The first positive Neumann eigenvalue has the mean-zero Rayleigh characterisation

The space $V$ is closed in $H^1(\mathrm{\Omega })$ and $L_0^2(\mathrm{\Omega })$ is closed in $L^2(\mathrm{\Omega })$ by [F1]; on $V$ the form $a$ is bounded and satisfies $a(u,u)\ge 0$ for every $u\in V$ by [F3]. The estimate of [F3] is exactly the coercivity statement $a(u,u)\ge \alpha \mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}^2,\alpha :=\frac{\theta }{1+C_W^2}>0,$ for all $u\in V$, obtained from Poincare--Wirtinger and ellipticity.

Solution operator. For each $f\in L_0^2(\mathrm{\Omega })$ the functional $v\mapsto (f,v{)}_{L^2}$ is bounded and conjugate-linear on $V$ by [F1], so by [F4] there is a unique $Sf\in V$ with $a(Sf,v)=(f,v{)}_{L^2}$ for every $v\in V$; the assignment $f\mapsto Sf$ is linear and bounded with the explicit estimate $\mathrm{\parallel }Sf{\mathrm{\parallel }}_{H^1}\le {\alpha }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}=(1+C_W^2){\theta }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}$, obtained by testing the defining identity at $v=Sf$ and using [F1]. By [F5] the operator $S:L_0^2(\mathrm{\Omega })\to L_0^2(\mathrm{\Omega })$ is self-adjoint, positive definite and injective, and by [F6] it is compact.

Spectral decomposition. By [F7] and [F8] the nonzero eigenvalues of $S$ are positive real numbers of finite multiplicity with no accumulation point except $0$; enumerate the distinct eigenvalues in decreasing order as ${\nu }_1>{\nu }_2>\cdots >0$ when the set is infinite (with ${\nu }_k\downarrow 0$), and let $P_k$ be the orthogonal projection onto the eigenspace $E_{{\nu }_k}$. Since $S$ is injective, [F7] gives $L_0^2(\mathrm{\Omega })=\stackrel{}{\mathrm{span}}{\bigcup }_k{E}_{{\nu }_k}$ with orthogonal summands, so for every $f\in L_0^2(\mathrm{\Omega })$ the net of partial sums $f_n:={\sum }_{k\le n}{P}_{k}f$ converges to $f$ in $L^2$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2={\sum }_k\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2$. Every eigenspace lies in $V$, because $Sg\in V$ and $g={\nu }^{-1}Sg$ for an eigenvector. Finally, for $g\in E_{{\nu }_k}$ and every $v\in V$ one has $a(Sg,v)=(g,v{)}_{L^2}$ by definition of $S$, that is $a(g,v)={\nu }_k^{-1}(g,v{)}_{L^2}$.

Partial sums in the form. Fix $f\in V$ and put $f_n={\sum }_{k\le n}{P}_{k}f$ as in step 3.1. Then step 3.1 gives, for every $v\in V$, $a(f_n,v)={\sum }_{k\le n}{\nu }_k^{-1}(P_{k}f,v{)}_{L^2}$; taking $v=f_n$ and $v=f$ and using orthogonality of the projections, $a(f_n,f_n)={\sum }_{k\le n}{\nu }_k^{-1}\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2=a(f_n,f),$ where the last identity is real. Hence $a(f-f_n,f-f_n)=a(f,f)-a(f_n,f_n)\ge 0$ by positivity of $a$ on $V$, and therefore ${\sum }_{k\le n}{\nu }_k^{-1}\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2\le a(f,f)$ for every $n$.

Form-norm expansion. The increasing partial sums of ${\sum }_k{\nu }_k^{-1}\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2$ are bounded by $a(f,f)$, so the series converges; consequently, for $m<n$, $a(f_n-f_m,f_n-f_m)={\sum }_{m<k\le n}{\nu }_k^{-1}\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2⟶0,$ so $(f_n)$ is Cauchy for the inner product $a$ on $V$. By the coercivity of step 1.1 it is Cauchy in $H^1(\mathrm{\Omega })$, hence converges in $H^1$ to some $u\in V$ (closedness of $V$). Since $H^1$ convergence implies $L^2$ convergence and $f_n\to f$ in $L^2$, we get $u=f$; continuity of $a$ in the $H^1$ norm then gives $a(f,f)={\lim }_{n}a(f_n,f_n)={\sum }_k{\nu }_k^{-1}\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2\mathrm{.}$

Rayleigh characterisation. Put ${\mu }_k:={\nu }_k^{-1}$, so that by [F8] ${\mu }_1={\nu }_1^{-1}=1\mathrm{/}\mathrm{\parallel }S\mathrm{\parallel }$ is the smallest of the ${\mu }_k$ and ${\mu }_k>0$. For $f\in V\setminus \{0\}$ step 5.1 and $\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2={\sum }_k\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2$ give $\frac{a(f,f)}{\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2}=\frac{{\sum }_k{\mu }_k\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2}{{\sum }_k\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2}\ge {\mu }_1,$ with equality precisely when $P_{k}f=0$ for every $k$ with ${\mu }_k>{\mu }_1$, that is $f\in E_{{\nu }_1}$. Hence ${\mu }_1>0$ is attained and the minimisers are exactly the nonzero elements of $E_{{\nu }_1}$. Moreover $f\in E_{{\nu }_1}\setminus \{0\}$ satisfies $a(f,v)={\mu }_1(f,v{)}_{L^2}$ for all $v\in V$ by step 3.1; conversely, if $0\mathrm{\ne }f\in V$ satisfies $a(f,v)={\mu }_1(f,v{)}_{L^2}$ for all $v\in V$, then the same expansion gives ${\sum }_k({\mu }_k-{\mu }_1)\mathrm{\parallel }{P}_{k}f{\mathrm{\parallel }}_{L^2}^2=0$ with all coefficients nonnegative, so $P_{k}f=0$ whenever ${\mu }_k>{\mu }_1$ and $f\in E_{{\nu }_1}$. Thus the eigenspace $\{u\in V:a(u,v)={\mu }_1(u,v{)}_{L^2}\mathrm{\forall }v\in V\}$ equals $E_{{\nu }_1}$, and no mean-zero weak Neumann eigenvalue $\lambda \in (0,{\mu }_1)$ exists, since it would give the same identity with a nonnegative combination vanishing.

Extension to $H^1(\mathrm{\Omega })$ and conclusions. Let $e_1\in E_{{\nu }_1}\setminus \{0\}$. By [F9] the constant $1$ has $a(e_1,1)=0$ and $(e_1,1{)}_{L^2}=0$ because $e_1\in V$; writing an arbitrary $v\in H^1(\mathrm{\Omega })$ as $v=(v-v_{\mathrm{\Omega }})+v_{\mathrm{\Omega }}$ with $v-v_{\mathrm{\Omega }}\in V$, the identity $a(e_1,w)={\mu }_1(e_1,w{)}_{L^2}$ for $w=v-v_{\mathrm{\Omega }}$ therefore extends to all $v\in H^1(\mathrm{\Omega })$, which is the weak Neumann eigenequation; the same argument extends the eigenspace description of step 6.1, showing that ${\mu }_1$ is the smallest positive weak Neumann eigenvalue on the mean-zero space. Together with ${\mu }_1=1\mathrm{/}\mathrm{\parallel }S\mathrm{\parallel }$ from step 6.1 this proves all the assertions; connectedness is used only through Poincare--Wirtinger [F2] (a disconnected domain admits the componentwise constants in $V$ with $a=0$, so the infimum would be $0$), and the extension-domain hypothesis is used only through [F2] and [F6].