Proof of Lipschitz Regularity of Bounded Continuous Viscosity Solutions of a Viscous Hamilton-Jacobi Equation with a Monotone Nonlinearity on a Hilbert Triple

Each result cited is universally quantified over the data in its own statement. Throughout, $h$ is the penalty function, $\sigma(f)$ is the sum of the square-summable sequence $f$ as in Square-Summable Sequences in the Form Space of a Hilbert Triple and the Trace of a Form along Them §square-summable (so $0\le\sigma(f)$), and we write $\ell=\ell_{g}$, $c=\tfrac{\ell}{\lambda_{0}}$ and $K=c+2C'+1$; thus $0\le c$, $1\le K$ and $\tfrac{\ell^{2}}{2\lambda_{0}^{2}\alpha}=\tfrac{c^{2}}{2\alpha}$ for every real $\alpha>1$. Elementary arithmetic and order in $\mathbb{R}$ are used freely by The Real Numbers: Standing Notation and Background §background.

Step 0 (standing facts about $u$ and $F$). By claim 6 of Properties of the Absolute Value in an Ordered Field, $|u(x)|\le C'$ gives $u(x)\le C'$ and $-C'\le u(x)$ for every $x\in H$, so by Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity §bound (with $U=H$) the function $u$ is bounded above and below near each point of $H$, and for every real $\delta>0$ the $\delta$-envelopes $u^{-}_{\delta}$, $u^{+}_{\delta}$ are defined on $V$ and satisfy

$$u^{-}_{\delta}(x)\le C'-\delta h(x)\le C'-\tfrac{\delta}{2}|x|_{H}^{2},\qquad -u^{+}_{\delta}(y)\le C'-\delta h(y)\le C'-\tfrac{\delta}{2}|y|_{H}^{2}\qquad(x,y\in V), \tag{0.1}$$

while Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity §semicontinuity gives

$$u(x)-\delta h(x)\le u^{-}_{\delta}(x),\qquad u^{+}_{\delta}(y)\le u(y)+\delta h(y)\qquad(x,y\in V). \tag{0.2}$$

Since $u$ is a viscosity solution of $F$ on $H$, it is both a viscosity subsolution and a viscosity supersolution of $F$ on $H$ by Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on a Hilbert Triple §solution.

The operator $F$ is the operator of A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses for the data $\lambda_{0},C_{g},g,B,\nu,f,\Gamma$ of the statement, the zero map $L$ of $H$ and $\omega_{g}(t)=\ell t$. Indeed, $L$ is Lipschitz with constant $0$ in the sense of Lipschitz Map Between Metric Spaces, because $d_{H}(L(x),L(x'))=d_{H}(0_{H},0_{H})=0$; its constant $c_{L}=|L(0_{H})|_{H}$ is $0$; and $\langle Ax+B(x)+L(x),p\rangle_{H}=\langle Ax+B(x),p\rangle_{H}$, so the two formulas for $F$ agree. The function $t\mapsto\ell t$ is a modulus of continuity by Linear Moduli of Continuity §modulus, as $0\le\ell$, and for $x,y\in V$ we have $x-y\in V$ and $|x-y|_{H}\le|x-y|_{V}$ by Hilbert Triples: Standing Notation and Background §triple, hence $|g(x)-g(y)|\le\ell|x-y|_{H}\le\ell|x-y|_{V}=\omega_{g}(|x-y|_{V})$. Consequently A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §operator, A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §proper, A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §structure, A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §tail and A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §shift apply to $F$. Since $H$ is not finite-dimensional, An Orthonormal Basis of the Ambient Space Contained in the Form Space of a Hilbert Triple §basis provides an orthonormal basis $(e_{k})_{k\in\mathbb{N}}$ of $H$ with $e_{k}\in V$ for every $k$, fixed from now on; by A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §tail, $F$ is tail-insensitive along it.

Proof of claim 1.

Step 1 (the data that do not move). Fix a real $\alpha>1$ and $x_{0},y_{0}\in V$, and put

$$L_{0}=u(x_{0})-u(y_{0})-\tfrac{\alpha}{2}|x_{0}-y_{0}|_{H}^{2}.$$

We must show $L_{0}\le\tfrac{c^{2}}{2\alpha}$. If $L_{0}\le0$ this holds because $0\le\tfrac{c^{2}}{2\alpha}$; so assume $0<L_{0}$. Put

$$B_{0}=4C'+4,$$

so that $0\le C'\le B_{0}$ and $3B_{0}+2$ is positive, and let $G$ be the nonnegative square root of $\alpha(4C'+4)$. By A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §proper, $\lambda_{0}$ is a properness constant for $F$ at $3B_{0}+2$, and by A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §structure (where the Lipschitz constant of $L$ is $0$ and $c_{L}=0$) there is a second-order structure pair $(\omega_{1},\omega_{2})$ for $F$ at $3B_{0}+2$; fix it. None of $B_{0}$, $G$, $\lambda_{0}$, $(\omega_{1},\omega_{2})$ and the basis depends on the parameters chosen below. Now fix a real $\eta$ with $0<\eta\le1$; Steps 2 to 8 show

$$\lambda_{0}L_{0}\le\frac{\ell^{2}}{2\lambda_{0}\alpha}+(5+2\lambda_{0})\eta, \tag{1.1}$$

and the parameters $\delta$, $R$, $\omega$, $\tau^{(0)}$, $\Lambda$, $\gamma$, $p$, $q$, $\hat{x}$, $\hat{y}$, $\varepsilon$ are chosen there in this order, each after $\eta$.

Step 2 (the parameter $\delta$). Put $D=2\bigl(1+h(x_{0})+h(y_{0})+\nu\,\sigma(f)+(G+2\alpha)^{2}\bigr)$, which is at least $2$ because $h(x_{0}),h(y_{0})\ge0$ by The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds §nonneg and $\nu\sigma(f)\ge0$, and set $\delta=\tfrac{\eta}{D}$. Then $0<\delta\le\tfrac{\eta}{2}<1$ and

$$\delta\bigl(h(x_{0})+h(y_{0})\bigr)\le\eta,\qquad \nu\,\delta\,\sigma(f)\le\eta,\qquad \delta\,(G+2\alpha)^{2}\le\eta . \tag{2.1}$$

Let $\Phi:V\times V\to\mathbb{R}$ be given by

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

By (0.2) and (2.1),

$$\Phi(x_{0},y_{0})\ge L_{0}-\delta\bigl(h(x_{0})+h(y_{0})\bigr)\ge L_{0}-\eta>-1. \tag{2.2}$$

Step 3 (the level $R$, the shift modulus, and the perturbed maximum). Set

$$R=\frac{2B_{0}+1}{\delta}+3B_{0}+2+8\alpha+2+G+2\alpha+1,$$

so that $\tfrac{2B_{0}+1}{\delta}<R$, $3B_{0}+2<R$, $8\alpha+2<R$ and $G+2\alpha<R$, all summands being positive. By A Viscous Hamilton-Jacobi Operator with a Monotone Nonlinearity Satisfies the Second-Order Comparison Hypotheses §shift and The Shift-Continuity Condition on Admissible Test Data §continuity there is a shift modulus $\omega$ for $F$ at $(\delta,R)$; fix it, and by clause 2 of Modulus of Continuity choose a positive real $\tau^{(0)}$ such that every $t$ with $0\le t\le\tau^{(0)}$ satisfies $\omega(t)\le\eta$. Put

$$K_{0}=2C'+1+|x_{0}|_{H}+|y_{0}|_{H}+\tfrac{2}{\delta},\qquad \Lambda=K_{0}+\tfrac{2}{\delta},$$

and choose a positive real $\gamma$ with

$$\gamma\le1,\qquad \gamma\le\tfrac{\tau^{(0)}}{2},\qquad \gamma\bigl(|x_{0}|_{H}+|y_{0}|_{H}+\Lambda\bigr)\le\eta .$$

By Basic Properties of the $\delta$-Envelopes: Semicontinuity, Duality, Closed Superlevel Sets, Bounds and Monotonicity §closed-superlevel, $u^{-}_{\delta}$ and $-u^{+}_{\delta}$ have closed superlevel sets in $(H,d_{H})$, and by (0.1) they are bounded above by $C'$ and satisfy the coercive bounds there. Hence Closed Superlevel Sets of a Sum on a Product Space and of a Doubled Function §doubled, applied with $S_{1}=S_{2}=V$, $u_{1}=u^{-}_{\delta}$, $u_{2}=-u^{+}_{\delta}$, $C_{1}=C_{2}=C'$, the nonnegative $\alpha$ and $\kappa=\tfrac{\delta}{2}$, shows that $\Phi$ has closed superlevel sets in $H\times H$ and that $\Phi(x,y)\le2C'-\tfrac{\delta}{2}|(x,y)|^{2}$ on $V\times V$. 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 (it contains $(x_{0},y_{0})$). By A Linear Perturbation Producing a Sequentially Strict Maximum under a Coercive Bound, applied to $\Phi$ with the constant $2C'$, the positive number $\tfrac{\delta}{2}$ and the positive number $\gamma$, there are $(p,q)\in H\times H$ with $|(p,q)|\le\gamma$ and $(\hat{x},\hat{y})\in V\times V$ such that the function with value

$$\Phi(x,y)-\langle(p,q),(x,y)\rangle=u^{-}_{\delta}(x)-u^{+}_{\delta}(y)-\tfrac{\alpha}{2}|x-y|_{H}^{2}-\langle p,x\rangle_{H}-\langle q,y\rangle_{H}$$

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 §product, and Properties of the Product of Two Real Inner Product Spaces §norm gives $|p|_{H}\le\gamma$ and $|q|_{H}\le\gamma$.

