$f\colon \mathbb{R}^n \rightarrow \mathbb{R}$ continuous, $\phi (r) = \sup_{|x| =r} f(x)$ is continuous Let $f\colon \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous function, then $\phi (r) = \sup_{|x| =r} f(x)$ is continuous.
My attempt: 
Since the set $ R = \{ x | |x| = r\}$ is compact, for each $r \in [0, \infty)$, there is $ x_R \in R$ such that $\phi(r) = f(x_0)$
 A: Indeed, compactness of the spheres is essential.
Consider a convregent sequence $r_n\to \hat r$ with $r_n\ge 0$. We want to show that $\phi(r_n)\to \phi(\hat r)$. 
By compactness of $r_nS^{n-1}=\{\,x\in\Bbb R^n\mid |x|=r_n\,\}$ we find $x_n\in r_nS^{n-1}$ with $f(x_n)=\phi(r_n)$ and likewise $\hat x\in \hat rS^{n-1}$ with $f(\hat x)=\phi(\hat r)$. 
Then the sequence $y_n:=\frac{\hat r}{r_n}x_n$ (we may assume wlog. that $r_n\ne 0$ for all $n$) lives in $\hat rS^{n-1}$ and by compactness of that set has a convergent subsequence. So wlog. $y_n\to y\in \hat rS^{n-1}$. But then also $x_n\to y$ and $\phi(r_n)=f(x_n)\to f(y)$. Assume $f(y)<\phi(\hat r)$. Then $f(x)>\frac{f(\hat x)+f(y)}2$ in an open neighbourhood of $\hat x$ and consequently $\phi(r_n)>\frac{f(\hat x)+f(y)}2$ for $n$ large enough, contradicting $\phi(r_n)\to f(y)$.
As $f(y)$ cannot be $>\phi(\hat r)$, we conclude $f(y)=\phi(\hat r)$ and so $\phi(r_n)\to \phi(\hat r)$ as desired.
A: Another solution.
Take $R$ large, put $K=\overline{D(0,R)}$, this is a compact set, and as the function $f$ is continuous on $K$, it is uniformly continuous. Given $\varepsilon>0$, there exists $\delta>0$, such that if $x,y$ are in $K$ and $\|x-y\|<\delta$, we have $|f(x)-f(y)|<\varepsilon$. 
Now let $r,s$ in $]0,R]$, and suppose that $|r-s|<\delta$. Take $x=x_r$ such that $\phi(r)=|f(x_r)|$ Let $y=tx_r$, with $t\in ]0,+\infty[$ such that $tr=s$. We have $\|x_r-y\|=|t-1|\|x_r\|=|r-s|<\delta$. Hence $|f(x_r)-f(y)|<\varepsilon$. We get $\phi(r)=|f(x_r)|\leq |f(y)|+\varepsilon\leq \phi(s)+\varepsilon$. In the same way, we have $\phi(s)\leq \phi(r)+\varepsilon$, hence $|\phi(r)-\phi(s)| \leq \varepsilon$.
