Proof of A Comparison Principle on a Hilbert Triple under the First-Order Structure Condition

Each result cited is universally quantified over the data in its own statement. Replacing $C$ by $|C|$ if necessary, we may assume $0\le C$: indeed $u(x)\le C\le|C|$ and $-|C|\le-C\le v(x)$ for every $x\in H$.

Notation. Let $J=\{\delta\in\mathbb{R}:0<\delta<1\}$. For $\delta\in J$ the $\delta$-envelopes $u^{-}_{\delta}$ and $v^{+}_{\delta}$ are defined on $V$, and by Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity (iv)

$$u^{-}_{\delta}(x)\le C-\delta h(x)\le C-\tfrac{\delta}{2}|x|_{H}^{2},\qquad -v^{+}_{\delta}(y)\le C-\delta h(y)\le C-\tfrac{\delta}{2}|y|_{H}^{2}\qquad(x,y\in V), \tag{i}$$

and by Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity (iii) both $u^{-}_{\delta}$ and $-v^{+}_{\delta}$ have closed superlevel sets in $H$. The set $V$ is nonempty, since $D(A)\cap H$ is dense in $H$ by The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds (v); fix $x_{*}\in V$.

For $\delta\in J$ and real $\alpha>1$ let $\Phi_{\delta,\alpha}:V\times V\to\mathbb{R}$ be given by

$$\Phi_{\delta,\alpha}(x,y)=u^{-}_{\delta}(x)-v^{+}_{\delta}(y)-\tfrac{\alpha}{2}|x-y|_{H}^{2}.$$

By (i) and Closed Superlevel Sets of a Sum on a Product Space and of a Doubled Function (ii), applied with $u_{1}=u^{-}_{\delta}$, $u_{2}=-v^{+}_{\delta}$, $C_{1}=C_{2}=C$ and $\kappa=\tfrac{\delta}{2}$, the function $\Phi_{\delta,\alpha}$ has closed superlevel sets in $H\times H$ and

$$\Phi_{\delta,\alpha}(x,y)\le 2C-\tfrac{\delta}{2}|(x,y)|^{2}\qquad\text{for all }(x,y)\in V\times V. \tag{ii}$$

Let $M(\delta,\alpha)$ be the supremum of the values of $\Phi_{\delta,\alpha}$, and let $N(\alpha)$ be the supremum of the set of all numbers $\Phi_{\delta,\alpha}(x,y)$ with $\delta\in J$ and $(x,y)\in V\times V$. Both are real numbers, and the needed properties come from Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence: applied with $Z=V\times V$, $\psi(x,y)=u^{-}_{\delta}(x)-v^{+}_{\delta}(y)$, which is bounded above by $2C$ by (i), $D(x,y)=\tfrac12|x-y|_{H}^{2}$ and $z_{0}=(x_{*},x_{*})$, it gives claims 1 to 3 for the function $\alpha\mapsto M(\delta,\alpha)$; applied with $Z=J\times V\times V$, $\psi(\delta,x,y)=u^{-}_{\delta}(x)-v^{+}_{\delta}(y)$, $D(\delta,x,y)=\tfrac12|x-y|_{H}^{2}$ and $z_{0}=(\tfrac12,x_{*},x_{*})$, it gives claims 1, 2 and 4 for $N$. In particular, with $\alpha_{k}=2^{k+1}$ for $k\in\mathbb{N}$,

$$\text{the sequence whose }k\text{-th term is }N(\alpha_{k})-N(\alpha_{k+1})\text{ converges to }0, \tag{iii}$$

this being claim 4 with $\beta_{0}=2$. Moreover $M(\delta,\alpha)\le N(\alpha)$ for every $\delta\in J$, and by Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity (v) we have $u^{-}_{\delta'}\le u^{-}_{\delta}$ and $v^{+}_{\delta}\le v^{+}_{\delta'}$ on $V$ whenever $\delta\le\delta'$ in $J$, so

$$M(\delta',\alpha)\le M(\delta,\alpha)\qquad\text{for }\delta,\delta'\in J\text{ with }\delta\le\delta'. \tag{iv}$$

Proof of claim 1. Suppose the claim fails. Then there is a positive $\epsilon_{0}\in\mathbb{R}$ such that for every positive $\theta\in\mathbb{R}$ there are $x,y\in V$ with $|x-y|_{H}\le\theta$ and $\epsilon_{0}<u(x)-v(y)$.

