Dgdrxrt

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Continuous map on a compact set

No continuous function switches $mathbbQ$ and the irrationalsStone-Cech compactificationExistence of Continuous bijective functionA problem about a continuous iterated functionContinuous map from $l^infty$ to $l^2.$Does there exist any surjective continuous linear map $T:l^2 to l^1$?Path connected compact set with given propertyProof Verification: There exist a real number $alpha$ such that $alpha^2 =2$Show the set $x in M: f(x)<g(x)$ is an open set in metric space when $f,g$ are continuousContinuous map on $mathbbR^d$

1

$begingroup$

Let $f$ be a continuous map on $mathbbR^d$. We denote $A=minlimits_xin Kf(x).$

I want to prove that there exist $epsilon_1>0$ such that

$$f(x)le epsilon_1Rightarrow f(x)ge fracepsilon_02$$

Please help me to do so. Thanks

real-analysis functional-analysis functions

share|cite|improve this question

edited Mar 31 at 21:03

Aymen

asked Mar 29 at 21:17

AymenAymen

224

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is not true. But if you change $f(x)geq epsilon_0$ into $f(x)geq fracepsilon_02$ it will be true.
  $endgroup$
  – Yu Ding
  Mar 29 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes that is what i need to prove, can you please help me to prove it. I edited my post
  $endgroup$
  – Aymen
  Mar 29 at 21:26
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $f$ be a continuous map on $mathbbR^d$. We denote $A=minlimits_xin Kf(x).$

I want to prove that there exist $epsilon_1>0$ such that

$$f(x)le epsilon_1Rightarrow f(x)ge fracepsilon_02$$

Please help me to do so. Thanks

real-analysis functional-analysis functions

share|cite|improve this question

edited Mar 31 at 21:03

Aymen

asked Mar 29 at 21:17

AymenAymen

224

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is not true. But if you change $f(x)geq epsilon_0$ into $f(x)geq fracepsilon_02$ it will be true.
  $endgroup$
  – Yu Ding
  Mar 29 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes that is what i need to prove, can you please help me to prove it. I edited my post
  $endgroup$
  – Aymen
  Mar 29 at 21:26
  

  
  
  
  

add a comment | 

$begingroup$

Let $f$ be a continuous map on $mathbbR^d$. We denote $A=minlimits_xin Kf(x).$

I want to prove that there exist $epsilon_1>0$ such that

$$f(x)le epsilon_1Rightarrow f(x)ge fracepsilon_02$$

Please help me to do so. Thanks

real-analysis functional-analysis functions

share|cite|improve this question

edited Mar 31 at 21:03

Aymen

asked Mar 29 at 21:17

AymenAymen

224

$endgroup$

share|cite|improve this question

edited Mar 31 at 21:03

Aymen

asked Mar 29 at 21:17

AymenAymen

224

share|cite|improve this question

edited Mar 31 at 21:03

Aymen

asked Mar 29 at 21:17

AymenAymen

224

asked Mar 29 at 21:17

AymenAymen

224

asked Mar 29 at 21:17

AymenAymen

224

2 Answers
2

active

oldest

votes

0

$begingroup$

Since $f$ is continuous and $fracepsilon_02>0$, we know that for each $p in K$ the set $f^-1left(<fracepsilon_02right)$ is open in $mathbbR^d$. We also know that $undersetp in Kbigcup f^-1left(<fracepsilon_02right)$ contains $K$. Since the collection of open sets $big f^-1left(<fracepsilon_02right) big_p in K$ covers $K$, there exists $epsilon_1>0$ so that for every $p in K$ we have $B_d(,p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$ (Lebesgue's Number Lemma). Thus if $d(x,K)<epsilon_1$, then $|f(x)-f(q)|<fracepsilon_02$ for some $q in K$. And since $f(q)geq epsilon_0$, we therefore have $f(x)>f(q)-fracepsilon_02geq fracepsilon_02$.

