f is a function from R to R and J is bounded in interval in R
$begingroup$
Let $f(x)=1+x$ on $ J_{1}=[0,1]$ and $ J_{2}=[0,3]$
So, (A) ,(B) and (D) hold.
c be sigmun function with f(x)= 0 for x=0 , -1 for x<0 and 1 for x>0.
so on [0,1] $W(f,[0,1])=1 $
Please help !!
real-analysis
asked Dec 26 '18 at 16:03
sejysejy
1589
$endgroup$
|
show 5 more comments
$begingroup$
Let $f(x)=1+x$ on $ J_{1}=[0,1]$ and $ J_{2}=[0,3]$
So, (A) ,(B) and (D) hold.
c be sigmun function with f(x)= 0 for x=0 , -1 for x<0 and 1 for x>0.
so on [0,1] $W(f,[0,1])=1 $
Please help !!
real-analysis
asked Dec 26 '18 at 16:03
sejysejy
1589
$endgroup$
1
$begingroup$
It's not really clear what is the question? Do you want help showing these, or are you trying to determine if (C) is true?
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:07
$begingroup$
please help in determing c and am i correct so all options will hold .
$endgroup$
– sejy
Dec 26 '18 at 16:08
1
$begingroup$
Also perhaps show what you've tried, or thought.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:08
1
$begingroup$
(B) is not correct, your other two statements seem true
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:09
1
$begingroup$
So do you no longer need the explanation? Because Xiao showed you how to approach this, but not to actually do it.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:57
|
show 5 more comments
$begingroup$
Let $f(x)=1+x$ on $ J_{1}=[0,1]$ and $ J_{2}=[0,3]$
So, (A) ,(B) and (D) hold.
c be sigmun function with f(x)= 0 for x=0 , -1 for x<0 and 1 for x>0.
so on [0,1] $W(f,[0,1])=1 $
Please help !!
real-analysis
asked Dec 26 '18 at 16:03
sejysejy
1589
$endgroup$
Let $f(x)=1+x$ on $ J_{1}=[0,1]$ and $ J_{2}=[0,3]$
So, (A) ,(B) and (D) hold.
c be sigmun function with f(x)= 0 for x=0 , -1 for x<0 and 1 for x>0.
so on [0,1] $W(f,[0,1])=1 $
Please help !!
real-analysis
real-analysis
asked Dec 26 '18 at 16:03
sejysejy
1589
asked Dec 26 '18 at 16:03
sejysejy
1589
edited Dec 26 '18 at 16:13
sejy
asked Dec 26 '18 at 16:03
sejysejy
1589
asked Dec 26 '18 at 16:03
sejysejy
1589
asked Dec 26 '18 at 16:03
sejysejy
1589
1589
1
$begingroup$
It's not really clear what is the question? Do you want help showing these, or are you trying to determine if (C) is true?
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:07
$begingroup$
please help in determing c and am i correct so all options will hold .
$endgroup$
– sejy
Dec 26 '18 at 16:08
1
$begingroup$
Also perhaps show what you've tried, or thought.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:08
1
$begingroup$
(B) is not correct, your other two statements seem true
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:09
1
$begingroup$
So do you no longer need the explanation? Because Xiao showed you how to approach this, but not to actually do it.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:57
|
show 5 more comments
1
$begingroup$
It's not really clear what is the question? Do you want help showing these, or are you trying to determine if (C) is true?
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:07
$begingroup$
please help in determing c and am i correct so all options will hold .
$endgroup$
– sejy
Dec 26 '18 at 16:08
1
$begingroup$
Also perhaps show what you've tried, or thought.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:08
1
$begingroup$
(B) is not correct, your other two statements seem true
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:09
1
$begingroup$
So do you no longer need the explanation? Because Xiao showed you how to approach this, but not to actually do it.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:57
1
1
$begingroup$
It's not really clear what is the question? Do you want help showing these, or are you trying to determine if (C) is true?
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:07
$begingroup$
It's not really clear what is the question? Do you want help showing these, or are you trying to determine if (C) is true?
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:07
$begingroup$
please help in determing c and am i correct so all options will hold .
$endgroup$
– sejy
Dec 26 '18 at 16:08
$begingroup$
please help in determing c and am i correct so all options will hold .
$endgroup$
– sejy
Dec 26 '18 at 16:08
1
1
$begingroup$
Also perhaps show what you've tried, or thought.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:08
$begingroup$
Also perhaps show what you've tried, or thought.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:08
1
1
$begingroup$
(B) is not correct, your other two statements seem true
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:09
$begingroup$
(B) is not correct, your other two statements seem true
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:09
1
1
$begingroup$
So do you no longer need the explanation? Because Xiao showed you how to approach this, but not to actually do it.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:57
$begingroup$
So do you no longer need the explanation? Because Xiao showed you how to approach this, but not to actually do it.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:57
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
There is a very common misconception.
