Continuous linearized BVP operator and its conditioning

Let $y^{\ast }\in C^1([a,b],{\mathbb{R}}^d)$ solve a square BVP, with $f$ and $B$ continuously differentiable on open neighborhoods of the trajectory and endpoint pair. Put $A(t)=D_{y}f(t,y^{\ast }(t))$, $B_1=D_{1}B(y^{\ast }(a),y^{\ast }(b))$, and $B_2=D_{2}B(y^{\ast }(a),y^{\ast }(b))$. Let

$X=C^1([a,b],{\mathbb{R}}^d),\mathrm{\parallel }v{\mathrm{\parallel }}_X=\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }},$

and

$Z=C([a,b],{\mathbb{R}}^d)\times {\mathbb{R}}^d,\mathrm{\parallel }(g,q){\mathrm{\parallel }}_Z=\max \{\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }},\mathrm{\parallel }q{\mathrm{\parallel }}_2\}\mathrm{.}$

The continuous linearized BVP operator is the bounded linear map $\mathcal{L}:X\to Z$ given by

$\mathcal{L}v=(v^{\mathrm{\prime }}-Av,B_{1}v(a)+B_{2}v(b))\mathrm{.}$

It is boundedly invertible when it is bijective and its inverse ${\mathcal{L}}^{-1}:Z\to X$ has finite operator norm. That norm is the conditioning constant for the linearized BVP used in the local discretization theorem.