Invoking Lebesgue's Number Lemma allows us to get at the fact that $f$ is uniformly continuous on $K$.

share|cite|improve this answer

answered Mar 29 at 22:05

Matt A PeltoMatt A Pelto

2,657621

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Matt A Pelto, i didn't understand why "for every $p in K$ we have $B_d(,p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$ (Lebesgue's Number Lemma)" implies "if $d(x,K)<epsilon_1$, then $|f(x)-f(q)|<fracepsilon_02$ for some $q in K$." Can you please explain this more on details? thanks in advance
  $endgroup$
  – Aymen
  Mar 29 at 23:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $d(x,K)<epsilon_1$ implies $x in B_d(p, epsilon_1)$ for some $p in K$, and $B_d(p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$.
  $endgroup$
  – Matt A Pelto
  Mar 30 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that is clear now for me. I have an other question, if w consider additionaly a continuous map $g$ on $mathbbR^d$ such that $g(x)=0$ if and only if $x$ belongs to $K.$ How can us prove that there is $epsilon_2>0$ such that beginalign* d(x,K)ge epsilon_1Rightarrow |g(x)|ge epsilon_2;. endalign*Thanks in advance
  $endgroup$
  – Aymen
  Mar 30 at 0:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is something wrong with the way your followup question has been worded. Where does $epsilon_1$ come from?
  $endgroup$
  – Matt A Pelto
  Mar 30 at 2:05
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Use the following three facts (prove them!):

1. For points outside of $K$, we have $d(x,K) = d(x,partial K)$.


2. The boundary $partial K$ is again compact.


3. 
   $f$ is continuous, use the $epsilon$-$delta$ formulation of continuity.

share|cite|improve this answer

answered Mar 29 at 22:10

Daniel Robert-NicoudDaniel Robert-Nicoud

20.6k33897

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can I answer as follows: suppose that for all $epsilon_1>0$ there exists $xinmathbbR^d$ such that $d(x,K)le epsilon_1$ and $f(x)<fracepsilon_02.$ Then we could let $epsilon_1$ tends to zero and this would imply that $d(x,K)=0$ which in turn implies that $xin K$ since $K$ is compact. Thus $f(x)ge epsilon_0$ which is a contradiction.
  $endgroup$
  – Aymen
  Mar 29 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aymen This looks like a good argument. The one I had in mind was that for each of the points $xinpartial K$ there exists $r_x>0$ so that $fgetfracepsilon_02$ on $B(x,r_x)$. Since $partial K$ is compact, it is covered by a finite number of these balls $B(x_i,r_x_i)$. Then you want to prove that there exists $epsilon_1$ such that for all $xinpartial K$, the ball $B(x,epsilon_1)$ is contained in $bigcup_iB(x_i,r_x_i)$, and conclude.
  $endgroup$
  – Daniel Robert-Nicoud
  Mar 29 at 22:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i think that what i said is false. In fact when letting $epsilon_1$ tends to zero $d(x,K)$ do not tend to $d(x,K)$ since $x$ depends on $epsilon_1$
  $endgroup$
  – Aymen
  Mar 29 at 23:06
  

  
  
  
  

add a comment | 

Your Answer

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0

$begingroup$

Since $f$ is continuous and $fracepsilon_02>0$, we know that for each $p in K$ the set $f^-1left(<fracepsilon_02right)$ is open in $mathbbR^d$. We also know that $undersetp in Kbigcup f^-1left(<fracepsilon_02right)$ contains $K$. Since the collection of open sets $big f^-1left(<fracepsilon_02right) big_p in K$ covers $K$, there exists $epsilon_1>0$ so that for every $p in K$ we have $B_d(,p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$ (Lebesgue's Number Lemma). Thus if $d(x,K)<epsilon_1$, then $|f(x)-f(q)|<fracepsilon_02$ for some $q in K$. And since $f(q)geq epsilon_0$, we therefore have $f(x)>f(q)-fracepsilon_02geq fracepsilon_02$.

Invoking Lebesgue's Number Lemma allows us to get at the fact that $f$ is uniformly continuous on $K$.

