Prove the continuous function is not necessarily absolutely continuous
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Let $f$ be continuous on $I=[0,1]$, and absolutly continuous on $[epsilon,1]$ for any $0<epsilon<1$.
(i) Show that $f$ may not be absolutely continuous on $I.$
(ii) Show that $f$ is absolutely continuous on $I$ if it is increasing.
(ii)Show that $f(x)=sqrt{x}$ is absolutly continuous BUT not Lipschitz on $I$.
I am struggling with the first part, I tried to find a finite disjoint collection of open intervals in $I$ such that $sum_{i=1}^nl(I_i)<delta$ for any $delta>0$, but $$sum_{i=1}^n|f(b_i)-f(a_i)|geq epsilon$$
where $I_i=(a_i,b_i).$ So I would appreciate any help with that.
measure-theory lebesgue-measure
asked 8 hours ago
Ahmed
26819
add a comment |
up vote
1
down vote
favorite
Let $f$ be continuous on $I=[0,1]$, and absolutly continuous on $[epsilon,1]$ for any $0<epsilon<1$.
(i) Show that $f$ may not be absolutely continuous on $I.$
(ii) Show that $f$ is absolutely continuous on $I$ if it is increasing.
(ii)Show that $f(x)=sqrt{x}$ is absolutly continuous BUT not Lipschitz on $I$.
I am struggling with the first part, I tried to find a finite disjoint collection of open intervals in $I$ such that $sum_{i=1}^nl(I_i)<delta$ for any $delta>0$, but $$sum_{i=1}^n|f(b_i)-f(a_i)|geq epsilon$$
where $I_i=(a_i,b_i).$ So I would appreciate any help with that.
measure-theory lebesgue-measure
asked 8 hours ago
Ahmed
26819
3
Let $f(x) = xsin(x) $ for $xne 0$ and $f(0)=0$. $f$ is continuous but not absolutely continuous on any interval that contains $0$.
– Mark Viola
8 hours ago
Can you explain that more please.
– Ahmed
7 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f$ be continuous on $I=[0,1]$, and absolutly continuous on $[epsilon,1]$ for any $0<epsilon<1$.
(i) Show that $f$ may not be absolutely continuous on $I.$
(ii) Show that $f$ is absolutely continuous on $I$ if it is increasing.
(ii)Show that $f(x)=sqrt{x}$ is absolutly continuous BUT not Lipschitz on $I$.
I am struggling with the first part, I tried to find a finite disjoint collection of open intervals in $I$ such that $sum_{i=1}^nl(I_i)<delta$ for any $delta>0$, but $$sum_{i=1}^n|f(b_i)-f(a_i)|geq epsilon$$
where $I_i=(a_i,b_i).$ So I would appreciate any help with that.
measure-theory lebesgue-measure
asked 8 hours ago
Ahmed
26819
Let $f$ be continuous on $I=[0,1]$, and absolutly continuous on $[epsilon,1]$ for any $0<epsilon<1$.
(i) Show that $f$ may not be absolutely continuous on $I.$
(ii) Show that $f$ is absolutely continuous on $I$ if it is increasing.
(ii)Show that $f(x)=sqrt{x}$ is absolutly continuous BUT not Lipschitz on $I$.
I am struggling with the first part, I tried to find a finite disjoint collection of open intervals in $I$ such that $sum_{i=1}^nl(I_i)<delta$ for any $delta>0$, but $$sum_{i=1}^n|f(b_i)-f(a_i)|geq epsilon$$
where $I_i=(a_i,b_i).$ So I would appreciate any help with that.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked 8 hours ago
Ahmed
26819
asked 8 hours ago
Ahmed
26819
asked 8 hours ago
Ahmed
26819
asked 8 hours ago
Ahmed
26819
asked 8 hours ago
Ahmed
26819
26819
3
Let $f(x) = xsin(x) $ for $xne 0$ and $f(0)=0$. $f$ is continuous but not absolutely continuous on any interval that contains $0$.
– Mark Viola
8 hours ago
Can you explain that more please.
– Ahmed
7 hours ago
add a comment |
3
Let $f(x) = xsin(x) $ for $xne 0$ and $f(0)=0$. $f$ is continuous but not absolutely continuous on any interval that contains $0$.
– Mark Viola
8 hours ago
Can you explain that more please.
– Ahmed
7 hours ago
3
3
Let $f(x) = xsin(x) $ for $xne 0$ and $f(0)=0$. $f$ is continuous but not absolutely continuous on any interval that contains $0$.
– Mark Viola
8 hours ago
Let $f(x) = xsin(x) $ for $xne 0$ and $f(0)=0$. $f$ is continuous but not absolutely continuous on any interval that contains $0$.
