Mixing
Why normal convergence $implies $ uniform convergence.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
We say that $sum_k=0^infty f_k(x)$ converge normally in $I$ if there is $M_n$ s.t. $|f_n(x)|leq M_n$ for all $n$ and all $xin I$ with $sum_k=0^infty M_n$ converge. Why does it implies that the serie converge uniformly ? I set $f(x)=sum_k=0^infty f_k(x)$. Why $$sup_I|f(x)-sum_k=0^n f_k(x)|to 0$$
when $nto infty $ ?
sequences-and-series
asked Jul 22 at 8:51
user352653
354212
add a comment |Â
up vote
0
down vote
favorite
We say that $sum_k=0^infty f_k(x)$ converge normally in $I$ if there is $M_n$ s.t. $|f_n(x)|leq M_n$ for all $n$ and all $xin I$ with $sum_k=0^infty M_n$ converge. Why does it implies that the serie converge uniformly ? I set $f(x)=sum_k=0^infty f_k(x)$. Why $$sup_I|f(x)-sum_k=0^n f_k(x)|to 0$$
when $nto infty $ ?
sequences-and-series
asked Jul 22 at 8:51
user352653
354212
2
It's the Weierstrass M-test. The proof is short and can be found here, for example: en.wikipedia.org/wiki/Weierstrass_M-test
– Mark
Jul 22 at 8:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We say that $sum_k=0^infty f_k(x)$ converge normally in $I$ if there is $M_n$ s.t. $|f_n(x)|leq M_n$ for all $n$ and all $xin I$ with $sum_k=0^infty M_n$ converge. Why does it implies that the serie converge uniformly ? I set $f(x)=sum_k=0^infty f_k(x)$. Why $$sup_I|f(x)-sum_k=0^n f_k(x)|to 0$$
when $nto infty $ ?
sequences-and-series
asked Jul 22 at 8:51
user352653
354212
We say that $sum_k=0^infty f_k(x)$ converge normally in $I$ if there is $M_n$ s.t. $|f_n(x)|leq M_n$ for all $n$ and all $xin I$ with $sum_k=0^infty M_n$ converge. Why does it implies that the serie converge uniformly ? I set $f(x)=sum_k=0^infty f_k(x)$. Why $$sup_I|f(x)-sum_k=0^n f_k(x)|to 0$$
when $nto infty $ ?
sequences-and-series
asked Jul 22 at 8:51
user352653
354212
asked Jul 22 at 8:51
user352653
354212
asked Jul 22 at 8:51
user352653
354212
asked Jul 22 at 8:51
user352653
354212
354212
2
It's the Weierstrass M-test. The proof is short and can be found here, for example: en.wikipedia.org/wiki/Weierstrass_M-test
– Mark
Jul 22 at 8:56
add a comment |Â
2
It's the Weierstrass M-test. The proof is short and can be found here, for example: en.wikipedia.org/wiki/Weierstrass_M-test
– Mark
Jul 22 at 8:56
2
2
It's the Weierstrass M-test. The proof is short and can be found here, for example: en.wikipedia.org/wiki/Weierstrass_M-test
– Mark
Jul 22 at 8:56
It's the Weierstrass M-test. The proof is short and can be found here, for example: en.wikipedia.org/wiki/Weierstrass_M-test
– Mark
Jul 22 at 8:56
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
$$left|f(x)-sum_k=0^n f_k(x)right|=left|sum_k=0^infty f_k(x)-sum_k=0^n f_k(x)right|=left|sum_k=n+1^infty f_k(x) right|leq sum_k=n+1^infty M_n,$$
and thus
$$sup_xin Ileft|f(x)-sum_k=0^n f_k(x)right|leq sum_k=n+1^infty M_nundersetnto infty longrightarrow 0.$$
answered Jul 22 at 8:54
Surb
36.3k84274
add a comment |Â
up vote
0
down vote
It converges uniformly (and absolutely) because it satisfies the uniform Cauchy criterion:
if $N$ is such that
$$|M_n+M_n+1+dots +M_p|<ε qquadtextfor all : p>nge N,$$
we also have for all $xin I$
beginalign
|f_n(x)+f_n+1(x)+dots +f_p(x)|&le |f_n(x)|+|f_n+1(x)|+dots +|f_p(x)|\ &le|M_n+M_n+1+dots +M_p|< ε.