share|cite|improve this answer

answered Mar 29 at 22:05

Matt A PeltoMatt A Pelto

2,657621

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Matt A Pelto, i didn't understand why "for every $p in K$ we have $B_d(,p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$ (Lebesgue's Number Lemma)" implies "if $d(x,K)<epsilon_1$, then $|f(x)-f(q)|<fracepsilon_02$ for some $q in K$." Can you please explain this more on details? thanks in advance
  $endgroup$
  – Aymen
  Mar 29 at 23:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $d(x,K)<epsilon_1$ implies $x in B_d(p, epsilon_1)$ for some $p in K$, and $B_d(p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$.
  $endgroup$
  – Matt A Pelto
  Mar 30 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that is clear now for me. I have an other question, if w consider additionaly a continuous map $g$ on $mathbbR^d$ such that $g(x)=0$ if and only if $x$ belongs to $K.$ How can us prove that there is $epsilon_2>0$ such that beginalign* d(x,K)ge epsilon_1Rightarrow |g(x)|ge epsilon_2;. endalign*Thanks in advance
  $endgroup$
  – Aymen
  Mar 30 at 0:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is something wrong with the way your followup question has been worded. Where does $epsilon_1$ come from?
  $endgroup$
  – Matt A Pelto
  Mar 30 at 2:05
  

  
  
  
  

add a comment | 

0

$begingroup$

Since $f$ is continuous and $fracepsilon_02>0$, we know that for each $p in K$ the set $f^-1left(<fracepsilon_02right)$ is open in $mathbbR^d$. We also know that $undersetp in Kbigcup f^-1left(<fracepsilon_02right)$ contains $K$. Since the collection of open sets $big f^-1left(<fracepsilon_02right) big_p in K$ covers $K$, there exists $epsilon_1>0$ so that for every $p in K$ we have $B_d(,p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$ (Lebesgue's Number Lemma). Thus if $d(x,K)<epsilon_1$, then $|f(x)-f(q)|<fracepsilon_02$ for some $q in K$. And since $f(q)geq epsilon_0$, we therefore have $f(x)>f(q)-fracepsilon_02geq fracepsilon_02$.

Invoking Lebesgue's Number Lemma allows us to get at the fact that $f$ is uniformly continuous on $K$.

share|cite|improve this answer

answered Mar 29 at 22:05

Matt A PeltoMatt A Pelto

2,657621

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Matt A Pelto, i didn't understand why "for every $p in K$ we have $B_d(,p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$ (Lebesgue's Number Lemma)" implies "if $d(x,K)<epsilon_1$, then $|f(x)-f(q)|<fracepsilon_02$ for some $q in K$." Can you please explain this more on details? thanks in advance
  $endgroup$
  – Aymen
  Mar 29 at 23:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $d(x,K)<epsilon_1$ implies $x in B_d(p, epsilon_1)$ for some $p in K$, and $B_d(p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$.
  $endgroup$
  – Matt A Pelto
  Mar 30 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that is clear now for me. I have an other question, if w consider additionaly a continuous map $g$ on $mathbbR^d$ such that $g(x)=0$ if and only if $x$ belongs to $K.$ How can us prove that there is $epsilon_2>0$ such that beginalign* d(x,K)ge epsilon_1Rightarrow |g(x)|ge epsilon_2;. endalign*Thanks in advance
  $endgroup$
  – Aymen
  Mar 30 at 0:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is something wrong with the way your followup question has been worded. Where does $epsilon_1$ come from?
  $endgroup$
  – Matt A Pelto
  Mar 30 at 2:05
  

  
  
  
  

add a comment | 

$begingroup$

Since $f$ is continuous and $fracepsilon_02>0$, we know that for each $p in K$ the set $f^-1left(<fracepsilon_02right)$ is open in $mathbbR^d$. We also know that $undersetp in Kbigcup f^-1left(<fracepsilon_02right)$ contains $K$. Since the collection of open sets $big f^-1left(<fracepsilon_02right) big_p in K$ covers $K$, there exists $epsilon_1>0$ so that for every $p in K$ we have $B_d(,p, epsilon_1) subseteq f^-1left(<fracepsilon_02right)$ for some $q in K$ (Lebesgue's Number Lemma). Thus if $d(x,K)<epsilon_1$, then $|f(x)-f(q)|<fracepsilon_02$ for some $q in K$. And since $f(q)geq epsilon_0$, we therefore have $f(x)>f(q)-fracepsilon_02geq fracepsilon_02$.

