Checking the range in which the given function is uniformly continuous
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Problem
For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.
a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.
b) What is the largest interval in which the given function is uniformly continuous?
$Attempt$
a)
$4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$
$d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.
Hence f is uniformly continuous on [0,2].
Doubt
How to approach part (b) of the question?
real-analysis
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up vote
4
down vote
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Problem
For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.
a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.
b) What is the largest interval in which the given function is uniformly continuous?
$Attempt$
a)
$4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$
$d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.
Hence f is uniformly continuous on [0,2].
Doubt
How to approach part (b) of the question?
real-analysis
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Problem
For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.
a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.
b) What is the largest interval in which the given function is uniformly continuous?
$Attempt$
a)
$4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$
$d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.
Hence f is uniformly continuous on [0,2].
Doubt
How to approach part (b) of the question?
real-analysis
Problem
For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.
a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.
b) What is the largest interval in which the given function is uniformly continuous?
$Attempt$
a)
$4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$
$d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.
Hence f is uniformly continuous on [0,2].
Doubt
How to approach part (b) of the question?
real-analysis
asked Aug 20 at 10:26
blue boy
630311
630311
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1 Answer
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Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].
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up vote
2
down vote
Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].
Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].
answered Aug 20 at 10:32
Kavi Rama Murthy
23.3k2933
23.3k2933
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