Proof of A Comparison Principle on a Hilbert Triple under the Second-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, 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),

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

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 (vii) 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

$$\Phi_{\delta',\alpha}(x,y)+(\delta'-\delta)\bigl(h(x)+h(y)\bigr)\le\Phi_{\delta,\alpha}(x,y)\qquad\text{for }x,y\in V\text{ and }\delta,\delta'\in J\text{ with }\delta\le\delta'. \tag{iv$'$}$$

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 (i) 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

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

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 (i) 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 second-order structure pair for $F$ at $3B+2$; both exist by hypothesis. By the tail-insensitivity condition fix an orthonormal basis $(e_{k})_{k\in\mathbb{N}}$ of $H$ with $e_{k}\in V$ for every $k\in\mathbb{N}$ along which $F$ is tail-insensitive; this basis is fixed once and for all, before the parameters chosen in the steps below. 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

$$\theta\le1,\qquad 8\theta\le\epsilon_{0},\qquad 96\theta\le\tau^{(1)}.$$

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 Second-Order Structure Condition for an Equation Operator on a Hilbert Triple (ii), 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

$$\theta'\le\theta,\qquad 16\theta'\le\tau^{(2)}.$$

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 $8\theta'\le8\theta\le\epsilon_{0}$,

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

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

$$R=\frac{2B+1}{\delta}+3B+2+8\alpha+2+G+2\alpha+1 ,$$

so that the four inequalities $\tfrac{2B+1}{\delta}<R$, $3B+2<R$, $8\alpha+2<R$ 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 $(\hat{x},\hat{y})$; 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), $\theta'\le\theta$ and $8\theta\le\epsilon_{0}$ give

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

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

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

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

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

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

$$\varepsilon\le1,\qquad (2\alpha+1)\varepsilon\le\tfrac{\tau^{(0)}}{2},\qquad 32\alpha\varepsilon^{2}\le\tau^{(1)},\qquad (4+4G)\varepsilon\le\tfrac{\tau^{(2)}}{4},\qquad 16\alpha\varepsilon^{2}\le\tau^{(2)},\qquad 2\lambda\varepsilon+2\varepsilon\le\epsilon_{1}.$$

All the hypotheses of The Second-Order Test-Datum Estimate at a Sequentially Strict Maximum 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 basis $(e_{k})_{k\in\mathbb{N}}$, 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 Second-Order Test-Datum Estimate at a Sequentially Strict Maximum of the Doubled Function (i).

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 (i), $t\le|x_{1}-\hat{x}|_{H}+\rho_{0}+|\hat{y}-y_{1}|_{H}\le\rho_{0}+2\varepsilon$, 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

$$\alpha t^{2}\le2\alpha\rho_{0}^{2}+8\alpha\varepsilon^{2}\le24\theta+\tfrac{\tau^{(1)}}{4}\le\tfrac{\tau^{(1)}}{4}+\tfrac{\tau^{(1)}}{4}=\tfrac{\tau^{(1)}}{2},$$

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

$$\Phi_{\delta,\alpha}(x_{1},y_{1})+\tfrac{\delta}{2}\bigl(h(x_{1})+h(y_{1})\bigr)\le\Phi_{\delta/2,\alpha}(x_{1},y_{1})\le M\bigl(\tfrac{\delta}{2},\alpha\bigr)\le N(\alpha). \tag{x}$$

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 $\alpha\rho_{0}\le G$,

$$\tfrac{\alpha}{2}t^{2}\le\tfrac{\alpha}{2}(\rho_{0}+2\varepsilon)^{2}=\tfrac{\alpha}{2}\rho_{0}^{2}+2\varepsilon\,\alpha\rho_{0}+2\alpha\varepsilon^{2}\le\tfrac{\alpha}{2}\rho_{0}^{2}+2G\varepsilon+2\alpha\varepsilon^{2}.$$

Hence, by (vii),

$$\Phi_{\delta,\alpha}(x_{1},y_{1})\ \ge\ u^{-}_{\delta}(\hat{x})-v^{+}_{\delta}(\hat{y})-\tfrac{\alpha}{2}\rho_{0}^{2}-2\varepsilon-2G\varepsilon-2\alpha\varepsilon^{2}\ =\ \Phi_{\delta,\alpha}(\hat{x},\hat{y})-(2+2G)\varepsilon-2\alpha\varepsilon^{2}\ \ge\ M(\delta,\alpha)-\theta'-(2+2G)\varepsilon-2\alpha\varepsilon^{2}.$$

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

$$\tfrac{\delta}{2}\bigl(h(x_{1})+h(y_{1})\bigr)\le N(\alpha)-M(\delta,\alpha)+\theta'+(2+2G)\varepsilon+2\alpha\varepsilon^{2}\le2\theta'+(2+2G)\varepsilon+2\alpha\varepsilon^{2},$$

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

$$\delta\bigl(h(x_{1})+h(y_{1})+1\bigr)\le4\theta'+(4+4G)\varepsilon+4\alpha\varepsilon^{2}+\delta\le\tfrac{\tau^{(2)}}{4}+\tfrac{\tau^{(2)}}{4}+\tfrac{\tau^{(2)}}{4}+\tfrac{\tau^{(2)}}{4}=\tau^{(2)} .$$

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 (i), 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

$$\lambda\bigl(u^{-}_{\delta}(\hat{x})-v^{+}_{\delta}(\hat{y})\bigr)\le5\epsilon_{1}=\tfrac{\lambda\epsilon_{0}}{4}.$$

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 (i) gives $u(x)-v(x)\le0$, that is, $u(x)\le v(x)$.