Invoking Lebesgue's Number Lemma allows us to get at the fact that $f$ is uniformly continuous on $K$.

share|cite|improve this answer

answered Mar 29 at 22:05

Matt A PeltoMatt A Pelto

2,657621

$endgroup$

share|cite|improve this answer

answered Mar 29 at 22:05

Matt A PeltoMatt A Pelto

2,657621

answered Mar 29 at 22:05

Matt A PeltoMatt A Pelto

2,657621

answered Mar 29 at 22:05

Matt A PeltoMatt A Pelto

2,657621

0

$begingroup$

Hint: Use the following three facts (prove them!):

1. For points outside of $K$, we have $d(x,K) = d(x,partial K)$.


2. The boundary $partial K$ is again compact.


3. 
   $f$ is continuous, use the $epsilon$-$delta$ formulation of continuity.

share|cite|improve this answer

answered Mar 29 at 22:10

Daniel Robert-NicoudDaniel Robert-Nicoud

20.6k33897

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can I answer as follows: suppose that for all $epsilon_1>0$ there exists $xinmathbbR^d$ such that $d(x,K)le epsilon_1$ and $f(x)<fracepsilon_02.$ Then we could let $epsilon_1$ tends to zero and this would imply that $d(x,K)=0$ which in turn implies that $xin K$ since $K$ is compact. Thus $f(x)ge epsilon_0$ which is a contradiction.
  $endgroup$
  – Aymen
  Mar 29 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aymen This looks like a good argument. The one I had in mind was that for each of the points $xinpartial K$ there exists $r_x>0$ so that $fgetfracepsilon_02$ on $B(x,r_x)$. Since $partial K$ is compact, it is covered by a finite number of these balls $B(x_i,r_x_i)$. Then you want to prove that there exists $epsilon_1$ such that for all $xinpartial K$, the ball $B(x,epsilon_1)$ is contained in $bigcup_iB(x_i,r_x_i)$, and conclude.
  $endgroup$
  – Daniel Robert-Nicoud
  Mar 29 at 22:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i think that what i said is false. In fact when letting $epsilon_1$ tends to zero $d(x,K)$ do not tend to $d(x,K)$ since $x$ depends on $epsilon_1$
  $endgroup$
  – Aymen
  Mar 29 at 23:06
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Use the following three facts (prove them!):

1. For points outside of $K$, we have $d(x,K) = d(x,partial K)$.


2. The boundary $partial K$ is again compact.


3. 
   $f$ is continuous, use the $epsilon$-$delta$ formulation of continuity.

share|cite|improve this answer

answered Mar 29 at 22:10

Daniel Robert-NicoudDaniel Robert-Nicoud

20.6k33897

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can I answer as follows: suppose that for all $epsilon_1>0$ there exists $xinmathbbR^d$ such that $d(x,K)le epsilon_1$ and $f(x)<fracepsilon_02.$ Then we could let $epsilon_1$ tends to zero and this would imply that $d(x,K)=0$ which in turn implies that $xin K$ since $K$ is compact. Thus $f(x)ge epsilon_0$ which is a contradiction.
  $endgroup$
  – Aymen
  Mar 29 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aymen This looks like a good argument. The one I had in mind was that for each of the points $xinpartial K$ there exists $r_x>0$ so that $fgetfracepsilon_02$ on $B(x,r_x)$. Since $partial K$ is compact, it is covered by a finite number of these balls $B(x_i,r_x_i)$. Then you want to prove that there exists $epsilon_1$ such that for all $xinpartial K$, the ball $B(x,epsilon_1)$ is contained in $bigcup_iB(x_i,r_x_i)$, and conclude.
  $endgroup$
  – Daniel Robert-Nicoud
  Mar 29 at 22:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i think that what i said is false. In fact when letting $epsilon_1$ tends to zero $d(x,K)$ do not tend to $d(x,K)$ since $x$ depends on $epsilon_1$
  $endgroup$
  – Aymen
  Mar 29 at 23:06
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Use the following three facts (prove them!):

1. For points outside of $K$, we have $d(x,K) = d(x,partial K)$.


2. The boundary $partial K$ is again compact.


3. 
   $f$ is continuous, use the $epsilon$-$delta$ formulation of continuity.

share|cite|improve this answer

answered Mar 29 at 22:10

Daniel Robert-NicoudDaniel Robert-Nicoud

20.6k33897

$endgroup$

share|cite|improve this answer

answered Mar 29 at 22:10

Daniel Robert-NicoudDaniel Robert-Nicoud

20.6k33897

answered Mar 29 at 22:10

Daniel Robert-NicoudDaniel Robert-Nicoud

20.6k33897

answered Mar 29 at 22:10

Daniel Robert-NicoudDaniel Robert-Nicoud

20.6k33897