Step 4 (the maximum point nearly attains $L_{0}$). Write $r_{x}=|\hat{x}|_{H}$, $r_{y}=|\hat{y}|_{H}$, $d_{0}=|\hat{x}-\hat{y}|_{H}$. Comparing the maximum value with the value at $(x_{0},y_{0})$ and using The Cauchy-Schwarz Inequality in a Real Inner Product Space with $|p|_{H},|q|_{H}\le\gamma$,

$$\Phi(\hat{x},\hat{y})\ge\Phi(x_{0},y_{0})-\gamma\bigl(|x_{0}|_{H}+|y_{0}|_{H}\bigr)-\gamma\,(r_{x}+r_{y}). \tag{4.1}$$

By (0.1), dropping the nonpositive term $-\tfrac{\alpha}{2}d_{0}^{2}$, we have $\Phi(\hat{x},\hat{y})\le2C'-\tfrac{\delta}{2}(r_{x}^{2}+r_{y}^{2})$. Together with (4.1), (2.2) and $\gamma\le1$,

$$\tfrac{\delta}{2}(r_{x}^{2}+r_{y}^{2})\le2C'-\Phi(\hat{x},\hat{y})\le2C'+1+|x_{0}|_{H}+|y_{0}|_{H}+r_{x}+r_{y}.$$

For every real $r$ one has $r\le\tfrac{\delta}{4}r^{2}+\tfrac{1}{\delta}$, since $\tfrac{\delta}{4}r^{2}-r+\tfrac{1}{\delta}=\tfrac{\delta}{4}\bigl(r-\tfrac{2}{\delta}\bigr)^{2}\ge0$; hence $r_{x}+r_{y}\le\tfrac{\delta}{4}(r_{x}^{2}+r_{y}^{2})+\tfrac{2}{\delta}$, and the last display gives $\tfrac{\delta}{4}(r_{x}^{2}+r_{y}^{2})\le K_{0}$ and then $r_{x}+r_{y}\le K_{0}+\tfrac{2}{\delta}=\Lambda$. Substituting this and (2.2) into (4.1) and using the choice of $\gamma$,

$$\Phi(\hat{x},\hat{y})\ge L_{0}-\eta-\gamma\bigl(|x_{0}|_{H}+|y_{0}|_{H}+\Lambda\bigr)\ge L_{0}-2\eta>-2. \tag{4.2}$$

Step 5 (the bounds required by the test estimate). By (0.1), $\Phi(\hat{x},\hat{y})\le\bigl(C'-\delta h(\hat{x})\bigr)+\bigl(C'-\delta h(\hat{y})\bigr)-\tfrac{\alpha}{2}d_{0}^{2}$, so by (4.2)

$$\delta h(\hat{x})+\delta h(\hat{y})+\tfrac{\alpha}{2}d_{0}^{2}\le2C'-\Phi(\hat{x},\hat{y})<2C'+2 .$$

All three summands are nonnegative (The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds §nonneg), so $\delta h(\hat{x})\le2C'+2\le B_{0}$, $\delta h(\hat{y})\le B_{0}$ and $\alpha d_{0}^{2}\le4C'+4$. Hence $\alpha^{2}d_{0}^{2}\le\alpha(4C'+4)$ and, comparing nonnegative square roots by Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, $\alpha d_{0}\le G$. Next, (0.1) and (0.2) give

$$-3C'-2\le u(\hat{x})-\delta h(\hat{x})\le u^{-}_{\delta}(\hat{x})\le C',\qquad -C'\le u^{+}_{\delta}(\hat{y})\le u(\hat{y})+\delta h(\hat{y})\le3C'+2,$$

so $|u^{-}_{\delta}(\hat{x})|\le B_{0}$ and $|u^{+}_{\delta}(\hat{y})|\le B_{0}$ by claim 6 of Properties of the Absolute Value in an Ordered Field.

