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 §bound

and by Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity §closed-superlevel 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 §dense; 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

By (i) and Closed Superlevel Sets of a Sum on a Product Space and of a Doubled Function §doubled, 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

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, for every real $\beta_{0}>1$, with $\beta_{k}=2^{k}\beta_{0}$ for $k\in\mathbb{N}$ (so that $1<\beta_{0}\le\beta_{k}$ for every $k$, by claims 2 and 5 of Properties of Natural Number Powers in a Field, and $N(\beta_{k})$ is defined),

this being claim 4. Moreover $M(\delta,\alpha)\le N(\alpha)$ for every $\delta\in J$. By Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity §gap we have $u^{-}_{\delta'}(x)+(\delta'-\delta)h(x)\le u^{-}_{\delta}(x)$ and $v^{+}_{\delta}(y)+(\delta'-\delta)h(y)\le v^{+}_{\delta'}(y)$ for $x,y\in V$ whenever $\delta\le\delta'$ in $J$; adding the first inequality to the second multiplied by $-1$ (claim 3 of Elementary Order Arithmetic in an Ordered Field in its weak form, and claim 4 of that lemma) and then subtracting $\tfrac{\alpha}{2}|x-y|_{H}^{2}$ from both sides gives

Since $0\le(\delta'-\delta)(h(x)+h(y))$ by The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds §nonneg and claims 2 and 5 of Elementary Arithmetic in an Ordered Field, (iv$'$) gives $\Phi_{\delta',\alpha}\le\Phi_{\delta,\alpha}$ on $V\times V$, so

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 §nonneg 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 §semicontinuity, $u(x)-\delta h(x)\le u^{-}_{\delta}(x)$ and $v^{+}_{\delta}(y)\le v(y)+\delta h(y)$, whence

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 §slack-above 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 $(\omega_{1},\omega_{2})$ be a structure pair 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 a positive real $\tau^{(1)}$ such that every $t$ with $0\le t\le\tau^{(1)}$ satisfies $\omega_{1}(t)\le\epsilon_{1}$.

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

By The Archimedean Property of the Real Numbers choose $m_{0}\in\mathbb{N}$ with $\tfrac{2}{\tau^{(1)}}\le m_{0}$ and put $m=m_{0}+1$, so that $m\in\mathbb{N}$, $\tfrac{2}{\tau^{(1)}}\le m$ and $2\le m$ (as $1\le m_{0}$). For $k\in\mathbb{N}$ put $\beta_{k}=2^{k}m$, and let $\beta_{0}$ denote $m$ itself, as in claim 4 of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence; since $1<2\le m$, (iii) applies with $\beta_{0}=m$: the sequence whose $k$-th term is $N(\beta_{k})-N(\beta_{k+1})$ converges to $0$. Choose $k\in\mathbb{N}$ with $N(\beta_{k})-N(\beta_{k+1})\le\theta$ and set $\alpha=\beta_{k+1}$, so that $\tfrac{\alpha}{2}=\beta_{k}$. Since $1\le2^{k}$ and $1\le2^{k+1}$ by claims 2 and 5 of Properties of Natural Number Powers in a Field, we have $2\le m\le\beta_{k}=\tfrac{\alpha}{2}$ and $m\le\alpha$; hence $1<\tfrac{\alpha}{2}$ and $1<\alpha$, so that $M(\delta,\tfrac{\alpha}{2})$, $M(\delta,\alpha)$ and $N(\alpha)$ are defined, and $\tfrac{2}{\tau^{(1)}}\le\alpha$, that is, $\tfrac{1}{\alpha}\le\tfrac{\tau^{(1)}}{2}$ (claims 4 and 5 of Elementary Arithmetic in an Ordered Field).

