Proof of The Square-Integrable and Sobolev Spaces of the Torus Form a Hilbert Triple

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

Proof of claim 1. The measure space $(Q,\mathcal{B}_{Q},\lambda_{Q})$ is the one fixed in The Flat Torus: Standing Notation (iii). By L²(μ) Is a Real Hilbert Space (ii), applied to that measure space, $L^{2}(\mathbb{T}^{n})$ with the inner product $\langle\,\cdot\,,\cdot\,\rangle_{L^{2}}$ of L²(μ) Is a Real Hilbert Space (i) is a real Hilbert space; by that clause its norm and distance are those fixed in The Flat Torus: Standing Notation (iv).

By The Sobolev Space of Once Weakly Differentiable Square-Integrable Classes on the Torus (i), $H^{1}(\mathbb{T}^{n})$ is a linear subspace of $L^{2}(\mathbb{T}^{n})$, and by The Sobolev Space of Once Weakly Differentiable Square-Integrable Classes on the Torus (ii) the map $\langle\,\cdot\,,\cdot\,\rangle_{H^{1}}$ is an inner product on it, with distance $d_{H^{1}}$. By The Sobolev Space on the Torus is a Separable Hilbert Space Densely Embedded in the Square-Integrable Space (ii), $H^{1}(\mathbb{T}^{n})$ with that inner product is a real Hilbert space.

We check the two conditions of Hilbert Triple: a Densely and Continuously Embedded Hilbert Space and Its Form Operator (i) for $H=L^{2}(\mathbb{T}^{n})$ and $V=H^{1}(\mathbb{T}^{n})$. Condition Hilbert Triple: a Densely and Continuously Embedded Hilbert Space and Its Form Operator (ii) requires that the $L^{2}$ norm of a member of $H^{1}(\mathbb{T}^{n})$ be at most its $H^{1}$ norm; this is the first inequality of The Sobolev Space on the Torus is a Separable Hilbert Space Densely Embedded in the Square-Integrable Space (i). Condition Hilbert Triple: a Densely and Continuously Embedded Hilbert Space and Its Form Operator (iii) requires that $H^{1}(\mathbb{T}^{n})$ be dense in $L^{2}(\mathbb{T}^{n})$; this is The Sobolev Space on the Torus is a Separable Hilbert Space Densely Embedded in the Square-Integrable Space (ix).

Hence Hilbert Triple: a Densely and Continuously Embedded Hilbert Space and Its Form Operator (i) applies to these data, and Hilbert Triple: a Densely and Continuously Embedded Hilbert Space and Its Form Operator (iv) furnishes the set $D(A)\subseteq H^{1}(\mathbb{T}^{n})$ and the map $A:D(A)\to L^{2}(\mathbb{T}^{n})$ with $\langle X,Y\rangle_{H^{1}}=\langle AX,Y\rangle_{L^{2}}$ for all $X\in D(A)$ and $Y\in H^{1}(\mathbb{T}^{n})$, so that $L^{2}(\mathbb{T}^{n})$, $H^{1}(\mathbb{T}^{n})$ with $\langle\,\cdot\,,\cdot\,\rangle_{H^{1}}$, and $A$ form a Hilbert triple.

Finally $(H^{1}(\mathbb{T}^{n}),d_{H^{1}})$ is separable by The Sobolev Space on the Torus is a Separable Hilbert Space Densely Embedded in the Square-Integrable Space (viii). Since $V=H^{1}(\mathbb{T}^{n})$ and $d_{V}=d_{H^{1}}$ for this triple, that is precisely the standing hypothesis Hilbert Triples: Standing Notation and Background (ii).

Proof of claim 2. Let $u\in C^{2}_{\mathrm{per}}$. The memberships $u\in C^{1}_{\mathrm{per}}$, $u\in C_{\mathrm{per}}$, $\partial_{i}\partial_{i}u\in C_{\mathrm{per}}$ for $i\in[n]$, $\Delta u\in C_{\mathrm{per}}$, $u-\Delta u\in C_{\mathrm{per}}$, $(u-\Delta u)|_{Q}\in\mathcal{L}^{2}(\mathbb{T}^{n})$ and $[u|_{Q}]\in H^{1}(\mathbb{T}^{n})$ are justified in the statement by the references attached there; for $\Delta u$ we note in addition that $\Delta u$ is by The Laplacian of a Twice Continuously Differentiable Function (i) the map whose value at $x$ is the finite sum $\sum_{i=1}^{n}\partial_{i}\partial_{i}u(x)$, and that a finite sum of members of $C_{\mathrm{per}}$ lies in $C_{\mathrm{per}}$ by induction on the upper summation index from Elementary Properties of Lattice-Periodic Functions (i), the base case being a single member of $C_{\mathrm{per}}$.

Write $U=[u|_{Q}]$. Since $u\in C^{1}_{\mathrm{per}}$, Elementary Properties of the Weak Partial Derivative on the Torus (i) gives $\partial_{j}U=[(\partial_{j}u)|_{Q}]$ for every $j\in[n]$, this class lying in $L^{2}(\mathbb{T}^{n})$ by that clause.