Step 1. $\epsilon_{0}\le N(\alpha)$ for every real $\alpha>1$. Let $\alpha>1$ and let $\mu\in\mathbb{R}$ be positive. Choose a positive $\theta$ with $\tfrac{\alpha}{2}\theta^{2}\le\tfrac{\mu}{2}$ and then $x,y\in V$ with $|x-y|_{H}\le\theta$ and $\epsilon_{0}<u(x)-v(y)$. By The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds (i) and claim 2 of Elementary Arithmetic in an Ordered Field the number $h(x)+h(y)$ is nonnegative, so there is $\delta\in J$ with $\delta\bigl(h(x)+h(y)\bigr)\le\tfrac{\mu}{2}$: take $\delta=\tfrac12$ if $h(x)+h(y)=0$, and otherwise any positive $\delta$ below both $\tfrac12$ and $\tfrac{\mu}{2(h(x)+h(y))}$. By Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity (i), $u(x)-\delta h(x)\le u^{-}_{\delta}(x)$ and $v^{+}_{\delta}(y)\le v(y)+\delta h(y)$, whence

$$N(\alpha)\ge\Phi_{\delta,\alpha}(x,y)\ge u(x)-v(y)-\delta\bigl(h(x)+h(y)\bigr)-\tfrac{\alpha}{2}|x-y|_{H}^{2}>\epsilon_{0}-\mu ,$$

where $|x-y|_{H}^{2}\le\theta^{2}$ by Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. As $\mu$ was an arbitrary positive real, Comparison of Real Numbers with Arbitrary Positive Slack gives $\epsilon_{0}\le N(\alpha)$.

Step 2 (constants that do not move). Set $B=4C+4$, so that $0\le C\le B$ and $3B+2$ is positive. Let $\lambda$ be a properness constant for $F$ at $3B+2$ and let $(\omega_{1},\omega_{2},\omega_{3})$ be a structure triple for $F$ at $3B+2$; both exist by hypothesis. Set $\epsilon_{1}=\tfrac{\lambda\epsilon_{0}}{20}$, a positive real, and use clause 2 of Modulus of Continuity to choose positive reals $\tau^{(1)},\tau^{(2)},\tau^{(3)}$ such that, for $i\in\{1,2,3\}$, every $t$ with $0\le t\le\tau^{(i)}$ satisfies $\omega_{i}(t)\le\epsilon_{1}$.

Step 3 (the parameters $\theta$ and $\alpha$). Choose a positive $\theta\in\mathbb{R}$ with

$$\theta\le1,\qquad 8\theta\le\epsilon_{0},\qquad 48\theta\le\tau^{(2)},\qquad 48\theta\le\bigl(\tau^{(1)}\bigr)^{2}.$$

By (iii) choose $k\in\mathbb{N}$ with $N(\alpha_{k})-N(\alpha_{k+1})\le\theta$ and set $\alpha=\alpha_{k+1}$, so that $1<\alpha$ and $\tfrac{\alpha}{2}=\alpha_{k}$.

Step 4 (the parameter $\delta$). By the definition of $N(\alpha)$ as a supremum there are $\delta_{1}\in J$ and $(x,y)\in V\times V$ with $N(\alpha)-\theta<\Phi_{\delta_{1},\alpha}(x,y)\le M(\delta_{1},\alpha)$; by (iv), $N(\alpha)-\theta<M(\delta,\alpha)$ for every $\delta\in J$ with $\delta\le\delta_{1}$. Choose such a $\delta$ with in addition $\delta\le\tfrac{\tau^{(3)}}{\alpha\tau^{(2)}}$. By Step 1 and $8\theta\le\epsilon_{0}$,