Step 4 (the parameters $\theta'$ and $\delta$). The function $t\mapsto\omega_{2}(t,\alpha)$ on the nonnegative reals is a modulus of continuity by The First-Order Structure Condition for a Second-Order Equation Operator on a Hilbert Triple §pair, so by clause 2 of Modulus of Continuity there is a positive real $\tau^{(2)}$ such that every $t$ with $0\le t\le\tau^{(2)}$ satisfies $\omega_{2}(t,\alpha)\le\epsilon_{1}$. Choose a positive $\theta'\in\mathbb{R}$ with

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 $4\delta\le\tau^{(2)}$. By Step 1 and $8\theta'\le8\theta\le\epsilon_{0}$,

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 $\theta'\le1$,

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

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

The product $H\times H$ is a real Hilbert space by Properties of the Product of Two Real Inner Product Spaces §hilbert, 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

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 §product, and Properties of the Product of Two Real Inner Product Spaces §norm 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,

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

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 $\sigma(2\Gamma+2\Gamma')\le\tfrac{\theta'}{2}$,

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

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

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 §near-maximiser for the function $\alpha\mapsto M(\delta,\alpha)$ with the parameter $\alpha$ and the number $\theta'$, so

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

where 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

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 pair $(\omega_{1},\omega_{2})$ 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 $|x_{1}-\hat{x}|_{H}<\varepsilon$, $|y_{1}-\hat{y}|_{H}<\varepsilon$, $|u^{-}_{\delta}(x_{1})-u^{-}_{\delta}(\hat{x})|<\varepsilon$, $|v^{+}_{\delta}(y_{1})-v^{+}_{\delta}(\hat{y})|<\varepsilon$, $\tau_{1}\le(2\alpha+1)\varepsilon+\sigma\le\tau^{(0)}$, $\tau_{2}\le\tau^{(0)}$ and the estimate of The Test-Datum Estimate at a Maximum Point of the Doubled Function §estimate.

The first modulus. Write $t=|x_{1}-y_{1}|_{H}$ and $\rho_{0}=|\hat{x}-\hat{y}|_{H}$. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle, $t\le|x_{1}-\hat{x}|_{H}+\rho_{0}+|\hat{y}-y_{1}|_{H}\le\rho_{0}+2\varepsilon$, so $t^{2}\le(\rho_{0}+2\varepsilon)^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. 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

using (ix), $32\alpha\varepsilon^{2}\le\tau^{(1)}$ and $96\theta\le\tau^{(1)}$; together with $\tfrac{1}{\alpha}\le\tfrac{\tau^{(1)}}{2}$ from Step 3 this gives $0\le\alpha t^{2}+\tfrac{1}{\alpha}\le\tau^{(1)}$, hence $\omega_{1}\bigl(\alpha t^{2}+\tfrac{1}{\alpha}\bigr)\le\epsilon_{1}$ by Step 2.

The penalty at the test points. Since $\tfrac{\delta}{2}\in J$ and $\tfrac{\delta}{2}\le\delta$, (iv$'$) applied with $\tfrac{\delta}{2}$ and $\delta$ in the roles of $\delta$ and $\delta'$ gives

On the other hand, claim 9 of Properties of the Absolute Value in an Ordered Field turns the two envelope inequalities provided by the estimate into $u^{-}_{\delta}(x_{1})>u^{-}_{\delta}(\hat{x})-\varepsilon$ and $v^{+}_{\delta}(y_{1})<v^{+}_{\delta}(\hat{y})+\varepsilon$, while expanding the square and using (ix) and $\alpha\rho_{0}\le G$,

Hence, by (vii),

Combining this with (x) and with $N(\alpha)-\theta'<M(\delta,\alpha)$ from Step 4,

and therefore, by $16\theta'\le\tau^{(2)}$, $(4+4G)\varepsilon\le\tfrac{\tau^{(2)}}{4}$, $16\alpha\varepsilon^{2}\le\tau^{(2)}$ and $4\delta\le\tau^{(2)}$,

The left-hand side is nonnegative by The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds §nonneg, so $\omega_{2}\bigl(\delta(h(x_{1})+h(y_{1})+1),\alpha\bigr)\le\epsilon_{1}$ by the choice of $\tau^{(2)}$ in Step 4.

Conclusion. Each of $\omega(\tau_{1})$, $\omega(\tau_{2})$, $\omega_{1}\bigl(\alpha t^{2}+\tfrac{1}{\alpha}\bigr)$ and $\omega_{2}\bigl(\delta(h(x_{1})+h(y_{1})+1),\alpha\bigr)$ is at most $\epsilon_{1}$, and $2\lambda\varepsilon+2\varepsilon\le\epsilon_{1}$, so the estimate gives

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 $\tfrac{3}{4}\le\tfrac{1}{4}$, 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 §slack-above gives $u(x)-v(x)\le0$, that is, $u(x)\le v(x)$.