Proof of Pairing an Integrable Function on the Torus with a Continuous Periodic Function

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 statement of this lemma. We use throughout that a map $v:Q\to\mathbb{R}$ belongs to $\mathcal{L}^{1}(\mathbb{T}^{n})$ exactly when it is measurable with respect to $\mathcal{B}_{Q}$ and $\int_{Q}|v|\,d\lambda_{Q}$ is finite, and that it is then integrable with respect to $\lambda_{Q}$: membership requires by Power-Integrable Functions and the p-Seminorm §space that $v$ be measurable and that the integral of $|v|^{1}$ be finite, and $|v|^{1}=|v|$ by Properties of Real Powers of Nonnegative Real Numbers §agreement.

Step 1. Proof of claim 1.

Let $w\in\mathcal{L}^{1}(\mathbb{T}^{n})$ and let $M$ be a real number with $0\le M$ and $|\varphi(x)|\le M$ for every $x\in\mathbb{R}^{n}$. The product $w\,(\varphi|_{Q})$ is measurable with respect to $\mathcal{B}_{Q}$ by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, since $w$ and $\varphi|_{Q}$ are; and $|w\,(\varphi|_{Q})|$ and $|w|$ are measurable by claim 4 of that lemma. For every $y\in Q$,

the equality by claim 4 of Properties of the Absolute Value in an Ordered Field and the inequality by claim 5 of Elementary Arithmetic in an Ordered Field, applied with the nonnegative factor $|w(y)|$. Since $w$ is integrable, $\int_{Q}|w|\,d\lambda_{Q}$ is a real number, so claim 1 of Linearity and Monotonicity of the Lebesgue Integral, used for monotonicity and for the scalar multiple, gives

Hence $w\,(\varphi|_{Q})\in\mathcal{L}^{1}(\mathbb{T}^{n})$, as recorded above.

For the bound, put $f=w\,(\varphi|_{Q})$, an integrable map. By claim 3 of Properties of the Absolute Value in an Ordered Field we have $f(y)\le|f(y)|$ and $-f(y)\le|f(y)|$ for every $y\in Q$. The maps $|f|$ and $-f$ are integrable, the second by claim 2 of Linearity and Monotonicity of the Lebesgue Integral with the scalar $-1$, so that claim, used for monotonicity and again for the scalar $-1$, gives

By claim 6 of Properties of the Absolute Value in an Ordered Field the two inequalities together give $\bigl|\int_{Q}f\,d\lambda_{Q}\bigr|\le\int_{Q}|f|\,d\lambda_{Q}$, and combining this with the previous display proves the stated bound.

Step 2. Proof of claim 2.

The memberships $w+c\,w'\in\mathcal{L}^{1}(\mathbb{T}^{n})$ and $\varphi+c\,\varphi'\in C_{\mathrm{per}}$ are the references recorded in the claim. At every $y\in Q$,

by distributivity and commutativity in the field of real numbers, the restriction to $Q$ of a pointwise sum or scalar multiple being the corresponding combination of the restrictions. By claim 1, applied to $w$ and to $w'$ with the map $\varphi$, and to $w$ with the map $\varphi'$, every product appearing here is integrable. Claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied to each of the two identities, therefore yields the two displayed equations of claim 2.

Step 3. Proof of claim 3.

By hypothesis and The Lebesgue Space of Power-Integrable Functions §equivalence there is a $\lambda_{Q}$-null set $N$ such that $w(y)=w'(y)$ for every $y\in Q$ with $y\notin N$. For such $y$ the products satisfy $w(y)\,\varphi(y)=w'(y)\,\varphi(y)$, so $w\,(\varphi|_{Q})=w'\,(\varphi|_{Q})$ $\lambda_{Q}$-almost everywhere on $Q$, with the same null set $N$. Both products are measurable and integrable by claim 1, so The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, applied to the pair of them, gives that their integrals over $\mathbb{T}^{n}$ coincide.

Step 4. Proof of claim 4.

Put $v=w-w'$, a member of $\mathcal{L}^{1}(\mathbb{T}^{n})$ by Minkowski's Inequality and the Seminormed Space of Power-Integrable Functions §vector-space, used with the scalar $-1$, and hence integrable with respect to $\lambda_{Q}$. Every $\varphi\in C^{\infty}_{\mathrm{per}}$ lies in $C_{\mathrm{per}}$, as recorded in the statement of this lemma, so claim 2, used with the scalar $-1$, together with the hypothesis gives

By The Fundamental Lemma of the Calculus of Variations on the Torus §vanishing it follows that $v=0$ $\lambda_{Q}$-almost everywhere on $Q$, that is, $w=w'$ $\lambda_{Q}$-almost everywhere on $Q$. Hence $[w]=[w']$ in $L^{1}(\mathbb{T}^{n})$ by The Lebesgue Space of Power-Integrable Functions §equivalence; and if $w$ and $w'$ lie in $\mathcal{L}^{q}(\mathbb{T}^{n})$, the same almost-everywhere equality gives $[w]=[w']$ in $L^{q}(\mathbb{T}^{n})$ by that same clause, applied to the exponent $q$. $\blacksquare$