s
x

{\displaystyle f:[a,b]\to \mathbb{R}, f(x)=e^{-sx}}

, and

ϕ
:
[
a
,
b
]
→

R

,
ϕ
(
x
# )

{

a
1

a
≤
x
<

λ
1

,

a
1

a
2

λ
1

≤
x
<

λ
2

,

…

…

a
1

a
2

…

a

m
−
1

λ

m
−
1

≤
x
<

λ
m

,

a
1

a
2

…

a

m

λ

m

≤
x
≤
b

.

{\displaystyle \phi:[a,b]\to \mathbb{R}, \phi(x)=\left\{
\begin{array}{lcl}
a\_1 & & {a \leq x < \lambda\_1 },\\
a\_1+a\_2 & & {\lambda\_1 \leq x < \lambda\_2},\\
\ldots & & {\ldots}\\
a\_1+a\_2+\ldots+a\_{m-1} & & {\lambda\_{m-1}\leq x < \lambda\_m},\\
a\_1+a\_2+\ldots+a\_{m} & & {\lambda\_{m}\leq x \leq b}.
\end{array} \right.}

Consider a partition

Γ
n

{\displaystyle \Gamma\_n}

 of the interval

[
a
,
b
]

{\displaystyle [a, b]}

 such that

Γ
n

=
{

x
0

,

x
1

,
…
,

x
n

}

{\displaystyle \Gamma\_n=\{x\_0,x\_1,\ldots,x\_n\}}

 with

x
i

=
a
i

b
−
a

n

{\displaystyle x\_i=a+i\frac{b-a}{n}}

,

n
>
m

{\displaystyle n > m}

 and there exists a subsequence

{

x

l

i

}

{\displaystyle \left \{ x\_{l\_{i}} \right \} }

 such that

λ
i

∈
[

x

l
i

−
1

,

x

l
i

]

{\displaystyle \lambda\_i \in [x\_{l\_i-1},x\_{l\_i}]}

, for every

λ
i

,
# i
1
,
2
,
…
,
m

{\displaystyle \lambda\_i,i=1,2,\ldots,m}

, .

Then we can write the Riemann–Stieltjes sum as

R

Γ
n

=

∑

# i
1

n

f
(

ξ
i

)
(
ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

)
)

=

∑

# k
1

m

f
(

ξ

l
k

)
(
ϕ
(

x

l
k

)
−
ϕ
(

x

l
k

−
1

)
)

=

∑

# k
1

m

e

−
s

ξ

l
k

a
k

,

{\displaystyle \begin{aligned}
R\_{\Gamma\_n}
&= \sum\_{i=1}^{n}f(\xi\_i)(\phi(x\_{i})-\phi(x\_{i-1}))\\
&= \sum\_{k=1}^{m} f(\xi\_{l\_k})(\phi(x\_{l\_k})-\phi(x\_{l\_k-1}))\\
&= \sum\_{k=1}^{m} e^{-s\xi\_{l\_k}}a\_k,
\end{aligned}}

where the

ξ

l

k

{\displaystyle \xi\_{l\_{k}}}

 are chosen sample points such that

ξ

l

k

∈
[

x

l

k

−
1

,

x

l

k

]

{\displaystyle \xi\_{l\_{k}} \in [x\_{l\_{k}-1},x\_{l\_{k}}]}

.(In this case,

λ

k

∈
[

x

l

k

−
1

,

x

l

k

]

{\displaystyle \lambda\_{k} \in [x\_{l\_{k}-1},x\_{l\_{k}}]}

, so

ϕ
(

x

l
k

)
−
ϕ
(

x

l
k

−
1

# )

a

k

{\displaystyle \phi(x\_{l\_k})-\phi(x\_{l\_k-1})=a\_{k}}

)

If we let

n
→
∞

{\displaystyle n \to \infty}

, then the gap of

Γ
n

{\displaystyle \Gamma\_n}

 converges to 0.

In addition, now, we can only choose points

{

ξ

l
k

}

{\displaystyle \{\xi\_{l\_k}\}}

 as points

{

λ
k

}
,
# k
1
,
2
,
…
,
m

{\displaystyle \{\lambda\_k\}, k=1,2,\ldots,m}

.

That is,

lim

n
→
∞

R

Γ
n

=

∑
k

a
k

e

−
s

λ
k

{\displaystyle \lim\_{n\to \infty}R\_{\Gamma\_n}=\sum\_k a\_k e^{-s \lambda\_k} }

, which is exactly the Riemann–Stieltjes integral of

f

{\displaystyle f}

 with respect to

ϕ

{\displaystyle \phi}

 on

[
a
,
b
]

{\displaystyle [a, b]}

.
Polska

**Polska** – kraj w Europie Środkowej, położony nad Morzem Bałtyckim. Sąsiaduje z Niemcami, Czechami, Słowacją, Ukrainą, Białorusią, Litwą i Rosją.

