Mixing
Continuous function on the open unit disc in $mathbb R^2.$
$begingroup$
Suppose $U=(0,frac{1}{2})times(0,frac{1}{2}),V=(-frac{1}{2},0)times (-frac{1}{2},0)$ and $D$ be the open unit disc with centre at the origin of $mathbb R^2.$ Let $f$ be a real valued continuous function on $D$ such that $f(U)=0.$ Then is follows that
$f(v)=0$ for every $v$ in $V$
$f(v)neq 0$ for every $v$ in $V$
$f(v)=0$ for some $v$ in $V$
$f$ can assume any real value on $V$.
If we take $f(x,y)=sqrt{x^2+y^2}$ for all $(x,y)in V$ then $(1)$ and $(3)$ are not true.
If we take $fequiv0$ on $mathbb R^2$ then $(2)$ is not true.
So I think $(4)$ is true. Am I correct?
Any help is appreciated. Thank you.
general-topology
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
$endgroup$
add a comment |
$begingroup$
Suppose $U=(0,frac{1}{2})times(0,frac{1}{2}),V=(-frac{1}{2},0)times (-frac{1}{2},0)$ and $D$ be the open unit disc with centre at the origin of $mathbb R^2.$ Let $f$ be a real valued continuous function on $D$ such that $f(U)=0.$ Then is follows that
$f(v)=0$ for every $v$ in $V$
$f(v)neq 0$ for every $v$ in $V$
$f(v)=0$ for some $v$ in $V$
$f$ can assume any real value on $V$.
If we take $f(x,y)=sqrt{x^2+y^2}$ for all $(x,y)in V$ then $(1)$ and $(3)$ are not true.
If we take $fequiv0$ on $mathbb R^2$ then $(2)$ is not true.
So I think $(4)$ is true. Am I correct?
Any help is appreciated. Thank you.
general-topology
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
$endgroup$
add a comment |
$begingroup$
Suppose $U=(0,frac{1}{2})times(0,frac{1}{2}),V=(-frac{1}{2},0)times (-frac{1}{2},0)$ and $D$ be the open unit disc with centre at the origin of $mathbb R^2.$ Let $f$ be a real valued continuous function on $D$ such that $f(U)=0.$ Then is follows that
$f(v)=0$ for every $v$ in $V$
$f(v)neq 0$ for every $v$ in $V$
$f(v)=0$ for some $v$ in $V$
$f$ can assume any real value on $V$.
If we take $f(x,y)=sqrt{x^2+y^2}$ for all $(x,y)in V$ then $(1)$ and $(3)$ are not true.
If we take $fequiv0$ on $mathbb R^2$ then $(2)$ is not true.
So I think $(4)$ is true. Am I correct?
Any help is appreciated. Thank you.
general-topology
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
$endgroup$
Suppose $U=(0,frac{1}{2})times(0,frac{1}{2}),V=(-frac{1}{2},0)times (-frac{1}{2},0)$ and $D$ be the open unit disc with centre at the origin of $mathbb R^2.$ Let $f$ be a real valued continuous function on $D$ such that $f(U)=0.$ Then is follows that
$f(v)=0$ for every $v$ in $V$
$f(v)neq 0$ for every $v$ in $V$
$f(v)=0$ for some $v$ in $V$
$f$ can assume any real value on $V$.
If we take $f(x,y)=sqrt{x^2+y^2}$ for all $(x,y)in V$ then $(1)$ and $(3)$ are not true.
If we take $fequiv0$ on $mathbb R^2$ then $(2)$ is not true.
So I think $(4)$ is true. Am I correct?
Any help is appreciated. Thank you.
general-topology
general-topology
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
asked Jan 29 at 16:49
nurun neshanurun nesha
1,0462623
1,0462623
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Counterexamples 1. $f(x,y) = 0,xge 0,$ $f(x,y) = x,xle 0.$ 2. $fequiv 0.$ 3. See 1.
