Mixing
Darboux continuity of the function $f(x) = limsup_{n to infty} frac{(x_{1}+…+x_{n})^{2}}{n^{2}}$
$begingroup$
Let $f : [0,1] to [0,1]$ be a function that assigns to each $x in [0,1]$ the following value:
$$ x = 0.x_{1}x_{2}x_{3} ... hspace{0.3cm} text{be the binary expansion of }x $$
define
$$ f(x): = limsup_{n to infty} frac{(x_{1}+...+x_{n})^{2}}{n^{2}}$$
Prove that $f$ is Darboux continuous but not continuous.
Can someone give hint on this problem?
Thank you.
real-analysis analysis continuity binary average
edited Jan 25 at 11:12
Hayk
2,6271214
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
$endgroup$
add a comment |
$begingroup$
Let $f : [0,1] to [0,1]$ be a function that assigns to each $x in [0,1]$ the following value:
$$ x = 0.x_{1}x_{2}x_{3} ... hspace{0.3cm} text{be the binary expansion of }x $$
define
$$ f(x): = limsup_{n to infty} frac{(x_{1}+...+x_{n})^{2}}{n^{2}}$$
Prove that $f$ is Darboux continuous but not continuous.
Can someone give hint on this problem?
Thank you.
real-analysis analysis continuity binary average
edited Jan 25 at 11:12
Hayk
2,6271214
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
$endgroup$
$begingroup$
how do you define Darboux continuous?
$endgroup$
– Hayk
Jan 20 at 6:54
1
$begingroup$
@Hayk I guess what it's meant is that the function has Darboux's intermediate value property.
$endgroup$
– Matteo
Jan 20 at 7:18
$begingroup$
@Hayk . A function $f$ is darboux continuous if it has intermediate value property or the image of an interval is an interval.
$endgroup$
– Amirhossein
Jan 20 at 7:58
$begingroup$
FYI, this is a variation of the Cesàro-Vietoris function (see also here and here).
$endgroup$
– Dave L. Renfro
Jan 20 at 9:39
$begingroup$
How are different binary expansions dealt with? For example $x = frac{1}{2}$ have the two expansions $(0.01111cdots)_2 = (0.1000cdots)_2$ giving rise to two different values $1$ and $0$ for $f(x)$. I assume you always choose the first one that doesn't end in ($000cdots$) ? (atleast that seems to work)
$endgroup$
– Winther
Jan 20 at 15:18
add a comment |
$begingroup$
Let $f : [0,1] to [0,1]$ be a function that assigns to each $x in [0,1]$ the following value:
$$ x = 0.x_{1}x_{2}x_{3} ... hspace{0.3cm} text{be the binary expansion of }x $$
define
$$ f(x): = limsup_{n to infty} frac{(x_{1}+...+x_{n})^{2}}{n^{2}}$$
Prove that $f$ is Darboux continuous but not continuous.
Can someone give hint on this problem?
Thank you.
real-analysis analysis continuity binary average
edited Jan 25 at 11:12
Hayk
2,6271214
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
$endgroup$
Let $f : [0,1] to [0,1]$ be a function that assigns to each $x in [0,1]$ the following value:
$$ x = 0.x_{1}x_{2}x_{3} ... hspace{0.3cm} text{be the binary expansion of }x $$
define
$$ f(x): = limsup_{n to infty} frac{(x_{1}+...+x_{n})^{2}}{n^{2}}$$
Prove that $f$ is Darboux continuous but not continuous.
Can someone give hint on this problem?
Thank you.
real-analysis analysis continuity binary average
real-analysis analysis continuity binary average
edited Jan 25 at 11:12
Hayk
2,6271214
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
edited Jan 25 at 11:12
Hayk
2,6271214
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
edited Jan 25 at 11:12
Hayk
2,6271214
edited Jan 25 at 11:12
Hayk
2,6271214
edited Jan 25 at 11:12
Hayk
2,6271214
2,6271214
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
asked Jan 20 at 4:09
AmirhosseinAmirhossein
797
797
$begingroup$
how do you define Darboux continuous?
