Show that inverse of this continuous function is continuous
This question is from baby Rudin, Ch 5, exer 2:
Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.
The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'
Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.
So, how to show that $g$ indeed continuous on $(a,b)$?
real-analysis derivatives continuity
asked Nov 25 at 13:45
Silent
2,64932050
add a comment |
This question is from baby Rudin, Ch 5, exer 2:
Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.
The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'
Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.
So, how to show that $g$ indeed continuous on $(a,b)$?
real-analysis derivatives continuity
asked Nov 25 at 13:45
Silent
2,64932050
Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58
As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05
@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18
@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44
add a comment |
This question is from baby Rudin, Ch 5, exer 2:
Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.
The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'
Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.
So, how to show that $g$ indeed continuous on $(a,b)$?
real-analysis derivatives continuity
asked Nov 25 at 13:45
Silent
2,64932050
This question is from baby Rudin, Ch 5, exer 2:
Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.
The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'
Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.
So, how to show that $g$ indeed continuous on $(a,b)$?
real-analysis derivatives continuity
real-analysis derivatives continuity
asked Nov 25 at 13:45
Silent
2,64932050
asked Nov 25 at 13:45
Silent
2,64932050
asked Nov 25 at 13:45
Silent
2,64932050
asked Nov 25 at 13:45
Silent
2,64932050
asked Nov 25 at 13:45
Silent
2,64932050
2,64932050
Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58
As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05
@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18
@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44
add a comment |
Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58
As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05
@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18
@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44
Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58
Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58
As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05
As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05
@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18
@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18
@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44
@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44
add a comment |
1 Answer
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It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.
answered Nov 25 at 16:03
KnobbyWan
195110
add a comment |
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It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.
answered Nov 25 at 16:03
KnobbyWan
195110
add a comment |
It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.
answered Nov 25 at 16:03
KnobbyWan
195110
add a comment |
It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.
answered Nov 25 at 16:03
KnobbyWan
195110
It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.
answered Nov 25 at 16:03
KnobbyWan
195110
answered Nov 25 at 16:03
KnobbyWan
195110
answered Nov 25 at 16:03
KnobbyWan
195110
answered Nov 25 at 16:03
KnobbyWan
195110
195110
add a comment |
add a comment |
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Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58
As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05
@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18
@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44