$$0\le\epsilon_{0}-\theta\le N(\alpha)-\theta<M(\delta,\alpha). \tag{v}$$

Step 5 (the maximum point). Choose $(x_{2},y_{2})\in V\times V$ with $M(\delta,\alpha)-\tfrac{\theta}{2}\le\Phi_{\delta,\alpha}(x_{2},y_{2})$. By (ii), (v) and $\theta\le1$,

$$\tfrac{\delta}{2}|(x_{2},y_{2})|^{2}\le2C-\Phi_{\delta,\alpha}(x_{2},y_{2})\le2C-M(\delta,\alpha)+\tfrac{\theta}{2}\le2C+1,$$

so $|(x_{2},y_{2})|\le\Gamma$, where $\Gamma$ denotes the nonnegative square root of $\tfrac{4C+2}{\delta}$. Let $\Gamma'$ denote the nonnegative square root of $\tfrac{4}{\delta}\bigl(2C+\tfrac12+2\Gamma+\tfrac{4}{\delta}\bigr)$, let $G$ denote the nonnegative square root of $12\alpha\theta$, and set

$$R=\frac{2B+1}{\delta}+3B+2+\alpha+1+G+2\alpha+1 ,$$

so that the four inequalities $\tfrac{2B+1}{\delta}<R$, $3B+2<R$, $\alpha+1<R$ and $G+2\alpha<R$ hold, all summands being positive. Let $\omega$ be a shift modulus for $F$ at $(\delta,R)$, which exists by hypothesis, and choose a positive $\tau^{(0)}$ such that every $t$ with $0\le t\le\tau^{(0)}$ satisfies $\omega(t)\le\epsilon_{1}$.

Now choose a positive $\sigma\in\mathbb{R}$ with

$$\sigma\le1,\qquad \sigma\le\tfrac{\tau^{(0)}}{2},\qquad \sigma\,(2\Gamma+2\Gamma')\le\tfrac{\theta}{2}.$$

The product $H\times H$ is a real Hilbert space by Properties of the Product of Two Real Inner Product Spaces (vi), and $V\times V$ is a nonempty subset of it. By (ii) and A Linear Perturbation Producing a Sequentially Strict Maximum under a Coercive Bound, applied to $\Phi_{\delta,\alpha}$ with the constant $2C$, the positive number $\tfrac{\delta}{2}$ and the positive number $\sigma$, there are $(p,q)\in H\times H$ with $|(p,q)|\le\sigma$ and $(\hat{x},\hat{y})\in V\times V$ such that the function with value

$$\Phi_{\delta,\alpha}(x,y)-\langle(p,q),(x,y)\rangle=\Phi_{\delta,\alpha}(x,y)-\langle p,x\rangle_{H}-\langle q,y\rangle_{H}$$

at $(x,y)$ attains a sequentially strict maximum on $V\times V$ at $(\hat{x},\hat{y})$, and in particular a maximum there, by the definition of a sequentially strict maximum; the displayed identity is the formula for the inner product of the product in The Product of Two Real Inner Product Spaces (i), and Properties of the Product of Two Real Inner Product Spaces (ii) gives $|p|_{H}\le\sigma$ and $|q|_{H}\le\sigma$.

Step 6 (the maximum point nearly maximises $\Phi_{\delta,\alpha}$). Write $\rho=|(\hat{x},\hat{y})|$, so that $|\hat{x}|_{H}+|\hat{y}|_{H}\le2\rho$ and $|x_{2}|_{H}+|y_{2}|_{H}\le2\Gamma$. By the maximum property and The Cauchy–Schwarz Inequality in a Real Inner Product Space,

$$\Phi_{\delta,\alpha}(\hat{x},\hat{y})\ \ge\ \Phi_{\delta,\alpha}(x_{2},y_{2})-\sigma\bigl(|x_{2}|_{H}+|y_{2}|_{H}\bigr)-\sigma\bigl(|\hat{x}|_{H}+|\hat{y}|_{H}\bigr)\ \ge\ M(\delta,\alpha)-\tfrac{\theta}{2}-2\sigma\Gamma-2\sigma\rho . \tag{vi}$$

