Are $F$ and $F^{-1}$ continuous in this given metric space?
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The full question is below:
Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?
I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.
By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.
real-analysis convergence metric-spaces
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
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add a comment |
$begingroup$
The full question is below:
Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?
I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.
By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.
real-analysis convergence metric-spaces
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
$endgroup$
add a comment |
$begingroup$
The full question is below:
Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?
I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.
By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.
real-analysis convergence metric-spaces
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
$endgroup$
The full question is below:
Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?
I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.
By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.
real-analysis convergence metric-spaces
real-analysis convergence metric-spaces
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
asked Dec 18 '18 at 22:09
Mohammed ShahidMohammed Shahid
1457
1457
add a comment |
add a comment |
1 Answer
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$begingroup$
Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.
For $F$ it's easy to see that it's Lipschitz:
$$d_1(f,g) le d_infty(f,g)$$
(we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
$begingroup$
I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:45
1
$begingroup$
@MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:46
$begingroup$
oh, I see. Thank you so much.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:51
1
$begingroup$
@MohammedShahid you're welcome. Good luck with the exam.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:53
add a comment |
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1 Answer
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$begingroup$
Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.
For $F$ it's easy to see that it's Lipschitz:
$$d_1(f,g) le d_infty(f,g)$$
(we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
$begingroup$
I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:45
1
$begingroup$
@MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:46
$begingroup$
oh, I see. Thank you so much.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:51
1
$begingroup$
@MohammedShahid you're welcome. Good luck with the exam.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:53
add a comment |
$begingroup$
Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.
For $F$ it's easy to see that it's Lipschitz:
$$d_1(f,g) le d_infty(f,g)$$
(we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
$begingroup$
I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:45
1
$begingroup$
@MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:46
$begingroup$
oh, I see. Thank you so much.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:51
1
$begingroup$
@MohammedShahid you're welcome. Good luck with the exam.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:53
add a comment |
$begingroup$
Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.
For $F$ it's easy to see that it's Lipschitz:
$$d_1(f,g) le d_infty(f,g)$$
(we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.
For $F$ it's easy to see that it's Lipschitz:
$$d_1(f,g) le d_infty(f,g)$$
(we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
answered Dec 18 '18 at 22:33
Henno BrandsmaHenno Brandsma
115k348124
115k348124
$begingroup$
I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:45
1
$begingroup$
@MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:46
$begingroup$
oh, I see. Thank you so much.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:51
1
$begingroup$
@MohammedShahid you're welcome. Good luck with the exam.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:53
add a comment |
$begingroup$
I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:45
1
$begingroup$
@MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:46
$begingroup$
oh, I see. Thank you so much.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:51
1
$begingroup$
@MohammedShahid you're welcome. Good luck with the exam.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:53
$begingroup$
I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:45
$begingroup$
I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:45
1
1
$begingroup$
@MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:46
$begingroup$
@MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:46
$begingroup$
oh, I see. Thank you so much.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:51
$begingroup$
oh, I see. Thank you so much.
$endgroup$
– Mohammed Shahid
Dec 18 '18 at 22:51
1
1
$begingroup$
@MohammedShahid you're welcome. Good luck with the exam.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:53
$begingroup$
@MohammedShahid you're welcome. Good luck with the exam.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:53
add a comment |
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