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Prove uniform continuity of the composition of two uniformly continuous functions.

0

$begingroup$

Given two functions that are uniformly continuous on $mathbbR$ prove that their composition is uniformly continuous. Here's my attempt.

Proof

Let $f,g$ be our given uniformly continuous functions. We need to show that for any $epsilon >0 $ there exists a $delta$ such that

$$left| (fcirc g)left(xright) - (fcirc g)left(yright) right| < epsilon$$

Since we are given $$left|x' - y' right| < delta_f implies left| fleft(x'right) - fleft(y'right) right| < epsilon$$

and $$left|x - y right| < delta implies left| gleft(xright) - gleft(yright) right| < delta_f$$

defining $g(a) = a'$ for $ain mathbbR$ completes the proof.

real-analysis uniform-continuity

share|cite|improve this question

edited Feb 22 '18 at 4:26

Eric Wofsey

193k14221352

asked Feb 22 '18 at 4:15

ZduffZduff

1,7501021

$endgroup$

add a comment | 

0

$begingroup$

Given two functions that are uniformly continuous on $mathbbR$ prove that their composition is uniformly continuous. Here's my attempt.

Proof

Let $f,g$ be our given uniformly continuous functions. We need to show that for any $epsilon >0 $ there exists a $delta$ such that

$$left| (fcirc g)left(xright) - (fcirc g)left(yright) right| < epsilon$$

Since we are given $$left|x' - y' right| < delta_f implies left| fleft(x'right) - fleft(y'right) right| < epsilon$$

and $$left|x - y right| < delta implies left| gleft(xright) - gleft(yright) right| < delta_f$$

defining $g(a) = a'$ for $ain mathbbR$ completes the proof.

real-analysis uniform-continuity

share|cite|improve this question

edited Feb 22 '18 at 4:26

Eric Wofsey

193k14221352

asked Feb 22 '18 at 4:15

ZduffZduff

1,7501021

$endgroup$

add a comment | 

$begingroup$

Given two functions that are uniformly continuous on $mathbbR$ prove that their composition is uniformly continuous. Here's my attempt.

Proof

Let $f,g$ be our given uniformly continuous functions. We need to show that for any $epsilon >0 $ there exists a $delta$ such that

$$left| (fcirc g)left(xright) - (fcirc g)left(yright) right| < epsilon$$

Since we are given $$left|x' - y' right| < delta_f implies left| fleft(x'right) - fleft(y'right) right| < epsilon$$

and $$left|x - y right| < delta implies left| gleft(xright) - gleft(yright) right| < delta_f$$

defining $g(a) = a'$ for $ain mathbbR$ completes the proof.

real-analysis uniform-continuity

share|cite|improve this question

edited Feb 22 '18 at 4:26

Eric Wofsey

193k14221352

asked Feb 22 '18 at 4:15

ZduffZduff

1,7501021

$endgroup$

share|cite|improve this question

edited Feb 22 '18 at 4:26

Eric Wofsey

193k14221352

asked Feb 22 '18 at 4:15

ZduffZduff

1,7501021

share|cite|improve this question

edited Feb 22 '18 at 4:26

Eric Wofsey

193k14221352

asked Feb 22 '18 at 4:15

ZduffZduff

1,7501021

edited Feb 22 '18 at 4:26

Eric Wofsey

193k14221352

edited Feb 22 '18 at 4:26

Eric Wofsey

193k14221352

asked Feb 22 '18 at 4:15

ZduffZduff

1,7501021

asked Feb 22 '18 at 4:15

ZduffZduff

1,7501021

1 Answer
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3

$begingroup$

This is the right idea but what you have written is nonsense if taken literally. It is extremely important to clearly state the hypotheses and quantifiers in each statement you make. For instance, we are not "given" $$left|x' - y' right| < delta_f implies left| fleft(x'right) - fleft(y'right) right| < epsilon.$$ Rather, we are given that there exists $delta_f>0$ such that for all $x',y'inmathbbR$, this implication is true. Clean up all your statements similarly and you'll have a great proof.

share|cite|improve this answer

answered Feb 22 '18 at 4:26

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

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3

$begingroup$

This is the right idea but what you have written is nonsense if taken literally. It is extremely important to clearly state the hypotheses and quantifiers in each statement you make. For instance, we are not "given" $$left|x' - y' right| < delta_f implies left| fleft(x'right) - fleft(y'right) right| < epsilon.$$ Rather, we are given that there exists $delta_f>0$ such that for all $x',y'inmathbbR$, this implication is true. Clean up all your statements similarly and you'll have a great proof.

share|cite|improve this answer

answered Feb 22 '18 at 4:26

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

3

$begingroup$

This is the right idea but what you have written is nonsense if taken literally. It is extremely important to clearly state the hypotheses and quantifiers in each statement you make. For instance, we are not "given" $$left|x' - y' right| < delta_f implies left| fleft(x'right) - fleft(y'right) right| < epsilon.$$ Rather, we are given that there exists $delta_f>0$ such that for all $x',y'inmathbbR$, this implication is true. Clean up all your statements similarly and you'll have a great proof.

share|cite|improve this answer

answered Feb 22 '18 at 4:26

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

$begingroup$

This is the right idea but what you have written is nonsense if taken literally. It is extremely important to clearly state the hypotheses and quantifiers in each statement you make. For instance, we are not "given" $$left|x' - y' right| < delta_f implies left| fleft(x'right) - fleft(y'right) right| < epsilon.$$ Rather, we are given that there exists $delta_f>0$ such that for all $x',y'inmathbbR$, this implication is true. Clean up all your statements similarly and you'll have a great proof.

share|cite|improve this answer

answered Feb 22 '18 at 4:26

Eric WofseyEric Wofsey

193k14221352

$endgroup$

share|cite|improve this answer

answered Feb 22 '18 at 4:26

Eric WofseyEric Wofsey

193k14221352

answered Feb 22 '18 at 4:26

Eric WofseyEric Wofsey

193k14221352

answered Feb 22 '18 at 4:26

Eric WofseyEric Wofsey

193k14221352