= Solution
Let $u$ be the actual <displacement field>, let $u'$ be an admissible trial <displacement>, and set $v=u'-u$. On the displacement-prescribed part of the boundary, $v=0$. With actual <stress> $\sigma=D_\varepsilon W(\varepsilon)$, the supporting-plane inequality for the convex <strain energy density> gives
$$
W(\varepsilon')-W(\varepsilon)\ge\sigma:(\varepsilon'-\varepsilon).
$$
Both $\sigma$ and $\sigma_0$ satisfy <static elastic equilibrium> with the same <body force>, so $\nabla\cdot(\sigma-\sigma_0)=0$. On the traction-prescribed boundary their <tractions> agree. Since $\varepsilon'-\varepsilon=\operatorname{sym}\nabla v$, <integration by parts> gives
$$
\int_\Omega(\sigma-\sigma_0):(\varepsilon'-\varepsilon)\,dx
=\int_{\partial\Omega}[(\sigma-\sigma_0)n]\cdot v\,dS=0.
$$
Subtracting the prescribed external-work representative proves the <minimum potential energy principle in elasticity>:
$$
\boxed{\int_\Omega\bigl(W(\varepsilon)-\sigma_0:\varepsilon\bigr)\,dx
\le\int_\Omega\bigl(W(\varepsilon')-\sigma_0:\varepsilon'\bigr)\,dx.}
$$
Strict convexity gives uniqueness of the minimizing <strain>, although rigid <displacements> can remain when the boundary conditions allow them.
For the comparison bound, put $F(e,x)=W(e,x)-W_0(e)$ and use the <lower conjugate>, with its infimum sign convention,
$$
F_*(\tau,x)=\inf_e\{\tau:e-F(e,x)\}.
$$
Its defining inequality is $F_*(\tau,x)\le\tau:e-F(e,x)$, hence $W(e,x)\le W_0(e)+\tau:e-F_*(\tau,x)$ for every $e$. Integrate this pointwise majorant for each compatible <strain>, then take the infimum over the same admissible set. This proves
$$
\boxed{W^{\mathrm{eff}}(E)\le\inf_{e\in K}\int_\Omega
\bigl[W_0(e)+\tau:e-F_*(\tau,x)\bigr],dx.}
$$
No convexity of $F$ is used for this inequality. If $F_*(\tau,x)=-\infty$, it yields only the extended-real upper bound $+\infty$. For example, $W_0=0$ and $W(e)=|e|^2/2$ give $F_*(\tau)=-\infty$ for every finite $\tau$. A useful finite bound requires a suitable comparison energy; choosing a sufficiently stiff quadratic comparison makes the usual excess energies concave. The derivative used below additionally presupposes finite differentiable duals, or an appropriate supergradient formulation.
Now normalize $|\Omega|=1$, take positive constant $C^0$, and write $F_r(e)=W_r(e)-e:C^0e/2$. For phasewise constant <elastic polarization> $\tau=\sum_r\chi_r\tau^r$, its <lower conjugate> contribution is constant within each phase. The remaining minimization is
$$
J_0(E,\tau)=\min_{e\in K}\int_\Omega\left(\tfrac12e:C^0e+\tau:e\right)dx.
$$
Write the minimizing <strain> as $\widehat e=E+\eta$, where $\eta$ comes from a zero-boundary <displacement field>. The <static elastic equilibrium> equation is $\nabla\cdot(C^0\eta+\tau)=0$, so the supplied <elastic strain Green operator> gives $\eta=-\Gamma\tau$. Its zero mean removes the cross term with the constant $E$. Using its reciprocity and projection property $\Gamma C^0\Gamma=\Gamma$, we obtain
$$
\begin{aligned}
J_0(E,\tau)
&=\tfrac12E:C^0E+\overline\tau:E
+\tfrac12\langle\Gamma\tau,C^0\Gamma\tau\rangle-\langle\tau,\Gamma\tau\rangle\\
&=\tfrac12E:C^0E+\overline\tau:E-\tfrac12\langle\tau,\Gamma\tau\rangle.
\end{aligned}
$$
Here the brackets in the last two terms denote the spatial integral and <tensor> contraction, not just a phase average.
Let $c_r=\int_\Omega\chi_r$. The zero output mean and reciprocity of $\Gamma$ imply that it annihilates constant <elastic polarizations> on either side. Thus subtracting $c_r$ and $c_s$ from the phase indicators does not change their bilinear integral. Define the fourth-order block <tensors>
$$
A_{rs}=\int_\Omega\!\int_\Omega
(\chi_r(x)-c_r)\Gamma(x,x')(\chi_s(x')-c_s)\,dx'\,dx.
$$
Consequently $A_{rs}=A_{sr}^T$, $\sum_s A_{rs}=0$, and
$$
\langle\tau,\Gamma\tau\rangle=\sum_{r,s}\tau^r:A_{rs}\tau^s.
$$
These identities follow from reciprocity and $\sum_s\chi_s=\sum_s c_s=1$; no phase independence assumption is needed. The <phase polarization comparison bound>, valid for every polarization with finite dual terms, is therefore
$$
\boxed{W^{\mathrm{eff}}(E)\le\mathcal B(E,\tau)
=\tfrac12E:C^0E+\sum_r c_r\tau^r:E
-\tfrac12\sum_{r,s}\tau^r:A_{rs}\tau^s
-\sum_r c_rF_{r*}(\tau^r).}
$$
In particular, it can be optimized over the phase <elastic polarizations> without changing the upper-bound direction.
To derive the final displayed form, impose the stationarity equations in the question. For $c_r>0$, put $e^r=DF_{r*}(\tau^r)$; these are dual <strains>, initially distinct from the comparison solution's phase averages. Reciprocity of $A$ gives
$$
D_{\tau^r}\mathcal B
=c_rE-\sum_sA_{rs}\tau^s-c_re^r.
$$
Thus stationarity is exactly
$$
\boxed{c_re^r+\sum_sA_{rs}\tau^s=c_rE.}
$$
The comparison solution has phase mean
$$
\frac1{c_r}\int_\Omega\chi_r\widehat e\,dx
=E-\frac1{c_r}\sum_sA_{rs}\tau^s,
$$
so \b[the dual strains become its phase means at stationarity]. Multiplying the stationarity equations by $\tau^r$ and summing gives $\sum_{r,s}\tau^r:A_{rs}\tau^s=\sum_r c_r\tau^r:(E-e^r)$.
For clarity, the <concave biconjugate> here uses another infimum: $F_{r**}(e)=\inf_s\{s:e-F_{r*}(s)\}$. The <lower conjugate> $F_{r*}$ is concave, since it is an infimum of affine functions. At a differentiable finite point, its supporting-plane inequality shows that $s=\tau^r$ minimizes this last expression when $e=e^r$. Hence
$$
F_{r*}(\tau^r)+F_{r**}(e^r)=\tau^r:e^r.
$$
Substitute both identities into $\mathcal B$ to obtain
$$
\boxed{W^{\mathrm{eff}}(E)\le\tfrac12E:C^0E
+\tfrac12\sum_r c_r\tau^r:(E-e^r)
+\sum_r c_rF_{r**}(e^r).}
$$
This is the requested final bound with the accompanying phase equations. It remains valid at any well-defined stationary choice; asserting existence of such a differentiable choice for arbitrary $W_0$ would require extra hypotheses. Without stationarity, the earlier $\mathcal B(E,\tau)$ is the correct general bound. In the concave closed excess-energy case $F_{r**}=F_r$; for general excess energies replacing the <concave biconjugate> by $F_r$ without proof would reverse the needed majorant logic.
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