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We use the homogeneous version of the Sobolev embedding theorem in three dimensions:Here homogeneous Sobolev space ${\dot{H}}^1$ is the completion of compactly supported smooth functions in the L2 norm of the gradient, identified with its $L^6$ representative. Apply this Sobolev inequality both to $\varphi$ and to its spatial derivatives.

LetThe Plancherel theorem identifies the L2 norm of the Hessian matrix with $\Vert \varphi {\Vert }_{{\dot{H}}^2}$, because ${\sum }_{i,j}{\xi }_i^2{\xi }_j^2=|\xi {|}^4$. In particular,Differentiating the defocusing semilinear wave equation gives $\square ({\partial }_i\varphi )=3{\varphi }^2{\partial }_i\varphi$. By the Holder inequality with exponents $3$ and $6$, the preceding Sobolev inequality, and conservation of the positive wave energy,Use the inhomogeneous wave energy estimate simultaneously for the three spatial derivatives. It givesThe Gronwall inequality therefore yields $B(t)\le D\exp (CEt)$. We may take the continuous, locally bounded functionThe constant is universal; the dependence on the initial data is only through $D$ and $E$. The H2 bound for the defocusing cubic wave equation holds on every existing smooth interval, without assuming the global conclusion. If $D=0$ or $E=0$, the zero solution satisfies the same estimate.

Under spatial reflection, a scalar field obeys $P\varphi (\mathbf{x})P=\varphi (-\mathbf{x})$, and hence each spatial derivative changes sign. The interaction contains three such derivatives, so changing the integration variable from $\mathbf{x}$ to $-\mathbf{x}$ givesThus this is a parity-odd scalar interaction.

Put $O({\mathbf{k}}_a)={\prod }_{a=1}^n{\varphi }_a({\mathbf{k}}_a)$ and $A({\mathbf{k}}_a)=⟨H_{\mathrm{int}}O({\mathbf{k}}_a)⟩$. Hermitian conjugation and commutativity of equal-time scalar fields giveThe parity-invariant vacuum and the odd parity of $H_{\mathrm{int}}$ imply $A(-{\mathbf{k}}_a)=-A({\mathbf{k}}_a)$. ThereforeThere is no conflict with the usual $2i\mathrm{Im}A$ rule for two Hermitian operators: a momentum-space product at fixed ${\mathbf{k}}_a$ is generally not itself Hermitian, since its adjoint carries momenta $-{\mathbf{k}}_a$.