One-dimensional P1 interpolation has an L2 error bound

Fix a cell $K=[a,b]$ of length $h_K$, and let $s_K=(u(b)-u(a))\mathrm{/}{h}_K$ be the slope of $I_{h}u$ on $K$. By the fundamental theorem of calculus, $s_K=h_K^{-1}{\int }_a^b{u}^{\mathrm{\prime }}(t)dt$. For every $x\in K$, the same theorem gives $\mathrm{\mid }{u}^{\mathrm{\prime }}(x)-u^{\mathrm{\prime }}(t)\mathrm{\mid }\le h_K\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$, so averaging in $t$ yields $\mathrm{\mid }(u-I_{h}u{)}^{\mathrm{\prime }}(x)\mathrm{\mid }=\mathrm{\mid }{u}^{\mathrm{\prime }}(x)-s_K\mathrm{\mid }\le h_K\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Since $(u-I_{h}u)(a)=0$, integrating the derivative bound from step 1.1 over $[a,x]$ gives $\mathrm{\mid }u(x)-I_{h}u(x)\mathrm{\mid }\le h_K^2\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $x\in K$. Therefore $\mathrm{\parallel }u-I_{h}u{\mathrm{\parallel }}_{L^2(K)}^2\le h_K^5\mathrm{\parallel }{u}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}^2$.

The cell interiors are disjoint, their lengths sum to $1$, and $h_K\le h$, so ${\sum }_K{h}_K^5\le h^4{\sum }_K{h}_K=h^4$. Summing step 2.1 over the cells and taking nonnegative square roots proves the estimate.