Is it necessary that either $f(x) leq 0$ for all $x$, or, $f(x) geq 0$ for all $x$ [duplicate]
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This question already has an answer here:
Properties of a continuous map $f : mathbb{R}^2rightarrow mathbb{R}$ with only finitely many zeroes
1 answer
Let $f : mathbb{R}^2 to mathbb{R}$ be a continuous map such that $f(x) = 0$ for only finitely
many values of $x$. Is it necessary that either $f(x) leq 0$ for all $x$, or, $f(x) geq 0$ for all $x$
I don't think its necessary because if $f(x,y)=x$ then $f$ is continous and it takes both positive and negative values.
real-analysis metric-spaces
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marked as duplicate by Community♦ Dec 22 '18 at 7:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Properties of a continuous map $f : mathbb{R}^2rightarrow mathbb{R}$ with only finitely many zeroes
1 answer
Let $f : mathbb{R}^2 to mathbb{R}$ be a continuous map such that $f(x) = 0$ for only finitely
many values of $x$. Is it necessary that either $f(x) leq 0$ for all $x$, or, $f(x) geq 0$ for all $x$
I don't think its necessary because if $f(x,y)=x$ then $f$ is continous and it takes both positive and negative values.
real-analysis metric-spaces
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marked as duplicate by Community♦ Dec 22 '18 at 7:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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how many pairs $(x,y)$ make your proposed function equal to $0$?
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– Mark S.
Dec 22 '18 at 3:02
2
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Your example $f(x,y)=x$ has infinitely many zeros -- $f(0,k)=0$ for any real $k.$
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– coffeemath
Dec 22 '18 at 3:02
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What does $f(x)$ mean if $f$ has domain $mathbb{R}^2$? Is $x$ a point in the plane?
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– Randall
Dec 22 '18 at 3:15
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@coffeemath, I see. Thanks
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– user408906
Dec 22 '18 at 4:47
add a comment |
$begingroup$
This question already has an answer here:
Properties of a continuous map $f : mathbb{R}^2rightarrow mathbb{R}$ with only finitely many zeroes
1 answer
Let $f : mathbb{R}^2 to mathbb{R}$ be a continuous map such that $f(x) = 0$ for only finitely
many values of $x$. Is it necessary that either $f(x) leq 0$ for all $x$, or, $f(x) geq 0$ for all $x$
I don't think its necessary because if $f(x,y)=x$ then $f$ is continous and it takes both positive and negative values.
real-analysis metric-spaces
$endgroup$
This question already has an answer here:
Properties of a continuous map $f : mathbb{R}^2rightarrow mathbb{R}$ with only finitely many zeroes
1 answer
Let $f : mathbb{R}^2 to mathbb{R}$ be a continuous map such that $f(x) = 0$ for only finitely
many values of $x$. Is it necessary that either $f(x) leq 0$ for all $x$, or, $f(x) geq 0$ for all $x$
I don't think its necessary because if $f(x,y)=x$ then $f$ is continous and it takes both positive and negative values.
This question already has an answer here:
Properties of a continuous map $f : mathbb{R}^2rightarrow mathbb{R}$ with only finitely many zeroes
1 answer
real-analysis metric-spaces
real-analysis metric-spaces
asked Dec 22 '18 at 2:56
user408906
marked as duplicate by Community♦ Dec 22 '18 at 7:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Community♦ Dec 22 '18 at 7:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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how many pairs $(x,y)$ make your proposed function equal to $0$?
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– Mark S.
Dec 22 '18 at 3:02
2
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Your example $f(x,y)=x$ has infinitely many zeros -- $f(0,k)=0$ for any real $k.$
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– coffeemath
Dec 22 '18 at 3:02
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What does $f(x)$ mean if $f$ has domain $mathbb{R}^2$? Is $x$ a point in the plane?
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– Randall
Dec 22 '18 at 3:15
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@coffeemath, I see. Thanks
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– user408906
Dec 22 '18 at 4:47
add a comment |
$begingroup$
how many pairs $(x,y)$ make your proposed function equal to $0$?
$endgroup$
– Mark S.
Dec 22 '18 at 3:02
2
$begingroup$
Your example $f(x,y)=x$ has infinitely many zeros -- $f(0,k)=0$ for any real $k.$
$endgroup$
– coffeemath
Dec 22 '18 at 3:02
$begingroup$
What does $f(x)$ mean if $f$ has domain $mathbb{R}^2$? Is $x$ a point in the plane?
$endgroup$
– Randall
Dec 22 '18 at 3:15
$begingroup$
@coffeemath, I see. Thanks
$endgroup$
– user408906
Dec 22 '18 at 4:47
$begingroup$
how many pairs $(x,y)$ make your proposed function equal to $0$?
$endgroup$
– Mark S.