$endgroup$
– Hayk
Jan 20 at 6:54
1
$begingroup$
@Hayk I guess what it's meant is that the function has Darboux's intermediate value property.
$endgroup$
– Matteo
Jan 20 at 7:18
$begingroup$
@Hayk . A function $f$ is darboux continuous if it has intermediate value property or the image of an interval is an interval.
$endgroup$
– Amirhossein
Jan 20 at 7:58
$begingroup$
FYI, this is a variation of the Cesàro-Vietoris function (see also here and here).
$endgroup$
– Dave L. Renfro
Jan 20 at 9:39
$begingroup$
How are different binary expansions dealt with? For example $x = frac{1}{2}$ have the two expansions $(0.01111cdots)_2 = (0.1000cdots)_2$ giving rise to two different values $1$ and $0$ for $f(x)$. I assume you always choose the first one that doesn't end in ($000cdots$) ? (atleast that seems to work)
$endgroup$
– Winther
Jan 20 at 15:18
add a comment |
$begingroup$
how do you define Darboux continuous?
$endgroup$
– Hayk
Jan 20 at 6:54
1
$begingroup$
@Hayk I guess what it's meant is that the function has Darboux's intermediate value property.
$endgroup$
– Matteo
Jan 20 at 7:18
$begingroup$
@Hayk . A function $f$ is darboux continuous if it has intermediate value property or the image of an interval is an interval.
$endgroup$
– Amirhossein
Jan 20 at 7:58
$begingroup$
FYI, this is a variation of the Cesàro-Vietoris function (see also here and here).
$endgroup$
– Dave L. Renfro
Jan 20 at 9:39
$begingroup$
How are different binary expansions dealt with? For example $x = frac{1}{2}$ have the two expansions $(0.01111cdots)_2 = (0.1000cdots)_2$ giving rise to two different values $1$ and $0$ for $f(x)$. I assume you always choose the first one that doesn't end in ($000cdots$) ? (atleast that seems to work)
$endgroup$
– Winther
Jan 20 at 15:18
$begingroup$
how do you define Darboux continuous?
$endgroup$
– Hayk
Jan 20 at 6:54
$begingroup$
how do you define Darboux continuous?
$endgroup$
– Hayk
Jan 20 at 6:54
1
1
$begingroup$
@Hayk I guess what it's meant is that the function has Darboux's intermediate value property.
$endgroup$
– Matteo
Jan 20 at 7:18
$begingroup$
@Hayk I guess what it's meant is that the function has Darboux's intermediate value property.
$endgroup$
– Matteo
Jan 20 at 7:18
$begingroup$
@Hayk . A function $f$ is darboux continuous if it has intermediate value property or the image of an interval is an interval.
$endgroup$
– Amirhossein
Jan 20 at 7:58
$begingroup$
@Hayk . A function $f$ is darboux continuous if it has intermediate value property or the image of an interval is an interval.
$endgroup$
– Amirhossein
Jan 20 at 7:58
$begingroup$
FYI, this is a variation of the Cesàro-Vietoris function (see also here and here).
$endgroup$
– Dave L. Renfro
Jan 20 at 9:39
$begingroup$
FYI, this is a variation of the Cesàro-Vietoris function (see also here and here).
$endgroup$
– Dave L. Renfro
Jan 20 at 9:39
$begingroup$
How are different binary expansions dealt with? For example $x = frac{1}{2}$ have the two expansions $(0.01111cdots)_2 = (0.1000cdots)_2$ giving rise to two different values $1$ and $0$ for $f(x)$. I assume you always choose the first one that doesn't end in ($000cdots$) ? (atleast that seems to work)
$endgroup$
– Winther
Jan 20 at 15:18
$begingroup$
How are different binary expansions dealt with? For example $x = frac{1}{2}$ have the two expansions $(0.01111cdots)_2 = (0.1000cdots)_2$ giving rise to two different values $1$ and $0$ for $f(x)$. I assume you always choose the first one that doesn't end in ($000cdots$) ? (atleast that seems to work)
$endgroup$
– Winther
Jan 20 at 15:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Thinking of the binary expansion of $xin [0,1]$ as infinite sequence of (fair) coin tosses implies the following:
Claim: $f(x) = frac{1}{4}$ almost everywhere on $[0,1]$.