If you want to show a statement is true, you have to show for all possible functions and all possible intervals that satisfy the assumptions in the statement, the conclusion in the statement holds. For example in $(A)$, you have to show for ANY function $f: mathbb{R} rightarrow mathbb{R}$, and ANY two bounded intervals $J_1 subset J_2 subset mathbb{R}$ (these are the only assumptions), we have (the conclusion)
$$W(f, J_1) leq W(f, J_2).$$
You can not pick your function $f(x)$ and your intervals $J_1, J_2$.
On the other hand, if you are trying to show a statement is false, it suffices to construct a function $f$ and a bounded interval (or intervals) that satisfy the assumption and yet the conclusion fails.
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
$endgroup$
add a comment |
$begingroup$
What you tried to give to prove (C) is actually an example and not a proof. Notice first that $W(f,J)geq 0$ always. Perhaps it would be more useful to see for a function $f$ on interval $J$, if $f(x_1)neq f(x_2)$ for $x_1,x_2in J$, what can you say about $W(f,J)$ (A lower estimate)?
As Xiao said to show that (B) is false, it suffices to find an function $f$ and a sequence of intervals $Jsupseteq I_1supseteq I_2supseteq ...$ with lengths tending to $0$ such that $W(f,J_n)not rightarrow0$. Try looking for a bounded function with a dicontinuity at $ain I_n$ for all $nin mathbb{N}$.
For (A), notice that if $Jsubseteq I$, then $sup{ f(x) :xin J } leq sup{ f(x) :xin I }$ and $inf{ f(x) :xin J } geq inf{ f(x) :xin I }$. Use these facts to show (A).
For (D) write what it means for $f$ to be continuous at $a$ in terms of $delta$ and $epsilon$. This should give an obvious sub-interval to work with.
If you still struggle after these hints, write in a comment to this answer and I'll explain further.
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a very common misconception.
If you want to show a statement is true, you have to show for all possible functions and all possible intervals that satisfy the assumptions in the statement, the conclusion in the statement holds. For example in $(A)$, you have to show for ANY function $f: mathbb{R} rightarrow mathbb{R}$, and ANY two bounded intervals $J_1 subset J_2 subset mathbb{R}$ (these are the only assumptions), we have (the conclusion)
$$W(f, J_1) leq W(f, J_2).$$
You can not pick your function $f(x)$ and your intervals $J_1, J_2$.
On the other hand, if you are trying to show a statement is false, it suffices to construct a function $f$ and a bounded interval (or intervals) that satisfy the assumption and yet the conclusion fails.
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
$endgroup$
add a comment |
$begingroup$
There is a very common misconception.
If you want to show a statement is true, you have to show for all possible functions and all possible intervals that satisfy the assumptions in the statement, the conclusion in the statement holds. For example in $(A)$, you have to show for ANY function $f: mathbb{R} rightarrow mathbb{R}$, and ANY two bounded intervals $J_1 subset J_2 subset mathbb{R}$ (these are the only assumptions), we have (the conclusion)
$$W(f, J_1) leq W(f, J_2).$$
You can not pick your function $f(x)$ and your intervals $J_1, J_2$.
On the other hand, if you are trying to show a statement is false, it suffices to construct a function $f$ and a bounded interval (or intervals) that satisfy the assumption and yet the conclusion fails.
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
$endgroup$
add a comment |
$begingroup$
There is a very common misconception.
If you want to show a statement is true, you have to show for all possible functions and all possible intervals that satisfy the assumptions in the statement, the conclusion in the statement holds. For example in $(A)$, you have to show for ANY function $f: mathbb{R} rightarrow mathbb{R}$, and ANY two bounded intervals $J_1 subset J_2 subset mathbb{R}$ (these are the only assumptions), we have (the conclusion)
$$W(f, J_1) leq W(f, J_2).$$
You can not pick your function $f(x)$ and your intervals $J_1, J_2$.
On the other hand, if you are trying to show a statement is false, it suffices to construct a function $f$ and a bounded interval (or intervals) that satisfy the assumption and yet the conclusion fails.
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
$endgroup$
There is a very common misconception.
If you want to show a statement is true, you have to show for all possible functions and all possible intervals that satisfy the assumptions in the statement, the conclusion in the statement holds. For example in $(A)$, you have to show for ANY function $f: mathbb{R} rightarrow mathbb{R}$, and ANY two bounded intervals $J_1 subset J_2 subset mathbb{R}$ (these are the only assumptions), we have (the conclusion)
$$W(f, J_1) leq W(f, J_2).$$
You can not pick your function $f(x)$ and your intervals $J_1, J_2$.
On the other hand, if you are trying to show a statement is false, it suffices to construct a function $f$ and a bounded interval (or intervals) that satisfy the assumption and yet the conclusion fails.
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
answered Dec 26 '18 at 16:40
XiaoXiao
4,87811636
4,87811636
add a comment |
add a comment |
$begingroup$
What you tried to give to prove (C) is actually an example and not a proof. Notice first that $W(f,J)geq 0$ always. Perhaps it would be more useful to see for a function $f$ on interval $J$, if $f(x_1)neq f(x_2)$ for $x_1,x_2in J$, what can you say about $W(f,J)$ (A lower estimate)?
