The method of continuity on a constant-coefficient one-dimensional path

The operators and the base point. For $u\in X$ one has $\mathrm{\parallel }{L}_{t}u{\mathrm{\parallel }}_{C^{0,\alpha }}\le \mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{C^{0,\alpha }}+tc\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{0,\alpha }}\le C(\alpha ,\pi )(1+c)\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{2,\alpha }}$, so $L_t$ maps $X$ boundedly into $Y$ for every $t$. For $t=0$, $L_0=-d^2\mathrm{/}d{x}^2$: if $-u^{\mathrm{\prime }\mathrm{\prime }}=0$ with $u(0)=u(\pi )=0$ then $u$ is affine and vanishes at both endpoints, so $u=0$ (injectivity); and for every $f\in Y$ the explicit function $Sf$ of [F3] satisfies $-S{f}^{\mathrm{\prime }\mathrm{\prime }}=f$, vanishes at the endpoints, and obeys $\mathrm{\parallel }Sf{\mathrm{\parallel }}_{C^{2,\alpha }}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$, so $L_{0}Sf=f$ (surjectivity). Hence $L_0$ is bijective.

The eigenvalue picture. By [F2], $L_t{e}_k=(k^2-tc)e_k$. For any $u\in X$, two integrations by parts, using $u=e_k=0$ at both endpoints, give ${\int }_0^{\pi }(L_{t}u)e_k=(k^2-tc){\int }_0^{\pi }u{e}_k$. If $L_{t}u=0$, all sine coefficients of $u$ vanish when $tc$ is not a square; if $tc=k_0^2$, all except the $k_0$th vanish. Since $u\in C^{0,\alpha }$, the Fourier identity in [F2] then gives $u=0$ in the first case and $u\in \mathrm{span}\{e_{k_0}\}$ in the second. Thus $L_t$ is injective exactly off the displayed spectrum, and its kernel at a spectral parameter is exactly $\mathrm{span}\{e_{k_0}\}$.

The Fourier solution away from resonance, including the coercive range. Fix $f\in Y$. If $tc=k_0^2$, assume $f_{k_0}=0$ and set $c_{k_0}=0$; for every $k\in J:=\{k\ge 1:k^2\mathrm{\ne }tc\}$ set $c_k=f_k\mathrm{/}(k^2-tc)$. In the non-resonant case this defines every $c_k$. In either case $m:={\inf }_{k\in J}\mathrm{\mid }{k}^2-tc\mathrm{\mid }\mathrm{/}{k}^2>0$, because the ratios tend to $1$ and none of the finitely many remaining ratios is zero. Put $u_N={\sum }_{k\le N}{c}_k\sin (kx)$. The coefficient bound in [F2] gives $\mathrm{\mid }{c}_k\mathrm{\mid }\le C_{\alpha }{m}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}{k}^{-2-\alpha }$ and $k\mathrm{\mid }{c}_k\mathrm{\mid }\le C_{\alpha }{m}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}{k}^{-1-\alpha }$. Thus only the series for $u_N$ and $u_N^{\mathrm{\prime }}$ are asserted to converge absolutely and uniformly; [F4] gives a limit $u\in C^1([0,\pi ])$ with zero endpoint values. For $tc<1$ one has $m\ge 1-tc$, giving the stated coercive-range bound. To see $u\in C^{0,\alpha }$, write $d=\mathrm{\mid }x-y\mathrm{\mid }$. For $0<d<1$, split ${\sum }_k\mathrm{\mid }{c}_k\mathrm{\mid }\min (2,kd)$ at $k\le d^{-1}$: the low-frequency part is at most $C{m}^{-1}\mathrm{\parallel }f\mathrm{\parallel }d{\sum }_{k\le d^{-1}}{k}^{-1-\alpha }\le C{m}^{-1}\mathrm{\parallel }f\mathrm{\parallel }d$, and the high-frequency part is at most $C{m}^{-1}\mathrm{\parallel }f\mathrm{\parallel }{\sum }_{k>d^{-1}}{k}^{-2-\alpha }\le C{m}^{-1}\mathrm{\parallel }f\mathrm{\parallel }{d}^{1+\alpha }$. For $1\le d\le \pi$, the supremum bound gives the same $C{m}^{-1}\mathrm{\parallel }f\mathrm{\parallel }{d}^{\alpha }$ control. Hence $[u{]}_{0,\alpha }\le C{m}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$.

