A boundary flux control uses the H1 trace and a different control operator

By [F2], $v$ has an absolutely continuous representative. By [F3], the hypotheses of [F4] hold. Applying [F4] to $F(x)=x$ and $G(x)=v(x)$ gives $v(1)={\int }_0^{1}v(x)dx+{\int }_0^{1}x{v}^{\mathrm{\prime }}(x)dx$. Cauchy--Schwarz [F5] yields $\mathrm{\mid }v(1)\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_2+\mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le \sqrt{2}\mathrm{\parallel }v{\mathrm{\parallel }}_V$. Applying [F4] to $F=u^{\mathrm{\prime }}$ and $G=v$ for a smooth state gives ${\int }_0^1(u^{\mathrm{\prime }}{v}^{\mathrm{\prime }}+uv)dx={\int }_0^{1}fvdx+gv(1)$. Thus the boundary control operator $B_{\mathrm{\partial }}g:v\mapsto gv(1)$ is bounded on $V$.

Let $v_n(x)=\max \{1-n(1-x),0\}$. Then $v_n\in H^1(0,1)$, $v_n(1)=1$, and $\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2=(3n{)}^{-1\mathrm{/}2}$. If $g\mathrm{\ne }0$ and $gv(1)={\int }_0^{1}qv$ for every $v\in V$ with some $q\in L^2(0,1)$, Cauchy--Schwarz [F5] gives $\mathrm{\mid }g\mathrm{\mid }=\mathrm{\mid }{\int }_0^{1}q{v}_n\mathrm{\mid }\le \mathrm{\parallel }q{\mathrm{\parallel }}_2(3n{)}^{-1\mathrm{/}2}\to 0$, a contradiction. For $g=0$, $q=0$ represents the same zero functional. The boundary input space is nevertheless the scalar $L^2(\{1\})$ space, while the distributed input space in [F1] is $L^2(0,1)$; their operators and control sets must be specified separately.