Mixing
Convergence of a sequence of functions to a continuous limit
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Suppose that $f_n$ is a sequence of continuous functions on $X$ s.t. $f_nto f$ for all $xin D$, where $D$ is a dense subset of $X$ and $f$ is continuous. Is it true that $f_nto f$ for all $xin X$.
Let $ynotin D$ and suppose that $f_n(y)to g(y)$. Then $g(y)$ must be equal $f(y)$. How does one show that $f_n(y)$ converges?
real-analysis convergence continuity
edited Jul 29 at 16:17
José Carlos Santos
112k1696173
asked Jul 29 at 16:13
Robert W.
35818
add a comment |Â
up vote
0
down vote
favorite
Suppose that $f_n$ is a sequence of continuous functions on $X$ s.t. $f_nto f$ for all $xin D$, where $D$ is a dense subset of $X$ and $f$ is continuous. Is it true that $f_nto f$ for all $xin X$.
Let $ynotin D$ and suppose that $f_n(y)to g(y)$. Then $g(y)$ must be equal $f(y)$. How does one show that $f_n(y)$ converges?
real-analysis convergence continuity
edited Jul 29 at 16:17
José Carlos Santos
112k1696173
asked Jul 29 at 16:13
Robert W.
35818
Are you assuming that $f$ is continuous too?
– José Carlos Santos
Jul 29 at 16:14
Yes. It is continuous
– Robert W.
Jul 29 at 16:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $f_n$ is a sequence of continuous functions on $X$ s.t. $f_nto f$ for all $xin D$, where $D$ is a dense subset of $X$ and $f$ is continuous. Is it true that $f_nto f$ for all $xin X$.
Let $ynotin D$ and suppose that $f_n(y)to g(y)$. Then $g(y)$ must be equal $f(y)$. How does one show that $f_n(y)$ converges?
real-analysis convergence continuity
edited Jul 29 at 16:17
José Carlos Santos
112k1696173
asked Jul 29 at 16:13
Robert W.
35818
Suppose that $f_n$ is a sequence of continuous functions on $X$ s.t. $f_nto f$ for all $xin D$, where $D$ is a dense subset of $X$ and $f$ is continuous. Is it true that $f_nto f$ for all $xin X$.
Let $ynotin D$ and suppose that $f_n(y)to g(y)$. Then $g(y)$ must be equal $f(y)$. How does one show that $f_n(y)$ converges?
real-analysis convergence continuity
edited Jul 29 at 16:17
José Carlos Santos
112k1696173
asked Jul 29 at 16:13
Robert W.
35818
edited Jul 29 at 16:17
José Carlos Santos
112k1696173
edited Jul 29 at 16:17
José Carlos Santos
112k1696173
edited Jul 29 at 16:17
José Carlos Santos
112k1696173
112k1696173
asked Jul 29 at 16:13
Robert W.
35818
asked Jul 29 at 16:13
Robert W.
35818
asked Jul 29 at 16:13
Robert W.
35818
35818
Are you assuming that $f$ is continuous too?
– José Carlos Santos
Jul 29 at 16:14
Yes. It is continuous
– Robert W.
Jul 29 at 16:14
add a comment |Â
Are you assuming that $f$ is continuous too?
– José Carlos Santos
Jul 29 at 16:14
Yes. It is continuous
– Robert W.
Jul 29 at 16:14
Are you assuming that $f$ is continuous too?
– José Carlos Santos
Jul 29 at 16:14
Are you assuming that $f$ is continuous too?
– José Carlos Santos
Jul 29 at 16:14
Yes. It is continuous
– Robert W.
Jul 29 at 16:14
Yes. It is continuous
– Robert W.
Jul 29 at 16:14
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
No, it is not true. Take $X=[0,1]$, $D=[0,1)$, $f(x)=0$ and $f_n(x)=x^n$. Then$$(forall xin D):lim_ntoinftyf_n(x)=f(x),$$but $lim_ntoinftyf_n(1)=1neq f(1)$.
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
No, it is not true. Take $X=[0,1]$, $D=[0,1)$, $f(x)=0$ and $f_n(x)=x^n$. Then$$(forall xin D):lim_ntoinftyf_n(x)=f(x),$$but $lim_ntoinftyf_n(1)=1neq f(1)$.
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
add a comment |Â
up vote
4
down vote
accepted
No, it is not true. Take $X=[0,1]$, $D=[0,1)$, $f(x)=0$ and $f_n(x)=x^n$. Then$$(forall xin D):lim_ntoinftyf_n(x)=f(x),$$but $lim_ntoinftyf_n(1)=1neq f(1)$.
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
No, it is not true. Take $X=[0,1]$, $D=[0,1)$, $f(x)=0$ and $f_n(x)=x^n$. Then$$(forall xin D):lim_ntoinftyf_n(x)=f(x),$$but $lim_ntoinftyf_n(1)=1neq f(1)$.
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
No, it is not true. Take $X=[0,1]$, $D=[0,1)$, $f(x)=0$ and $f_n(x)=x^n$. Then$$(forall xin D):lim_ntoinftyf_n(x)=f(x),$$but $lim_ntoinftyf_n(1)=1neq f(1)$.
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
answered Jul 29 at 16:17
José Carlos Santos
112k1696173
112k1696173
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866191%2fconvergence-of-a-sequence-of-functions-to-a-continuous-limit%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Are you assuming that $f$ is continuous too?
– José Carlos Santos
Jul 29 at 16:14
Yes. It is continuous
– Robert W.
Jul 29 at 16:14