## Polscy asasyni

### Średniowiecze

Znanymi polskimi asasynami z czasów średniowiecza są:

* Bogusław Czesławiak
* Lech Bogusławowicz

### Współczesność

Obecnie w Polsce znajdują się cztery komórki bractwa. Mają one swoje kryjówki na Śląsku, w Małopolsce, na Pomorzu i w Wielkopolsce.

#### Komórka Małopolska

* Andrzej Wiśniewski
* Karolina Kłos
* Michał Lato
* Grzegorz Walicki
* Anna Piotrowska
Bolesław Miłoszewicz

**Bolesław Miłoszewicz** (ur. ok. 1300 roku we wsi pod Poznaniem) – polski chłop, mąż nieznanej z imienia kobiety, ojciec Bastiena Duponta.

## Biografia

Bolesław Miłoszewicz urodził się około 1300 roku w jednej ze wsi pod Poznaniem, podczas rywalizacji Wacława II i Władysława Łokietka o polski tron. Około 1330 roku wyjechał do Francji, gdzie poznał nieznaną z imienia chłopkę, z którą niedługo potem się ożenił. Mieszkali oni w małej miejscowości pod Paryżem. W 1334 roku urodził im się syn, Bastien.

Bolesław wiódł wraz z żoną i synem spokojne chłopskie życie. Uczył również Bastiena podstaw języka polskiego, ponieważ chciał, aby jego syn chłonął dużo wiedzy i wyrósł na wykształconego człowieka. W 1348 roku szerząca się po Europie epidemia dżumy dotarła pod Paryż, a sąsiedzi Bolesława w bardzo szybkim tempie zarażali się i umierali. W najbliższej okolicy nie było również żadnego lekarza, który mógłby zająć się chorymi. Nie minęło dużo czasu, a dżuma dotarła do domu Bolesława, który zaraził się wraz z żoną od którychś sąsiadów. Wiedzieli oni, że nie zostało im zbyt wiele czasu, jednak ich syn miał jeszcze szansę ujść z życiem z miejscowości. Bolesław dał więc mu wszystkie swoje pieniądze i kazał uciekać do Polski w nadziei, że epidemia jeszcze tam nie dotarła. Bastien wysłuchał ojca i jak najszybciej opuścił miejscowość. Dalsze losy Bolesława Miłoszewicza są nieznane, lecz prawdopodobnie zmarł on niedługo później na skutek zarażenia się dżumą.
Ex2-28

Let

f

{\displaystyle f}

 be Lipschitz continuous on an interval

[
a
,
b
]

{\displaystyle [a, b]}

. Show that

V
[
f
;
a
,
b
]

{\displaystyle V[ f ; a, b]}

 is strictly less than the length of the graph of

f

{\displaystyle f}

 over

[
a
,
b
]

{\displaystyle [a, b]}

.

Proof:

Recall that

f

{\displaystyle f}

 is said to be Lipschitz continuous on

[
a
,
b
]

{\displaystyle [a, b]}

, or to be a Lipschitz function

[
a
,
b
]

{\displaystyle [a, b]}

, if there is a constant

C
≥
0

{\displaystyle C\geq 0}

 such that

f
(
x
)
−
f
(
y
)

≤
C

x
−
y

for all

x
,
y
∈
[
a
,
b
]
.

Consider a partition

# a

x
0

,

x
1

,
…
,

x
m

=
b
}
.

{\displaystyle \Gamma=\{a=x\_0,x\_1,\ldots,x\_m=b\}.}

 With each partition

Γ

{\displaystyle \Gamma}

, we have

S

Γ

[
f
;
a
,
b
]

=

∑

# i
1

m

f
(

x

i

)
−
f
(

x

i
−
1

)

=

f
(

x

1

)
−
f
(

x

0

)

f
(

x

2

)
−
f
(

x

1

)

…

f
(

x

m

)
−
f
(

x

m
−
1

)

≤
C

x

1

−

x

0

C

x

2

−

x

1

…
C

x

m

−

x

m
−
1

=
C
(

x

1

−

x

0

)
C
(

x

2

−

x

1

)
…
C
(

x

m

−

x

m
−
1

)

=
C
[

x

1

−

x

0

x

2

−

x

1

…

x

m

−

x

m
−
1

]

=
C
[

x

m

−

x

0

]

=
C
[
b
−
a
]
.