Dec 22 '18 at 3:02
$begingroup$
how many pairs $(x,y)$ make your proposed function equal to $0$?
$endgroup$
– Mark S.
Dec 22 '18 at 3:02
2
2
$begingroup$
Your example $f(x,y)=x$ has infinitely many zeros -- $f(0,k)=0$ for any real $k.$
$endgroup$
– coffeemath
Dec 22 '18 at 3:02
$begingroup$
Your example $f(x,y)=x$ has infinitely many zeros -- $f(0,k)=0$ for any real $k.$
$endgroup$
– coffeemath
Dec 22 '18 at 3:02
$begingroup$
What does $f(x)$ mean if $f$ has domain $mathbb{R}^2$? Is $x$ a point in the plane?
$endgroup$
– Randall
Dec 22 '18 at 3:15
$begingroup$
What does $f(x)$ mean if $f$ has domain $mathbb{R}^2$? Is $x$ a point in the plane?
$endgroup$
– Randall
Dec 22 '18 at 3:15
$begingroup$
@coffeemath, I see. Thanks
$endgroup$
– user408906
Dec 22 '18 at 4:47
$begingroup$
@coffeemath, I see. Thanks
$endgroup$
– user408906
Dec 22 '18 at 4:47
add a comment |
2 Answers
2
active
oldest
votes
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Yes, this statement is true. And one idea could be to prove it by contradiction.
So, assume that $f(mathbf{x})=0$ only for finitely many $mathbf{x}inmathbb{R}^2$, and yet it's not true that either $f(mathbf{x})le0$ for all $mathbf{x}inmathbb{R}^2$ or $f(mathbf{x})ge0$ for all $mathbf{x}inmathbb{R}^2$. The last part means that there exist at least one $mathbf{x}_1inmathbb{R}^2$ such that $f(mathbf{x}_1)<0$ and at least one $mathbf{x}_2inmathbb{R}^2$ such that $f(mathbf{x}_2)>0$. Now the key idea is that there are infinitely many paths connecting the points $mathbf{x}_1$ and $mathbf{x}_2$ in the plane, and since $f$ is continuous, there exist points at which $f$ is zero on each such path.
For example, let's say $f(-1,0)<0$ and $f(1,0)>0$. Consider $f$ restricted to the arcs of parabolas $y=a(x^2-1)$, $-1le xle1$, for all $a>0$. All these arcs connects these two points, and at the same time they don't have any common points other than the endpoints. Since $f$ is continuous along each arc and goes from being negative to being positive, $f$ is equal to zero at least once on each of them.
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
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1
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do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly?
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– user408906
Dec 22 '18 at 4:49
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@Suraj Yes, that's exactly what I mean.
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– zipirovich
Dec 22 '18 at 13:49
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I like this argument.
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– Randall
Dec 22 '18 at 14:53
add a comment |
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Given two points $A$ and $B$ in the plane, consider the family of all circular arcs including both points. Each such arc is the locus of $X$ such that $angle AXB = theta$ for some fixed $theta$ (as a signed angle), so they don't overlap at all.
There are infinitely many such arcs. If $f$ is continuous and $f(A)$ and $f(B)$ have opposite signs, then each arc must contain a point $X$ at which $f(X)=0$. As such, there are infinitely many points at which $f$ is zero.
There it is - the contrapositive of your statement. If a continuous function from $mathbb{R}^2$ to $mathbb{R}$ takes both positive and negative values, then it has infinitely many zeros. Done.
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this statement is true. And one idea could be to prove it by contradiction.
So, assume that $f(mathbf{x})=0$ only for finitely many $mathbf{x}inmathbb{R}^2$, and yet it's not true that either $f(mathbf{x})le0$ for all $mathbf{x}inmathbb{R}^2$ or $f(mathbf{x})ge0$ for all $mathbf{x}inmathbb{R}^2$. The last part means that there exist at least one $mathbf{x}_1inmathbb{R}^2$ such that $f(mathbf{x}_1)<0$ and at least one $mathbf{x}_2inmathbb{R}^2$ such that $f(mathbf{x}_2)>0$. Now the key idea is that there are infinitely many paths connecting the points $mathbf{x}_1$ and $mathbf{x}_2$ in the plane, and since $f$ is continuous, there exist points at which $f$ is zero on each such path.
For example, let's say $f(-1,0)<0$ and $f(1,0)>0$. Consider $f$ restricted to the arcs of parabolas $y=a(x^2-1)$, $-1le xle1$, for all $a>0$. All these arcs connects these two points, and at the same time they don't have any common points other than the endpoints. Since $f$ is continuous along each arc and goes from being negative to being positive, $f$ is equal to zero at least once on each of them.
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
$endgroup$
1
$begingroup$
do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly?
