Proof of The Fundamental Lemma of the Calculus of Variations on 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 statement of this lemma.

Step 1. Proof of claim 1.

Let $v:Q\to\mathbb{R}$ be integrable with respect to $\lambda_{Q}$ and satisfy the hypothesis. By Integrable Function and the Lebesgue Integral the map $v$ is measurable with respect to $\mathcal{B}_{Q}$ and $\int_{Q}|v|\,d\lambda_{Q}<\infty$. Since $|v|^{1}=|v|$ by Properties of Real Powers of Nonnegative Real Numbers §agreement, the integral of $|v|^{1}$ is finite as well, so $v\in\mathcal{L}^{1}(\mathbb{T}^{n})$ by Power-Integrable Functions and the p-Seminorm §space.

By Existence of Mollifier Kernels of Every Radius there is a mollifier kernel $\rho$ of radius $1$ on $\mathbb{R}^{n}$. Let $\varepsilon$ be a real number with $0<\varepsilon$ and let $\rho_{\varepsilon}$ be its rescaling, a mollifier kernel of radius $\varepsilon$ by that lemma. By Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ the map $\rho_{\varepsilon}$ is smooth on $\mathbb{R}^{n}$, hence continuous by claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous, and vanishes at every $y$ with $\varepsilon<\lVert y\rVert$. So The Periodised Kernel of a Periodic Convolution applies with the kernel $\rho_{\varepsilon}$ and the radius $\varepsilon$; let $\Psi$ be its periodised kernel and, for $x\in\mathbb{R}^{n}$, let $\Psi_{x}$ be the restriction of $\Psi(x,\cdot)$ to $Q$.

Fix $x\in\mathbb{R}^{n}$. Since $\rho_{\varepsilon}$ is smooth, The Periodised Kernel of a Periodic Convolution §regularity gives $\Psi(x,\cdot)\in C^{\infty}_{\mathrm{per}}$. Taking $\varphi=\Psi(x,\cdot)$ in the hypothesis, whose restriction to $Q$ is $\Psi_{x}$, we get

On the other hand The Periodised Kernel of a Periodic Convolution §representation, applied to $v\in\mathcal{L}^{1}(\mathbb{T}^{n})$, gives

which is the same number, the pointwise products $v\,\Psi_{x}$ and $\Psi_{x}\,v$ being equal. Hence $(\rho_{\varepsilon}\star v)(x)=0$. As $x$ was arbitrary, $(\rho_{\varepsilon}\star v)|_{Q}$ is the map on $Q$ with constant value $0$, and therefore

the last equality by Elementary Properties of the p-Seminorm §homogeneous with the scalar $-1$, together with claim 2 of Properties of the Absolute Value in an Ordered Field. This holds for every positive real $\varepsilon$.

Let $\eta$ be a real number with $0<\eta$. By The Lebesgue Bound for Periodic Convolution and Convergence of Mollifications §convergence, used with the exponent $1$, the kernel $\rho$ of radius $1$ and the function $v$, there is a positive real $\varepsilon_{0}$ such that $\lVert(\rho_{\varepsilon}\star v)|_{Q}-v\rVert_{1}\le\eta$ for every real $\varepsilon$ with $0<\varepsilon<\varepsilon_{0}$. Choosing one such $\varepsilon$ — for instance $\varepsilon_{0}/2$, positive and smaller than $\varepsilon_{0}$ by claim 8 of Elementary Order Arithmetic in an Ordered Field — and combining with the previous display gives $\lVert v\rVert_{1}\le\eta$.

Thus $\lVert v\rVert_{1}$ is a nonnegative real number not exceeding any positive real, so $\lVert v\rVert_{1}=0$ by Comparison of Real Numbers with Arbitrary Positive Slack §vanishing. By Power-Integrable Functions and the p-Seminorm §seminorm we have $\lVert v\rVert_{1}=\bigl(\int_{Q}|v|^{1}\,d\lambda_{Q}\bigr)^{1/1}$, and $|v|^{1}=|v|$ together with $t^{1}=t$ for nonnegative $t$ by Properties of Real Powers of Nonnegative Real Numbers §agreement; hence $\int_{Q}|v|\,d\lambda_{Q}=0$. Since $|v|$ is measurable and nonnegative, The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing gives $|v(y)|=0$ for almost every $y\in Q$, that is, $v=0$ $\lambda_{Q}$-almost everywhere on $Q$, by claim 1 of Properties of the Absolute Value in an Ordered Field.

Step 2. Proof of claim 2.

Let $u\in\mathcal{L}^{p}(\mathbb{T}^{n})$ satisfy the hypothesis of claim 2. By The Periodic Extension of a Function on the Unit Cell §finite-measure the map $u$ is integrable with respect to $\lambda_{Q}$, so claim 1 applies to $u$ and gives $u=0$ $\lambda_{Q}$-almost everywhere on $Q$.

The map on $Q$ with constant value $0$ lies in $\mathcal{L}^{p}(\mathbb{T}^{n})$ and its class is the zero element of $L^{p}(\mathbb{T}^{n})$, by The Lebesgue Space of Power-Integrable Functions §space. Since $u$ agrees with it almost everywhere, $u$ and it lie in the same class under almost-everywhere equality, so $[u]$ is that zero element. $\blacksquare$