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Suppose $f: mathbbR to mathbbR$ is function such that $f^3$ is continuous on $mathbbR$. Prove that $f$ is continuous on $mathbbR$
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Suppose $epsilon > 0$ and that $c in mathbbR$. Since $f^3$ is continuous, we can choose $delta > 0$ such that whenever $lvert x - c rvert < delta$ we have that $rvert f^3(x) - f^3(c) lvert < epsilon$.
To show that $f$ is continuous, I know I need some number $delta_1 > 0$ such that $lvert x - c rvert < delta_1$ implies that $lvert f(x) - f(c) rvert < epsilon$.
I've tried manipulating the expression $lvert f(x) - f(c) rvert < epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get
$$
lvert f(x) - f(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) + f^2(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) rvert.
$$
However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
real-analysis continuity
asked Jul 23 at 23:28
user100000000000000
381313
add a comment |Â
up vote
0
down vote
favorite
Suppose $epsilon > 0$ and that $c in mathbbR$. Since $f^3$ is continuous, we can choose $delta > 0$ such that whenever $lvert x - c rvert < delta$ we have that $rvert f^3(x) - f^3(c) lvert < epsilon$.
To show that $f$ is continuous, I know I need some number $delta_1 > 0$ such that $lvert x - c rvert < delta_1$ implies that $lvert f(x) - f(c) rvert < epsilon$.
I've tried manipulating the expression $lvert f(x) - f(c) rvert < epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get
$$
lvert f(x) - f(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) + f^2(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) rvert.
$$
However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
real-analysis continuity
asked Jul 23 at 23:28
user100000000000000
381313
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $epsilon > 0$ and that $c in mathbbR$. Since $f^3$ is continuous, we can choose $delta > 0$ such that whenever $lvert x - c rvert < delta$ we have that $rvert f^3(x) - f^3(c) lvert < epsilon$.
To show that $f$ is continuous, I know I need some number $delta_1 > 0$ such that $lvert x - c rvert < delta_1$ implies that $lvert f(x) - f(c) rvert < epsilon$.
I've tried manipulating the expression $lvert f(x) - f(c) rvert < epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get
$$
lvert f(x) - f(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) + f^2(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) rvert.
$$
However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
real-analysis continuity
asked Jul 23 at 23:28
user100000000000000
381313
Suppose $epsilon > 0$ and that $c in mathbbR$. Since $f^3$ is continuous, we can choose $delta > 0$ such that whenever $lvert x - c rvert < delta$ we have that $rvert f^3(x) - f^3(c) lvert < epsilon$.
To show that $f$ is continuous, I know I need some number $delta_1 > 0$ such that $lvert x - c rvert < delta_1$ implies that $lvert f(x) - f(c) rvert < epsilon$.
I've tried manipulating the expression $lvert f(x) - f(c) rvert < epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get
$$
lvert f(x) - f(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) + f^2(c) rvert < fraclvert f^3(x) - f^3(c)rvertlvert f^2(x) + f(x)f(c) rvert.
$$
However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
real-analysis continuity
asked Jul 23 at 23:28
user100000000000000
381313
asked Jul 23 at 23:28
user100000000000000
381313
asked Jul 23 at 23:28
user100000000000000
381313
asked Jul 23 at 23:28
user100000000000000
381313
381313
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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up vote
5
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accepted
I suppose that the best way is to use the fact that the function $xmapstosqrt[3]x$ is continuous. So,$$f^3text continuousimpliessqrt[3]f^3text continuousiff ftext continuous.$$
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
Thanks for the help!
– user100000000000000
Jul 24 at 0:01
I'm glad I could help.
– José Carlos Santos
Jul 24 at 0:02
1
Right. Note also that $f^2$ continuous does not imply that $f$ continuous: consider the function that is $1$ on the rationals and $0$ on the irrationals. So the issue of when $x^n$ has an inverse function seems to enter in crucially: blindly pushing around inequalities is probably not the way.
– Pete L. Clark
Jul 24 at 2:55
@PeteL.Clark I suppose that you meant that $f$ is $-1$ on the irrationals.
– José Carlos Santos
Jul 24 at 9:13
1
@JoséCarlosSantos: Yes, thank you.
– Pete L. Clark
Jul 24 at 15:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
I suppose that the best way is to use the fact that the function $xmapstosqrt[3]x$ is continuous. So,$$f^3text continuousimpliessqrt[3]f^3text continuousiff ftext continuous.$$
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
Thanks for the help!
– user100000000000000
Jul 24 at 0:01
I'm glad I could help.
– José Carlos Santos
Jul 24 at 0:02
1
Right. Note also that $f^2$ continuous does not imply that $f$ continuous: consider the function that is $1$ on the rationals and $0$ on the irrationals. So the issue of when $x^n$ has an inverse function seems to enter in crucially: blindly pushing around inequalities is probably not the way.