$endgroup$
– user408906
Dec 22 '18 at 4:49
$begingroup$
@Suraj Yes, that's exactly what I mean.
$endgroup$
– zipirovich
Dec 22 '18 at 13:49
$begingroup$
I like this argument.
$endgroup$
– Randall
Dec 22 '18 at 14:53
add a comment |
$begingroup$
Yes, this statement is true. And one idea could be to prove it by contradiction.
So, assume that $f(mathbf{x})=0$ only for finitely many $mathbf{x}inmathbb{R}^2$, and yet it's not true that either $f(mathbf{x})le0$ for all $mathbf{x}inmathbb{R}^2$ or $f(mathbf{x})ge0$ for all $mathbf{x}inmathbb{R}^2$. The last part means that there exist at least one $mathbf{x}_1inmathbb{R}^2$ such that $f(mathbf{x}_1)<0$ and at least one $mathbf{x}_2inmathbb{R}^2$ such that $f(mathbf{x}_2)>0$. Now the key idea is that there are infinitely many paths connecting the points $mathbf{x}_1$ and $mathbf{x}_2$ in the plane, and since $f$ is continuous, there exist points at which $f$ is zero on each such path.
For example, let's say $f(-1,0)<0$ and $f(1,0)>0$. Consider $f$ restricted to the arcs of parabolas $y=a(x^2-1)$, $-1le xle1$, for all $a>0$. All these arcs connects these two points, and at the same time they don't have any common points other than the endpoints. Since $f$ is continuous along each arc and goes from being negative to being positive, $f$ is equal to zero at least once on each of them.
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
$endgroup$
1
$begingroup$
do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly?
$endgroup$
– user408906
Dec 22 '18 at 4:49
$begingroup$
@Suraj Yes, that's exactly what I mean.
$endgroup$
– zipirovich
Dec 22 '18 at 13:49
$begingroup$
I like this argument.
$endgroup$
– Randall
Dec 22 '18 at 14:53
add a comment |
$begingroup$
Yes, this statement is true. And one idea could be to prove it by contradiction.
So, assume that $f(mathbf{x})=0$ only for finitely many $mathbf{x}inmathbb{R}^2$, and yet it's not true that either $f(mathbf{x})le0$ for all $mathbf{x}inmathbb{R}^2$ or $f(mathbf{x})ge0$ for all $mathbf{x}inmathbb{R}^2$. The last part means that there exist at least one $mathbf{x}_1inmathbb{R}^2$ such that $f(mathbf{x}_1)<0$ and at least one $mathbf{x}_2inmathbb{R}^2$ such that $f(mathbf{x}_2)>0$. Now the key idea is that there are infinitely many paths connecting the points $mathbf{x}_1$ and $mathbf{x}_2$ in the plane, and since $f$ is continuous, there exist points at which $f$ is zero on each such path.
For example, let's say $f(-1,0)<0$ and $f(1,0)>0$. Consider $f$ restricted to the arcs of parabolas $y=a(x^2-1)$, $-1le xle1$, for all $a>0$. All these arcs connects these two points, and at the same time they don't have any common points other than the endpoints. Since $f$ is continuous along each arc and goes from being negative to being positive, $f$ is equal to zero at least once on each of them.
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
$endgroup$
Yes, this statement is true. And one idea could be to prove it by contradiction.
So, assume that $f(mathbf{x})=0$ only for finitely many $mathbf{x}inmathbb{R}^2$, and yet it's not true that either $f(mathbf{x})le0$ for all $mathbf{x}inmathbb{R}^2$ or $f(mathbf{x})ge0$ for all $mathbf{x}inmathbb{R}^2$. The last part means that there exist at least one $mathbf{x}_1inmathbb{R}^2$ such that $f(mathbf{x}_1)<0$ and at least one $mathbf{x}_2inmathbb{R}^2$ such that $f(mathbf{x}_2)>0$. Now the key idea is that there are infinitely many paths connecting the points $mathbf{x}_1$ and $mathbf{x}_2$ in the plane, and since $f$ is continuous, there exist points at which $f$ is zero on each such path.
For example, let's say $f(-1,0)<0$ and $f(1,0)>0$. Consider $f$ restricted to the arcs of parabolas $y=a(x^2-1)$, $-1le xle1$, for all $a>0$. All these arcs connects these two points, and at the same time they don't have any common points other than the endpoints. Since $f$ is continuous along each arc and goes from being negative to being positive, $f$ is equal to zero at least once on each of them.
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
answered Dec 22 '18 at 3:37
zipirovichzipirovich
11.4k11731
11.4k11731
1
$begingroup$
do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly?
$endgroup$
– user408906
Dec 22 '18 at 4:49
$begingroup$
@Suraj Yes, that's exactly what I mean.
