Mixing
Prove that a sequence converges
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I am a little insecure about my proof because the book provide a longer one, can you check it? Thanks in advance
Let $Isubseteq mathbbR$ be an interval and $f,f_n$ functions from I to $mathbbR$ for $ninmathbbN$ such that $(f_n)$ uniformly converges to $f$.
Let $t_0 in I$ such that for every $n$ exists $l_n=lim_tto t_0f_n(t)$.
Prove that $(l_n)$ converges.
Let $epsilon >0$ and $n,m in mathbbN$; we have $|l_n-l_m|=|lim_t to t_0(f_n(t)-f_m(t))|=lim_t to t_0|f_n(t)-f_m(t)| le lim_t to t_0epsilon=epsilon$ if $n, m$ are sufficiently large because $(f_n)$ is uniformly convergent. So $(l_n)$ is a Cauchy sequence and therefore it converges.
real-analysis sequences-and-series proof-writing uniform-convergence cauchy-sequences
asked Jul 20 at 7:16
user
1369
add a comment |Â
up vote
1
down vote
favorite
I am a little insecure about my proof because the book provide a longer one, can you check it? Thanks in advance
Let $Isubseteq mathbbR$ be an interval and $f,f_n$ functions from I to $mathbbR$ for $ninmathbbN$ such that $(f_n)$ uniformly converges to $f$.
Let $t_0 in I$ such that for every $n$ exists $l_n=lim_tto t_0f_n(t)$.
Prove that $(l_n)$ converges.
Let $epsilon >0$ and $n,m in mathbbN$; we have $|l_n-l_m|=|lim_t to t_0(f_n(t)-f_m(t))|=lim_t to t_0|f_n(t)-f_m(t)| le lim_t to t_0epsilon=epsilon$ if $n, m$ are sufficiently large because $(f_n)$ is uniformly convergent. So $(l_n)$ is a Cauchy sequence and therefore it converges.
real-analysis sequences-and-series proof-writing uniform-convergence cauchy-sequences
asked Jul 20 at 7:16
user
1369
@KaviRamaMurthy I'm not sure of what you are saying, what if you take $f_n=f$ with f discontinuous in $t_0$? For example $f: [0,1] to mathbbR, x to 1_1(x)$ and $t_0 = 1$
– user
Jul 20 at 7:46
1
The proof is correct.
– Paul Frost
Jul 20 at 9:53
1
A mathematician would say "OK" but a marker on an exam for an elementary course might demand some more detail.
– DanielWainfleet
Jul 20 at 14:55
1
@user. $l_n$ does not necessarily converge to $f(t_0)$.
– DanielWainfleet
Jul 20 at 14:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am a little insecure about my proof because the book provide a longer one, can you check it? Thanks in advance
Let $Isubseteq mathbbR$ be an interval and $f,f_n$ functions from I to $mathbbR$ for $ninmathbbN$ such that $(f_n)$ uniformly converges to $f$.
Let $t_0 in I$ such that for every $n$ exists $l_n=lim_tto t_0f_n(t)$.
Prove that $(l_n)$ converges.
Let $epsilon >0$ and $n,m in mathbbN$; we have $|l_n-l_m|=|lim_t to t_0(f_n(t)-f_m(t))|=lim_t to t_0|f_n(t)-f_m(t)| le lim_t to t_0epsilon=epsilon$ if $n, m$ are sufficiently large because $(f_n)$ is uniformly convergent. So $(l_n)$ is a Cauchy sequence and therefore it converges.
real-analysis sequences-and-series proof-writing uniform-convergence cauchy-sequences
asked Jul 20 at 7:16
user
1369
I am a little insecure about my proof because the book provide a longer one, can you check it? Thanks in advance
Let $Isubseteq mathbbR$ be an interval and $f,f_n$ functions from I to $mathbbR$ for $ninmathbbN$ such that $(f_n)$ uniformly converges to $f$.