– Pete L. Clark
Jul 24 at 2:55
@PeteL.Clark I suppose that you meant that $f$ is $-1$ on the irrationals.
– José Carlos Santos
Jul 24 at 9:13
1
@JoséCarlosSantos: Yes, thank you.
– Pete L. Clark
Jul 24 at 15:55
add a comment |Â
up vote
5
down vote
accepted
I suppose that the best way is to use the fact that the function $xmapstosqrt[3]x$ is continuous. So,$$f^3text continuousimpliessqrt[3]f^3text continuousiff ftext continuous.$$
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
Thanks for the help!
– user100000000000000
Jul 24 at 0:01
I'm glad I could help.
– José Carlos Santos
Jul 24 at 0:02
1
Right. Note also that $f^2$ continuous does not imply that $f$ continuous: consider the function that is $1$ on the rationals and $0$ on the irrationals. So the issue of when $x^n$ has an inverse function seems to enter in crucially: blindly pushing around inequalities is probably not the way.
– Pete L. Clark
Jul 24 at 2:55
@PeteL.Clark I suppose that you meant that $f$ is $-1$ on the irrationals.
– José Carlos Santos
Jul 24 at 9:13
1
@JoséCarlosSantos: Yes, thank you.
– Pete L. Clark
Jul 24 at 15:55
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
I suppose that the best way is to use the fact that the function $xmapstosqrt[3]x$ is continuous. So,$$f^3text continuousimpliessqrt[3]f^3text continuousiff ftext continuous.$$
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
I suppose that the best way is to use the fact that the function $xmapstosqrt[3]x$ is continuous. So,$$f^3text continuousimpliessqrt[3]f^3text continuousiff ftext continuous.$$
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
edited Jul 24 at 0:02
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
answered Jul 23 at 23:36
José Carlos Santos
113k1698176
113k1698176
Thanks for the help!
– user100000000000000
Jul 24 at 0:01
I'm glad I could help.
– José Carlos Santos
Jul 24 at 0:02
1
Right. Note also that $f^2$ continuous does not imply that $f$ continuous: consider the function that is $1$ on the rationals and $0$ on the irrationals. So the issue of when $x^n$ has an inverse function seems to enter in crucially: blindly pushing around inequalities is probably not the way.
– Pete L. Clark
Jul 24 at 2:55
@PeteL.Clark I suppose that you meant that $f$ is $-1$ on the irrationals.
– José Carlos Santos
Jul 24 at 9:13
1
@JoséCarlosSantos: Yes, thank you.
– Pete L. Clark
Jul 24 at 15:55
add a comment |Â
Thanks for the help!
– user100000000000000
Jul 24 at 0:01
I'm glad I could help.
– José Carlos Santos
Jul 24 at 0:02
1
Right. Note also that $f^2$ continuous does not imply that $f$ continuous: consider the function that is $1$ on the rationals and $0$ on the irrationals. So the issue of when $x^n$ has an inverse function seems to enter in crucially: blindly pushing around inequalities is probably not the way.
– Pete L. Clark
Jul 24 at 2:55
@PeteL.Clark I suppose that you meant that $f$ is $-1$ on the irrationals.
– José Carlos Santos
Jul 24 at 9:13
1
@JoséCarlosSantos: Yes, thank you.
– Pete L. Clark
Jul 24 at 15:55
Thanks for the help!
– user100000000000000
Jul 24 at 0:01
Thanks for the help!
– user100000000000000
Jul 24 at 0:01
I'm glad I could help.
– José Carlos Santos
Jul 24 at 0:02
I'm glad I could help.
– José Carlos Santos
Jul 24 at 0:02
1
1
Right. Note also that $f^2$ continuous does not imply that $f$ continuous: consider the function that is $1$ on the rationals and $0$ on the irrationals. So the issue of when $x^n$ has an inverse function seems to enter in crucially: blindly pushing around inequalities is probably not the way.
– Pete L. Clark
Jul 24 at 2:55
Right. Note also that $f^2$ continuous does not imply that $f$ continuous: consider the function that is $1$ on the rationals and $0$ on the irrationals. So the issue of when $x^n$ has an inverse function seems to enter in crucially: blindly pushing around inequalities is probably not the way.
– Pete L. Clark
Jul 24 at 2:55
@PeteL.Clark I suppose that you meant that $f$ is $-1$ on the irrationals.
– José Carlos Santos
Jul 24 at 9:13
@PeteL.Clark I suppose that you meant that $f$ is $-1$ on the irrationals.
– José Carlos Santos
Jul 24 at 9:13
1
1
@JoséCarlosSantos: Yes, thank you.
– Pete L. Clark
Jul 24 at 15:55
@JoséCarlosSantos: Yes, thank you.
– Pete L. Clark
Jul 24 at 15:55
add a comment |Â
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