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Regarding continuous function not in the Disc algebra
$begingroup$
Let $D={zinmathbb{C}: |z|<1}$.
$C(bar{D})={f:bar Dlongrightarrow mathbb{C}: f ;text{is continuous on}; bar{D}}$
$A(D)={fin C(bar{D}): f ;text{is analytic in} ;D}$
Can you give me example of a function which is in $C(bar{D})$ but not in $A(D)$? A function that is not analytic exactly at the boundary but continuous.
functional-analysis banach-algebras
asked Jan 10 at 16:02
user31459user31459
284
$endgroup$
add a comment |
$begingroup$
Let $D={zinmathbb{C}: |z|<1}$.
$C(bar{D})={f:bar Dlongrightarrow mathbb{C}: f ;text{is continuous on}; bar{D}}$
$A(D)={fin C(bar{D}): f ;text{is analytic in} ;D}$
Can you give me example of a function which is in $C(bar{D})$ but not in $A(D)$? A function that is not analytic exactly at the boundary but continuous.
functional-analysis banach-algebras
asked Jan 10 at 16:02
user31459user31459
284
$endgroup$
1
$begingroup$
How about $zmapstooverline{z}$?
$endgroup$
– SmileyCraft
Jan 10 at 16:03
add a comment |
$begingroup$
Let $D={zinmathbb{C}: |z|<1}$.
$C(bar{D})={f:bar Dlongrightarrow mathbb{C}: f ;text{is continuous on}; bar{D}}$
$A(D)={fin C(bar{D}): f ;text{is analytic in} ;D}$
Can you give me example of a function which is in $C(bar{D})$ but not in $A(D)$? A function that is not analytic exactly at the boundary but continuous.
functional-analysis banach-algebras
asked Jan 10 at 16:02
user31459user31459
284
$endgroup$
Let $D={zinmathbb{C}: |z|<1}$.
$C(bar{D})={f:bar Dlongrightarrow mathbb{C}: f ;text{is continuous on}; bar{D}}$
$A(D)={fin C(bar{D}): f ;text{is analytic in} ;D}$
Can you give me example of a function which is in $C(bar{D})$ but not in $A(D)$? A function that is not analytic exactly at the boundary but continuous.
functional-analysis banach-algebras
functional-analysis banach-algebras
asked Jan 10 at 16:02
user31459user31459
284
asked Jan 10 at 16:02
user31459user31459
284
asked Jan 10 at 16:02
user31459user31459
284
asked Jan 10 at 16:02
user31459user31459
284
asked Jan 10 at 16:02
user31459user31459
284
284
1
$begingroup$
How about $zmapstooverline{z}$?
$endgroup$
– SmileyCraft
Jan 10 at 16:03
add a comment |
1
$begingroup$
How about $zmapstooverline{z}$?
$endgroup$
– SmileyCraft
Jan 10 at 16:03
1
1
$begingroup$
How about $zmapstooverline{z}$?
$endgroup$
– SmileyCraft
Jan 10 at 16:03
$begingroup$
How about $zmapstooverline{z}$?
$endgroup$
– SmileyCraft
Jan 10 at 16:03
add a comment |
1 Answer
1
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$begingroup$
Any function that is not differentiable will do. For instance $f(z)=|z|$.
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
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$begingroup$
Any function that is not differentiable will do. For instance $f(z)=|z|$.
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Any function that is not differentiable will do. For instance $f(z)=|z|$.
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Any function that is not differentiable will do. For instance $f(z)=|z|$.
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
Any function that is not differentiable will do. For instance $f(z)=|z|$.
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 16:55
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
add a comment |
add a comment |
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$begingroup$
How about $zmapstooverline{z}$?
$endgroup$
– SmileyCraft
Jan 10 at 16:03