# Integral representation of a limit

Let $$\tilde{\mathbb C}={\mathbb C}\smallsetminus (-\infty,0]$$ the complex plane without the negative real axis. Let $V$ denote the set of all holomorphic functions $f:\tilde{\mathbb C}\to{ \mathbb C}$ such that $$f^+(0)=\lim_{\lambda\searrow 0}f(\lambda)$$ exists. If $K\subset \{\mathrm{Re}(z)>0\}$ is a compact subset which is not finite, then a given $f\in V$ is uniquely determined by its restriction to $K$. Therefore it is reasonable to ask:

Does there exists a compact set $K$ and a complex valued Radon measure $\mu$ on $K$ such that $$f^+(0)=\int_K f(\lambda)\,d\mu(\lambda)$$ holds for every $f\in V$?

No. The existence of such a representation would imply that the linear functional $\Phi:V\to \mathbb C$, $f\mapsto f^+(0)$ is a continuous with respect to the topology of uniform convergence on compact subsets of $K$. By Hahn-Banach one could thus extend $\Phi$ to $H(\tilde{\mathbb C})$. This is certainly not true: The functions $f_n(z)= \frac{1}{z+1/n}$ are in $V$ and converge in $H(\tilde{\mathbb C})$. But $\Phi(f_n)=n$ does not converge.
• Where do we use HB in this argument? All these functions belong to $V$ itself and they are uniformly bounded on $K$ while their limits in 0 are not, is not it already a contradiction? Jul 13, 2016 at 8:43
• Right. But Hahn-Banach was my first idea because it is ''obvious'' (even without looking at examples) that $\Phi$ cannot extend continuously to $H(\tilde{\mathbb C})$ ''because'' $f^+(0)$ need not exist for $f\in H(\tilde{\mathbb C})$. Jul 13, 2016 at 8:53
Here is a simple proof without Hahn-Banach. I assume that the complex Radon measure is finite. Then your representation would imply $$|\lim_{\lambda\to 0}f(\lambda)|\leq C\max_{K}|f(z)|,$$ which is evidently not true.