Proof of The Cube Map is a Monotone Nonlinearity on the Sobolev Hilbert Triple of the Torus

Each result cited below is universally quantified over the data appearing in its own statement, and is applied here to the data named in the step in question.

Preliminary remarks.

(R1) For $y\in\mathbb{R}$ with $0\le y$ one has $|y|=y$: by claim 1 of Properties of the Absolute Value in an Ordered Field either $|y|=y$ or $|y|=-y$, and in the second case $0\le|y|=-y$ gives $y\le0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, whence $y=0$ by antisymmetry of the order and $|y|=-0=0=y$ by claim 4 of Additive Cancellation and Elementary Additive Identities in a Field.

(R2) For $y\in\mathbb{R}$ one has $0\le y^{2}$: by claim 1 of Properties of the Absolute Value in an Ordered Field either $|y|=y$ or $|y|=-y$, so in both cases $|y|\,|y|=y\,y=y^{2}$; and $0\le|y|$ by that claim, so claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative multiplier $|y|$ gives $0=0\,|y|\le|y|\,|y|$, the first equality by claim 1 of Zero Products and Elementary Identities in a Field.

(R3) For $a,b\in\mathbb{R}$ with $0\le a$ and $a\le b$ one has $a^{3}\le b^{3}$, by claim 5 of Properties of Natural Number Powers in a Field.

Claim 1. Let $U\in H^{1}(\mathbb{T}^{n})$ and let $u$ be a representative of $U$. By The Sobolev Embedding of the First Sobolev Space of the Torus into the Sixth Lebesgue Space in Dimensions at Most Three (ii), which applies because $n\le3$, we have $u\in\mathcal{L}^{6}(\mathbb{T}^{n})$ and $\lVert u\rVert_{6}\le5\,\lVert U\rVert_{H^{1}}$. Applying Elementary Properties of the p-Seminorm (vi) with $r=3$, $s=2$ and $h=u$, so that $0<r$, $1\le s$ and $1\le rs=6$, we conclude that $|u|^{3}$ is measurable, belongs to $\mathcal{L}^{2}(\mathbb{T}^{n})$, and satisfies

$$\bigl\lVert\,|u|^{3}\,\bigr\rVert_{2}=\bigl(\lVert u\rVert_{6}\bigr)^{3}.$$

The pointwise cube $u^{3}$ is measurable by claim 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, applied twice, and $|u^{3}(x)|=|u(x)|^{3}$ for every $x\in Q$ by claim 4 of Properties of the Absolute Value in an Ordered Field, applied twice. Hence Elementary Properties of the p-Seminorm (iii) gives $u^{3}\in\mathcal{L}^{2}(\mathbb{T}^{n})$ with $\lVert u^{3}\rVert_{2}\le\bigl\lVert|u|^{3}\bigr\rVert_{2}$. Since seminorms are nonnegative by Power-Integrable Functions and the p-Seminorm (iii), (R3) applied to $0\le\lVert u\rVert_{6}\le5\lVert U\rVert_{H^{1}}$ gives $(\lVert u\rVert_{6})^{3}\le(5\lVert U\rVert_{H^{1}})^{3}$, and claim 3 of Properties of Natural Number Powers in a Field together with $5^{3}=125$ gives $(5\lVert U\rVert_{H^{1}})^{3}=125\,(\lVert U\rVert_{H^{1}})^{3}$. Combining,

$$\lVert u^{3}\rVert_{2}\le125\,\bigl(\lVert U\rVert_{H^{1}}\bigr)^{3}.$$

If $\tilde{u}$ is a further representative of $U$ then $u$ and $\tilde{u}$ agree almost everywhere, hence so do $u^{3}$ and $\tilde{u}^{3}$, and $[u^{3}]=[\tilde{u}^{3}]$ by The Lebesgue Space Lᵖ(μ) (i). Since $L^{2}(\mathbb{T}^{n})$ is a real vector space by The Riesz–Fischer Theorem (i), the scalar multiple $b\,[u^{3}]$ lies in $L^{2}(\mathbb{T}^{n})$, so $B$ is a well-defined map from $H^{1}(\mathbb{T}^{n})$ to $L^{2}(\mathbb{T}^{n})$.

Claim 2. We first record a pointwise inequality. For $s,t\in\mathbb{R}$, expanding the products in the field $\mathbb{R}$,

$$(s-t)\,(s^{2}+st+t^{2})=s^{3}-t^{3},\qquad 4\,(s^{2}+st+t^{2})=(2s+t)^{2}+3\,t^{2},$$

where $2=1+1$, $3=2+1$ and $4=3+1$. By (R2) both $(2s+t)^{2}$ and $t^{2}$ are nonnegative, hence so is $3t^{2}$ and hence so is their sum, by claim 5 of Elementary Arithmetic in an Ordered Field and clause 2 of Ordered Field. Since $4$ is positive, its inverse $4^{-1}$ is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, so multiplying by $4^{-1}$ gives $0\le s^{2}+st+t^{2}$. Consequently, by (R2) and clause 2 of Ordered Field,

$$0\le(s-t)^{2}\,(s^{2}+st+t^{2})=(s-t)\,(s-t)\,(s^{2}+st+t^{2})=(s^{3}-t^{3})\,(s-t).$$

