Can a cube of discontinuous function be continuous?
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
asked 4 hours ago
J. AbrahamJ. Abraham
473314
add a comment |
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
asked 4 hours ago
J. AbrahamJ. Abraham
473314
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago
What is your domain? It matters really quite a lot.
– user3482749
4 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago
add a comment |
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
asked 4 hours ago
J. AbrahamJ. Abraham
473314
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
continuity
asked 4 hours ago
J. AbrahamJ. Abraham
473314
asked 4 hours ago
J. AbrahamJ. Abraham
473314
edited 3 hours ago
J. Abraham
asked 4 hours ago
J. AbrahamJ. Abraham
473314
asked 4 hours ago
J. AbrahamJ. Abraham
473314
asked 4 hours ago
J. AbrahamJ. Abraham
473314
473314
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago
What is your domain? It matters really quite a lot.
– user3482749
4 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago
add a comment |
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago
What is your domain? It matters really quite a lot.
– user3482749
4 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago
3
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago
What is your domain? It matters really quite a lot.
– user3482749
4 hours ago
What is your domain? It matters really quite a lot.
– user3482749
4 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago
add a comment |
1 Answer
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Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
add a comment |
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Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
add a comment |
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
add a comment |
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
edited 3 hours ago
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
answered 3 hours ago
Paul FrostPaul Frost
9,5002631
9,5002631
add a comment |
add a comment |
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3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago
What is your domain? It matters really quite a lot.
– user3482749
4 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago