Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 math.ubc.ca/~feldman/m511/analytic.pdf which instead uses the uniform boundedness principle. –  Jeff Sep 29 '12 at 8:30
add comment

1 Answer

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 $f$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.