$\lim_{\alpha \to 0} U_\alpha u(x)$ is in $C_0$ if $\sup_{\alpha>0}U_\alpha u(x)=c_d \int_{\mathbb{R}^d} |y|^{2-d} u(x-y)dy$. Let $u\ge 0$ be a $C_0(\mathbb{R}^d)$ function, i.e., continuous function vanishing at infinity and $d\ge 3$. We have, for any $u\in C_0^+(\mathbb{R}^d)$, $$\sup_{\alpha>0}U_\alpha u(x)=c_d \int_{\mathbb{R}^d} |y|^{2-d} u(x-y)dy$$
with $c_d =\pi^{-d/2}\Gamma(d/2)/(d-2).$  We define $U_0u(x):= \lim_{\alpha \to 0} U_\alpha u(x)$ where the limit is taken in the sup norm. $U_\alpha$ is the $\alpha$-potential operator of the Brownian semigroup here, i.e. 
$$U_\alpha u(x):= \int_0^\infty e^{-\alpha t}P_t u(x) dt$$ where $P_t(u(x))=\int_{\mathbb{R}^d} \frac{1}{(2\pi t)^{d/2}} e^{-|y|^2/2t} u(x-y)dy dt$.
Hence the identity for $\sup_{\alpha>0}U_\alpha u(x)$ follows using monotone convergence and Tonelli's theorem. 
My question is about the next statement below.
Then since $\int_{|y|\le 1} |y|^{2-d} dy < \infty$, we see with dominated convergence that the function $U_0 u(x)$ is in $C_0(\mathbb{R}^d)$ for all $u \in C_0^+(\mathbb{R}^d)\cap L^1(dy).$
I don't understand how we get this last statement, $U_0 u(x)$ is in $C_0(\mathbb{R}^d)$ for all $u \in C_0^+(\mathbb{R}^d)\cap L^1(dy)$, from $\int_{|y|\le 1} |y|^{2-d} dy < \infty$ and dominated convergence. I would greatly appreciate an explanation of this statement.
 A: Continuity: Choose a cut-off function $\chi \in C_c(\mathbb{R}^d)$ such that $1_{B(0,1)} \leq \chi \leq 1_{B(0,2)}$. Clearly, $$U_0(x) = c_d \int \chi(y) |y|^{2-d} u(x-y) \, dy + c_d\int (1-\chi(y))|y|^{2-d} u(x-y) \, dy.$$
Since $u$ is bounded and continuous and $\int_{|y| \leq 2} |y|^{2-d} \, dy < \infty$, a straight-forward application of the dominated convergence theorem shows that the first term on the right-hand side is continuous. For the second term write
$$\int (1-\chi(y)) |y|^{2-d} u(x-y) \, dy = \int (1-\chi(x-z)) \frac{1}{|x-z|^{2-d}} u(z) \, dz. \tag{1}$$
As $\chi(x-z)=1$ for $|x-z| < 1$ and $0 \leq \chi \leq 1$, we have
$$\frac{1}{|z-x|^{2-d}} u(z) (1-\chi(x-z)) \leq u(z) \in L^1,$$
and therefore we may apply again the dominated convergence theorem to show that $(1)$ is a continuous function of $x$.
Vanishing at infinity: Fix $\epsilon>0$ and write
$$U_0(x) = c_d \int_{|y| \leq R} |y|^{2-d} u(x-y) \, dy + c_d \int_{|y|>R} |y|^{2-d} u(x-y) \, dy =: I_1+I_2$$
for fixed $R>0$. Clearly, as $d \geq 3$, 
$$I_2 \leq \frac{c_d}{R} \int u(z) \, dz \leq c' \frac{1}{R},$$
and therefore we can choose $R>0$ sufficiently large such that $I_2 \leq \epsilon$ for all $x \in \mathbb{R}^d$. Since $u(x-y) \to 0$ as $|x| \to \infty$ it follows from the dominated convergence theorem that we can choose $r>0$ such that $I_1 \leq \epsilon$ for all $|x| \geq r$.  Hence,
$$|U_0(x)| \leq 2 \epsilon  \quad \text{for all $|x| \geq r$},$$
and this finishes the proof.
