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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.

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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 $\lambda_n=-n^2\pi^2$.

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  • $\begingroup$ For me it is not clear. If $z>0$ you need to show that $\sum_{n=0}^{\infty}\frac{z}{z+n^2}$ is bounded in $z$ which is clearly not. $\endgroup$
    – Marcin Malogrosz
    Commented Aug 25, 2014 at 16:21
  • $\begingroup$ You mean for $z\to\infty$? I didn't consider this case; I focused on $z$ between eigenvalues (as mentioned in your question). $\endgroup$
    – Christian Remling
    Commented Aug 25, 2014 at 16:43
  • $\begingroup$ If you want to analyze $|z|\to\infty$, I would try to work with the explicit formula for $G$ and use facts about Fourier transforms. $\endgroup$
    – Christian Remling
    Commented Aug 25, 2014 at 16:55
  • $\begingroup$ Which explicit formula do You have in mind? From your reasoning $||R(z,A_p)||_{\mathcal{L}(L_p)}\leq ||||G(x,t;z)||_{L_p(dx)}||_{L_{p'}(dt)}$. Now if $p=2$ one has $||||G(x,t;z)||_{L_p(dx)}||_{L_{p'}(dt)}=||G(x,t;z)||_{L_2(dx\otimes dt)}=(\sum_{n=0}^{\infty}1/|\lambda_n-z|^2)^{1/2}$, so the estimate seems to be a bit to coarse even for $p=2$. $\endgroup$
    – Marcin Malogrosz
    Commented Aug 25, 2014 at 17:23
  • $\begingroup$ You can of course solve $y''-zy=f$, $y'(0)=y'(1)=0$ explicitly. This gives a (simple) formula for $G$. As for the estimate being coarse, I'm sorry to learn that you don't like it. I was addressing the question you asked. $\endgroup$
    – Christian Remling
    Commented Aug 25, 2014 at 17:57

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