{\displaystyle \begin{align}
S\_{\Gamma}[f;a,b]&=\sum\_{i=1}^m |f(x\_{i})-f(x\_{i-1})|\\
&=C(x\_{1}-x\_{0})+C(x\_{2}-x\_{1})+\ldots+C(x\_{m}-x\_{m-1})\\
&=C[x\_{1}-x\_{0}+x\_{2}-x\_{1}+\ldots+x\_{m}-x\_{m-1}]\\
&=C[x\_{m}-x\_{0}]\\
&=C[b-a].
\end{align}}

Since this inequality holds for any

Γ

{\displaystyle \Gamma}

, thus

V
[
f
;
a
,
b
]

=

s
u
p

Γ

S

Γ

[
f
;
a
,
b
]

≤
C
[
b
−
a
]
.

{\displaystyle \begin{align}
V[f;a,b]&=\underset{\Gamma}{sup}~S\_{\Gamma}[f;a,b]\\
&\leq C[b-a].
\end{align}}

Also, the length of the graph of

f

{\displaystyle f}

 over

[
a
,
b
]

{\displaystyle [a, b]}

 is defined as

∫
a
b

1

f
′

(
x

)
2

d
x
.

{\displaystyle \int\_a^b \sqrt{1+f'(x)^2}~dx.}

 From Corollary 2.10,

V

=

∫
a
b

f
′

(
x
)

d
x

=

∫
a
b

[

f
′

(
x
)

]
2

d
x

<

∫
a
b

1

f
′

(
x

)
2

d
x
.

{\displaystyle \begin{align}
V&=\int\_a^b |f'(x)|~dx\\
&=\int\_a^b \sqrt{[f'(x)]^2}~dx\\
&<\int\_a^b \sqrt{1+f'(x)^2}~dx.
\end{align}}
9-5

**Question 9.5**. Let G and G1 be bounded open subsets of Rn such that

G
1

¯

⊂
G

{\displaystyle \bar{G\_1} \subset G}

 .Construct a function

h
∈

∁
0
∞

{\displaystyle h\in\complement\_0^\infty}

 such that h =1 in G1 and h =0 outside G.

Solutions:

We can choose an open G2 such that

G
1

¯

⊂

G
2

{\displaystyle \bar{G\_1} \subset {G\_2}}

 ,and

G
2

¯

⊂

G

{\displaystyle \bar{G\_2} \subset {G}}

.

**Proof.**

We use the fact that Rn is a normal space: every two disjoint closed sets of Rn have disjoint open neighborhoods.

Note that

G
1

¯

{\displaystyle \bar{G\_1} }

and Gc (complement of G) are disjoint closed sets. Thus there are disjoint open sets G2, G3 such that

G
1

¯

⊂

G
2

{\displaystyle \bar{G\_1} \subset {G\_2}}

and

G
c

¯

⊂

G
3

{\displaystyle \bar{G^c} \subset {G\_3}}

. Note that G2 ∩ G3 = ∅ implies

G
2

⊂

G
3
c

{\displaystyle G\_2 \subset G\_3^c}

 and

G
c

⊂

G
3

{\displaystyle G^c \subset G\_3}

 implies

G
3
c

⊂
G

{\displaystyle G\_3^c \subset G}

 . Further- more

G
3
c

{\displaystyle G\_3^c }

 is closed. Then

G
1

¯

⊂

G
2

⊂

G
2

¯

⊂

G
3
c

⊂

G

{\displaystyle \bar{G\_1}\subset{G\_2}\subset\bar{G\_2}\subset{G\_3^c}\subset{G}}

Define

ϵ
1

=
i
n
f

x
−
y

:
x
∈

G
1

,
y
∈

G
2
c

=
d
i
s
t
(

G
1

,
δ

G
2

)

{\displaystyle \epsilon\_1 =inf {|x-y|:x\in G\_1,y\in G\_2^c}= dist(G\_1,\delta G\_2)}

ϵ
2

=
i
n
f

x
−
y

:
x
∈

G
2

,
y
∈

G
c

=
d
i
s
t
(

G
2

,
δ
G
)

{\displaystyle \epsilon\_2 =inf {|x-y|:x\in G\_2,y\in G^c}= dist(G\_2,\delta G)}

And let

# ϵ
m
i
n
{

ϵ
1

,

ϵ
2

}

{\displaystyle \epsilon= min\{\epsilon\_1,\epsilon\_2\}}

By Question 4(c), we can choose

K
∈

C
0
∞

{\displaystyle K\in C\_0^\infty}

 with sup(K) =

B
ϵ

(
0
)

¯

{\displaystyle \bar{B\_\epsilon(0)}}

. Since K is continuous with compact support, K is integrable. By multiplying with a suitable constant,we can further assume that

∫
# K
1

{\displaystyle \int\limits K=1}

.

