The existence, characterization and essential uniqueness of solutions of $L^{\infty }$ extremal problems

Abstract:

Let $I = (a,b)$ be an interval in R and let ${H^{n,\infty }}$ consist of those real-valued functions f such that ${f^{(n - 1)}}$ is absolutely continuous on I and ${f^{(n)}} \in {L^\infty }(I)$. Let L be a linear differential operator of order n with leading coefficient $1,a = {x_1} < \cdots < {x_m} = b$ be a partition of I and let the linear functionals ${L_{ij}}$ on ${H^{n,\infty }}$ be given by \[ {L_{ij}}f = \sum \limits _{v = 0}^{n - 1} {a_{ij}^{(v)}{f^{(v)}}({x_i}),\quad j = 1, \cdots ,{k_i},i = 1, \cdots ,m,} \] where $1 \leq {k_i} \leq n$ and the ${k_i}$ n-tuples $(a_{ij}^{(0)}, \cdots ,a_{ij}^{(n - 1)})$ are linearly inde pendent. Let ${r_{ij}}$ be prescribed real numbers and let $U = \{ f \in {H^{n,\infty }}:{L_{ij}}f = {r_{ij}},j = 1, \cdots ,{k_i},i = 1, \cdots ,m\}$. In this paper we consider the extremal problem \begin{equation}\tag {$\ast $}{\left \| {Ls} \right \|_{{L^\infty }}} = \alpha = \inf \{ {\left \| {Lf} \right \|_{{L^\infty }}}:f \in U\} .\end{equation} We show that there are, in general, many solutions to $( \ast )$ but that there is, under certain consistency assumptions on L and the ${L_{ij}}$, a fundamental (or core) interval of the form $({x_i},{x_{i + {n_0}}})$ on which all solutions to $( \ast )$ agree; ${n_0}$ is determined by the ${k_i}$ and satisfies ${n_0} \geq 1$. Further, if s is any solution to $( \ast )$ then on $({x_i},{x_{i + {n_0}}}),|Ls| = \alpha$ a.e. Further, we show that there is a uniquely determined solution ${s_ \ast }$ to $( \ast )$, found by minimizing ${\left \| {Lf} \right \|_{{L^\infty }}}$ over all subintervals $({x_j},{x_{j + 1}}),j = 1, \cdots ,m - 1$, with the property that $|L{s_ \ast }|$ is constant on each subinterval $({x_j},{x_{j + 1}})$ and $L{s_ \ast }$ is a step function with at most $n - 1$ discontinuities on $({x_j},{x_{j + 1}})$. When $L = {D^n},{s_ \ast }$ is a piecewise perfect spline. Examples show that the results are essentially best possible.