Now let $U,W\in H^{1}(\mathbb{T}^{n})$ with representatives $u$ and $w$. By the vector space operations of The Lebesgue Space Lᵖ(μ) (ii), $u-w$ is a representative of $U-W$ and $b\,(u^{3}-w^{3})$ is a representative of $B(U)-B(W)$. By claim 1 of L²(μ) Is a Real Hilbert Space (i) the pointwise product of two members of $\mathcal{L}^{2}(\mathbb{T}^{n})$ is integrable and

$$\bigl\langle B(U)-B(W),U-W\bigr\rangle_{L^{2}}=\int_{\mathbb{T}^{n}}b\,\bigl(u^{3}-w^{3}\bigr)\,(u-w)\,dx .$$

By the pointwise inequality just proved and $0\le b$, the integrand is nonnegative at every point of $Q$, by clause 2 of Ordered Field. Comparing it with the zero function, claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives that the integral is nonnegative. Thus $B$ is monotone.

Claim 3. Let $R$ and $U$ be as in the claim and let $u$ be a representative of $U$. By The Riesz–Fischer Theorem (i) the norm of $L^{2}(\mathbb{T}^{n})$ is absolutely homogeneous and $\lVert[v]\rVert_{L^{2}}=\lVert v\rVert_{2}$, so, using $|b|=b$ from (R1),

$$\lVert B(U)\rVert_{L^{2}}=b\,\lVert u^{3}\rVert_{2}\le b\cdot125\,\bigl(\lVert U\rVert_{H^{1}}\bigr)^{3},$$

the inequality by claim 1 together with claim 5 of Elementary Arithmetic in an Ordered Field and the nonnegative multiplier $b$. Norms are nonnegative by Real Inner Product Spaces (ii), so (R3) applied to $0\le\lVert U\rVert_{H^{1}}\le R$ gives $(\lVert U\rVert_{H^{1}})^{3}\le R^{3}$, and multiplying by the nonnegative number $125\,b$ yields $\lVert B(U)\rVert_{L^{2}}\le125\,b\,R^{3}$. As $R$ was an arbitrary positive real number and $125bR^{3}$ does not depend on $U$, $B$ is bounded on $V$-bounded sets.

Claim 4. Let $U\in D(A)$. By Hilbert Triples: Standing Notation and Background (iii) we have $U\in H^{1}(\mathbb{T}^{n})$, $AU\in L^{2}(\mathbb{T}^{n})$ and

$$\bigl\langle AU,Y\bigr\rangle_{L^{2}}=\bigl\langle U,Y\bigr\rangle_{H^{1}}\qquad\text{for every }Y\in H^{1}(\mathbb{T}^{n}).$$

Let $u$ be a representative of $U$, let $g_{j}$ be a representative of $\partial_{j}U$ for $j\in[n]$, and let $z$ be a representative of $AU$.

By Existence of a Sequence of Positive Real Numbers with Limit Zero there is a sequence $(\varepsilon_{m})_{m\in\mathbb{N}}$ of positive real numbers converging to $0$. Fix $m\in\mathbb{N}$ and let $\phi_{\varepsilon_{m}}$ be the function of A Bounded-Derivative Truncation of the Cube Map on the Real Line (i) for the parameter $\varepsilon_{m}$. By A Bounded-Derivative Truncation of the Cube Map on the Real Line (ii) it is of class $C^{1}$ on $\mathbb{R}^{1}$ with $\partial_{1}\phi_{\varepsilon_{m}}=\phi_{\varepsilon_{m}}'$, and by A Bounded-Derivative Truncation of the Cube Map on the Real Line (iii) together with (R1) its derivative satisfies $|\phi_{\varepsilon_{m}}'(s)|\le4\,\varepsilon_{m}^{-1}$ for every $s\in\mathbb{R}$. Hence The Chain Rule for Weak Derivatives on the Torus applies with $\phi=\phi_{\varepsilon_{m}}$ and $M=4\varepsilon_{m}^{-1}$, which is nonnegative by claim 7 of Elementary Order Arithmetic in an Ordered Field: by The Chain Rule for Weak Derivatives on the Torus (ii) the class $\phi_{\varepsilon_{m}}(U)=[\phi_{\varepsilon_{m}}\circ u]$ lies in $L^{2}(\mathbb{T}^{n})$, and by The Chain Rule for Weak Derivatives on the Torus (iii) it lies in $H^{1}(\mathbb{T}^{n})$ with $\partial_{j}\phi_{\varepsilon_{m}}(U)=[(\phi_{\varepsilon_{m}}'\circ u)\,g_{j}]$ for every $j\in[n]$.