Indeed, consider the probability space $([0,1], mathcal{B}([0,1]), mu )$ where $mathcal{B}([0,1])$ is the Borel $sigma$-algebra of $[0,1]$ and $mu$ is the Lebesgue measure of $[0,1]$. For each $nin mathbb{N}$, let $X_n :[0,1] to {0,1}$ be the $n$-th bit (coordinate) of the binary expansion of $xin [0,1]$. Notice that $X_n$ is not well-defined on dyadic rationals, however they form a set of measure (probability) $0$, thus we will ignore it without loss of generality.
Notice that ${X_n}_{n=1}^infty$ is a sequence of i.i.d. random variables with $mu(X_n = 1) = mu(X_n = 0) = 1/2$ for all $nin mathbb{N}$. To see this simply follow the structure of the set $X_n = a$ where $a in {0,1}$, precisely if $[0,1]$ is partitioned into intervals of the form $Delta_i: = [i2^{-n}, (i+1)2^{-n})$, $i=0,1,...,2^n-1$, then if the set $X_n = a $ in each $Delta_i$ is either the left half if $a=0$ or the right half if $a=1$. Using this structure the independence follows easily.
Now, the function $f$, in our new notation becomes
$$
f(omega) = limsuplimits_{nto infty} left(frac{X_1(omega)+...+X_n(omega)}{n}right)^2, text{ where } omega in [0,1].
$$
In view of the i.i.d. condition of $(X_n)$, from the strong law of large numbers we get that $f(omega) = (mathbb{E}X_1)^2 = 1/4$ almost surely, i.e. almost everywhere on $[0,1]$.
We next prove that the image of any dyadic interval under $f$ is the $[0,1]$. This will prove that $f$ is Darboux continuous, and coupled with the fact that $f=1/4$ a.e. will show that $f$ is nowhere continuous.
Let $Delta_k^i: = [i/2^k,(i+1)/2^k]$ where $0leq i leq 2^k - 1 $ and $k=0,1,...$ . Then $f(Delta_k^i) = [0,1]$.
To prove this, take any $0<a<1$ and consider its decimal expansion
$$
a = sumlimits_{n=1}^infty frac{a_n}{10^n}, text{ where } a_n in {0,1,...,9}.
$$
Let $S_n = sumlimits_{k=1}^n frac{a_k}{10^k} := frac{p_n}{10^n}$ be the $n$-th partial sum of the expansion. Here $p_n$ is a non-negative integer, and
$$
frac{p_n}{10^n} to a, tag{1}
$$
hence for $nin mathbb{N}$ large enough, we have
$$
frac a2 10^n leq p_n leq 10^n tag{2}
$$
Take a rapidly growing sequence of integers $n_1<n_2<...$ and define $xin (0,1) $ with binary expansion of the form
$$
x = (b_1,...,b_{10^{n_1}}, b_{10^{n_1} + 1}, ..., b_{10^{n_2}}, ....)
$$
as follows:
in each block $(b_{10^{n_k} + 1}, ..., b_{10^{n_{k+1}}})$ take precisely $p_{n_{k+1}} -10^{n_k} $ number of $1$s and put the rest of the bits as $0$. The length of the block equals $10^{n_{k+1}} - 10^{n_k} $, hence using $(2)$, for all $k=1,2,...$ we get
$$
frac{p_{n_k}}{10^{n_k}} - frac{1}{10^{n_k - n_{k-1}} } leq frac{b_1+...+b_{10^{n_k}}}{10^{n_k}} leq frac{1}{10^{n_k - n_{k-1}} } + frac{p_{n_k} - 10^{n_{k-1}}}{10^{n_k}} = frac{p_{n_k}}{10^{n_k}}.
$$
The latter combined with $(1)$ implies that $limsuplimits_{nto infty} frac{b_1+...+b_n}{n} = a $, i.e. $f(x) = a^2$. Since $0<a<1$ is any, we get that $f([0,1]) = [0,1]$.
