Proof of Integrability of Products of Four Sobolev Classes on the Torus in Dimensions at Most Three

Each result cited is universally quantified over the data in its own statement and is applied here to the data at hand; the notation is that of the statement, and the measure space throughout is $(Q,\mathcal{B}_{Q},\lambda_{Q})$ of The Flat Torus: Standing Notation §measure. Finite products of real numbers are the finite products of that definition, so that $\prod_{k=1}^{4}a_{k}$ is the iterated product $((a_{1}a_{2})a_{3})a_{4}$ by the recursion defining it.

Step 0 (A product comparison). Let $\alpha,\beta,\gamma,\delta\in\mathbb{R}$ satisfy $0\le\alpha$, $\alpha\le\beta$, $0\le\gamma$ and $\gamma\le\delta$. Then $\alpha\gamma\le\beta\delta$. Indeed, claim 5 of Elementary Arithmetic in an Ordered Field, applied to $\alpha\le\beta$ with the nonnegative multiplier $\gamma$, gives $\gamma\alpha\le\gamma\beta$; the number $\beta$ is nonnegative by transitivity of the order of The Real Numbers: Standing Notation and Background §numbers applied to $0\le\alpha$ and $\alpha\le\beta$, so the same claim applied to $\gamma\le\delta$ with the multiplier $\beta$ gives $\beta\gamma\le\beta\delta$; and transitivity together with commutativity of multiplication gives $\alpha\gamma\le\beta\delta$. We record also that a product of two nonnegative real numbers is nonnegative: if $0\le\alpha$ and $0\le\gamma$, then $\alpha\cdot0\le\alpha\gamma$ by claim 5 of Elementary Arithmetic in an Ordered Field, while $\alpha\cdot0=0\cdot\alpha=0$ by commutativity and claim 1 of Zero Products and Elementary Identities in a Field, so $0\le\alpha\gamma$.

Step 1 (Membership in $\mathcal{L}^{1}(\mathbb{T}^{n})$ and integrability). Let $f:Q\to\mathbb{R}$ be measurable. By Properties of Real Powers of Nonnegative Real Numbers §agreement the power of a nonnegative real number with exponent $1$ is that number itself, so the map $|f|^{1}$ of Power-Integrable Functions and the p-Seminorm §measurable-power is $|f|$. Hence, by Power-Integrable Functions and the p-Seminorm §space, $f\in\mathcal{L}^{1}(\mathbb{T}^{n})$ if and only if $\int_{\mathbb{T}^{n}}|f|\,dx<\infty$, which by the last sentence of Integrable Function and the Lebesgue Integral holds if and only if $f$ is integrable; and in that case $\lVert f\rVert_{1}=\int_{\mathbb{T}^{n}}|f|\,dx$ by Power-Integrable Functions and the p-Seminorm §seminorm, since $1/1=1$ and the power with exponent $1$ is the identity by the clause just cited.

Claim 1. Let $U_{1},U_{2},U_{3},U_{4}$ and their representatives $u_{1},u_{2},u_{3},u_{4}$ be as in the claim.

(a) Since $1\le n$ and $n\le3$, The Sobolev Embedding of the First Sobolev Space of the Torus into the Sixth Lebesgue Space in Dimensions at Most Three §embedding, applied for each $k$ to the class $U_{k}$ and its representative $u_{k}$, gives $u_{k}\in\mathcal{L}^{6}(\mathbb{T}^{n})$ and $\lVert u_{k}\rVert_{6}\le5\,\lVert U_{k}\rVert_{H^{1}}$.

(b) The real numbers $4$ and $6$ satisfy $1\le4$ and $4\le6$, so Comparison of the Lebesgue Seminorms on the Torus §comparison, applied with $r=4$, $s=6$ and $v=u_{k}$, gives $u_{k}\in\mathcal{L}^{4}(\mathbb{T}^{n})$ and $\lVert u_{k}\rVert_{4}\le\lVert u_{k}\rVert_{6}$. With (a) and transitivity of the order, $\lVert u_{k}\rVert_{4}\le5\,\lVert U_{k}\rVert_{H^{1}}$ for each $k$.

(c) The real number $4$ is positive, hence nonzero and invertible, by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. Repeated use of the recursion in claim 1 of Properties of Finite Sums together with claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field gives $\sum_{k=1}^{4}1=4$, whence by claim 3 of the former

Therefore Hoelder's Inequality, for Two and for Finitely Many Factors §several, applied to the measure space $(Q,\mathcal{B}_{Q},\lambda_{Q})$ with the number of factors — written $n$ in that clause, and not to be confused with the dimension $n$ of the present statement — taken to be $4$, with $p_{k}=4$ and $f_{k}=u_{k}$ for $k\in[4]$, shows that the pointwise product $F=u_{1}u_{2}u_{3}u_{4}$ lies in $\mathcal{L}^{1}(\mathbb{T}^{n})$ and

(d) Seminorms are nonnegative by Power-Integrable Functions and the p-Seminorm §seminorm, norms are nonnegative by Real Inner Product Space §norm, and $5=1+1+1+1+1$ is nonnegative by claims 1 and 2 of Elementary Arithmetic in an Ordered Field; hence each $5\lVert U_{k}\rVert_{H^{1}}$ is nonnegative by the product remark of Step 0, and so is each partial product of the numbers $\lVert u_{k}\rVert_{4}$ and each partial product of the numbers $5\lVert U_{k}\rVert_{H^{1}}$. Three applications of Step 0 to the four comparisons of (b), the hypotheses $0\le\alpha$ and $0\le\gamma$ being supplied at each application by these nonnegativity facts, give

and the right-hand side equals $625\,\lVert U_{1}\rVert_{H^{1}}\lVert U_{2}\rVert_{H^{1}}\lVert U_{3}\rVert_{H^{1}}\lVert U_{4}\rVert_{H^{1}}$ by associativity and commutativity of multiplication together with claim 1 of Properties of Natural Number Powers in a Field, which gives $((5\cdot5)\cdot5)\cdot5=5^{4}=625$.