endalign
answered Jul 22 at 9:14
Bernard
110k635103
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$left|f(x)-sum_k=0^n f_k(x)right|=left|sum_k=0^infty f_k(x)-sum_k=0^n f_k(x)right|=left|sum_k=n+1^infty f_k(x) right|leq sum_k=n+1^infty M_n,$$
and thus
$$sup_xin Ileft|f(x)-sum_k=0^n f_k(x)right|leq sum_k=n+1^infty M_nundersetnto infty longrightarrow 0.$$
answered Jul 22 at 8:54
Surb
36.3k84274
add a comment |Â
up vote
2
down vote
accepted
$$left|f(x)-sum_k=0^n f_k(x)right|=left|sum_k=0^infty f_k(x)-sum_k=0^n f_k(x)right|=left|sum_k=n+1^infty f_k(x) right|leq sum_k=n+1^infty M_n,$$
and thus
$$sup_xin Ileft|f(x)-sum_k=0^n f_k(x)right|leq sum_k=n+1^infty M_nundersetnto infty longrightarrow 0.$$
answered Jul 22 at 8:54
Surb
36.3k84274
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$left|f(x)-sum_k=0^n f_k(x)right|=left|sum_k=0^infty f_k(x)-sum_k=0^n f_k(x)right|=left|sum_k=n+1^infty f_k(x) right|leq sum_k=n+1^infty M_n,$$
and thus
$$sup_xin Ileft|f(x)-sum_k=0^n f_k(x)right|leq sum_k=n+1^infty M_nundersetnto infty longrightarrow 0.$$
answered Jul 22 at 8:54
Surb
36.3k84274
$$left|f(x)-sum_k=0^n f_k(x)right|=left|sum_k=0^infty f_k(x)-sum_k=0^n f_k(x)right|=left|sum_k=n+1^infty f_k(x) right|leq sum_k=n+1^infty M_n,$$
and thus
$$sup_xin Ileft|f(x)-sum_k=0^n f_k(x)right|leq sum_k=n+1^infty M_nundersetnto infty longrightarrow 0.$$
answered Jul 22 at 8:54
Surb
36.3k84274
answered Jul 22 at 8:54
Surb
36.3k84274
answered Jul 22 at 8:54
Surb
36.3k84274
answered Jul 22 at 8:54
Surb
36.3k84274
36.3k84274
add a comment |Â
add a comment |Â
up vote
0
down vote
It converges uniformly (and absolutely) because it satisfies the uniform Cauchy criterion:
if $N$ is such that
$$|M_n+M_n+1+dots +M_p|<ε qquadtextfor all : p>nge N,$$
we also have for all $xin I$
beginalign
|f_n(x)+f_n+1(x)+dots +f_p(x)|&le |f_n(x)|+|f_n+1(x)|+dots +|f_p(x)|\ &le|M_n+M_n+1+dots +M_p|< ε.
endalign
answered Jul 22 at 9:14
Bernard
110k635103
add a comment |Â
up vote
0
down vote
It converges uniformly (and absolutely) because it satisfies the uniform Cauchy criterion:
if $N$ is such that
$$|M_n+M_n+1+dots +M_p|<ε qquadtextfor all : p>nge N,$$
we also have for all $xin I$
beginalign
|f_n(x)+f_n+1(x)+dots +f_p(x)|&le |f_n(x)|+|f_n+1(x)|+dots +|f_p(x)|\ &le|M_n+M_n+1+dots +M_p|< ε.
endalign
answered Jul 22 at 9:14
Bernard
110k635103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It converges uniformly (and absolutely) because it satisfies the uniform Cauchy criterion:
if $N$ is such that
$$|M_n+M_n+1+dots +M_p|<ε qquadtextfor all : p>nge N,$$
we also have for all $xin I$
beginalign
|f_n(x)+f_n+1(x)+dots +f_p(x)|&le |f_n(x)|+|f_n+1(x)|+dots +|f_p(x)|\ &le|M_n+M_n+1+dots +M_p|< ε.
endalign
answered Jul 22 at 9:14
Bernard
110k635103
It converges uniformly (and absolutely) because it satisfies the uniform Cauchy criterion:
if $N$ is such that
$$|M_n+M_n+1+dots +M_p|<ε qquadtextfor all : p>nge N,$$
we also have for all $xin I$
beginalign
|f_n(x)+f_n+1(x)+dots +f_p(x)|&le |f_n(x)|+|f_n+1(x)|+dots +|f_p(x)|\ &le|M_n+M_n+1+dots +M_p|< ε.
endalign
answered Jul 22 at 9:14
Bernard
110k635103
answered Jul 22 at 9:14
Bernard
110k635103
answered Jul 22 at 9:14
Bernard
110k635103
answered Jul 22 at 9:14
Bernard
110k635103
110k635103
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%2f2859215%2fwhy-normal-convergence-implies-uniform-convergence%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
2
It's the Weierstrass M-test. The proof is short and can be found here, for example: en.wikipedia.org/wiki/Weierstrass_M-test
– Mark
Jul 22 at 8:56