Choose the real zero-trace Sobolev space
The Poincaré inequality makes this an equivalent Sobolev norm. Since the dimension is five, the Sobolev conjugate exponent is . The Sobolev embedding theorem gives continuously. Boundedness of and Lp inclusion on a finite measure space then give continuously, say .
For , the weak formulation is
The zero Dirichlet boundary condition is imposed through the Sobolev trace. Hölder's inequality verifies every term:
Thus , while , and all three pairings are finite bounded linear functionals in . The choice is sufficient; it is not intended to be the largest possible space of data.
For the spaces in part (b)(i) and fixed , define by
For each , this is a linear functional in , and the estimates in part (b)(i) show
Hence belongs to the continuous dual space , and sends bounded sets to bounded sets. The weak formulation is exactly
The forcing term is included in so that its right side is zero, as requested.
For , the forcing terms cancel, giving
The scalar function is strictly increasing on all of , so the second integrand is nonnegative. For the first one, use the quartic gradient monotonicity inequality. Put and for real vectors . Expanding gives
Consequently
The last strict inequality uses the Poincaré inequality: if the weak gradient of vanishes, then . This proves that is a strictly monotone operator.

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