As Xiao said to show that (B) is false, it suffices to find an function $f$ and a sequence of intervals $Jsupseteq I_1supseteq I_2supseteq ...$ with lengths tending to $0$ such that $W(f,J_n)not rightarrow0$. Try looking for a bounded function with a dicontinuity at $ain I_n$ for all $nin mathbb{N}$.
For (A), notice that if $Jsubseteq I$, then $sup{ f(x) :xin J } leq sup{ f(x) :xin I }$ and $inf{ f(x) :xin J } geq inf{ f(x) :xin I }$. Use these facts to show (A).
For (D) write what it means for $f$ to be continuous at $a$ in terms of $delta$ and $epsilon$. This should give an obvious sub-interval to work with.
If you still struggle after these hints, write in a comment to this answer and I'll explain further.
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
$endgroup$
add a comment |
$begingroup$
What you tried to give to prove (C) is actually an example and not a proof. Notice first that $W(f,J)geq 0$ always. Perhaps it would be more useful to see for a function $f$ on interval $J$, if $f(x_1)neq f(x_2)$ for $x_1,x_2in J$, what can you say about $W(f,J)$ (A lower estimate)?
As Xiao said to show that (B) is false, it suffices to find an function $f$ and a sequence of intervals $Jsupseteq I_1supseteq I_2supseteq ...$ with lengths tending to $0$ such that $W(f,J_n)not rightarrow0$. Try looking for a bounded function with a dicontinuity at $ain I_n$ for all $nin mathbb{N}$.
For (A), notice that if $Jsubseteq I$, then $sup{ f(x) :xin J } leq sup{ f(x) :xin I }$ and $inf{ f(x) :xin J } geq inf{ f(x) :xin I }$. Use these facts to show (A).
For (D) write what it means for $f$ to be continuous at $a$ in terms of $delta$ and $epsilon$. This should give an obvious sub-interval to work with.
If you still struggle after these hints, write in a comment to this answer and I'll explain further.
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
$endgroup$
add a comment |
$begingroup$
What you tried to give to prove (C) is actually an example and not a proof. Notice first that $W(f,J)geq 0$ always. Perhaps it would be more useful to see for a function $f$ on interval $J$, if $f(x_1)neq f(x_2)$ for $x_1,x_2in J$, what can you say about $W(f,J)$ (A lower estimate)?
As Xiao said to show that (B) is false, it suffices to find an function $f$ and a sequence of intervals $Jsupseteq I_1supseteq I_2supseteq ...$ with lengths tending to $0$ such that $W(f,J_n)not rightarrow0$. Try looking for a bounded function with a dicontinuity at $ain I_n$ for all $nin mathbb{N}$.
For (A), notice that if $Jsubseteq I$, then $sup{ f(x) :xin J } leq sup{ f(x) :xin I }$ and $inf{ f(x) :xin J } geq inf{ f(x) :xin I }$. Use these facts to show (A).
For (D) write what it means for $f$ to be continuous at $a$ in terms of $delta$ and $epsilon$. This should give an obvious sub-interval to work with.
If you still struggle after these hints, write in a comment to this answer and I'll explain further.
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
$endgroup$
What you tried to give to prove (C) is actually an example and not a proof. Notice first that $W(f,J)geq 0$ always. Perhaps it would be more useful to see for a function $f$ on interval $J$, if $f(x_1)neq f(x_2)$ for $x_1,x_2in J$, what can you say about $W(f,J)$ (A lower estimate)?
As Xiao said to show that (B) is false, it suffices to find an function $f$ and a sequence of intervals $Jsupseteq I_1supseteq I_2supseteq ...$ with lengths tending to $0$ such that $W(f,J_n)not rightarrow0$. Try looking for a bounded function with a dicontinuity at $ain I_n$ for all $nin mathbb{N}$.
For (A), notice that if $Jsubseteq I$, then $sup{ f(x) :xin J } leq sup{ f(x) :xin I }$ and $inf{ f(x) :xin J } geq inf{ f(x) :xin I }$. Use these facts to show (A).
For (D) write what it means for $f$ to be continuous at $a$ in terms of $delta$ and $epsilon$. This should give an obvious sub-interval to work with.
If you still struggle after these hints, write in a comment to this answer and I'll explain further.
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
answered Dec 26 '18 at 18:15
Keen-ameteurKeen-ameteur
1,550516
1,550516
add a comment |
add a comment |
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1
$begingroup$
It's not really clear what is the question? Do you want help showing these, or are you trying to determine if (C) is true?
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:07
$begingroup$
please help in determing c and am i correct so all options will hold .
$endgroup$
– sejy
Dec 26 '18 at 16:08
1
$begingroup$
Also perhaps show what you've tried, or thought.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:08
1
$begingroup$
(B) is not correct, your other two statements seem true
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:09
1
$begingroup$
So do you no longer need the explanation? Because Xiao showed you how to approach this, but not to actually do it.
$endgroup$
– Keen-ameteur
Dec 26 '18 at 16:57