To get the result for the dyadic interval $Delta_k^i$, observe that changing finitely many bits in $x$ does not affect the value of $f(x)$. Thus, if $i=(b_1...b_k)$ is the binary representation of the integer $0leq ileq 2^k-1$, we simply append these bits to beginning of the binary expansion of $x$ constructed above for the given $a$ to make $xin Delta_k^i$, completing the proof of the claim $f(Delta_k^i) = [0,1]$.
answered Jan 25 at 11:09
HaykHayk
2,6271214
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1 Answer
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$begingroup$
Thinking of the binary expansion of $xin [0,1]$ as infinite sequence of (fair) coin tosses implies the following:
Claim: $f(x) = frac{1}{4}$ almost everywhere on $[0,1]$.
Indeed, consider the probability space $([0,1], mathcal{B}([0,1]), mu )$ where $mathcal{B}([0,1])$ is the Borel $sigma$-algebra of $[0,1]$ and $mu$ is the Lebesgue measure of $[0,1]$. For each $nin mathbb{N}$, let $X_n :[0,1] to {0,1}$ be the $n$-th bit (coordinate) of the binary expansion of $xin [0,1]$. Notice that $X_n$ is not well-defined on dyadic rationals, however they form a set of measure (probability) $0$, thus we will ignore it without loss of generality.
Notice that ${X_n}_{n=1}^infty$ is a sequence of i.i.d. random variables with $mu(X_n = 1) = mu(X_n = 0) = 1/2$ for all $nin mathbb{N}$. To see this simply follow the structure of the set $X_n = a$ where $a in {0,1}$, precisely if $[0,1]$ is partitioned into intervals of the form $Delta_i: = [i2^{-n}, (i+1)2^{-n})$, $i=0,1,...,2^n-1$, then if the set $X_n = a $ in each $Delta_i$ is either the left half if $a=0$ or the right half if $a=1$. Using this structure the independence follows easily.
Now, the function $f$, in our new notation becomes
$$
f(omega) = limsuplimits_{nto infty} left(frac{X_1(omega)+...+X_n(omega)}{n}right)^2, text{ where } omega in [0,1].
$$
In view of the i.i.d. condition of $(X_n)$, from the strong law of large numbers we get that $f(omega) = (mathbb{E}X_1)^2 = 1/4$ almost surely, i.e. almost everywhere on $[0,1]$.
We next prove that the image of any dyadic interval under $f$ is the $[0,1]$. This will prove that $f$ is Darboux continuous, and coupled with the fact that $f=1/4$ a.e. will show that $f$ is nowhere continuous.
Let $Delta_k^i: = [i/2^k,(i+1)/2^k]$ where $0leq i leq 2^k - 1 $ and $k=0,1,...$ . Then $f(Delta_k^i) = [0,1]$.
To prove this, take any $0<a<1$ and consider its decimal expansion
$$
a = sumlimits_{n=1}^infty frac{a_n}{10^n}, text{ where } a_n in {0,1,...,9}.
$$
Let $S_n = sumlimits_{k=1}^n frac{a_k}{10^k} := frac{p_n}{10^n}$ be the $n$-th partial sum of the expansion. Here $p_n$ is a non-negative integer, and
$$
frac{p_n}{10^n} to a, tag{1}
$$
hence for $nin mathbb{N}$ large enough, we have
$$
frac a2 10^n leq p_n leq 10^n tag{2}
$$
Take a rapidly growing sequence of integers $n_1<n_2<...$ and define $xin (0,1) $ with binary expansion of the form
$$
x = (b_1,...,b_{10^{n_1}}, b_{10^{n_1} + 1}, ..., b_{10^{n_2}}, ....)
$$
as follows:
in each block $(b_{10^{n_k} + 1}, ..., b_{10^{n_{k+1}}})$ take precisely $p_{n_{k+1}} -10^{n_k} $ number of $1$s and put the rest of the bits as $0$. The length of the block equals $10^{n_{k+1}} - 10^{n_k} $, hence using $(2)$, for all $k=1,2,...$ we get
$$
frac{p_{n_k}}{10^{n_k}} - frac{1}{10^{n_k - n_{k-1}} } leq frac{b_1+...+b_{10^{n_k}}}{10^{n_k}} leq frac{1}{10^{n_k - n_{k-1}} } + frac{p_{n_k} - 10^{n_{k-1}}}{10^{n_k}} = frac{p_{n_k}}{10^{n_k}}.