Let

h
(
x
# )
(
χ

G
2

∗
K
)
(
x
)
.

{\displaystyle h(x)=(\chi G\_2\*K)(x).}

Theorem 9.3 tells us that h ∈ C ∞

h
∈

C
∞

{\displaystyle h\in C^\infty}

, since

χ

G
2

∈

L
1

{\displaystyle \chi G\_2 \in L^1}

 as G2 is a bounded set.

Let

x
∈

G
1

{\displaystyle x\in G\_1}

.Then

h
(
x
# )

∫

R
n

χ

G
2

(
x
−
t
)
K
(
t
)
d
# t

∫

B
ϵ

(
0
)

χ

G
2

(
x
−
t
)
K
(
t
)
d
# t

∫

B
ϵ

(
0
)

K
(
t
)
d
# t
1

{\displaystyle h(x)=\int\limits\_{R^n} \chi G\_2(x-t)K(t)dt
=\int\limits\_{B\_\epsilon(0)} \chi G\_2(x-t)K(t)dt
=\int\limits\_{B\_\epsilon(0)} K(t)dt
=1}

Finally, if

x
∈

G
c

{\displaystyle x\in G\_c}

 , then

h
(
x
# )

∫

R
n

χ

G
2

(
x
−
t
)
K
(
t
)
d
# t

∫

B
ϵ

(
0
)

χ

G
2

(
x
−
t
)
K
(
t
)
d
# t
0

{\displaystyle h(x)=\int\limits\_{R^n} \chi G\_2(x-t)K(t)dt
=\int\limits\_{B\_\epsilon(0)} \chi G\_2(x-t)K(t)dt
=0}

since

x
−
t
∉

G
2

{\displaystyle x-t \not\in G\_2}

 for

t
<
ϵ

{\displaystyle t<\epsilon}

.(proved)
Chapter 2

Here are the homework exercises for Chapter 2: Functions of Bounded Variation and the Riemann-Stieltjes Integral

* Ex2-1

✓

{\displaystyle \checkmark}
* Ex2-2

✓

{\displaystyle \checkmark}
* Ex2-3 *✓

{\displaystyle \checkmark}*
* Ex2-4

✓

{\displaystyle \checkmark}
* Ex2-5✓

{\displaystyle \checkmark}
* Ex2-6✓

{\displaystyle \checkmark}
* Ex2-7*✓

{\displaystyle \checkmark}*
* Ex2-8 *✓

{\displaystyle \checkmark}*
* Ex2-9*✓

{\displaystyle \checkmark}*
* Ex2-10 *✓

{\displaystyle \checkmark}*
* Ex2-11 *✓

{\displaystyle \checkmark}*
* Ex2-12
* Ex2-13 *✓

{\displaystyle \checkmark}*
* Ex2-14

✓

{\displaystyle \checkmark}
* Ex2-15
* Ex2-16

✓

{\displaystyle \checkmark}
* Ex2-17
* Ex2-18
* Ex2-19
* Ex2-20
* Ex2-21
* Ex2-22
* Ex2-23
* Ex2-24
* Ex2-25
* Ex2-26
* Ex2-27
* Ex2-28
* Ex2-29
* Ex2-30 *(part (a) is proven)*
* Ex2-31 *✓

{\displaystyle \checkmark}*
The First Episode

The First Episode is an episode of NumberDecynary

## Characters

* Bluebuzz
* Alya
* Greenie
* Sunny
* Zay
* NDN-34 (mentioned in story)

## Story

On a hot rising sunny day. Nothing was there, but when fun came, Bluebuzz appeared in a lair, Bluebuzz walks down the 34 stairs. He finds so much fun, When he saw he was lonely, he spins around and makes Alya. Alya showed Bluebuzz how fun it is in a couple. She lighted her blocks saying “One two, one two!”, She lighted too much and she bumped right into Bluebuzz making Greenie. Greenie wanted to make a square. But He added up to Bluebuzz making Sunny. Despite his name, he was playing in the sun. Making squares and show how they are fun. Bluebuzz fell down as a prank making Zay, Zay showed Alya and Greenie prime numbers. Then Bluebuzz and Zay added up and made an explosion for a sign that says “Next episode for Six!” The episode ended. Credits happened
Winterrowd

**Winterrowd.com** was a website that was hosted by Jeremy Winterrowd, This website consist of screamers that can be often used to trick people, The type of screamers are videos, animations, games and more.

