## Dual space of $W^{\frac{1}{p’},p}(\partial\Omega)$

Let $$u\in W^{1,p}(\Omega)$$ verifying $$\begin{eqnarray}\int_{\Omega}|D\widehat{u}|^{p-2}(D\widehat{u},Dh)_{R^N}dz+\widehat{\xi}\int_{\Omega}|\widehat{u}|^{p-2}\widehat{u}hdz+\int_{\partial\Omega}\beta(Z)|\widehat{u}|^{p-2}\widehat{u}hd\sigma=\int_{\Omega}ghdz \label{gh} \end{eqnarray}$$ $$\forall h\in W^{1,p}(\Omega)$$. Let $$\left\langle A_p(u),y\right\rangle =\int_{\Omega}|Du|^{p-2}(Du,Dy)_{\mathbb{R}^N}dz\quad \forall u,y\in W^{1,p}(\Omega)$$ The green identity gives us $$\begin{eqnarray}\left\langle A(\widehat{u}),h\right\rangle =\left\langle \Delta_p\widehat{u},h\right\rangle +\left\langle \frac{\partial \widehat{u}}{\partial n_p},h\right\rangle _{\partial \Omega}\quad \forall h\in W^{1,p}(\Omega)\label{hi}\end{eqnarray}$$ We note by $$\left\langle .,.\right\rangle_{\partial\Omega}$$ the duality bracket $$(W^{-\frac{1}{p’},p’}(\partial\Omega),W^{\frac{1}{p’},p}(\partial\Omega))$$ $$(\frac{1}{p}+\frac{1}{p}=1)$$. The representation theorem of the dual space $$W^{1,p}(\Omega)^*=W^{-1,p’}(\Omega)$$, imply $$\Delta_p\widehat{u}\in W^{-1,p’}(\Omega)$$. Choosing $$h\in W^{1,p}_0(\Omega)\subseteq W^{1,p}(\Omega)$$ In the the last equality, we get $$\begin{eqnarray*} \left\langle \Delta_p\widehat{u},h\right\rangle++\widehat{\xi}\int_{\Omega}|\widehat{u}|^{p-2}\widehat{u}hdz=\int_{\Omega}ghdz\quad \forall h\in W^{1,p} \end{eqnarray*}$$ $$\Rightarrow-\Delta_{p}\widehat{u}(z)+\xi |\widehat{u}(z)|^{p-2}\widehat{u}(z)=g(z)\quad a.e\Omega$$ So from above we get $$\left\langle \frac{\partial \widehat{u}}{\partial n_p},h\right\rangle _{\partial \Omega}+\beta(z)|\widehat{u}|^{p-2}\widehat{u}=0\quad \forall h\in W^{1,p}(\Omega)$$ Thus $$\frac{\partial \widehat{u}}{\partial n_p}+\beta(z)|\widehat{u}|^{p-2}\widehat{u}=0 \text{a.e}\partial\Omega$$ my question is why do we have this.last eqaulity?is it some kind o f identification?