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$f_n:Mmapsto M'$ uniformly converges iff $d'(f_{n_k}(x_k),f(x_k))rightarrow 0inBbb Rforall$ subsequence...
$begingroup$
Let $(M,d)$ and $(M',d')$ be two metric spaces and $f_n:Mmapsto M'$ a sequence of functions and $f:Mmapsto M'$ is a function.
$f_nrightarrow f$ uniformly $iff foralltext{ subsequence } (f_{n_k}),forall text{ sequence }(x_k)$ in $M$ $d'(f_{n_k}(x_k),f(x_k))rightarrow 0$ in $(Bbb R,|.|)$
Here is how we prove the direction $big[~f_n$ does not uniformly converge towards $f$ $big]impliesexists (n_k)_{kinBbb N}$ and $(x_k)subset M$ such that $sup d'(f_{n_k}(x_k),f(x_k))>delta>0~forall kinBbb N$ for some $delta$:
Since $f_n$ does not uniformly converge towards $f$ $$existsdelta>0 text{ such that }forall NinBbb N~exists nge Ntext{ with }suplimits_{xin M}d'(f_n(x),f(x))>delta$$
Then we can recursively construct two sequences $n_k$ and $x_k$ such that $n_{k+1}>n_k$ (or $n_{k+1}ge n_k+1$) and $d'(f_{n_k}(x_k),f(x_k))>delta/2$
I get why we can find the appropriate $n_k$ but why shoud the $x_k$ exist? We're working under the assumption that $forall N~exists nge N$ such that $suplimits_{xin M}d'(f_n(x),f(x))>delta$.
How I think I understand it is that for each $n_k$ $sup$ has the property of there being a sequence $(y_l)subset M$ such that $limlimits_{lrightarrowinfty}d'(f_{n_k}(y_l),f(y_l))>delta$ and so for $l$ sufficiently large we can find an actual $x_k:=y_lin M$ such that $d'(f_n(y_l),f(y_l))>delta/2$.
So for each $n_k$ the $sup$ is $>delta$ so there exists $x_kin M$ such that $d'(f_{n_k}(x_k),f(x_k))>delta/2$ and the result follows.
Is there any subtlety that I didn't get?
sequences-and-series metric-spaces
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
Let $(M,d)$ and $(M',d')$ be two metric spaces and $f_n:Mmapsto M'$ a sequence of functions and $f:Mmapsto M'$ is a function.
$f_nrightarrow f$ uniformly $iff foralltext{ subsequence } (f_{n_k}),forall text{ sequence }(x_k)$ in $M$ $d'(f_{n_k}(x_k),f(x_k))rightarrow 0$ in $(Bbb R,|.|)$
Here is how we prove the direction $big[~f_n$ does not uniformly converge towards $f$ $big]impliesexists (n_k)_{kinBbb N}$ and $(x_k)subset M$ such that $sup d'(f_{n_k}(x_k),f(x_k))>delta>0~forall kinBbb N$ for some $delta$:
Since $f_n$ does not uniformly converge towards $f$ $$existsdelta>0 text{ such that }forall NinBbb N~exists nge Ntext{ with }suplimits_{xin M}d'(f_n(x),f(x))>delta$$
Then we can recursively construct two sequences $n_k$ and $x_k$ such that $n_{k+1}>n_k$ (or $n_{k+1}ge n_k+1$) and $d'(f_{n_k}(x_k),f(x_k))>delta/2$
I get why we can find the appropriate $n_k$ but why shoud the $x_k$ exist? We're working under the assumption that $forall N~exists nge N$ such that $suplimits_{xin M}d'(f_n(x),f(x))>delta$.
How I think I understand it is that for each $n_k$ $sup$ has the property of there being a sequence $(y_l)subset M$ such that $limlimits_{lrightarrowinfty}d'(f_{n_k}(y_l),f(y_l))>delta$ and so for $l$ sufficiently large we can find an actual $x_k:=y_lin M$ such that $d'(f_n(y_l),f(y_l))>delta/2$.
So for each $n_k$ the $sup$ is $>delta$ so there exists $x_kin M$ such that $d'(f_{n_k}(x_k),f(x_k))>delta/2$ and the result follows.
