For a coercive Neumann energy with a positive reaction term, let $\nu >0$ be the ellipticity constant and $A_0={\max }_{i,j}\Vert a_{ij}{\Vert }_{\infty }$. Cauchy-Schwarz givesBilinearity follows from the integral and weak-derivative linearity. Ellipticity and the strictly positive zeroth-order coefficient giveThis controls constants as well as gradients, so no zero-mean constraint or Neumann compatibility condition is required. Symmetry of the coefficients makes $B$ symmetric.

Write $F(v)={\int }_{\Omega }fv$. This is a bounded linear functional on $H^1$, with $|F(v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Consequentlyas the norm tends to infinity. The infimum is finite, and every minimizing sequence is bounded. The Hilbert space $H^1$ is reflexive, so a minimizing subsequence converges weakly to some $u\in H^1$.

To verify lower semicontinuity directly, put $w_j=v_j-u$. Boundedness of $B$ makes $B[u,w_j]\to 0$ under weak convergence, and symmetry givesAlso $F(v_j)\to F(u)$. Therefore $I(u)\le \mathrm{liminf}I(v_j)$, so the absolute minimum is attained.

For every $h\in H^1$ the real function $t\mapsto I(u+th)$ has a minimum at zero. Its derivative there is $B[u,h]-F(h)$, yielding the required weak equation for every $h$. If two weak solutions exist, their difference $z$ satisfies $B[z,h]=0$ for all $h$; choose $h=z$ and use coercivity to obtain $z=0$. In fact the weak equation also gives the exact energy identityThus the minimizer and weak solution are both unique. The natural boundary condition is encoded by testing in the whole $H^1$ space, without needing pointwise conormal derivatives for rough coefficients.

Suppose $\sigma \ne e$. Choose a symbol $i$ moved by $\sigma$ and write $j=\sigma (i)\ne i$. Since $n\ge 3$, choose a third symbol $k$ distinct from $i,j$. Let $\tau =(jk)$, a transposition that fixes $i$. ThenThus ${\tau }^{-1}\sigma \tau \ne \sigma$, proving that the conjugacy class of any nonidentity permutation has more than one element. By part (a)(iii), no such permutation lies in the center of a group. The identity always does, soThe restriction on $n$ is necessary: $S_2$ is an abelian group and equals its own center.

For the given symmetric matrix, $(1,1{)}^T$ is an eigenvector with eigenvalue $1$, and $(1,-1{)}^T$ is an eigenvector with eigenvalue $27$. Normalize them to obtainTake the real cube roots of the two diagonal entries and transform back:It is a real symmetric matrix, and $C^3=P\mathrm{diag}(1,27)P^T=A$ because $P^{T}P=I$. Direct multiplication also gives $C^2=\left(\begin{array}{cc}5& -4\\ -4& 5\end{array}\right)$ and then $C^3=\left(\begin{array}{cc}14& -13\\ -13& 14\end{array}\right)$.

Suppose $A^T=A$. For eigenvectors $v_i,v_j$ with distinct real eigenvalues, symmetry givesThus $({\lambda }_i-{\lambda }_j)v_i^T{v}_j=0$, and the eigenvectors are orthogonal. Divide each by its nonzero Euclidean norm to obtain an orthonormal basis. The matrix $P$ having these normalized columns satisfies $P^{T}P=I$, so it is an orthogonal matrix and $P^{-1}=P^T$. As in part (a),This proves the needed orthogonal diagonalization from symmetry and distinct eigenvalues, without invoking an unproved spectral theorem.