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Put $\delta =w-u$, which vanishes on the boundary. Expanding the Dirichlet energy givesThe cross term vanishes by integration by parts: it is ${\int }_{\partial S}\delta {\partial }_{n}uds-{\int }_S\delta \Delta udA=0$, using both the zero boundary values of $\delta$ and Laplace's equation for $u$. Thus the Dirichlet principle gives the stronger identityWith the usual regular boundary, equality forces $w=u$.

In the unit disc, take boundary data $\cos \theta$. The harmonic function $u=r\cos \theta =x$ realizes these data and has constant gradient of length one, so its energy is $\pi$. For $w=g(r)\cos \theta$, the polar gradient givesThe energy principle therefore yields the sharp radial Dirichlet energy inequality

There is a small regularity point at the centre: finite energy need not make the polar expression classically smooth there. The bound still holds under the stated finite-energy condition. Indeed for $0<s<t$,The tails tend to zero, so $g^2$ has a limit at zero. That limit must be zero, since a positive limit would make $\int g^2/r$ diverge. Completing the square on $[\epsilon ,1]$ and letting $\epsilon \to 0$ givesThis proves the finite-energy extension directly, and also shows that equality holds exactly for $g(r)=r$.