# Decouple distributional limit from function limit

Problem:
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.