= Solution
{c}
Test with $v_h=u_h^{m+1}$ and apply $2(a-b,a)=\|a\|^2-\|b\|^2+\|a-b\|^2$. The exact energy identity is
$$
\frac{\|u_h^{m+1}\|_2^2-\|u_h^m\|_2^2+\|u_h^{m+1}-u_h^m\|_2^2}{2\Delta t}
+\|(u_h^{m+1})'\|_2^2+\|u_h^{m+1}\|_2^2=0.
$$
The printed identity omits the last reaction contribution while explicitly describing its last term as the $H^1$ seminorm. \b[It requires the full $H^1$ <norm> instead of that seminorm.] With the printed seminorm its left side equals $-\|u_h^{m+1}\|_2^2$, rather than zero. The source reaction term forces this correction.
All terms other than the difference of successive squared <norms> are nonnegative. Consequently
$$
\boxed{\|u_h^m\|_2\leq\|u_h^0\|_2\leq\|u_0\|_2.}
$$
The second inequality uses contractivity of the initial <orthogonal projection>. No restriction relates $\Delta t$ to $h$, giving unconditional stability. The retained reaction term strengthens dissipation.
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