Identify the equation and upgrade regularity. Put $f_N={\sum }_{k\le N}{f}_k\sin (kx)$. At a resonance the omitted coefficient is zero by hypothesis, so for every sufficiently large $N$ the partial-sum identity is still $u_N^{\mathrm{\prime }\mathrm{\prime }}=-f_N-tc{u}_N$. By [F2], $f_N\to f$ in $L^2(0,\pi )$, while $u_N\to u$ uniformly; hence $u_N^{\mathrm{\prime }\mathrm{\prime }}\to g:=-f-tcu$ in $L^2$. Since also $u_N\to u$ in $L^2$, continuity of distributional differentiation and the regular-function embedding in [F7--F8] show $u^{\mathrm{\prime }\mathrm{\prime }}=g$ distributionally on $(0,\pi )$. The function $g$ is continuous. Set $G(x)={\int }_0^{x}g(s)ds$; then the distributional derivative of the continuous function $u^{\mathrm{\prime }}-G$ is zero. By [F9], $u^{\mathrm{\prime }}-G$ is a constant distribution, hence equals that constant pointwise; therefore $u\in C^2([0,\pi ])$ (with one-sided endpoint derivatives) and $u^{\mathrm{\prime }\mathrm{\prime }}=g$. Since $u\in C^{0,\alpha }$ by step 1.3 and $f\in C^{0,\alpha }$, $g=-f-tcu\in C^{0,\alpha }$, so $u\in X$ and $L_{t}u=f$.

The range at a spectral parameter. If $tc=k_0^2$, integration by parts as in step 1.2 gives ${\int }_0^{\pi }(L_{t}u)e_{k_0}=0$ for every $u\in X$, so the range lies in the proper closed hyperplane $\{f\in Y:f_{k_0}=0\}$. Conversely, for any $f$ in that hyperplane, steps 1.3 and 2.1 construct $u\in X$ with $L_{t}u=f$; thus this hyperplane is exactly the range. Moreover, the inverse norm of $L_t$ on its bijective parameters blows up near $t_0:=k_0^2\mathrm{/}c$: for $t\mathrm{\ne }{t}_0$, testing on $e_{k_0}$ gives $\mathrm{\parallel }{L}_t{e}_{k_0}{\mathrm{\parallel }}_{C^{0,\alpha }}=\mathrm{\mid }{k}_0^2-tc\mathrm{\mid }\mathrm{\parallel }{e}_{k_0}{\mathrm{\parallel }}_{C^{0,\alpha }}$ and hence $\mathrm{\parallel }{L}_t^{-1}\mathrm{\parallel }\ge \mathrm{\parallel }{e}_{k_0}{\mathrm{\parallel }}_X\mathrm{/}(\mathrm{\mid }{k}_0^2-tc\mathrm{\mid }\mathrm{\parallel }{e}_{k_0}{\mathrm{\parallel }}_Y)\to \mathrm{\infty }$ as $t\to t_0$.

Estimate in the coercive range. For $tc<1$, $m\ge 1-tc$ in step 1.3, so the sup and Hölder bounds there control $\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{0,\alpha }}$ and $\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$ by $C(1-tc{)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$. From step 2.1, $u^{\mathrm{\prime }\mathrm{\prime }}=-f-tcu$, hence $\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{C^{0,\alpha }}\le C(1-tc{)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$. Thus $\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{2,\alpha }}\le C(1-tc{)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$. Uniqueness follows from step 1.2; in particular $L_t$ is bijective for every non-spectral parameter, while this is the stated quantitative estimate on the coercive range.

The two continuity properties of the bijectivity set. By steps 1.2 and 2.1, $I=[0,1]\setminus \{k^2\mathrm{/}c:k^2\le c\}$ is exactly the bijectivity set. Its complement is finite, so $I$ is open in $[0,1]$. Every subinterval $J\subseteq [0,1]$ on which all $L_t$ are injective contains no spectral parameter by step 1.2, hence $I\cap J=J$ is relatively closed in $J$. These are the two properties inspected in the abstract method of continuity [F5]. At a spectral parameter the inverse norms on neighboring bijective parameters blow up as in step 3.1, so no a priori estimate uniform across that parameter can hold.

Conclusion. The model family $L_{t}u=-u^{\mathrm{\prime }\mathrm{\prime }}-tcu$ on $(0,\pi )$ is bounded $X\to Y$ for every $t$, has the bijective base point $L_0=-d^2\mathrm{/}d{x}^2$, and has bijectivity set $I=\{t:tc\mathrm{\notin }\{k^2:k\ge 1\}\}$. For $tc<1$ the eigenfunction expansion gives the inverse bound $C(1-tc{)}^{-1}$; at $tc=k_0^2$ the kernel is $\mathrm{span}\{\sin (k_{0}x)\}$ and the range is the closed hyperplane orthogonal to it. Openness and the relative-closedness property hold by direct inspection of the finite exceptional set. This one-dimensional example illustrates the abstract method of continuity [F5] and its PDE-level application [F6].