If $f$ is in a weighted Bergman space for the upper half plane, then $\forall\varepsilon>0, z\mapsto f(z+i\varepsilon)$ is in the Hardy space. Let $U:=\{z\in\mathbb{C}\ |\ \operatorname{Im}(z)>0\}$ and denote with $H(U)$ the set of holomorphic functions on $U$. Define:
$$H^2(U):=\{f\in H(U) \ |\ \sup_{y>0}\int_\mathbb{R} |f(x+iy)|^2\operatorname{d}x<+\infty \}$$
and if $\nu>-1$ define:
$$A^2_\nu(U):=\{f\in H(U)\ |\ \int_0^\infty\int_\mathbb{R} |f(x+iy)|^2y^\nu\operatorname{d}x\operatorname{d}y <+\infty \}.$$
In my lecture notes it is claimed without proof that if $f\in A^2_\nu(U)$ then for every $\varepsilon>0$, the function
$$f_\varepsilon:U\rightarrow\mathbb{C}, z\mapsto f(z+i\varepsilon)$$
is in $H^2(U)$ (actually, I implicitly assumed that this must be the case because in there it is used the Paley-Wiener theorem for $H^2(U)$ on $f_\varepsilon)$. I managed to prove this claim for $\nu\ge0$ but I've no idea how to prove it for $-1<\nu<0$. Any suggestion?
 A: Assume $-1<\nu<0$.
Denoting with $\mu$ the Lebesgue measure, if $x+iy\in U$, then by the mean value theorem and Holder inequality:
$$|f(x+iy)|=\left|\frac{4}{\pi y^2}\int_{D_{\frac{y}{2}}(x+iy)}f(w)\operatorname{d}\mu(w)\right|\le\frac{4}{\pi y^2}\int_{D_{\frac{y}{2}}(x+iy)}|f(w)|\operatorname{d}\mu(w) \\ \le \left(\frac{4}{\pi y^2}\int_{D_{\frac{y}{2}}(x+iy)}|f(w)|^2\operatorname{d}\mu(w)\right)^{1/2} = \frac{2}{\sqrt{\pi}y}\left(\int_{D_{\frac{y}{2}}(x+iy)}|f(w)|^2\operatorname{d}\mu(w)\right)^{1/2} \\ \le \frac{2}{\sqrt{\pi}y}\left(\int_{x-y/2}^{x+y/2}\int_{y/2}^{3y/2}|f(s+it)|^2\operatorname{d}t\operatorname{d}s\right)^{1/2} \le \frac{2}{\sqrt{\pi}y}\left(\int_{x-y/2}^{x+y/2}\int_{y/2}^{3y/2}|f(s+it)|^2 \left(\frac{2t}{3y}\right)^{\nu}\operatorname{d}t\operatorname{d}s\right)^{1/2} \\ = \frac{2^{1+\nu/2}}{3^{\nu/2}\sqrt{\pi}y^{1+\nu/2}}\left(\int_{x-y/2}^{x+y/2}\int_{y/2}^{3y/2}|f(s+it)|^2 t^{\nu}\operatorname{d}t\operatorname{d}s\right)^{1/2}.$$
So, if $x+iy\in U$:
$$|f(x+iy)|^2\le \frac{2^{2+\nu}}{3^{\nu}\pi y^{2+\nu}}\int_{x-y/2}^{x+y/2}\int_{y/2}^{3y/2}|f(s+it)|^2 t^{\nu}\operatorname{d}t\operatorname{d}s.$$
Then, if $y>0$
$$\int_\mathbb{R} |f(x+iy)|^2\operatorname{d}x \le \int_\mathbb{R} \frac{2^{2+\nu}}{3^{\nu}\pi y^{2+\nu}}\int_{x-y/2}^{x+y/2}\int_{y/2}^{3y/2}|f(s+it)|^2 t^{\nu}\operatorname{d}t\operatorname{d}s\operatorname{d}x \\ = \frac{2^{2+\nu}}{3^{\nu}\pi y^{2+\nu}} \int_\mathbb{R} \int_\mathbb{R}\int_{y/2}^{3y/2}\chi_{[x-\frac{y}{2},x+\frac{y}{2}]}(s)|f(s+it)|^2 t^{\nu}\operatorname{d}t\operatorname{d}s\operatorname{d}x \\ = \frac{2^{2+\nu}}{3^{\nu}\pi y^{2+\nu}} \int_{y/2}^{3y/2} \int_\mathbb{R} \int_\mathbb{R}\chi_{[x-\frac{y}{2},x+\frac{y}{2}]}(s)|f(s+it)|^2 t^{\nu}\operatorname{d}x\operatorname{d}s\operatorname{d}t \\ =\frac{2^{2+\nu}}{3^{\nu}\pi y^{2+\nu}} \int_{y/2}^{3y/2} \int_\mathbb{R} \int_\mathbb{R}\chi_{[s-\frac{y}{2},s+\frac{y}{2}]}(x)\operatorname{d}x|f(s+it)|^2t^{\nu}\operatorname{d}s\operatorname{d}t \\ = \frac{2^{2+\nu}}{3^{\nu}\pi y^{2+\nu}} \int_{y/2}^{3y/2} \int_\mathbb{R} y|f(s+it)|^2t^{\nu}\operatorname{d}s\operatorname{d}t \\ \le \frac{2^{2+\nu}}{3^{\nu}\pi y^{1+\nu}} \int_0^{\infty} \int_\mathbb{R} |f(s+it)|^2t^{\nu}\operatorname{d}s\operatorname{d}t$$
and so if $\varepsilon>0$:
$$\sup_{y>0}\int_\mathbb{R} |f_\varepsilon(x+iy)|^2\operatorname{d}x \le \frac{2^{2+\nu}}{3^{\nu}\pi \varepsilon^{1+\nu}} \int_\mathbb{R} |f(s+it)|^2t^{\nu}\operatorname{d}s\operatorname{d}t<+\infty.$$