- Is it possible for $f(V)=mathbb R?$ No, because $V$ is contained in a compact subset of $D$ (namely $[-1/2,0]times [-1/2,0].$) Hence $f(V)$ is bounded. Is it possible, given an arbitrary $ain mathbb R,$ to construct an $f$ satisfying the given hypotheses such that $f=a$ somewhere in $V?$ Sure: Let $f$ be as in 1. above. Then the function $-4af$ takes on the value $a$ at $(-1/4,-1/4)in V.$
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
$endgroup$
$begingroup$
I don't understand your reasoning regarding $f(V)$ being necessarily bounded. Here $D$ is not the closed, but open disc.
$endgroup$
– Robert Thingum
Jan 30 at 1:07
$begingroup$
@RobertThingum That doesn't matter. $Vsubset K=[-1/2,0]times [-1/2,0] subset D.$ Therefore $f(V)subset f(K).$ And $f$ is continuous on $K,$ a compact set.
$endgroup$
– zhw.
Jan 30 at 5:06
$begingroup$
Ah, I see, thank you.
$endgroup$
– Robert Thingum
Jan 30 at 5:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Counterexamples 1. $f(x,y) = 0,xge 0,$ $f(x,y) = x,xle 0.$ 2. $fequiv 0.$ 3. See 1.
- Is it possible for $f(V)=mathbb R?$ No, because $V$ is contained in a compact subset of $D$ (namely $[-1/2,0]times [-1/2,0].$) Hence $f(V)$ is bounded. Is it possible, given an arbitrary $ain mathbb R,$ to construct an $f$ satisfying the given hypotheses such that $f=a$ somewhere in $V?$ Sure: Let $f$ be as in 1. above. Then the function $-4af$ takes on the value $a$ at $(-1/4,-1/4)in V.$
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
$endgroup$
$begingroup$
I don't understand your reasoning regarding $f(V)$ being necessarily bounded. Here $D$ is not the closed, but open disc.
$endgroup$
– Robert Thingum
Jan 30 at 1:07
$begingroup$
@RobertThingum That doesn't matter. $Vsubset K=[-1/2,0]times [-1/2,0] subset D.$ Therefore $f(V)subset f(K).$ And $f$ is continuous on $K,$ a compact set.
$endgroup$
– zhw.
Jan 30 at 5:06
$begingroup$
Ah, I see, thank you.
$endgroup$
– Robert Thingum
Jan 30 at 5:09
add a comment |
$begingroup$
Counterexamples 1. $f(x,y) = 0,xge 0,$ $f(x,y) = x,xle 0.$ 2. $fequiv 0.$ 3. See 1.
- Is it possible for $f(V)=mathbb R?$ No, because $V$ is contained in a compact subset of $D$ (namely $[-1/2,0]times [-1/2,0].$) Hence $f(V)$ is bounded. Is it possible, given an arbitrary $ain mathbb R,$ to construct an $f$ satisfying the given hypotheses such that $f=a$ somewhere in $V?$ Sure: Let $f$ be as in 1. above. Then the function $-4af$ takes on the value $a$ at $(-1/4,-1/4)in V.$
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
$endgroup$
$begingroup$
I don't understand your reasoning regarding $f(V)$ being necessarily bounded. Here $D$ is not the closed, but open disc.
$endgroup$
– Robert Thingum
Jan 30 at 1:07
$begingroup$
@RobertThingum That doesn't matter. $Vsubset K=[-1/2,0]times [-1/2,0] subset D.$ Therefore $f(V)subset f(K).$ And $f$ is continuous on $K,$ a compact set.
$endgroup$
– zhw.
Jan 30 at 5:06
$begingroup$
Ah, I see, thank you.
$endgroup$
– Robert Thingum
Jan 30 at 5:09
add a comment |
$begingroup$
Counterexamples 1. $f(x,y) = 0,xge 0,$ $f(x,y) = x,xle 0.$ 2. $fequiv 0.$ 3. See 1.
