Can a cube of discontinuous function be continuous?
Clash Royale CLAN TAG#URR8PPP
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Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbbR$ ($D$ is subset of $mathbbR$) 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 Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
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show 2 more comments
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Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbbR$ ($D$ is subset of $mathbbR$) 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 Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
$endgroup$
11
$begingroup$
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^1/3$ is continuous.
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– Mindlack
Jan 8 at 1:03
3
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What is your domain? It matters really quite a lot.
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– user3482749
Jan 8 at 1:04
2
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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.
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– MJD
Jan 8 at 1:29
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No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
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– Eric Duminil
Jan 8 at 7:53
3
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I misspoke: it's the range that matters. If it's $mathbbC$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
$endgroup$
– user3482749
Jan 8 at 15:19
|
show 2 more comments
$begingroup$
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbbR$ ($D$ is subset of $mathbbR$) 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 Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
$endgroup$
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbbR$ ($D$ is subset of $mathbbR$) 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 Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
asked Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
edited Jan 8 at 1:07
J. Abraham
asked Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
asked Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
asked Jan 8 at 0:58
J. AbrahamJ. Abraham
528316
528316
11
$begingroup$
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^1/3$ is continuous.
$endgroup$
– Mindlack
Jan 8 at 1:03
3
$begingroup$
What is your domain? It matters really quite a lot.
$endgroup$
– user3482749
Jan 8 at 1:04
2
$begingroup$
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.
$endgroup$
– MJD
Jan 8 at 1:29
$begingroup$
No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
$endgroup$
– Eric Duminil
Jan 8 at 7:53
3
$begingroup$
I misspoke: it's the range that matters. If it's $mathbbC$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
$endgroup$
– user3482749
Jan 8 at 15:19
|
show 2 more comments
11
$begingroup$
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^1/3$ is continuous.
$endgroup$
– Mindlack
Jan 8 at 1:03
3
$begingroup$
What is your domain? It matters really quite a lot.
$endgroup$
– user3482749
Jan 8 at 1:04
2
$begingroup$
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.
$endgroup$
– MJD
Jan 8 at 1:29
$begingroup$
No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
$endgroup$
– Eric Duminil
Jan 8 at 7:53
3
$begingroup$
I misspoke: it's the range that matters. If it's $mathbbC$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
$endgroup$
– user3482749
Jan 8 at 15:19
11
11
$begingroup$
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^1/3$ is continuous.
$endgroup$
– Mindlack
Jan 8 at 1:03
$begingroup$
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^1/3$ is continuous.
$endgroup$
– Mindlack
Jan 8 at 1:03
3
3
$begingroup$
What is your domain? It matters really quite a lot.
$endgroup$
– user3482749
Jan 8 at 1:04
$begingroup$
What is your domain? It matters really quite a lot.
$endgroup$
– user3482749
Jan 8 at 1:04
2
2
$begingroup$
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.
$endgroup$
– MJD
Jan 8 at 1:29
$begingroup$
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.
$endgroup$
– MJD
Jan 8 at 1:29
$begingroup$
No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
$endgroup$
– Eric Duminil
Jan 8 at 7:53
$begingroup$
No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
$endgroup$
– Eric Duminil
Jan 8 at 7:53
3
3
$begingroup$
I misspoke: it's the range that matters. If it's $mathbbC$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
$endgroup$
– user3482749
Jan 8 at 15:19
$begingroup$
I misspoke: it's the range that matters. If it's $mathbbC$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
$endgroup$
– user3482749
Jan 8 at 15:19
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^1/3$ is also continuous.
So the contrapositive is also true, which is:
If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.
(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^1/3$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
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add a comment |
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Since $phi : mathbbR to mathbbR, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
$endgroup$
9
$begingroup$
To lower the level of this answer, note that $phi^-1(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^-1circ phi circ f$.
$endgroup$
– Paul Sinclair
Jan 8 at 18:09
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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active
oldest
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$begingroup$
If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^1/3$ is also continuous.
So the contrapositive is also true, which is:
If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.
(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^1/3$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
$endgroup$
add a comment |
$begingroup$
If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^1/3$ is also continuous.
So the contrapositive is also true, which is:
If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.
(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^1/3$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
$endgroup$
add a comment |
$begingroup$
If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^1/3$ is also continuous.
So the contrapositive is also true, which is:
If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.
(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^1/3$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
$endgroup$
If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^1/3$ is also continuous.
So the contrapositive is also true, which is:
If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.
(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^1/3$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
answered Jan 8 at 5:51
Tanner SwettTanner Swett
4,2241639
4,2241639
add a comment |
add a comment |
$begingroup$
Since $phi : mathbbR to mathbbR, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
$endgroup$
9
$begingroup$
To lower the level of this answer, note that $phi^-1(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^-1circ phi circ f$.
$endgroup$
– Paul Sinclair
Jan 8 at 18:09
add a comment |
$begingroup$
Since $phi : mathbbR to mathbbR, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
$endgroup$
9
$begingroup$
To lower the level of this answer, note that $phi^-1(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^-1circ phi circ f$.
$endgroup$
– Paul Sinclair
Jan 8 at 18:09
add a comment |
$begingroup$
Since $phi : mathbbR to mathbbR, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
$endgroup$
Since $phi : mathbbR to mathbbR, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
edited Jan 8 at 1:19
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
answered Jan 8 at 1:05
Paul FrostPaul Frost
10.1k3933
10.1k3933
9
$begingroup$
To lower the level of this answer, note that $phi^-1(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^-1circ phi circ f$.
$endgroup$
– Paul Sinclair
Jan 8 at 18:09
add a comment |
9
$begingroup$
To lower the level of this answer, note that $phi^-1(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^-1circ phi circ f$.
$endgroup$
– Paul Sinclair
Jan 8 at 18:09
9
9
$begingroup$
To lower the level of this answer, note that $phi^-1(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^-1circ phi circ f$.
$endgroup$
– Paul Sinclair
Jan 8 at 18:09
$begingroup$
To lower the level of this answer, note that $phi^-1(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^-1circ phi circ f$.
$endgroup$
– Paul Sinclair
Jan 8 at 18:09
add a comment |
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11
$begingroup$
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^1/3$ is continuous.
$endgroup$
– Mindlack
Jan 8 at 1:03
3
$begingroup$
What is your domain? It matters really quite a lot.
$endgroup$
– user3482749
Jan 8 at 1:04
2
$begingroup$
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.
$endgroup$
– MJD
Jan 8 at 1:29
$begingroup$
No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
$endgroup$
– Eric Duminil
Jan 8 at 7:53
3
$begingroup$
I misspoke: it's the range that matters. If it's $mathbbC$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
$endgroup$
– user3482749
Jan 8 at 15:19