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I am struggling with the following exercise:

Let $f:B(0,1) \rightarrow \mathbb{C}$ be a holomorphic function and we have $\forall n \in \mathbb{N}_{\ge 2}: f'(\frac{1}{n})=f(\frac{1}{n})$ then f can be extended to a holomorphic function on the whole complex plane.

Now my idea was the following: Obviously this is a property that only the exponential function on $\mathbb{R}$ fulfills, therefore f has to be the exponential function, as they coincide on countably many points and have the same limit point, they are equal as given by the identity theorem.

Does this approach work or does anybody of you have a better idea?

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I think this is correct reasoning. –  Cameron Williams Jun 27 '13 at 18:01
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I don't think that the exponential function the only (non-constant, differentiable) function on $\mathbb{R}$ to have this property. Take $f(x) = 0$ on $[0,\frac12]$ and $f(x)=(x-\frac12)^2$ on $[\frac12, 1]$ and extend differentiably on $\mathbb{R}$ as you like. For all of the points $\frac1n$ for the given range of $n$, $f(\frac1n)=f'(\frac1n)=0$ and the function is differentiable on $\mathbb{R}$. –  Tom Oldfield Jun 27 '13 at 18:06
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@TomOldfield: But the question was about holomorphic functions. –  Martin R Jun 27 '13 at 18:09
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Identity theorem is correct. $f - f'$ has $0$ as a limit point of zeros, so you know $f' = f$. Thus $f(z) = c\cdot e^z$. –  Daniel Fischer Jun 27 '13 at 18:10
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@MartinR sure, but in the proof, the op says "obviously this is a property that only the exponential function on $\mathbb{R}$ fulfills" and then makes conclusions. My point is that this claim is not true when restricted to $\mathbb{R}$, so a proof shouldn't be based on it. –  Tom Oldfield Jun 27 '13 at 18:11
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up vote 3 down vote accepted

Now my idea was the following: Obviously this is a property that only the exponential function on $\mathbb{R}$ fulfills, therefore f has to be the exponential function, as they coincide on countably many points and have the same limit point, they are equal as given by the identity theorem.

The first part of that is rather hand-wavy, and as it stands not correct, since any multiple of the exponential function has that property.

That these are the only such functions can be seen with the help of the identity theorem.

Since $f'(\frac1n) = f(\frac1n)$ for $n\in \mathbb{N}_{\geqslant 2}$ is the premise, it is immediate that $g = f - f' \in \mathcal{O}(\mathbb{D})$ has $\frac1n$ as zeros for $n\in \mathbb{N}_{\geqslant 2}$, thus $0$ is a limit point of zeros of $g$, and by the identity theorem $g \equiv 0$, or in other words $f' = f$.

The only solutions of the differential equation $f' = f$ are the multiples of the exponential function (consider $g(z) = f(z)\cdot e^{-z}$ for a solution $f$, differentiation yields $g'(z) = f'(z) e^{-z} - f(z)e^{-z} = 0$), hence $f(z) = f(0)\cdot e^z$, and that is well known to be extensible to the entire plane.

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