Lp estimate for resolvent of Laplace operator Consider for $1<p<\infty$ operator $A_p:L_p(0,1)\to L_p(0,1), \ D(A_p)=\{u\in W^2_p(0,1): u'(0)=u'(1)=0\}, \ A_pu=u''$, i.e. $L_p$-realisation of the Laplace operator with Neumann bcd on the unit interval. I need reference (or proof) for the following result:
\begin{align}
||R(\lambda,A_p)||_{\mathcal{L}(L_p)}\leq\frac{C}{dist(\lambda,\sigma(A_p))}, \ \lambda\in\rho(A_p).
\end{align}
I am interested especially in the case when $\lambda$ lies between two consecutive eigenvalues of $A_p$. When $p=2$ the result follows with $C=1$ from the spectral theorem.
 A: We can use the integral kernel of $R=(A-z)^{-1}$ and nothing much changes when $p\not= 2$. By the variation of constants formula, $(Rf)(x)=\int_0^1 G(x,t;z) f(t)\, dt$. Here $G$ could of course be worked out explicitly, but I'll just use that
$$
G(x,t;z) =\sum_{n=1}^{\infty} \frac{u_n(x)u_n(t)}{\lambda_n-z} ,
$$
with $u_n(x)=\sqrt{2} \cos n\pi x$ (the $n$th eigenfunction) and $\lambda_n=-n^2\pi^2$ is the corresponding eigenvalue. It doesn't matter here what $p$ is because we always need to solve the same ODE with the same boundary conditions (and, if we want to, we can assume that $f\in C_0^{\infty}(0,1)$).
Now clearly
$$
|(Rf)(x)|\le \sum \frac{|u_n(x)|}{|\lambda_n-z|} \|u_n\|_{p'}\|f\|_p ,
$$
and $\|u_n\|_p, \|u_n\|_{p'}$ are uniformly bounded, so to obtain the desired bound, I only need to show that
$$
\sum_{n=1}^{\infty} \frac{1}{|\lambda_n-z|} \lesssim \frac{1}{\textrm{dist}(z,\{\lambda_n\})} .
$$
This is clear for $z$ between eigenvalues (or $z<1$, say) since $\lambda_n=-n^2\pi^2$.