Using $\sigma\le1$, $M(\delta,\alpha)\ge0$, $\theta\le1$ and (ii),

$$\tfrac{\delta}{2}\rho^{2}\le2C-\Phi_{\delta,\alpha}(\hat{x},\hat{y})\le2C+\tfrac12+2\Gamma+2\rho ,$$

and since $\tfrac{\delta}{4}\bigl(\rho-\tfrac{4}{\delta}\bigr)^{2}\ge0$ gives $2\rho\le\tfrac{\delta}{4}\rho^{2}+\tfrac{4}{\delta}$, we obtain $\tfrac{\delta}{4}\rho^{2}\le2C+\tfrac12+2\Gamma+\tfrac{4}{\delta}$, that is, $\rho\le\Gamma'$. Substituting $\rho\le\Gamma'$ into (vi) and using $\sigma(2\Gamma+2\Gamma')\le\tfrac{\theta}{2}$,

$$M(\delta,\alpha)-\theta\ \le\ \Phi_{\delta,\alpha}(\hat{x},\hat{y}). \tag{vii}$$

Step 7 (the bounds required by the estimate). Since $\tfrac{\alpha}{2}|\hat{x}-\hat{y}|_{H}^{2}\ge0$, (vii), (v) and $8\theta\le\epsilon_{0}$ give

$$u^{-}_{\delta}(\hat{x})-v^{+}_{\delta}(\hat{y})\ \ge\ \Phi_{\delta,\alpha}(\hat{x},\hat{y})\ \ge\ M(\delta,\alpha)-\theta\ \ge\ \epsilon_{0}-2\theta\ \ge\ \tfrac{3}{4}\epsilon_{0}\ >\ 0. \tag{viii}$$

By (i) we have $u^{-}_{\delta}(\hat{x})\le C$ and $-C\le v^{+}_{\delta}(\hat{y})$; combined with (viii) this yields $u^{-}_{\delta}(\hat{x})\ge v^{+}_{\delta}(\hat{y})\ge-C$ and $v^{+}_{\delta}(\hat{y})\le u^{-}_{\delta}(\hat{x})\le C$, so

$$|u^{-}_{\delta}(\hat{x})|\le C\le B,\qquad |v^{+}_{\delta}(\hat{y})|\le C\le B .$$

By (i) again, $\delta h(\hat{x})\le C-u^{-}_{\delta}(\hat{x})\le2C\le B$ and $\delta h(\hat{y})\le v^{+}_{\delta}(\hat{y})+C\le2C\le B$.

By (vii) the point $(\hat{x},\hat{y})$ satisfies the hypothesis of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence (iii) for the function $\alpha\mapsto M(\delta,\alpha)$ with the parameter $\alpha$ and the number $\theta$, so

$$\tfrac{\alpha}{4}|\hat{x}-\hat{y}|_{H}^{2}=\tfrac{\alpha}{2}\cdot\tfrac12|\hat{x}-\hat{y}|_{H}^{2}\le M\bigl(\delta,\tfrac{\alpha}{2}\bigr)-M(\delta,\alpha)+\theta\le N(\alpha_{k})-\bigl(N(\alpha_{k+1})-\theta\bigr)+\theta\le3\theta ,$$

using $\tfrac{\alpha}{2}=\alpha_{k}$, $M(\delta,\alpha_{k})\le N(\alpha_{k})$, Step 4 and Step 3. Hence

$$\alpha|\hat{x}-\hat{y}|_{H}^{2}\le12\theta,\qquad |\hat{x}-\hat{y}|_{H}\le\sqrt{12\theta}\le\tfrac{\tau^{(1)}}{2},\qquad \alpha|\hat{x}-\hat{y}|_{H}=\sqrt{\alpha\cdot\alpha|\hat{x}-\hat{y}|_{H}^{2}}\le G , \tag{ix}$$

where the second inequality uses $1<\alpha$ and $48\theta\le(\tau^{(1)})^{2}$, and square roots are compared by Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field.