– Mark Viola
8 hours ago
Can you explain that more please.
– Ahmed
7 hours ago
Can you explain that more please.
– Ahmed
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
HINT:
Let $f$ be the function given by
$$f(x)=begin{cases}
xsin(pi/2x)&,xne0\\
0&,x=0
end{cases}$$
Take $epsilon=1$. Let $delta >0$ be given.
Then, take $x_k =frac1{Nk}$ and $y_k=frac1{N(k+1)}$ for $N$ and odd integer and $1/delta <N$.
Show that the sum $sum_{k=1}^N|x_k-y_k|<delta$, but $sum_{k=1}^N|f(x_k)-f(y_k) |ge1$.
answered 6 hours ago
Mark Viola
129k1273170
@ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
4 hours ago
And please feel free to up vote and accept an answer as you see fit of course. ;-)
– Mark Viola
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
HINT:
Let $f$ be the function given by
$$f(x)=begin{cases}
xsin(pi/2x)&,xne0\\
0&,x=0
end{cases}$$
Take $epsilon=1$. Let $delta >0$ be given.
Then, take $x_k =frac1{Nk}$ and $y_k=frac1{N(k+1)}$ for $N$ and odd integer and $1/delta <N$.
Show that the sum $sum_{k=1}^N|x_k-y_k|<delta$, but $sum_{k=1}^N|f(x_k)-f(y_k) |ge1$.
answered 6 hours ago
Mark Viola
129k1273170
@ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
4 hours ago
And please feel free to up vote and accept an answer as you see fit of course. ;-)
– Mark Viola
4 hours ago
add a comment |
up vote
0
down vote
HINT:
Let $f$ be the function given by
$$f(x)=begin{cases}
xsin(pi/2x)&,xne0\\
0&,x=0
end{cases}$$
Take $epsilon=1$. Let $delta >0$ be given.
Then, take $x_k =frac1{Nk}$ and $y_k=frac1{N(k+1)}$ for $N$ and odd integer and $1/delta <N$.
Show that the sum $sum_{k=1}^N|x_k-y_k|<delta$, but $sum_{k=1}^N|f(x_k)-f(y_k) |ge1$.
answered 6 hours ago
Mark Viola
129k1273170
@ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
4 hours ago
And please feel free to up vote and accept an answer as you see fit of course. ;-)
– Mark Viola
4 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT:
Let $f$ be the function given by
$$f(x)=begin{cases}
xsin(pi/2x)&,xne0\\
0&,x=0
end{cases}$$
Take $epsilon=1$. Let $delta >0$ be given.
Then, take $x_k =frac1{Nk}$ and $y_k=frac1{N(k+1)}$ for $N$ and odd integer and $1/delta <N$.
Show that the sum $sum_{k=1}^N|x_k-y_k|<delta$, but $sum_{k=1}^N|f(x_k)-f(y_k) |ge1$.
answered 6 hours ago
Mark Viola
129k1273170
HINT:
Let $f$ be the function given by
$$f(x)=begin{cases}
xsin(pi/2x)&,xne0\\
0&,x=0
end{cases}$$
Take $epsilon=1$. Let $delta >0$ be given.
Then, take $x_k =frac1{Nk}$ and $y_k=frac1{N(k+1)}$ for $N$ and odd integer and $1/delta <N$.
Show that the sum $sum_{k=1}^N|x_k-y_k|<delta$, but $sum_{k=1}^N|f(x_k)-f(y_k) |ge1$.
answered 6 hours ago
Mark Viola
129k1273170
edited 6 hours ago
answered 6 hours ago
Mark Viola
129k1273170
answered 6 hours ago
Mark Viola
129k1273170
answered 6 hours ago
Mark Viola
129k1273170
129k1273170
@ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
4 hours ago
And please feel free to up vote and accept an answer as you see fit of course. ;-)
– Mark Viola
4 hours ago
add a comment |
@ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
4 hours ago
And please feel free to up vote and accept an answer as you see fit of course. ;-)
– Mark Viola
4 hours ago
@ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
4 hours ago
@ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
4 hours ago
And please feel free to up vote and accept an answer as you see fit of course. ;-)
– Mark Viola
4 hours ago
And please feel free to up vote and accept an answer as you see fit of course. ;-)
– Mark Viola
4 hours ago
add a comment |
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3
Let $f(x) = xsin(x) $ for $xne 0$ and $f(0)=0$. $f$ is continuous but not absolutely continuous on any interval that contains $0$.
– Mark Viola
8 hours ago
Can you explain that more please.
– Ahmed
7 hours ago