$endgroup$
– zipirovich
Dec 22 '18 at 13:49
$begingroup$
I like this argument.
$endgroup$
– Randall
Dec 22 '18 at 14:53
add a comment |
1
$begingroup$
do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly?
$endgroup$
– user408906
Dec 22 '18 at 4:49
$begingroup$
@Suraj Yes, that's exactly what I mean.
$endgroup$
– zipirovich
Dec 22 '18 at 13:49
$begingroup$
I like this argument.
$endgroup$
– Randall
Dec 22 '18 at 14:53
1
1
$begingroup$
do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly?
$endgroup$
– user408906
Dec 22 '18 at 4:49
$begingroup$
do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly?
$endgroup$
– user408906
Dec 22 '18 at 4:49
$begingroup$
@Suraj Yes, that's exactly what I mean.
$endgroup$
– zipirovich
Dec 22 '18 at 13:49
$begingroup$
@Suraj Yes, that's exactly what I mean.
$endgroup$
– zipirovich
Dec 22 '18 at 13:49
$begingroup$
I like this argument.
$endgroup$
– Randall
Dec 22 '18 at 14:53
$begingroup$
I like this argument.
$endgroup$
– Randall
Dec 22 '18 at 14:53
add a comment |
$begingroup$
Given two points $A$ and $B$ in the plane, consider the family of all circular arcs including both points. Each such arc is the locus of $X$ such that $angle AXB = theta$ for some fixed $theta$ (as a signed angle), so they don't overlap at all.
There are infinitely many such arcs. If $f$ is continuous and $f(A)$ and $f(B)$ have opposite signs, then each arc must contain a point $X$ at which $f(X)=0$. As such, there are infinitely many points at which $f$ is zero.
There it is - the contrapositive of your statement. If a continuous function from $mathbb{R}^2$ to $mathbb{R}$ takes both positive and negative values, then it has infinitely many zeros. Done.
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
$endgroup$
add a comment |
$begingroup$
Given two points $A$ and $B$ in the plane, consider the family of all circular arcs including both points. Each such arc is the locus of $X$ such that $angle AXB = theta$ for some fixed $theta$ (as a signed angle), so they don't overlap at all.
There are infinitely many such arcs. If $f$ is continuous and $f(A)$ and $f(B)$ have opposite signs, then each arc must contain a point $X$ at which $f(X)=0$. As such, there are infinitely many points at which $f$ is zero.
There it is - the contrapositive of your statement. If a continuous function from $mathbb{R}^2$ to $mathbb{R}$ takes both positive and negative values, then it has infinitely many zeros. Done.
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
$endgroup$
add a comment |
$begingroup$
Given two points $A$ and $B$ in the plane, consider the family of all circular arcs including both points. Each such arc is the locus of $X$ such that $angle AXB = theta$ for some fixed $theta$ (as a signed angle), so they don't overlap at all.
There are infinitely many such arcs. If $f$ is continuous and $f(A)$ and $f(B)$ have opposite signs, then each arc must contain a point $X$ at which $f(X)=0$. As such, there are infinitely many points at which $f$ is zero.
There it is - the contrapositive of your statement. If a continuous function from $mathbb{R}^2$ to $mathbb{R}$ takes both positive and negative values, then it has infinitely many zeros. Done.
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
$endgroup$
Given two points $A$ and $B$ in the plane, consider the family of all circular arcs including both points. Each such arc is the locus of $X$ such that $angle AXB = theta$ for some fixed $theta$ (as a signed angle), so they don't overlap at all.
There are infinitely many such arcs. If $f$ is continuous and $f(A)$ and $f(B)$ have opposite signs, then each arc must contain a point $X$ at which $f(X)=0$. As such, there are infinitely many points at which $f$ is zero.
There it is - the contrapositive of your statement. If a continuous function from $mathbb{R}^2$ to $mathbb{R}$ takes both positive and negative values, then it has infinitely many zeros. Done.
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
answered Dec 22 '18 at 3:37
jmerryjmerry
16.8k11633
16.8k11633
add a comment |
add a comment |
$begingroup$
how many pairs $(x,y)$ make your proposed function equal to $0$?
$endgroup$
– Mark S.
Dec 22 '18 at 3:02
2
$begingroup$
Your example $f(x,y)=x$ has infinitely many zeros -- $f(0,k)=0$ for any real $k.$
$endgroup$
– coffeemath
Dec 22 '18 at 3:02
$begingroup$
What does $f(x)$ mean if $f$ has domain $mathbb{R}^2$? Is $x$ a point in the plane?
$endgroup$
– Randall
Dec 22 '18 at 3:15
$begingroup$
@coffeemath, I see. Thanks
$endgroup$
– user408906
Dec 22 '18 at 4:47