Step 8 (the contradiction). Choose a positive $\varepsilon\in\mathbb{R}$ with

$$\varepsilon\le1,\qquad (2\alpha+1)\varepsilon\le\tfrac{\tau^{(0)}}{2},\qquad 2\varepsilon\le\tfrac{\tau^{(1)}}{2},\qquad 8\alpha\varepsilon^{2}\le\tfrac{\tau^{(2)}}{2},\qquad 2\lambda\varepsilon+2\varepsilon\le\epsilon_{1}.$$

All the hypotheses of The Test-Datum Estimate at a Maximum Point of the Doubled Function now hold for the data $\delta,\alpha,\sigma,\varepsilon,B,G,R$, the vectors $p,q$, the point $(\hat{x},\hat{y})$, the constant $\lambda$, the triple $(\omega_{1},\omega_{2},\omega_{3})$ and the modulus $\omega$, by Steps 2 to 7. It provides $x_{1},y_{1}\in D(A)$ and nonnegative $\tau_{1},\tau_{2}$ with $|x_{1}-\hat{x}|_{H}<\varepsilon$, $|y_{1}-\hat{y}|_{H}<\varepsilon$, $\tau_{1}\le(2\alpha+1)\varepsilon+\sigma\le\tau^{(0)}$, $\tau_{2}\le\tau^{(0)}$ and the estimate of The Test-Datum Estimate at a Maximum Point of the Doubled Function (i).

Write $t=|x_{1}-y_{1}|_{H}$. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity (i) and (ix), $t\le|\hat{x}-\hat{y}|_{H}+2\varepsilon\le\tfrac{\tau^{(1)}}{2}+\tfrac{\tau^{(1)}}{2}=\tau^{(1)}$. Since $(a+b)^{2}\le2a^{2}+2b^{2}$ for all real $a,b$, because $2a^{2}+2b^{2}-(a+b)^{2}=(a-b)^{2}\ge0$, we get

$$\alpha t^{2}\le2\alpha|\hat{x}-\hat{y}|_{H}^{2}+8\alpha\varepsilon^{2}\le24\theta+\tfrac{\tau^{(2)}}{2}\le\tau^{(2)},$$

using (ix) and $48\theta\le\tau^{(2)}$, and consequently, by the choice of $\delta$ in Step 4,

$$\delta\alpha^{2}t^{2}=\delta\alpha\,(\alpha t^{2})\le\delta\alpha\,\tau^{(2)}\le\tau^{(3)} .$$

Therefore each of $\omega(\tau_{1})$, $\omega(\tau_{2})$, $\omega_{1}(t)$, $\omega_{2}(\alpha t^{2})$ and $\omega_{3}(\delta\alpha^{2}t^{2})$ is at most $\epsilon_{1}$, and $2\lambda\varepsilon+2\varepsilon\le\epsilon_{1}$, so the estimate gives

$$\lambda\bigl(u^{-}_{\delta}(\hat{x})-v^{+}_{\delta}(\hat{y})\bigr)\le6\epsilon_{1}=\tfrac{6\lambda\epsilon_{0}}{20}.$$

On the other hand (viii) and $\lambda>0$ give $\lambda\bigl(u^{-}_{\delta}(\hat{x})-v^{+}_{\delta}(\hat{y})\bigr)\ge\tfrac{3}{4}\lambda\epsilon_{0}$. Dividing by the positive number $\lambda\epsilon_{0}$ yields $\tfrac{3}{4}\le\tfrac{3}{10}$, which is false. This contradiction proves claim 1.

Proof of claim 2. Let $x\in V$ and let $\epsilon\in\mathbb{R}$ be positive. By claim 1 there is a positive $\theta$ such that all $x',y'\in V$ with $|x'-y'|_{H}\le\theta$ satisfy $u(x')-v(y')\le\epsilon$; taking $x'=y'=x$, which is legitimate because $|x-x|_{H}=0\le\theta$, gives $u(x)-v(x)\le\epsilon$. As $\epsilon$ was an arbitrary positive real, Comparison of Real Numbers with Arbitrary Positive Slack gives $u(x)-v(x)\le0$, that is, $u(x)\le v(x)$.