$$
The latter combined with $(1)$ implies that $limsuplimits_{nto infty} frac{b_1+...+b_n}{n} = a $, i.e. $f(x) = a^2$. Since $0<a<1$ is any, we get that $f([0,1]) = [0,1]$.
To get the result for the dyadic interval $Delta_k^i$, observe that changing finitely many bits in $x$ does not affect the value of $f(x)$. Thus, if $i=(b_1...b_k)$ is the binary representation of the integer $0leq ileq 2^k-1$, we simply append these bits to beginning of the binary expansion of $x$ constructed above for the given $a$ to make $xin Delta_k^i$, completing the proof of the claim $f(Delta_k^i) = [0,1]$.
answered Jan 25 at 11:09
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Thinking of the binary expansion of $xin [0,1]$ as infinite sequence of (fair) coin tosses implies the following:
Claim: $f(x) = frac{1}{4}$ almost everywhere on $[0,1]$.
Indeed, consider the probability space $([0,1], mathcal{B}([0,1]), mu )$ where $mathcal{B}([0,1])$ is the Borel $sigma$-algebra of $[0,1]$ and $mu$ is the Lebesgue measure of $[0,1]$. For each $nin mathbb{N}$, let $X_n :[0,1] to {0,1}$ be the $n$-th bit (coordinate) of the binary expansion of $xin [0,1]$. Notice that $X_n$ is not well-defined on dyadic rationals, however they form a set of measure (probability) $0$, thus we will ignore it without loss of generality.
Notice that ${X_n}_{n=1}^infty$ is a sequence of i.i.d. random variables with $mu(X_n = 1) = mu(X_n = 0) = 1/2$ for all $nin mathbb{N}$. To see this simply follow the structure of the set $X_n = a$ where $a in {0,1}$, precisely if $[0,1]$ is partitioned into intervals of the form $Delta_i: = [i2^{-n}, (i+1)2^{-n})$, $i=0,1,...,2^n-1$, then if the set $X_n = a $ in each $Delta_i$ is either the left half if $a=0$ or the right half if $a=1$. Using this structure the independence follows easily.
Now, the function $f$, in our new notation becomes
$$
f(omega) = limsuplimits_{nto infty} left(frac{X_1(omega)+...+X_n(omega)}{n}right)^2, text{ where } omega in [0,1].
$$
In view of the i.i.d. condition of $(X_n)$, from the strong law of large numbers we get that $f(omega) = (mathbb{E}X_1)^2 = 1/4$ almost surely, i.e. almost everywhere on $[0,1]$.
We next prove that the image of any dyadic interval under $f$ is the $[0,1]$. This will prove that $f$ is Darboux continuous, and coupled with the fact that $f=1/4$ a.e. will show that $f$ is nowhere continuous.
Let $Delta_k^i: = [i/2^k,(i+1)/2^k]$ where $0leq i leq 2^k - 1 $ and $k=0,1,...$ . Then $f(Delta_k^i) = [0,1]$.
To prove this, take any $0<a<1$ and consider its decimal expansion
$$
a = sumlimits_{n=1}^infty frac{a_n}{10^n}, text{ where } a_n in {0,1,...,9}.
$$
Let $S_n = sumlimits_{k=1}^n frac{a_k}{10^k} := frac{p_n}{10^n}$ be the $n$-th partial sum of the expansion. Here $p_n$ is a non-negative integer, and
$$
frac{p_n}{10^n} to a, tag{1}
$$
hence for $nin mathbb{N}$ large enough, we have
$$
frac a2 10^n leq p_n leq 10^n tag{2}
$$
Take a rapidly growing sequence of integers $n_1<n_2<...$ and define $xin (0,1) $ with binary expansion of the form
$$
x = (b_1,...,b_{10^{n_1}}, b_{10^{n_1} + 1}, ..., b_{10^{n_2}}, ....)