The website used to host his resume and did not provide any screamers (at the time when his game didn't attract an amount of popularity), However, in 2007, he provided of his screamer creations such as Optical Illusion, The Maze, or Ghost Photography and many more

The creator also created a contest since 2007 called: "Winterrowd Prank Contest", the objective of the contest is to send a reaction video to Jeremy, if someone wins the contest, they are rewarded with 100$ dollars, and will be uploaded to a channel called: "prankcontest", this channel also provides some screamers despite the reaction.
Solid Sharkey's Polybius Page

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**Solid Sharkey's Polybius Page** is a website that is created by Solid Sharkeys on October 2003.

When opened, the animation next to a text that states: "Polybius" is a flashing screenshot of a Zombie from Fatal Frame, it is the same picture used from Kikia.

Despite the screamer, the page actually provides some information that summarizes the story of Polybius expect for the part that states: "Interrupting the game with a screaming horror like the one on the top of this page (which probably just cost me a few friends.)"
Say It

The screenshot.

**Say It** is a simple screamer animation that was uploaded to Ebaumsworld in 2002.

This animation consist of a background and a music of Say It by Bumbai Kings plays on the lyrics of the "verse" part of the song scrolls up on the screen,

This continues normally until 22 seconds, right after the line: "Say those three little words I can't wait to hear" is shown, a picture of Regan MacNeil from The Exorcist appears along with a loud scream, which later to be used in Toy Story Secrets. Then the animation returns back to its original state.

There's another version that was created by joshtheblade on 2006, but the excerpt of the song is than the original screamer itself, and the animation uses another Regan picture (where she has no eyes.) with the K-Fee scream.. It appears before the scream. This version can be found here.
Scary Maze Game

The Level 3 in the game.

**Scary Maze Game** or known as **The Maze**, **Maze Game** is a popular flash game that was developed by Jeremy Winterrowd in 2004, the purpose of the game is to scare off people or it can be often used as a bait and switch by sending a link to someone.

The game begins with a start menu that consist of a button, a turquoise background, and a note that informs the player to guide the dot all the way of 4 levels without hitting the walls or else it leads back to the menu and the player has to start over, in addition to its note, it also request them to adjust the volume, However, there are no sounds during the gameplay expect for one part.

The player controls the dot with the use of arrow keys, then, after passing the first 2 levels, the third one, of course, contains thin walls that makes the player stare at the computer screen while trying to not touch it, However, if the player almost completes it, as a result, a picture of Regan MacNeil from The Exorcist appears with 2 loud screams, in conclusion, the replay button is displayed in the corner of the screen as well as a link to the now-defunct website: "Winterrowd.com".

## **Versions**

There is also an alternate version that uses the face from What's Wrong instead of Regan, Additionally, there are some minor changes of the game, the path and the start menu's background is yellow instead of turquoise, this version was also used in America's Funniest Home Videos.

### Level 4

Since the game only consist of 3 levels, there are also a video that demonstrates how to get to Level 4 in the game, However, the viewer can see some differences on the game, the text are in italic font (after the completion of 4 level), there are some cuts when the player gets close on completing the level 4 (you can't clearly notice since the creator stops there at the moment.) As of 2021, the video has accumulated over 75,000 views.

## External Link

* The game itself.
* The Alternative version.
* Another version that is playable in HTML5.
Car Page

Car Page is a fictional social network website used by Cars in Fernanxa's Universe. It is based off of Facebook and Instagram. Almost all the Piston Cup Racers have this website.

## Trivia (Fernanxa's Universe)

* Cruz has an account, her first post was when she won the 2017 Florida 500
* Car Page was used in Lightning starts to race again (Fernanxa's Universe)
* Lightning McQueen has over 10 million followers
* Cruz Ramirez has 5 million followers
* Doc also had the website. He installed it on his computer in 2013. He had 4 million followers. His account was closed after he died. Doc only followed the cars that were residents in Radiator Springs.
Screamer Database Wiki

**Welcome to the Screamer Database Wiki!**
his wiki is another clone of Screamer Wiki, this wiki is an encyclopedia of screamers, meaning that you can read the discovery of screamers, everyone can get to edit, add informations, galleries, images, and more, you are welcome to do so.

## Navigation

**Solid Sharkey's Polybius Page** is a website that is created by Solid Sharkeys on October 2003, the page provides some information that summarizes the story of Polybius, and there is a screamer...