Let $t_0 in I$ such that for every $n$ exists $l_n=lim_tto t_0f_n(t)$.
Prove that $(l_n)$ converges.
Let $epsilon >0$ and $n,m in mathbbN$; we have $|l_n-l_m|=|lim_t to t_0(f_n(t)-f_m(t))|=lim_t to t_0|f_n(t)-f_m(t)| le lim_t to t_0epsilon=epsilon$ if $n, m$ are sufficiently large because $(f_n)$ is uniformly convergent. So $(l_n)$ is a Cauchy sequence and therefore it converges.
real-analysis sequences-and-series proof-writing uniform-convergence cauchy-sequences
asked Jul 20 at 7:16
user
1369
asked Jul 20 at 7:16
user
1369
asked Jul 20 at 7:16
user
1369
asked Jul 20 at 7:16
user
1369
1369
@KaviRamaMurthy I'm not sure of what you are saying, what if you take $f_n=f$ with f discontinuous in $t_0$? For example $f: [0,1] to mathbbR, x to 1_1(x)$ and $t_0 = 1$
– user
Jul 20 at 7:46
1
The proof is correct.
– Paul Frost
Jul 20 at 9:53
1
A mathematician would say "OK" but a marker on an exam for an elementary course might demand some more detail.
– DanielWainfleet
Jul 20 at 14:55
1
@user. $l_n$ does not necessarily converge to $f(t_0)$.
– DanielWainfleet
Jul 20 at 14:58
add a comment |Â
@KaviRamaMurthy I'm not sure of what you are saying, what if you take $f_n=f$ with f discontinuous in $t_0$? For example $f: [0,1] to mathbbR, x to 1_1(x)$ and $t_0 = 1$
– user
Jul 20 at 7:46
1
The proof is correct.
– Paul Frost
Jul 20 at 9:53
1
A mathematician would say "OK" but a marker on an exam for an elementary course might demand some more detail.
– DanielWainfleet
Jul 20 at 14:55
1
@user. $l_n$ does not necessarily converge to $f(t_0)$.
– DanielWainfleet
Jul 20 at 14:58
@KaviRamaMurthy I'm not sure of what you are saying, what if you take $f_n=f$ with f discontinuous in $t_0$? For example $f: [0,1] to mathbbR, x to 1_1(x)$ and $t_0 = 1$
– user
Jul 20 at 7:46
@KaviRamaMurthy I'm not sure of what you are saying, what if you take $f_n=f$ with f discontinuous in $t_0$? For example $f: [0,1] to mathbbR, x to 1_1(x)$ and $t_0 = 1$
– user
Jul 20 at 7:46
1
1
The proof is correct.
– Paul Frost
Jul 20 at 9:53
The proof is correct.
– Paul Frost
Jul 20 at 9:53
1
1
A mathematician would say "OK" but a marker on an exam for an elementary course might demand some more detail.
– DanielWainfleet
Jul 20 at 14:55
A mathematician would say "OK" but a marker on an exam for an elementary course might demand some more detail.
– DanielWainfleet
Jul 20 at 14:55
1
1
@user. $l_n$ does not necessarily converge to $f(t_0)$.
– DanielWainfleet
Jul 20 at 14:58
@user. $l_n$ does not necessarily converge to $f(t_0)$.
– DanielWainfleet
Jul 20 at 14:58
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2857370%2fprove-that-a-sequence-converges%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
@KaviRamaMurthy I'm not sure of what you are saying, what if you take $f_n=f$ with f discontinuous in $t_0$? For example $f: [0,1] to mathbbR, x to 1_1(x)$ and $t_0 = 1$
– user
Jul 20 at 7:46
1
The proof is correct.
– Paul Frost
Jul 20 at 9:53
1
A mathematician would say "OK" but a marker on an exam for an elementary course might demand some more detail.
– DanielWainfleet
Jul 20 at 14:55
1
@user. $l_n$ does not necessarily converge to $f(t_0)$.
– DanielWainfleet
Jul 20 at 14:58