Let $T \in \mathcal F'$ be a distribution on some function space $\mathcal F := \{f \in \mathcal C^\infty ~:~ ||f||_{\mathcal F} < \infty \}$ with a certain fall-off behavior defined by the norm (I can elaborate if necessary).
$T$ defined as the distributional limit of an analytic function $T(z) : \mathbb R + i\mathbb R_{>0} \to \mathbb C$ s.t. $$T(g) := \lim_{\epsilon \to 0} \int dx T(x + i\epsilon) g(x) \leq C ||g||_{\mathcal F}$$ Now, let $f(\cdot + i\lambda) \in \mathcal F$ analytic for $\lambda \in [0,1]$. I want to show that: $$T(g) = \lim_{\epsilon \to 0} \int dx~ T(x + i\epsilon) f(x) \overset{!}{=} \lim_{\epsilon \to 0} \int dx~ T(x + i\epsilon) f(x + i\epsilon). $$ My thoughts:
If $T$ was a tempered distribution, I could use the Bros-Epstein-Glaser lemma to write $T(z) = P(\partial_z) G(z)$ for some analytic function $G$ and partially integrate. This is not possible because I have a generalized function space and would have to prove a similar property first.
While this would be a valid approach, I ask myself whether there is a more straightforward way to prove the equality by estimating against the norm $$\Big|\lim_{\epsilon \to 0} \int dx~ T(x + i\epsilon) (f(x) - f(x + i\epsilon)) \Big| \leq \cdots \leq C \lim_{\epsilon\to 0} ||f(\cdot) - f(\cdot + i\epsilon) ||_{\mathcal F}.$$ But it is unclear to me how to decouple the $\epsilon$ in $T$ and $f$ to obtain this estimate.



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