Let A be the disk algebra, of continuous functions on the closed disk holomorphic on the interior, with sup-norm denoted || . || . Let x be an interior point of the disk. Does there exist a proper A-ideal J such that the sup over nonzero elements of J of | f(x) | / || f || is 1 ?
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$\begingroup$ Sorry, I keep mixing the question. This seems true for any ideal containing an $f$ with $f(x)\ne0$. By multiplying the germ by $1/f$, you can make it constant about $x$; then, multiply it by a bell-like function to make it $0$ far from $x$. For this $g\in J$, you have $\|g\|=g(x)$. $\endgroup$– Alex DegtyarevCommented Mar 26, 2014 at 20:06
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$\begingroup$ @AlexDegtyarev The current version of the question asks that the function be holomorphic in the interior. $\endgroup$– Andreas BlassCommented Mar 26, 2014 at 21:34
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$\begingroup$ I think if such a $J$ exists, its hull will have to lie inside the unit circle. Otherwise it would be contained in a maximal ideal corresponding to a point of the open unit disk and then (if I've not made a daft mistake) a Schwarz lemma argument shows the ratio in your question is bounded away from 1. $\endgroup$– Yemon ChoiCommented Mar 26, 2014 at 22:31
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$\begingroup$ @Yemon Choi : I'm not sure ($1$ beeing a logarithmic branch point) that for $n\geq 2,$ your $f_n$ is continuous. Take a real sequence $\theta_m$ with $0<\theta_m<\pi /2$ and $\lim\limits_{m~\infty}\theta_m=0,$ consider $z_m=\exp(2i\theta_m);$ $z_m$ belongs to the closed unit disk, and $\lim\limits_{m~\infty}z_m=1.$ Now $~1-z_m=2\sin(\theta_m)\exp(i(\theta_m-\pi/2)),$ so $~\ln(1-z_m)=\ln|2\sin(\theta_m)|+i(\theta_m-\pi/2),$ and $\lim\limits_{m~\infty}f_n(z_m)=\exp(-i\pi/2n).$ But if $~w~$ is real in $~(-1,1)~~~\lim\limits_{w\rightarrow1^-}f_n(w)=0.$ Am I wrong ? $\endgroup$– user45639Commented Mar 28, 2014 at 15:49
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$\begingroup$ @Smaug: Thanks, but isn't $\lim \ln |2\sin(\theta_m)| = -\infty$? (And you should probably leave your comment as a comment on my answer, if that's possible) $\endgroup$– Yemon ChoiCommented Mar 28, 2014 at 21:35
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How about the following: let $M_1=\{f\in A : f(1)=0\}$, which is a maximal ideal in $A$, and let $f_n(z) = (1-z)^{1/n} = \exp (n^{-1}\log(1-z))$ where we take $(-\infty,0]$ as our branch cut for $\log$ (the point being that $z\mapsto 1-z$ maps the open unit disc into the open right-half-plane).
Then $(f_n)$ is a sequence in $M_1$, $\Vert f_n\Vert_\infty = 2^{1/n}$, and $f_n(0)=1$ for all $n$, so $$ 1\geq \sup \{ |f(0)| : f\in M_1, \Vert f\Vert_\infty=1\} \geq \sup_n 2^{-1/n} |f_n(0)| = 1$$ which seems to be what you wanted.
(Of course there was nothing special about $0$ as a choice of interior point, since we can always post-compose with an appropriate Möbius transformation to move any prescribed interior point to $0$ while not affecting norms in the disk algebra.)
answered Mar 26, 2014 at 23:12
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$\begingroup$ you're perfectly right, my comment was totally stupid and misplaced... Please accept my apologies. $\endgroup$– user45639Commented Mar 29, 2014 at 6:58
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$\begingroup$ @Smaug that's fine; no need to apologize $\endgroup$ Commented Mar 29, 2014 at 15:08