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With the standard Sobolev norm, the printed constant one is not valid for arbitrary $a$. The constant function $u=1$ has supremum norm one and $H^1$ norm $\sqrt{2a}$, which is smaller when $a<1/2$.

For the continuous one-dimensional Sobolev representative, let $\bar{u}=(2a{)}^{-1}{\int }_{I}u$. The Cauchy-Schwarz inequality gives $|\bar{u}|\le (2a{)}^{-1/2}\Vert u{\Vert }_2$. Using the preceding Hölder seminorm estimate,Consequently the corrected interval Sobolev supremum estimate isOne may first prove this for smooth approximations and pass on the full-measure convergence set. Averaging differences also handles complex-valued functions without assuming that a function takes its complex average at some point. At the fixed interval used below the constant is absolute, which suffices for the intended interior estimate.

Choose a fixed smooth cutoff function with $\Vert {\zeta }^{\prime }{\Vert }_{\infty }=C_0$ and set $J=(-1/2,1/2)$. The Caccioppoli inequality gives $\Vert u^{\prime }{\Vert }_{L^2(J)}\le 2R{C}_0\Vert u{\Vert }_{L^2(-1,1)}$. Since the length of $J$ is one, the corrected interval Sobolev supremum estimate yields $\Vert u{\Vert }_{L^{\infty }(J)}\le \Vert u{\Vert }_{L^2(J)}+\Vert u^{\prime }{\Vert }_{L^2(J)}$, and the Hölder seminorm is at most $\Vert u^{\prime }{\Vert }_{L^2(J)}$. ThereforeFor locally $H^1$ weak solutions with $u\in L^2(-1,1)$, density of smooth functions in a Sobolev space justifies the test ${\zeta }^{2}u$, and the previous representative estimates already apply to $H^1$. Mollifying a solution need not preserve the equation with the same measurable coefficient, so approximation is used for admissible tests and Sobolev estimates, not to assert that the mollified function solves the original equation. The required constant depends only on $R$.