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Let $X$ be a (complex) banach space, $U$ be an open subset of $\mathbb{C}$ and $f: U \to X$ be a function that is completely arbitrary except that it satisfies the property that for any continuous linear functional $l$ on $X$, $l \circ f$ is complex analytic in the usual sense. Is it possible to deduce from this that $f$ is continuous? What about strongly analytic? (This means that the usual limit of the difference quotient exists in the norm of $X$.)

Can strong analyticity be concluded if I assume the weak analyticity condition plus continuity?

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It follows that $f$ is strongly analytic conditional on Hahn-Banach. I don't think you need to assume continuity. – Qiaochu Yuan Sep 29 '12 at 6:35
@QiaochuYuan looked at it slightly inefficiently, and I think I see how to change this as you suggest. Prove the Cauchy integral formula by just applying a general continuous linear functional, and then undo this with Hahn banach. This works when f is continuous so that we have the integral, but what if f is not continuous? – Jeff Sep 29 '12 at 8:17
It seems this approach deadends at precisely where I stopped. But I found this which instead uses the uniform boundedness principle. – Jeff Sep 29 '12 at 8:30
up vote 1 down vote accepted

This may be a variation on the reference you linked above. If you shift so that $f(0) = 0$, we can check continuity at $0$. Apply linear functionals $l$ to the values of $f(z)/z$, which up to removable singularities is holomorphic, apply Cauchy's integral formula in the classical case, and then use the Cauchy estimate $|l \circ f(z)/z)| \le C/r$, with $C$ dependent on $l$. This means the set of values $f(z)/z$ is weakly bounded, so by Uniform Boundedness Principle or Banach-Alaoglu, the set is bounded, so $f(z)$ must be continuous at $0$. Thus indeed $f$ is continuous.

Then, as suggested by Qiaochu, there are no worries about verifying the Cauchy integral formula on $f$.

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There are two useful tricks here that I did not notice, so I was confused when classmates suggested as a hint that I use the uniform boundedness principle. The first problem I encountered was the issue that $X$ may not be a banach space of operators, but you deal with that by embedding in $X^{**}$ And then a set being bounded didn't seem all that useful, but you deal with that by dividing and multiplying by $z$. The more you learn... Also the Cauchy estimate part confused me, since I did not say $U$ was a disk, but you can get away with a dumb supnorm estimate on a compact subdisk. – Jeff Sep 29 '12 at 17:13
It's quite possible the Cauchy estimate makes sense, but I just don't know what "$r$" is referring to. If it's the modulus of $z$, I don't think this is quite right. In a Cauchy estimate, you can choose any radius that fits in the open set, but the disk has to be centered on $z$, so I think your estimate just works at z=0. – Jeff Sep 29 '12 at 17:17
The disc does not have to be centered at zero. It's centered at whichever point you wish. In general we just get $|l \circ f(z)/(z - z_0)| \le C/r$, with $r$ the radius of the disc $D(z_0,r)$. – Christopher A. Wong Sep 29 '12 at 22:48
Oh okay, if r is not a universal constant, then of course that works, but is still unnecessary, and perhaps also insufficient, but I'm sure it suffices to use just the usual compactness argument. – Jeff Sep 30 '12 at 0:37
Well, if I just want to prove continuity, then everything is local. – Christopher A. Wong Sep 30 '12 at 0:54

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