Lemma 1 (Euler-Poisson-Darboux equation). Fix $x \in \mathbb{R}^n$ and let $u$ satisfy $(11)$. Then $U \in C^m(\bar{\mathbb{R}}_+ \times [0,\infty))$ and \begin{cases} U_{tt} - U_{rr}-\frac{n-1}{r}U_r=0 & \text{in }\mathbb{R}_+ \times (0,\infty) \\ U=G,U_t=H & \text{on } \mathbb{R}_+ \times \{t=0\}\end{cases}
In case $(11)$ is needed: \begin{cases} u_{tt}-\Delta u=0 &\text{in }\mathbb{R}^n \times (0,\infty) \\ u=g,u_t=h & \text{on } \mathbb{R}^n \times \{t=0\} \tag{11}\end{cases}
The book also computed $U_{rr}(x;r,t)$, which is $$U_{rr}(x;r,t)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} \Delta u \, dS + \left(\frac{1}{n}-1 \right)\frac{1}{\alpha(n)r^n}\int_{B(x,r)} \Delta u \, dy $$ Thus, $\lim_{r \rightarrow 0^+} U_{rr}(x;r,t) = \frac{1}{n} \Delta u(x,t)$.
But I am unable to derive $\lim_{r \rightarrow 0^+} U_{rr}(x;r,t)$ by myself. I just need help on this part here. (I understand all other derivations up to this point, though.)
Let's fix $x$ and $t$. Our $u$ is sufficiently smooth so that $\Delta u$ is continuous. So, for every $\epsilon>0$ we have, for sufficiently small $r$, $$|\Delta u(y,t) - \Delta u(x,t)|<\epsilon ,\quad \forall \ y\in B(x,r)$$ Therefore, $$ (\Delta u(x,t)-\epsilon) |\partial B(x,r)| \le \int_{\partial B(x,r)} \Delta u \, dS \le (\Delta u(x,t)+\epsilon) |\partial B(x,r)| $$ and $$ (\Delta u(x,t)-\epsilon) | B(x,r)| \le \int_{\partial B(x,r)} \Delta u\,dy \le (\Delta u(x,t)+\epsilon) | B(x,r)| $$ Since $ |\partial B(x,r)| = n\alpha(n)r^{n-1}$ and $ | B(x,r)| = \alpha(n)r^{n}$, we end up with
$$(\Delta u(x,t)- \epsilon)\left(1 + \frac1n -1\right) \le U_{rr} \le (\Delta u(x,t)+ \epsilon)\left(1 + \frac1n -1\right)$$