(e) By Step 1 the function $F$ is integrable, so claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives $\bigl|\int_{\mathbb{T}^{n}}F\,dx\bigr|\le\int_{\mathbb{T}^{n}}|F|\,dx$. Combining this with (c) and (d) and using transitivity of the order yields the asserted bound.

(f) Let $u'_{1},u'_{2},u'_{3},u'_{4}$ be any representatives of $U_{1},U_{2},U_{3},U_{4}$. For each $k$ one has $u'_{k}=u_{k}$ almost everywhere by The Lebesgue Space of Power-Integrable Functions §equivalence. Applying the last sentence of claim 1 of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere to the sequence of properties $Q_{m}$ ($m\in\mathbb{N}$), where $Q_{m}$ is the property $u'_{k}(y)=u_{k}(y)$ with $k$ the lesser of $m$ and $4$, we conclude that almost every $y\in Q$ satisfies $u'_{k}(y)=u_{k}(y)$ for all $k\in[4]$ simultaneously; at every such $y$ the pointwise products $u'_{1}u'_{2}u'_{3}u'_{4}$ and $F$ agree. The product $u'_{1}u'_{2}u'_{3}u'_{4}$ is measurable by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Hence the last sentence of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, applied with the integrable $F$ and this product, shows that the latter is integrable with the same integral. This proves claim 1.

Claim 2. Let $U$ and $u$ be as in the claim. By claim 1 of Properties of Natural Number Powers in a Field, for every $y\in Q$,

so the pointwise fourth power $u^{4}$ is the pointwise product of the four maps $u,u,u,u$. Claim 1, applied with $U_{1}=U_{2}=U_{3}=U_{4}=U$ and $u_{1}=u_{2}=u_{3}=u_{4}=u$, therefore gives $u^{4}\in\mathcal{L}^{1}(\mathbb{T}^{n})$, the independence of $\int_{\mathbb{T}^{n}}u^{4}\,dx$ from the representative chosen, and

the last equality again by claim 1 of Properties of Natural Number Powers in a Field.

Claim 3. Let $U,W$ and their representatives $u,w$ be as in the claim.

(a) By (a) of claim 1, $u\in\mathcal{L}^{6}(\mathbb{T}^{n})$. The real numbers $r=3$ and $s=2$ satisfy $0<3$, $1\le2$ and $1\le6$, and $rs=6$; so claim 6 of Elementary Properties of the p-Seminorm, applied to the measurable map $u$ on $(Q,\mathcal{B}_{Q},\lambda_{Q})$, gives that the map $|u|^{3}:y\mapsto(|u(y)|)^{3}$ is measurable and lies in $\mathcal{L}^{2}(\mathbb{T}^{n})$.

(b) The pointwise cube $u^{3}$ is measurable by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and for every $y\in Q$

by claim 1 of Properties of Natural Number Powers in a Field and two applications of claim 4 of Properties of the Absolute Value in an Ordered Field. Hence claim 3 of Elementary Properties of the p-Seminorm, applied with $p=2$, with $f=u^{3}$ and with $g=|u|^{3}$, gives $u^{3}\in\mathcal{L}^{2}(\mathbb{T}^{n})$.

(c) If $u'$ is any representative of $U$, then $u'=u$ almost everywhere by The Lebesgue Space of Power-Integrable Functions §equivalence, so $(u')^{3}=u^{3}$ at every point where $u'=u$, hence almost everywhere; by (b) applied to $u'$ and by the same clause, $(u')^{3}\in\mathcal{L}^{2}(\mathbb{T}^{n})$ and $[(u')^{3}]=[u^{3}]$. Thus $[u^{3}]$ does not depend on the representative.

(d) $H^{1}(\mathbb{T}^{n})$ is a subset of $L^{2}(\mathbb{T}^{n})$ by The Sobolev Space of Once Weakly Differentiable Square-Integrable Classes on the Torus §space, so $W\in L^{2}(\mathbb{T}^{n})$ and its representative $w$ lies in $\mathcal{L}^{2}(\mathbb{T}^{n})$ with $[w]=W$. The real numbers $2$ and $2$ are conjugate exponents, since $1<2$ and $\tfrac{1}{2}+\tfrac{1}{2}=1$; hence Hoelder's Inequality, for Two and for Finitely Many Factors §holder, applied with $p=q=2$, $f=u^{3}$ and $g=w$, gives $u^{3}w\in\mathcal{L}^{1}(\mathbb{T}^{n})$.

(e) Finally The Lebesgue Space of Square-Integrable Functions is a Real Hilbert Space §inner-product, applied to the measure space $(Q,\mathcal{B}_{Q},\lambda_{Q})$ and to $u^{3},w\in\mathcal{L}^{2}(\mathbb{T}^{n})$, gives

which is claim 3.