statistical tool
Factsheet
$$\mathbb{R}_+=[0,+\infty)\qquad\bar{\mathbb{R}}_{+}=\mathbb R_+\cup\{+\infty\}$$
I think you got confused by calculating like this:
We have known that $$ \frac{\partial}{\partial r}\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} u\,\mathrm{d}S = \frac{r}{n}\frac{1}{\alpha(n)r^n}\int_{B(x,r)} \Delta u\,\mathrm{d}x $$ and $$ \frac{\partial}{\partial r}\frac{1}{\alpha(n)r^n}\int_{B(x,r)} u\,\mathrm{d}x = \frac{n}{r}\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} u\,\mathrm{d}S - \frac{1}{r}\frac{1}{\alpha(n)r^n}\int_{B(x,r)} u\,\mathrm{d}x $$ Thus $$ U_{rrr}(x;r,t) = \frac{n-1}{nr}\frac{1}{\alpha(n)r^n}\int_{B(x,r)} \Delta u\,\mathrm{d}x - \frac{n-1}{r}\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} \Delta u\,\mathrm{d}x + \frac{r}{n}\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} \Delta(\Delta u)\,\mathrm{d}x $$
If $u\in C^3(\mathbb{R}^n\times [0,\infty))$ but not $C^{4}$, we can't compute $\Delta(\Delta u)$, let alone $U_{rrr}(x;r,t)\in C^3(\bar{\mathbb{R}}_+\times [0,\infty))$.
But we don't have to use the Laplace operator. To avoid confusion I'd like to remind that $u\in C^m$ is wrt. $x$, while $U\in C^m$ is wrt. $r$. Actually $$ U_r = \frac{r}{n}\frac{1}{\alpha(n)r^n}\int_{B(x,r)} \Delta u\,\mathrm{d}x $$ requires $u$ is $C^2$. If we don't know this we have $$ U_r = \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} Du\cdot \vec{\nu}\,\mathrm{d}x $$ which needs only $u\in C^1$.
Use \bar for short lines and \overline for longer lines.
A separate comment: In fine math typography, a \cdot is generally preferred over * to denote multiplication.
\documentclass[11pt]{report}
\usepackage[utf8]{inputenc}
\usepackage{mathtools,array}
\begin{document}
\begin{align*}
C_f
&= 100 - 100\cdot\frac{\overline{R_{j(pr)}}}{\overline{R_{j(obs)}}} \\
&= 100 - 100\cdot\frac{\bar{R}_{j(pr)}}{\bar{R}_{j(obs)}}
\end{align*}
\end{document}
Like this? Then you just messed around with the curly brackets. (Edited the over-frac-issue)
\documentclass[11pt]{report}
\usepackage[utf8]{inputenc}
\usepackage{mathtools,array,dcolumn}
\begin{document}
\begin{equation}
Cf = 100 - (100* \frac{\bar{R}_{j(pr)}}{\bar{R}_{j(obs)}})
\end{equation}
\end{document}
When using the bar-command the following brackets include the barred content. Because you just wanted the 'R' to be barred - only include the R into the brackets.
Here is a command \xoverline[width percent]{symb} that will do it. Note that it will not scale inside sub or superscripts. If you need that, everthing has to go through a \mathchoice resulting in a lot more complex code.
\documentclass{article}
\usepackage{amsmath,amssymb}
\makeatletter
\newsavebox\myboxA
\newsavebox\myboxB
\newlength\mylenA
\newcommand*\xoverline[2][0.75]{%
\sbox{\myboxA}{$\m@th#2$}%
\setbox\myboxB\null% Phantom box
\ht\myboxB=\ht\myboxA%
\dp\myboxB=\dp\myboxA%
\wd\myboxB=#1\wd\myboxA% Scale phantom
\sbox\myboxB{$\m@th\overline{\copy\myboxB}$}% Overlined phantom
\setlength\mylenA{\the\wd\myboxA}% calc width diff
\addtolength\mylenA{-\the\wd\myboxB}%
\ifdim\wd\myboxB<\wd\myboxA%
\rlap{\hskip 0.5\mylenA\usebox\myboxB}{\usebox\myboxA}%
\else
\hskip -0.5\mylenA\rlap{\usebox\myboxA}{\hskip 0.5\mylenA\usebox\myboxB}%
\fi}
\makeatother
\begin{document}
$|\xoverline{W}|~~
|\xoverline{i}|~~
|\xoverline[3.0]{i}|$
\bigskip
$\bar{\mathbb{R}}~~\overline{\mathbb{R}}~~\xoverline{\mathbb{R}}$
\end{document}
I usually define a command \overbar, which reduced the width of \overline by 1.5mu on each side.
\documentclass{article}
\usepackage{amsfonts}
\newcommand{\overbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}
\begin{document}
$\bar{\mathbb{R}}$ $\overbar{\mathbb{R}}$ $\overline{\mathbb{R}}$
\end{document}