Let $h(\phi):\mathbb{R}\to\mathbb{R}$ be a bounded piecewise continuous function on the real line. Define a function on the upper half-plane by the formula $\tilde{h}(s+it):=\int_{-\infty}^{\infty} C_t(s-\phi)h(\phi)d\phi$ where $C_t(s)=\frac{1}{\pi}\frac{t}{s^2+t^2}$.
I need to show that $\tilde{h}$ is an extension of $h$ in the sense that if $\phi_0\in \mathbb{R}$ and there is a sequence of $\lbrace z_n \rbrace_{n=1}^\infty \to \phi_0$ then $\lbrace \tilde{h}(z_n) \rbrace_{n=1}^\infty \to h(\phi_0)$ when $h$ is continuous at $\phi_0$.
I have already shown that $\tilde{h}$ is harmonic and bounded in $\mathbb{H}$. I am trying to simplify the computation by saying $\lbrace z_n \rbrace_{n=1}^\infty \to \phi_0$ $\implies$ $\lbrace Re(z_n) \rbrace_{n=1}^\infty \to \phi_0$. Can anyone help me with this limit?
Write $z_n = s_n + it_n$. We have $z_n \to \phi_0 \iff (s_n\to\phi_0 \land t_n\to 0)$. Further, write
$$\tilde{h}(s+it) = \int_{-\infty}^\infty C_t(\phi)h(s-\phi)\,d\phi.$$
That is not important, but IMO more convenient. Next, note that $\int_{-\infty}^\infty C_t(\phi)\,d\phi = 1$ for all, $t > 0$, so we can write
$$\tilde{h}(s+it) - h(\phi_0) = \int_{-\infty}^\infty C_t(\phi)\left(h(s-\phi) - h(\phi_0)\right)\,d\phi.\tag{1}$$
To obtain the desired conclusion, we split $(1)$ into two integrals, one part where $\lvert h(s-\phi) - h(\phi_0)\rvert$ is small, and one part where $\int C_t(\phi)\,d\phi$ becomes small as $t\to 0$.
Let $K = \sup \{\lvert h(\phi)\rvert : \phi \in\mathbb{R}\}$. Given $\varepsilon > 0$, since $h$ is supposed to be continuous in $\phi_0$, there is a $\delta > 0$ such that $\lvert h(\phi) - h(\phi_0)\rvert < \varepsilon/2$ for $\lvert \phi-\phi_0\rvert < 2\delta$. Choose $n_0$ such that $\lvert s_n - \phi_0\rvert < \delta$ for all $n \geqslant n_0$. Then, for $n \geqslant n_0$ we have
$$\begin{align} \lvert \tilde{h}(s_n+it_n) - h(\phi_0)\rvert &= \left\lvert \int_{-\infty}^\infty C_{t_n}(\phi)\left(h(s_n-\phi) - h(\phi_0)\right)\,d\phi\right\rvert\\ &\leqslant \int_{-\infty}^\infty\left\lvert C_{t_n}(\phi)\left(h(s_n-\phi) - h(\phi_0)\right)\right\rvert\,d\phi\\ &= \underbrace{\int_{-\delta}^\delta C_{t_n}(\phi)\lvert h(s_n-\phi)-h(\phi_0)\rvert\,d\phi}_{I} + \underbrace{\int_{\lvert\phi\rvert\geqslant\delta} C_{t_n}(\phi)\lvert h(s_n-\phi)-h(\phi_0)\rvert\,d\phi}_{II}. \end{align}$$
We have $\lvert s_n-\phi-\phi_0\rvert \leqslant \lvert s_n-\phi_0\rvert + \lvert\phi\rvert < \delta+\delta$ in $I$, hence
$$I \leqslant \int_{-\delta}^\delta C_{t_n}(\phi)\frac{\varepsilon}{2}\,d\phi < \frac{\varepsilon}{2}\int_{-\infty}^\infty C_{t_n}(\phi)\,d\phi = \frac{\varepsilon}{2}$$
for $n\geqslant n_0$. In $II$, we can estimate $\lvert h(s_n-\phi) - h(\phi)\rvert$ by $2K$, so
$$\begin{align} II &\leqslant 2K\int_{\lvert\phi\rvert\geqslant\delta} C_{t_n}(\phi)\,d\phi\\ &= 4K\int_{\delta}^\infty C_{t_n}(\phi)\,d\phi\\ &= \frac{4K}{\pi}\int_\delta^\infty \frac{t_n}{t_n^2+\phi^2}\,d\phi\\ &= \frac{4K}{\pi}\int_{\delta/t_n}^\infty \frac{d\psi}{1+\psi^2}\\ &= \frac{4K}{\pi}\int_0^{t_n/\delta} \frac{d\theta}{1+\theta^2}\\ &= \frac{4K}{\pi} \arctan \frac{t_n}{\delta}. \end{align}$$
So choosing $n_1 \geqslant n_0$ such that for $n\geqslant n_1$ we have
$$t_n < \delta\cdot\tan \frac{\varepsilon\pi}{8K},$$
it follows that
$$\lvert \tilde{h}(s_n+it_n) - h(\phi_0)\rvert < \varepsilon$$
for $n \geqslant n_1$.