$$
as follows:
in each block $(b_{10^{n_k} + 1}, ..., b_{10^{n_{k+1}}})$ take precisely $p_{n_{k+1}} -10^{n_k} $ number of $1$s and put the rest of the bits as $0$. The length of the block equals $10^{n_{k+1}} - 10^{n_k} $, hence using $(2)$, for all $k=1,2,...$ we get
$$
frac{p_{n_k}}{10^{n_k}} - frac{1}{10^{n_k - n_{k-1}} } leq frac{b_1+...+b_{10^{n_k}}}{10^{n_k}} leq frac{1}{10^{n_k - n_{k-1}} } + frac{p_{n_k} - 10^{n_{k-1}}}{10^{n_k}} = frac{p_{n_k}}{10^{n_k}}.
$$
The latter combined with $(1)$ implies that $limsuplimits_{nto infty} frac{b_1+...+b_n}{n} = a $, i.e. $f(x) = a^2$. Since $0<a<1$ is any, we get that $f([0,1]) = [0,1]$.
To get the result for the dyadic interval $Delta_k^i$, observe that changing finitely many bits in $x$ does not affect the value of $f(x)$. Thus, if $i=(b_1...b_k)$ is the binary representation of the integer $0leq ileq 2^k-1$, we simply append these bits to beginning of the binary expansion of $x$ constructed above for the given $a$ to make $xin Delta_k^i$, completing the proof of the claim $f(Delta_k^i) = [0,1]$.
answered Jan 25 at 11:09
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Thinking of the binary expansion of $xin [0,1]$ as infinite sequence of (fair) coin tosses implies the following:
Claim: $f(x) = frac{1}{4}$ almost everywhere on $[0,1]$.
Indeed, consider the probability space $([0,1], mathcal{B}([0,1]), mu )$ where $mathcal{B}([0,1])$ is the Borel $sigma$-algebra of $[0,1]$ and $mu$ is the Lebesgue measure of $[0,1]$. For each $nin mathbb{N}$, let $X_n :[0,1] to {0,1}$ be the $n$-th bit (coordinate) of the binary expansion of $xin [0,1]$. Notice that $X_n$ is not well-defined on dyadic rationals, however they form a set of measure (probability) $0$, thus we will ignore it without loss of generality.
Notice that ${X_n}_{n=1}^infty$ is a sequence of i.i.d. random variables with $mu(X_n = 1) = mu(X_n = 0) = 1/2$ for all $nin mathbb{N}$. To see this simply follow the structure of the set $X_n = a$ where $a in {0,1}$, precisely if $[0,1]$ is partitioned into intervals of the form $Delta_i: = [i2^{-n}, (i+1)2^{-n})$, $i=0,1,...,2^n-1$, then if the set $X_n = a $ in each $Delta_i$ is either the left half if $a=0$ or the right half if $a=1$. Using this structure the independence follows easily.
Now, the function $f$, in our new notation becomes
$$
f(omega) = limsuplimits_{nto infty} left(frac{X_1(omega)+...+X_n(omega)}{n}right)^2, text{ where } omega in [0,1].
$$
In view of the i.i.d. condition of $(X_n)$, from the strong law of large numbers we get that $f(omega) = (mathbb{E}X_1)^2 = 1/4$ almost surely, i.e. almost everywhere on $[0,1]$.
We next prove that the image of any dyadic interval under $f$ is the $[0,1]$. This will prove that $f$ is Darboux continuous, and coupled with the fact that $f=1/4$ a.e. will show that $f$ is nowhere continuous.
Let $Delta_k^i: = [i/2^k,(i+1)/2^k]$ where $0leq i leq 2^k - 1 $ and $k=0,1,...$ . Then $f(Delta_k^i) = [0,1]$.
