Endpoint Poincaré inequality ($\Vert v{\Vert }_{L^2(b,c)}^2\le \frac{(c-b{)}^2}{2}\Vert v^{\prime }{\Vert }_{L^2(b,c)}^2,v(c)=0$)

For the one-dimensional Sobolev representative with $v(c)=0$, the fundamental theorem of calculus gives $v(x)=-{\int }_x^c{v}^{\prime }(t)dt$. The Cauchy-Schwarz inequality then gives $|v(x){|}^2\le (c-x)\Vert v^{\prime }{\Vert }_2^2$. Integrating proves the displayed, generally nonoptimal Poincaré inequality. In particular $\Vert v{\Vert }_{H^1}^2\le [1+(c-b{)}^2/2]\Vert v^{\prime }{\Vert }_2^2$. Only one endpoint must be fixed, which is useful for mixed boundary conditions and for controlling a negative potential.