Taking $Y=\phi_{\varepsilon_{m}}(U)$ in the displayed identity and expanding the Sobolev inner product by The Sobolev Space of Once Weakly Differentiable Square-Integrable Classes on the Torus (ii) and then each $L^{2}$ inner product by claim 1 of L²(μ) Is a Real Hilbert Space (i),

$$\bigl\langle AU,\phi_{\varepsilon_{m}}(U)\bigr\rangle_{L^{2}} =\int_{\mathbb{T}^{n}}u\,\bigl(\phi_{\varepsilon_{m}}\circ u\bigr)\,dx +\sum_{j=1}^{n}\int_{\mathbb{T}^{n}}\bigl(\phi_{\varepsilon_{m}}'\circ u\bigr)\,g_{j}\,g_{j}\,dx .$$

Every integrand here is nonnegative at every point of $Q$: the first by A Bounded-Derivative Truncation of the Cube Map on the Real Line (iv), and each of the others because $0\le\phi_{\varepsilon_{m}}'$ by A Bounded-Derivative Truncation of the Cube Map on the Real Line (iii) and $0\le g_{j}(x)^{2}$ by (R2), so that clause 2 of Ordered Field applies. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion each integral is therefore nonnegative, and by Finite Sums of Real Numbers: Nonnegativity, Domination by the Sum, and Limits (i) so is the finite sum. Hence

$$0\le\bigl\langle AU,\phi_{\varepsilon_{m}}(U)\bigr\rangle_{L^{2}}=\int_{\mathbb{T}^{n}}z\,\bigl(\phi_{\varepsilon_{m}}\circ u\bigr)\,dx ,$$

the equality again by claim 1 of L²(μ) Is a Real Hilbert Space (i).

We now let $m$ grow. Let $x\in Q$. By A Bounded-Derivative Truncation of the Cube Map on the Real Line (v),

$$\bigl|\phi_{\varepsilon_{m}}(u(x))-u(x)^{3}\bigr|\le\varepsilon_{m}\,|u(x)|^{5},$$

and the right-hand side converges to $0$ by Arithmetic of Limits of Real Sequences, being the constant $|u(x)|^{5}$ times a sequence converging to $0$; so claim 3 of Order Properties of Limits of Real Sequences shows that $(\phi_{\varepsilon_{m}}(u(x)))_{m\in\mathbb{N}}$ converges to $u(x)^{3}$, and Arithmetic of Limits of Real Sequences then shows that $(z(x)\,\phi_{\varepsilon_{m}}(u(x)))_{m\in\mathbb{N}}$ converges to $z(x)\,u(x)^{3}$. Moreover, by A Bounded-Derivative Truncation of the Cube Map on the Real Line (iv) and claim 4 of Properties of the Absolute Value in an Ordered Field,

$$\bigl|z(x)\,\phi_{\varepsilon_{m}}(u(x))\bigr|=|z(x)|\,\bigl|\phi_{\varepsilon_{m}}(u(x))\bigr|\le|z(x)|\,|u(x)|^{3}$$

for every $m$, using claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative multiplier $|z(x)|$. The functions $|z|$ and $|u|^{3}$ lie in $\mathcal{L}^{2}(\mathbb{T}^{n})$, the first by Elementary Properties of the p-Seminorm (iii) and the second by claim 1, so their pointwise product is integrable by claim 1 of L²(μ) Is a Real Hilbert Space (i). Each function $z\,(\phi_{\varepsilon_{m}}\circ u)$ is measurable, being a product of measurable functions by claim 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and so is $z\,u^{3}$. By Dominated Convergence Theorem,

$$\int_{\mathbb{T}^{n}}z\,\bigl(\phi_{\varepsilon_{m}}\circ u\bigr)\,dx\longrightarrow\int_{\mathbb{T}^{n}}z\,u^{3}\,dx .$$

Every term of the left-hand sequence is nonnegative, so comparing with the constant sequence $0$ and applying claim 1 of Order Properties of Limits of Real Sequences gives

$$0\le\int_{\mathbb{T}^{n}}z\,u^{3}\,dx=\bigl\langle AU,[u^{3}]\bigr\rangle_{L^{2}} .$$

Finally, by the symmetry and bilinearity of the inner product, Real Inner Product Spaces (i), and $0\le b$,

$$\bigl\langle B(U),AU\bigr\rangle_{L^{2}}=b\,\bigl\langle[u^{3}],AU\bigr\rangle_{L^{2}}=b\int_{\mathbb{T}^{n}}z\,u^{3}\,dx\ge0 ,$$

using claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative multiplier $b$. Thus $B$ is $A$-monotone.

Claim 5. By claim 1 the map $B$ sends $V=H^{1}(\mathbb{T}^{n})$ into $H=L^{2}(\mathbb{T}^{n})$; by claims 2, 3 and 4 it is monotone, bounded on $V$-bounded sets and $A$-monotone. By Monotone, $A$-Monotone and Locally Bounded Nonlinearities on a Hilbert Triple (iv), $B$ is a monotone nonlinearity for $(H,V,A)$.