Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly on $(a, b)$ then is $f$ analytic on...
$begingroup$
Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?
Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?
real-analysis convergence uniform-convergence analytic-functions sequence-of-function
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
asked Dec 14 '18 at 15:24
josephjoseph
496111
$endgroup$
add a comment |
$begingroup$
Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?
Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?
real-analysis convergence uniform-convergence analytic-functions sequence-of-function
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
asked Dec 14 '18 at 15:24
josephjoseph
496111
$endgroup$
2
$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30
add a comment |
$begingroup$
Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?
Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?
real-analysis convergence uniform-convergence analytic-functions sequence-of-function
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
asked Dec 14 '18 at 15:24
josephjoseph
496111
$endgroup$
Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?
Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?
real-analysis convergence uniform-convergence analytic-functions sequence-of-function
real-analysis convergence uniform-convergence analytic-functions sequence-of-function
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
asked Dec 14 '18 at 15:24
josephjoseph
496111
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
asked Dec 14 '18 at 15:24
josephjoseph
496111
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
edited Dec 14 '18 at 15:32
José Carlos Santos
165k22132235
165k22132235
asked Dec 14 '18 at 15:24
josephjoseph
496111
asked Dec 14 '18 at 15:24
josephjoseph
496111
asked Dec 14 '18 at 15:24
josephjoseph
496111
496111
2
$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30
add a comment |
2
$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30
2
2
$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30
$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
Nice counterexample
$endgroup$
– joseph
Dec 14 '18 at 15:33
$begingroup$
@joseph If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 16:59
add a comment |
$begingroup$
Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
Nice counterexample
$endgroup$
– joseph
Dec 14 '18 at 15:33
$begingroup$
@joseph If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 16:59
add a comment |
$begingroup$
Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
Nice counterexample
$endgroup$
– joseph
Dec 14 '18 at 15:33
$begingroup$
@joseph If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 16:59
add a comment |
$begingroup$
Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
answered Dec 14 '18 at 15:32
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
$begingroup$
Nice counterexample
$endgroup$
– joseph
Dec 14 '18 at 15:33
$begingroup$
@joseph If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 16:59
add a comment |
$begingroup$
Nice counterexample
$endgroup$
– joseph
Dec 14 '18 at 15:33
$begingroup$
@joseph If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 16:59
$begingroup$
Nice counterexample
$endgroup$
– joseph
Dec 14 '18 at 15:33
$begingroup$
Nice counterexample
$endgroup$
– joseph
Dec 14 '18 at 15:33
$begingroup$
@joseph If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 16:59
$begingroup$
@joseph If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 16:59
add a comment |
$begingroup$
Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
$begingroup$
Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
$begingroup$
Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
$endgroup$
Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
answered Dec 14 '18 at 16:46
BigbearZzzBigbearZzz
8,88821652
8,88821652
add a comment |
add a comment |
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$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30