Let $V\in H^{1}(\mathbb{T}^{n})$, let $v$ be a representative of $V$ and, for $j\in[n]$, let $h_{j}$ be a representative of $\partial_{j}V$. By The Sobolev Space of Once Weakly Differentiable Square-Integrable Classes on the Torus (ii),

$$\langle U,V\rangle_{H^{1}}=\langle U,V\rangle_{L^{2}}+\sum_{j=1}^{n}\bigl\langle[(\partial_{j}u)|_{Q}],\partial_{j}V\bigr\rangle_{L^{2}} .$$

Fix $j\in[n]$. Since $u\in C^{2}_{\mathrm{per}}$, we have $\partial_{j}u\in C^{1}_{\mathrm{per}}$ by Elementary Properties of Lattice-Periodic Functions (ii). Applying Elementary Properties of the Weak Partial Derivative on the Torus (iii) to the class $V$, its $j$-th weak partial derivative $\partial_{j}V$, the representatives $v$ and $h_{j}$, and the test function $\varphi=\partial_{j}u\in C^{1}_{\mathrm{per}}$,

$$\int_{\mathbb{T}^{n}}v\,\bigl((\partial_{j}\partial_{j}u)|_{Q}\bigr)\,dx=-\int_{\mathbb{T}^{n}}h_{j}\,\bigl((\partial_{j}u)|_{Q}\bigr)\,dx .$$

By L²(μ) Is a Real Hilbert Space (i) the right-hand integral is $\bigl\langle[(\partial_{j}u)|_{Q}],\partial_{j}V\bigr\rangle_{L^{2}}$, the pointwise product being commutative, so

$$\bigl\langle[(\partial_{j}u)|_{Q}],\partial_{j}V\bigr\rangle_{L^{2}}=-\int_{\mathbb{T}^{n}}v\,\bigl((\partial_{j}\partial_{j}u)|_{Q}\bigr)\,dx .$$

For each $j\in[n]$ put $f_{j}=v\cdot\bigl((\partial_{j}\partial_{j}u)|_{Q}\bigr)$. Since $\partial_{j}\partial_{j}u\in C_{\mathrm{per}}$, its restriction to $Q$ lies in $\mathcal{L}^{2}(\mathbb{T}^{n})$ by Continuous Periodic Functions are Power-Integrable and Dense on the Torus (i), and $v\in\mathcal{L}^{2}(\mathbb{T}^{n})$; so $f_{j}$ is integrable by L²(μ) Is a Real Hilbert Space (i). Applying Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions (i) and Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions (ii) with $m=n$, these maps $f_{j}$ and the coefficients $c_{j}=-1$,

$$\sum_{j=1}^{n}\Bigl(-\int_{\mathbb{T}^{n}}f_{j}\,dx\Bigr)=\int_{\mathbb{T}^{n}}\Bigl(\sum_{j=1}^{n}(-1)\,f_{j}\Bigr)dx ,$$

the map $x\mapsto\sum_{j=1}^{n}(-1)f_{j}(x)$ being integrable. By claims 1 and 4 of Properties of a Sum over a Finite Index Set, applied to the map $j\mapsto\partial_{j}\partial_{j}u(x)$ on $[n]$ with the scalar $-v(x)$, we have for every $x\in Q$

$$\sum_{j=1}^{n}(-1)f_{j}(x)=\bigl(-v(x)\bigr)\sum_{j=1}^{n}\partial_{j}\partial_{j}u(x)=-v(x)\,\Delta u(x),$$

the last equality by The Laplacian of a Twice Continuously Differentiable Function (i). Since the finite sums of two families with the same terms coincide, combining the last three displays gives

$$\sum_{j=1}^{n}\bigl\langle[(\partial_{j}u)|_{Q}],\partial_{j}V\bigr\rangle_{L^{2}}=-\int_{\mathbb{T}^{n}}v\,\bigl((\Delta u)|_{Q}\bigr)\,dx=-\bigl\langle[(\Delta u)|_{Q}],V\bigr\rangle_{L^{2}} ,$$

the last equality again by L²(μ) Is a Real Hilbert Space (i), the class $[(\Delta u)|_{Q}]$ lying in $L^{2}(\mathbb{T}^{n})$ by Continuous Periodic Functions are Power-Integrable and Dense on the Torus (i).

Therefore

$$\langle U,V\rangle_{H^{1}}=\langle U,V\rangle_{L^{2}}-\bigl\langle[(\Delta u)|_{Q}],V\bigr\rangle_{L^{2}} .$$

By L²(μ) Is a Real Hilbert Space (i) the right-hand side is $\int_{\mathbb{T}^{n}}\bigl(u|_{Q}\bigr)v\,dx-\int_{\mathbb{T}^{n}}\bigl((\Delta u)|_{Q}\bigr)v\,dx$, which by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, applied with the coefficients $1$ and $-1$, equals

$$\int_{\mathbb{T}^{n}}\bigl(u|_{Q}-(\Delta u)|_{Q}\bigr)v\,dx=\int_{\mathbb{T}^{n}}\bigl((u-\Delta u)|_{Q}\bigr)v\,dx=\bigl\langle[(u-\Delta u)|_{Q}],V\bigr\rangle_{L^{2}} ,$$

since $(u-\Delta u)|_{Q}$ and $u|_{Q}-(\Delta u)|_{Q}$ are the same map on $Q$, the difference being pointwise. This is the identity asserted in the claim.

Since $V\in H^{1}(\mathbb{T}^{n})$ was arbitrary, the class $Z=[(u-\Delta u)|_{Q}]\in L^{2}(\mathbb{T}^{n})$ satisfies $\langle U,Y\rangle_{H^{1}}=\langle Z,Y\rangle_{L^{2}}$ for every $Y\in H^{1}(\mathbb{T}^{n})$. By Hilbert Triple: a Densely and Continuously Embedded Hilbert Space and Its Form Operator (iv) this says exactly that $U\in D(A)$ and $AU=Z$, that is, $[u|_{Q}]\in D(A)$ and $A[u|_{Q}]=[(u-\Delta u)|_{Q}]$.