Is a bounded continuous function always have a supremum?
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My problem is related to Proving that a bounded, continuous function has a supremum
- A continous function $f(R)$ defined on $Rin[0,1]$ with $f(0)=0$ and $f(1)=1$.
- Assuming that we have $f(R)<R$ on the open interval $(0,R_1)$ and $f(R_1)=R_1$. $R_1in(0,1)$ is fixed.
- How can I prove there will always be a supremum $lambda$ so that $sup{f(R)/R}=lambda<1$，$Rin(0,R_1)$
real-analysis sequences-and-series
asked Nov 21 at 3:44
QZHua
112
add a comment |
up vote
-1
down vote
favorite
My problem is related to Proving that a bounded, continuous function has a supremum
- A continous function $f(R)$ defined on $Rin[0,1]$ with $f(0)=0$ and $f(1)=1$.
- Assuming that we have $f(R)<R$ on the open interval $(0,R_1)$ and $f(R_1)=R_1$. $R_1in(0,1)$ is fixed.
- How can I prove there will always be a supremum $lambda$ so that $sup{f(R)/R}=lambda<1$，$Rin(0,R_1)$
real-analysis sequences-and-series
asked Nov 21 at 3:44
QZHua
112
Is $R_1$ fixed?
– Offlaw
Nov 21 at 3:53
With the given assumptions you can construct a function $f$ such that $f(R)=1$ for $R = R_1 + epsilon < 1$ for some small $epsilon$, yielding $f(R)/R = frac{1}{R_1 + epsilon} > 1$, a contradiction. So something in your post is stated incorrectly (or I have misunderstood something).
– angryavian
Nov 21 at 3:58
Looks like the supremum will be $1$.
– Lord Shark the Unknown
Nov 21 at 4:07
I think @ Lord Shark the Unknown is right, then it makes the problem no-sense.
– QZHua
Dec 4 at 3:34
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
My problem is related to Proving that a bounded, continuous function has a supremum
- A continous function $f(R)$ defined on $Rin[0,1]$ with $f(0)=0$ and $f(1)=1$.
- Assuming that we have $f(R)<R$ on the open interval $(0,R_1)$ and $f(R_1)=R_1$. $R_1in(0,1)$ is fixed.
- How can I prove there will always be a supremum $lambda$ so that $sup{f(R)/R}=lambda<1$，$Rin(0,R_1)$
real-analysis sequences-and-series
asked Nov 21 at 3:44
QZHua
112
My problem is related to Proving that a bounded, continuous function has a supremum
- A continous function $f(R)$ defined on $Rin[0,1]$ with $f(0)=0$ and $f(1)=1$.
- Assuming that we have $f(R)<R$ on the open interval $(0,R_1)$ and $f(R_1)=R_1$. $R_1in(0,1)$ is fixed.
- How can I prove there will always be a supremum $lambda$ so that $sup{f(R)/R}=lambda<1$，$Rin(0,R_1)$
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Nov 21 at 3:44
QZHua
112
asked Nov 21 at 3:44
QZHua
112
edited Nov 22 at 5:17
asked Nov 21 at 3:44
QZHua
112
asked Nov 21 at 3:44
QZHua
112
asked Nov 21 at 3:44
QZHua
112
112
Is $R_1$ fixed?
– Offlaw
Nov 21 at 3:53
With the given assumptions you can construct a function $f$ such that $f(R)=1$ for $R = R_1 + epsilon < 1$ for some small $epsilon$, yielding $f(R)/R = frac{1}{R_1 + epsilon} > 1$, a contradiction. So something in your post is stated incorrectly (or I have misunderstood something).
– angryavian
Nov 21 at 3:58
Looks like the supremum will be $1$.
– Lord Shark the Unknown
Nov 21 at 4:07
I think @ Lord Shark the Unknown is right, then it makes the problem no-sense.
– QZHua
Dec 4 at 3:34
add a comment |
Is $R_1$ fixed?
– Offlaw
Nov 21 at 3:53
With the given assumptions you can construct a function $f$ such that $f(R)=1$ for $R = R_1 + epsilon < 1$ for some small $epsilon$, yielding $f(R)/R = frac{1}{R_1 + epsilon} > 1$, a contradiction. So something in your post is stated incorrectly (or I have misunderstood something).
– angryavian
Nov 21 at 3:58
Looks like the supremum will be $1$.
– Lord Shark the Unknown
Nov 21 at 4:07
I think @ Lord Shark the Unknown is right, then it makes the problem no-sense.
– QZHua
Dec 4 at 3:34
Is $R_1$ fixed?
– Offlaw
Nov 21 at 3:53
Is $R_1$ fixed?
– Offlaw
Nov 21 at 3:53
With the given assumptions you can construct a function $f$ such that $f(R)=1$ for $R = R_1 + epsilon < 1$ for some small $epsilon$, yielding $f(R)/R = frac{1}{R_1 + epsilon} > 1$, a contradiction. So something in your post is stated incorrectly (or I have misunderstood something).
– angryavian
Nov 21 at 3:58
With the given assumptions you can construct a function $f$ such that $f(R)=1$ for $R = R_1 + epsilon < 1$ for some small $epsilon$, yielding $f(R)/R = frac{1}{R_1 + epsilon} > 1$, a contradiction. So something in your post is stated incorrectly (or I have misunderstood something).
– angryavian
Nov 21 at 3:58
Looks like the supremum will be $1$.
– Lord Shark the Unknown
Nov 21 at 4:07
Looks like the supremum will be $1$.
– Lord Shark the Unknown
Nov 21 at 4:07
I think @ Lord Shark the Unknown is right, then it makes the problem no-sense.
– QZHua
Dec 4 at 3:34
I think @ Lord Shark the Unknown is right, then it makes the problem no-sense.
– QZHua
Dec 4 at 3:34
add a comment |
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Is $R_1$ fixed?
– Offlaw
Nov 21 at 3:53
With the given assumptions you can construct a function $f$ such that $f(R)=1$ for $R = R_1 + epsilon < 1$ for some small $epsilon$, yielding $f(R)/R = frac{1}{R_1 + epsilon} > 1$, a contradiction. So something in your post is stated incorrectly (or I have misunderstood something).
– angryavian
Nov 21 at 3:58
Looks like the supremum will be $1$.
– Lord Shark the Unknown
Nov 21 at 4:07
I think @ Lord Shark the Unknown is right, then it makes the problem no-sense.
– QZHua
Dec 4 at 3:34