Mixing
Continuous function:
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Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$
I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.
And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.
real-analysis analysis
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Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$
I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.
And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.
real-analysis analysis
edited yesterday
idea
2,07121023
asked yesterday
Guy
133
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$
I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.
And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.
real-analysis analysis
edited yesterday
idea
2,07121023
asked yesterday
Guy
133
Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$
I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.
And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.
real-analysis analysis
real-analysis analysis
edited yesterday
idea
2,07121023
asked yesterday
Guy
133
edited yesterday
idea
2,07121023
asked yesterday
Guy
133
edited yesterday
idea
2,07121023
edited yesterday
idea
2,07121023
edited yesterday
idea
2,07121023
2,07121023
asked yesterday
Guy
133
asked yesterday
Guy
133
asked yesterday
Guy
133
133
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add a comment |
3 Answers
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A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.
answered yesterday
Sorin Tirc
5039
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HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.
answered yesterday
Y. Forman
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Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:
$lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
$l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
answered yesterday
José Carlos Santos
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.
answered yesterday
Sorin Tirc
5039
add a comment |
up vote
3
down vote
A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.
answered yesterday
Sorin Tirc
5039
add a comment |
up vote
3
down vote
up vote
3
down vote
A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.
answered yesterday
Sorin Tirc
5039
A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
5039
add a comment |
add a comment |
up vote
0
down vote
HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.
answered yesterday
Y. Forman
11.3k423
add a comment |
up vote
0
down vote
HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.
answered yesterday
Y. Forman
11.3k423
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.
answered yesterday
Y. Forman
11.3k423
HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
11.3k423
add a comment |
add a comment |
up vote
0
down vote
Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:
$lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
$l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
answered yesterday
José Carlos Santos
139k18111203
add a comment |
up vote
0
down vote
Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:
$lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
$l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
answered yesterday
José Carlos Santos
139k18111203
add a comment |
up vote
0
down vote
up vote
0
down vote
Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:
$lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
$l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
answered yesterday
José Carlos Santos
139k18111203
Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:
$lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
$l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
139k18111203
add a comment |
add a comment |
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