To prove this, take any $0<a<1$ and consider its decimal expansion
$$
a = sumlimits_{n=1}^infty frac{a_n}{10^n}, text{ where } a_n in {0,1,...,9}.
$$
Let $S_n = sumlimits_{k=1}^n frac{a_k}{10^k} := frac{p_n}{10^n}$ be the $n$-th partial sum of the expansion. Here $p_n$ is a non-negative integer, and
$$
frac{p_n}{10^n} to a, tag{1}
$$
hence for $nin mathbb{N}$ large enough, we have
$$
frac a2 10^n leq p_n leq 10^n tag{2}
$$
Take a rapidly growing sequence of integers $n_1<n_2<...$ and define $xin (0,1) $ with binary expansion of the form
$$
x = (b_1,...,b_{10^{n_1}}, b_{10^{n_1} + 1}, ..., b_{10^{n_2}}, ....)
$$
as follows:
in each block $(b_{10^{n_k} + 1}, ..., b_{10^{n_{k+1}}})$ take precisely $p_{n_{k+1}} -10^{n_k} $ number of $1$s and put the rest of the bits as $0$. The length of the block equals $10^{n_{k+1}} - 10^{n_k} $, hence using $(2)$, for all $k=1,2,...$ we get
$$
frac{p_{n_k}}{10^{n_k}} - frac{1}{10^{n_k - n_{k-1}} } leq frac{b_1+...+b_{10^{n_k}}}{10^{n_k}} leq frac{1}{10^{n_k - n_{k-1}} } + frac{p_{n_k} - 10^{n_{k-1}}}{10^{n_k}} = frac{p_{n_k}}{10^{n_k}}.
$$
The latter combined with $(1)$ implies that $limsuplimits_{nto infty} frac{b_1+...+b_n}{n} = a $, i.e. $f(x) = a^2$. Since $0<a<1$ is any, we get that $f([0,1]) = [0,1]$.
To get the result for the dyadic interval $Delta_k^i$, observe that changing finitely many bits in $x$ does not affect the value of $f(x)$. Thus, if $i=(b_1...b_k)$ is the binary representation of the integer $0leq ileq 2^k-1$, we simply append these bits to beginning of the binary expansion of $x$ constructed above for the given $a$ to make $xin Delta_k^i$, completing the proof of the claim $f(Delta_k^i) = [0,1]$.
answered Jan 25 at 11:09
HaykHayk
2,6271214
$endgroup$
Thinking of the binary expansion of $xin [0,1]$ as infinite sequence of (fair) coin tosses implies the following:
Claim: $f(x) = frac{1}{4}$ almost everywhere on $[0,1]$.
Indeed, consider the probability space $([0,1], mathcal{B}([0,1]), mu )$ where $mathcal{B}([0,1])$ is the Borel $sigma$-algebra of $[0,1]$ and $mu$ is the Lebesgue measure of $[0,1]$. For each $nin mathbb{N}$, let $X_n :[0,1] to {0,1}$ be the $n$-th bit (coordinate) of the binary expansion of $xin [0,1]$. Notice that $X_n$ is not well-defined on dyadic rationals, however they form a set of measure (probability) $0$, thus we will ignore it without loss of generality.
Notice that ${X_n}_{n=1}^infty$ is a sequence of i.i.d. random variables with $mu(X_n = 1) = mu(X_n = 0) = 1/2$ for all $nin mathbb{N}$. To see this simply follow the structure of the set $X_n = a$ where $a in {0,1}$, precisely if $[0,1]$ is partitioned into intervals of the form $Delta_i: = [i2^{-n}, (i+1)2^{-n})$, $i=0,1,...,2^n-1$, then if the set $X_n = a $ in each $Delta_i$ is either the left half if $a=0$ or the right half if $a=1$. Using this structure the independence follows easily.
Now, the function $f$, in our new notation becomes
$$
f(omega) = limsuplimits_{nto infty} left(frac{X_1(omega)+...+X_n(omega)}{n}right)^2, text{ where } omega in [0,1].
$$
In view of the i.i.d. condition of $(X_n)$, from the strong law of large numbers we get that $f(omega) = (mathbb{E}X_1)^2 = 1/4$ almost surely, i.e. almost everywhere on $[0,1]$.
