= Solution
The convergence is in the <strong operator topology>: for each fixed $f\in E$, one must prove $\|\lambda R_\lambda f-f\|\to0$. First, the <contraction semigroup> bound gives
$$
\|\lambda R_\lambda f\|
\leq\int_0^\infty\lambda e^{-\lambda t}\|P_tf\|\,dt
\leq\|f\|.
$$
Thus \b[$\lambda R_\lambda$ is a contraction]. This is the useful scaling of the contraction estimate in the source's hint. Since $\int_0^\infty\lambda e^{-\lambda t}\,dt=1$,
$$
\lambda R_\lambda f-f
=\int_0^\infty\lambda e^{-\lambda t}(P_tf-f)\,dt.
$$
Fix $\varepsilon>0$. By <strong continuity>, choose $\delta>0$ so that $\|P_tf-f\|<\varepsilon$ for $0\leq t\leq\delta$. Split the integral there and use $\|P_tf-f\|\leq2\|f\|$ for the tail:
$$
\begin{aligned}
\|\lambda R_\lambda f-f\|
&\leq\varepsilon\int_0^\delta\lambda e^{-\lambda t}\,dt
+2\|f\|\int_\delta^\infty\lambda e^{-\lambda t}\,dt\\
&\leq\varepsilon+2\|f\|e^{-\lambda\delta}.
\end{aligned}
$$
Let $\lambda\to\infty$ and then $\varepsilon\downarrow0$. Therefore
$$
\boxed{\lambda R_\lambda f\longrightarrow f\text{ in }E\quad\text{for every }f\in E.}
$$
The distinction from <operator norm> convergence is essential. For example, the <multiplication semigroup> on $\ell^2(\mathbb N)$ defined by $(P_tf)_n=e^{-nt}f_n$ is a <strongly continuous semigroup> of contractions: coordinatewise convergence and <dominated convergence> applied to $\sum_n|f_n|^2$ prove <strong continuity>. Its <resolvent operator> is $(R_\lambda f)_n=f_n/(\lambda+n)$, and
$$
\|\lambda R_\lambda-I\|=\sup_{n\geq1}\frac n{\lambda+n}=1.
$$
Thus the claimed convergence cannot in general be read as convergence in <operator norm>.
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