Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.
Let $A\in\mathcal{B}(F)$. The following quantities $$\|A\|=\displaystyle\sup_{\substack{x\in F,\\ \|x\|= 1}}\|Ax\|,$$ $$r(A)=\lim_{n\to\infty}\|A^n\|^{1/n},$$ and $$\omega(A)=\sup\left\{|\langle Ax\mid x\rangle|\,;\;x\in F,\;\|x\|= 1\right\},$$ denote respectively the norm, the spectral radius and numerical radius of $A$.
I want to prove that $$r(A)=\|A\|\Longleftrightarrow \omega(A)=\|A\|.$$
It is well know that
$$r(A)\leq\omega(A)\leq\|A\|,$$
for every $A\in\mathcal{B}(F)$. So clearly if $r(A)=\|A\|$ then $\omega(A)=\|A\|$.
In infinite dimension this is not true. If you consider the $n\times n$ "shift", i.e. $S_n=\sum_{j=1}^nE_{j,j+1}^{(n)}$, then $\omega(S_n)=\cos \pi/(n+1)$ and $\|S_n\|=1$ and $r(S_n)=0$. If we form $$ A=\bigoplus_{n\in\mathbb N} S_n, $$ then $$ \|A\|=\sup\{\|S_n\|:\ n\}=1, $$ $$ \sigma(A)=\overline{\bigcup_n\sigma(S_n)\}}=\{0\} $$ so $r(A)=0$, and $$ \omega(A)=1 $$ since the numerical range is the closed convex hull of the union of the numerical ranges, which in this case amounts to an increasing union of concentric balls.