We next prove that the image of any dyadic interval under $f$ is the $[0,1]$. This will prove that $f$ is Darboux continuous, and coupled with the fact that $f=1/4$ a.e. will show that $f$ is nowhere continuous.
Let $Delta_k^i: = [i/2^k,(i+1)/2^k]$ where $0leq i leq 2^k - 1 $ and $k=0,1,...$ . Then $f(Delta_k^i) = [0,1]$.
To prove this, take any $0<a<1$ and consider its decimal expansion
$$
a = sumlimits_{n=1}^infty frac{a_n}{10^n}, text{ where } a_n in {0,1,...,9}.
$$
Let $S_n = sumlimits_{k=1}^n frac{a_k}{10^k} := frac{p_n}{10^n}$ be the $n$-th partial sum of the expansion. Here $p_n$ is a non-negative integer, and
$$
frac{p_n}{10^n} to a, tag{1}
$$
hence for $nin mathbb{N}$ large enough, we have
$$
frac a2 10^n leq p_n leq 10^n tag{2}
$$
Take a rapidly growing sequence of integers $n_1<n_2<...$ and define $xin (0,1) $ with binary expansion of the form
$$
x = (b_1,...,b_{10^{n_1}}, b_{10^{n_1} + 1}, ..., b_{10^{n_2}}, ....)
$$
as follows:
in each block $(b_{10^{n_k} + 1}, ..., b_{10^{n_{k+1}}})$ take precisely $p_{n_{k+1}} -10^{n_k} $ number of $1$s and put the rest of the bits as $0$. The length of the block equals $10^{n_{k+1}} - 10^{n_k} $, hence using $(2)$, for all $k=1,2,...$ we get
$$
frac{p_{n_k}}{10^{n_k}} - frac{1}{10^{n_k - n_{k-1}} } leq frac{b_1+...+b_{10^{n_k}}}{10^{n_k}} leq frac{1}{10^{n_k - n_{k-1}} } + frac{p_{n_k} - 10^{n_{k-1}}}{10^{n_k}} = frac{p_{n_k}}{10^{n_k}}.
$$
The latter combined with $(1)$ implies that $limsuplimits_{nto infty} frac{b_1+...+b_n}{n} = a $, i.e. $f(x) = a^2$. Since $0<a<1$ is any, we get that $f([0,1]) = [0,1]$.
To get the result for the dyadic interval $Delta_k^i$, observe that changing finitely many bits in $x$ does not affect the value of $f(x)$. Thus, if $i=(b_1...b_k)$ is the binary representation of the integer $0leq ileq 2^k-1$, we simply append these bits to beginning of the binary expansion of $x$ constructed above for the given $a$ to make $xin Delta_k^i$, completing the proof of the claim $f(Delta_k^i) = [0,1]$.
answered Jan 25 at 11:09
HaykHayk
2,6271214
edited Jan 25 at 11:32
answered Jan 25 at 11:09
HaykHayk
2,6271214
answered Jan 25 at 11:09
HaykHayk
2,6271214
answered Jan 25 at 11:09
HaykHayk
2,6271214
2,6271214
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$begingroup$
how do you define Darboux continuous?
$endgroup$
– Hayk
Jan 20 at 6:54
1
$begingroup$
@Hayk I guess what it's meant is that the function has Darboux's intermediate value property.
$endgroup$
– Matteo
Jan 20 at 7:18
$begingroup$
@Hayk . A function $f$ is darboux continuous if it has intermediate value property or the image of an interval is an interval.
$endgroup$
– Amirhossein
Jan 20 at 7:58
$begingroup$
FYI, this is a variation of the Cesàro-Vietoris function (see also here and here).
$endgroup$
– Dave L. Renfro
Jan 20 at 9:39
$begingroup$
How are different binary expansions dealt with? For example $x = frac{1}{2}$ have the two expansions $(0.01111cdots)_2 = (0.1000cdots)_2$ giving rise to two different values $1$ and $0$ for $f(x)$. I assume you always choose the first one that doesn't end in ($000cdots$) ? (atleast that seems to work)
$endgroup$
– Winther
Jan 20 at 15:18