One-dimensional P1 interpolation has a first-derivative error bound

Fix a cell $K=[a,b]=[x_j,x_{j+1}]$ of length $h_K$. Its interpolant slope is $s_K=(u(b)-u(a))\mathrm{/}{h}_K=h_K^{-1}{\int }_a^b{u}^{\mathrm{\prime }}(t)dt$ by [F1].

For every $x\in K$, $u^{\mathrm{\prime }}(x)-s_K=h_K^{-1}{\int }_a^b(u^{\mathrm{\prime }}(x)-u^{\mathrm{\prime }}(t))dt$. A second use of [F1] and then [F2] gives $\mathrm{\mid }{u}^{\mathrm{\prime }}(x)-u^{\mathrm{\prime }}(t)\mathrm{\mid }\le \mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }x-t\mathrm{\mid }\le h_K\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$, hence $\mathrm{\mid }{u}^{\mathrm{\prime }}(x)-s_K\mathrm{\mid }\le h_K\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$.

The continuous function $\mathrm{\mid }{u}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }$ has a finite maximum by [F4]. On each open cell $K^{\circ }=(x_j,x_{j+1})$, $q=\mathrm{\mid }{u}^{\mathrm{\prime }}-s_K{\mathrm{\mid }}^2$ is continuous and hence measurable; define $q=0$ at the finitely many nodes. This gives a nonnegative measurable representative of the squared derivative error, bounded by the finite cell bounds in step 2.1. The Lebesgue interval-measure and monotonicity facts in [F3] give ${\int }_K\mathrm{\mid }(u-I_{h}u{)}^{\mathrm{\prime }}{\mathrm{\mid }}^2\le h_K^3\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}^2$. The open cells are disjoint and their complement consists of finitely many nodes of Lebesgue measure zero. Finite additivity of the measure $E\mapsto {\int }_{E}q$ from [F3], with zero contribution from the nodes where $q=0$, therefore identifies the global Lebesgue squared norm with the sum of these cell integrals. Using $h_K\le h$ and ${\sum }_K{h}_K=1$ gives $\mathrm{\parallel }(u-I_{h}u{)}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\le h^2\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}^2$. Taking nonnegative square roots proves the estimate; endpoint values follow from the nodal definition.