Is there any subtlety that I didn't get?
sequences-and-series metric-spaces
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
Let $(M,d)$ and $(M',d')$ be two metric spaces and $f_n:Mmapsto M'$ a sequence of functions and $f:Mmapsto M'$ is a function.
$f_nrightarrow f$ uniformly $iff foralltext{ subsequence } (f_{n_k}),forall text{ sequence }(x_k)$ in $M$ $d'(f_{n_k}(x_k),f(x_k))rightarrow 0$ in $(Bbb R,|.|)$
Here is how we prove the direction $big[~f_n$ does not uniformly converge towards $f$ $big]impliesexists (n_k)_{kinBbb N}$ and $(x_k)subset M$ such that $sup d'(f_{n_k}(x_k),f(x_k))>delta>0~forall kinBbb N$ for some $delta$:
Since $f_n$ does not uniformly converge towards $f$ $$existsdelta>0 text{ such that }forall NinBbb N~exists nge Ntext{ with }suplimits_{xin M}d'(f_n(x),f(x))>delta$$
Then we can recursively construct two sequences $n_k$ and $x_k$ such that $n_{k+1}>n_k$ (or $n_{k+1}ge n_k+1$) and $d'(f_{n_k}(x_k),f(x_k))>delta/2$
I get why we can find the appropriate $n_k$ but why shoud the $x_k$ exist? We're working under the assumption that $forall N~exists nge N$ such that $suplimits_{xin M}d'(f_n(x),f(x))>delta$.
How I think I understand it is that for each $n_k$ $sup$ has the property of there being a sequence $(y_l)subset M$ such that $limlimits_{lrightarrowinfty}d'(f_{n_k}(y_l),f(y_l))>delta$ and so for $l$ sufficiently large we can find an actual $x_k:=y_lin M$ such that $d'(f_n(y_l),f(y_l))>delta/2$.
So for each $n_k$ the $sup$ is $>delta$ so there exists $x_kin M$ such that $d'(f_{n_k}(x_k),f(x_k))>delta/2$ and the result follows.
Is there any subtlety that I didn't get?
sequences-and-series metric-spaces
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
$endgroup$
Let $(M,d)$ and $(M',d')$ be two metric spaces and $f_n:Mmapsto M'$ a sequence of functions and $f:Mmapsto M'$ is a function.
$f_nrightarrow f$ uniformly $iff foralltext{ subsequence } (f_{n_k}),forall text{ sequence }(x_k)$ in $M$ $d'(f_{n_k}(x_k),f(x_k))rightarrow 0$ in $(Bbb R,|.|)$
Here is how we prove the direction $big[~f_n$ does not uniformly converge towards $f$ $big]impliesexists (n_k)_{kinBbb N}$ and $(x_k)subset M$ such that $sup d'(f_{n_k}(x_k),f(x_k))>delta>0~forall kinBbb N$ for some $delta$:
Since $f_n$ does not uniformly converge towards $f$ $$existsdelta>0 text{ such that }forall NinBbb N~exists nge Ntext{ with }suplimits_{xin M}d'(f_n(x),f(x))>delta$$
Then we can recursively construct two sequences $n_k$ and $x_k$ such that $n_{k+1}>n_k$ (or $n_{k+1}ge n_k+1$) and $d'(f_{n_k}(x_k),f(x_k))>delta/2$
I get why we can find the appropriate $n_k$ but why shoud the $x_k$ exist? We're working under the assumption that $forall N~exists nge N$ such that $suplimits_{xin M}d'(f_n(x),f(x))>delta$.
How I think I understand it is that for each $n_k$ $sup$ has the property of there being a sequence $(y_l)subset M$ such that $limlimits_{lrightarrowinfty}d'(f_{n_k}(y_l),f(y_l))>delta$ and so for $l$ sufficiently large we can find an actual $x_k:=y_lin M$ such that $d'(f_n(y_l),f(y_l))>delta/2$.
So for each $n_k$ the $sup$ is $>delta$ so there exists $x_kin M$ such that $d'(f_{n_k}(x_k),f(x_k))>delta/2$ and the result follows.
Is there any subtlety that I didn't get?
sequences-and-series metric-spaces
sequences-and-series metric-spaces
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
edited Jan 18 at 18:00
John Cataldo
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
asked Jan 18 at 17:53
John CataldoJohn Cataldo
1,1881316
1,1881316
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