- Is it possible for $f(V)=mathbb R?$ No, because $V$ is contained in a compact subset of $D$ (namely $[-1/2,0]times [-1/2,0].$) Hence $f(V)$ is bounded. Is it possible, given an arbitrary $ain mathbb R,$ to construct an $f$ satisfying the given hypotheses such that $f=a$ somewhere in $V?$ Sure: Let $f$ be as in 1. above. Then the function $-4af$ takes on the value $a$ at $(-1/4,-1/4)in V.$
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
$endgroup$
Counterexamples 1. $f(x,y) = 0,xge 0,$ $f(x,y) = x,xle 0.$ 2. $fequiv 0.$ 3. See 1.
- Is it possible for $f(V)=mathbb R?$ No, because $V$ is contained in a compact subset of $D$ (namely $[-1/2,0]times [-1/2,0].$) Hence $f(V)$ is bounded. Is it possible, given an arbitrary $ain mathbb R,$ to construct an $f$ satisfying the given hypotheses such that $f=a$ somewhere in $V?$ Sure: Let $f$ be as in 1. above. Then the function $-4af$ takes on the value $a$ at $(-1/4,-1/4)in V.$
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
answered Jan 29 at 20:24
zhw.zhw.
74.7k43275
74.7k43275
$begingroup$
I don't understand your reasoning regarding $f(V)$ being necessarily bounded. Here $D$ is not the closed, but open disc.
$endgroup$
– Robert Thingum
Jan 30 at 1:07
$begingroup$
@RobertThingum That doesn't matter. $Vsubset K=[-1/2,0]times [-1/2,0] subset D.$ Therefore $f(V)subset f(K).$ And $f$ is continuous on $K,$ a compact set.
$endgroup$
– zhw.
Jan 30 at 5:06
$begingroup$
Ah, I see, thank you.
$endgroup$
– Robert Thingum
Jan 30 at 5:09
add a comment |
$begingroup$
I don't understand your reasoning regarding $f(V)$ being necessarily bounded. Here $D$ is not the closed, but open disc.
$endgroup$
– Robert Thingum
Jan 30 at 1:07
$begingroup$
@RobertThingum That doesn't matter. $Vsubset K=[-1/2,0]times [-1/2,0] subset D.$ Therefore $f(V)subset f(K).$ And $f$ is continuous on $K,$ a compact set.
$endgroup$
– zhw.
Jan 30 at 5:06
$begingroup$
Ah, I see, thank you.
$endgroup$
– Robert Thingum
Jan 30 at 5:09
$begingroup$
I don't understand your reasoning regarding $f(V)$ being necessarily bounded. Here $D$ is not the closed, but open disc.
$endgroup$
– Robert Thingum
Jan 30 at 1:07
$begingroup$
I don't understand your reasoning regarding $f(V)$ being necessarily bounded. Here $D$ is not the closed, but open disc.
$endgroup$
– Robert Thingum
Jan 30 at 1:07
$begingroup$
@RobertThingum That doesn't matter. $Vsubset K=[-1/2,0]times [-1/2,0] subset D.$ Therefore $f(V)subset f(K).$ And $f$ is continuous on $K,$ a compact set.
$endgroup$
– zhw.
Jan 30 at 5:06
$begingroup$
@RobertThingum That doesn't matter. $Vsubset K=[-1/2,0]times [-1/2,0] subset D.$ Therefore $f(V)subset f(K).$ And $f$ is continuous on $K,$ a compact set.
$endgroup$
– zhw.
Jan 30 at 5:06
$begingroup$
Ah, I see, thank you.
$endgroup$
– Robert Thingum
Jan 30 at 5:09
$begingroup$
Ah, I see, thank you.
$endgroup$
– Robert Thingum
Jan 30 at 5:09
add a comment |
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