Separated-box degeneracy

[Orthonormality and completeness] For $n,m,n^{\mathrm{\prime }},m^{\mathrm{\prime }}$, $⟨e_{nm},e_{n^{\mathrm{\prime }}{m}^{\mathrm{\prime }}}⟩=\left({\int }_0^L{e}_n\stackrel{‾}{e_{n^{\mathrm{\prime }}}}\right)\left({\int }_0^L{e}_m\stackrel{‾}{e_{m^{\mathrm{\prime }}}}\right)={\delta }_{n{n}^{\mathrm{\prime }}}{\delta }_{m{m}^{\mathrm{\prime }}}$ by [F1] and Fubini. If $g\in L^2((0,L{)}^2)$ is orthogonal to all $e_{nm}$, then for every $n$ the function $h_n(y):={\int }_0^{L}g(x,y)e_n(x)dx$ (defined as zero on exceptional sections) is measurable and satisfies $\mathrm{\mid }{h}_n(y){\mathrm{\mid }}^2\le {\int }_0^L\mathrm{\mid }g(x,y){\mathrm{\mid }}^{2}dx$ a.e. by Cauchy-Schwarz, hence $h_n\in L^2(0,L)$ by Tonelli; moreover $g(x,y)e_n(x)e_m(y)$ is absolutely integrable by Cauchy-Schwarz on the square, so ${\int }_0^L{h}_n\overline{e_m}=0$ for all $m$ by Fubini, so $h_n=0$ a.e. by completeness of $(e_m)$. Fubini also gives a full-measure set $Y_0$ on which $g(\cdot ,y)\in L^2(0,L)$. For each $n$, choose a full-measure set $Y_n$ on which $h_n(y)=0$; Countable Choice permits choosing these sets simultaneously, and countable subadditivity makes $Y_0\cap {\bigcap }_n{Y}_n$ full measure. For every $y$ in this intersection, all sine coefficients of $g(\cdot ,y)$ vanish, so completeness of $(e_n)$ gives $g(\cdot ,y)=0$ a.e. in $x$. Fubini then yields $g=0$ a.e. on the square. This is (a).

[Diagonal operator and eigenspace] By [F2] applied to the complete orthonormal family $(e_{nm})$ and the real numbers $({\mu }_{nm})$ (indexed by ${\mathbb{N}}^2$, which is countable), define $D({\mathrm{\Delta }}_D):=\{g\in H:{\sum }_{n,m\ge 1}{\mu }_{nm}^2\mathrm{\mid }⟨g,e_{nm}⟩{\mathrm{\mid }}^2<\mathrm{\infty }\}$ and ${\mathrm{\Delta }}_{D}g:={\sum }_{n,m\ge 1}{\mu }_{nm}⟨g,e_{nm}⟩e_{nm}$. To match the supplier's indexing by $\mathbb{N}=\{0,1,\dots \}$, enumerate $(n,m)$ by increasing $n+m\ge 2$, and within each sum by increasing $n$; this lists every positive pair once. The supplier proves convergence on this domain and self-adjointness, with ${\mathrm{\Delta }}_D{e}_{nm}={\mu }_{nm}{e}_{nm}$; a vector $x={\sum }_{n,m}{c}_{nm}{e}_{nm}\in D({\mathrm{\Delta }}_D)$ lies in $\ker ({\mathrm{\Delta }}_D-\mu )$ exactly when ${\sum }_{n,m}({\mu }_{nm}-\mu )c_{nm}{e}_{nm}=0$, which by uniqueness of coefficients forces $c_{nm}=0$ whenever ${\mu }_{nm}\mathrm{\ne }\mu$. The equation $n^2+m^2=5$ with $n,m\ge 1$ has exactly the solutions $(1,2)$ and $(2,1)$, so $\ker ({\mathrm{\Delta }}_D-\mu )=\mathbb{C}{e}_{12}\oplus \mathbb{C}{e}_{21}$ is two-dimensional. This proves (b) and the displayed eigenspace assertion in (c).

[One-dimensional comparison] The operator $A$ in the comparison has precisely the domain and action required by [F3]; the hypotheses $L>0$, $\alpha ,\beta \in \mathbb{R}$, $V\in C([0,L];\mathbb{R})$ and both choice assumptions are explicit. Thus its spectrum consists of eigenvalues with one-dimensional eigenspaces. By step 2.1, the square operator instead has a two-dimensional eigenspace at $5{\pi }^2\mathrm{/}{L}^2$. This establishes the stated contrast and completes (c).