Step 6 (the parameter $\varepsilon$ and the test data). Choose a positive real $\varepsilon$ with

$$\varepsilon\le1,\qquad (2\alpha+1)\varepsilon\le\tfrac{\tau^{(0)}}{2},\qquad (2+2\lambda_{0}+2\ell)\varepsilon\le\eta .$$

All hypotheses of The Second-Order Test-Datum Estimate at a Sequentially Strict Maximum of the Doubled Function now hold for $u$ and $v=u$, the constant $C'$ in the role of $C$ (by Step 0, with $0\le C'$), the basis $(e_{k})_{k\in\mathbb{N}}$ (Step 0), the numbers $\delta,\alpha,\varepsilon,G,R$, the number $\gamma$ in the role of $\sigma$, $B_{0}$ in the role of $B$, the vectors $p,q$ and the point $(\hat{x},\hat{y})$ (Steps 2, 3 and 5), the properness constant $\lambda_{0}$ and the structure pair $(\omega_{1},\omega_{2})$ at $3B_{0}+2$ (Step 1), and the shift modulus $\omega$ at $(\delta,R)$ (Step 3). By The Second-Order Test-Datum Estimate at a Sequentially Strict Maximum of the Doubled Function §nearby there are $x_{1},y_{1}\in D(A)$ and $\tau_{1},\tau_{2}\in\mathbb{R}$ with $|x_{1}-\hat{x}|_{H}<\varepsilon$, $|y_{1}-\hat{y}|_{H}<\varepsilon$ and $0\le\tau_{i}\le(2\alpha+1)\varepsilon+\gamma\le\tau^{(0)}$ for $i=1,2$, so that $\omega(\tau_{1})\le\eta$ and $\omega(\tau_{2})\le\eta$; and by The Second-Order Test-Datum Estimate at a Sequentially Strict Maximum of the Doubled Function §raw there are $s,t\in\mathbb{R}$ and a pair $(X,Y)$ of members of $\mathrm{Sym}(H)$ admitted at $\alpha$ with $|s-u^{-}_{\delta}(\hat{x})|<\varepsilon$, $|t-u^{+}_{\delta}(\hat{y})|<\varepsilon$ and, writing $P_{1}=\alpha(x_{1}-y_{1})$,

$$F^{-}_{\delta}(x_{1},s,P_{1},X)\le\varepsilon+\omega(\tau_{1}),\qquad -\varepsilon-\omega(\tau_{2})\le F^{+}_{\delta}(y_{1},t,P_{1},Y). \tag{6.1}$$

Since $(X,Y)$ is admitted at $\alpha$, $X\preceq Y$ by The Second-Order Structure Condition for an Equation Operator on a Hilbert Triple §admitted. Write $d_{1}=|x_{1}-y_{1}|_{H}$. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle, $d_{1}\le|x_{1}-\hat{x}|_{H}+d_{0}+|\hat{y}-y_{1}|_{H}<d_{0}+2\varepsilon$, so $|P_{1}|_{H}=\alpha d_{1}\le\alpha d_{0}+2\alpha\varepsilon\le G+2\alpha$ (Step 5, $\varepsilon\le1$), and by (2.1)

$$\delta|P_{1}|_{H}^{2}\le\delta(G+2\alpha)^{2}\le\eta . \tag{6.2}$$

Step 7 (expanding the $\delta$-shifts). We have $x_{1},y_{1}\in D(A)\subseteq V$ by Hilbert Triples: Standing Notation and Background §operator; put $a_{1}=Ax_{1}$ and $b_{1}=Ay_{1}$, elements of $H$. By Second-Order Equation Operator on an Open Subset of a Hilbert Triple and Its $\delta$-Shifts §shifted and the formula for $F$,

$$F^{-}_{\delta}(x_{1},s,P_{1},X)=\lambda_{0}\bigl(s+\delta h(x_{1})\bigr)-\tfrac{\nu}{2}\mathrm{Tr}_{f}\bigl(X|_{V}+\delta I_{V}\bigr)+\tfrac12\Gamma(P_{1}+\delta a_{1},P_{1}+\delta a_{1})+\langle a_{1}+B(x_{1}),P_{1}+\delta a_{1}\rangle_{H}-g(x_{1}),$$ $$F^{+}_{\delta}(y_{1},t,P_{1},Y)=\lambda_{0}\bigl(t-\delta h(y_{1})\bigr)-\tfrac{\nu}{2}\mathrm{Tr}_{f}\bigl(Y|_{V}-\delta I_{V}\bigr)+\tfrac12\Gamma(P_{1}-\delta b_{1},P_{1}-\delta b_{1})+\langle b_{1}+B(y_{1}),P_{1}-\delta b_{1}\rangle_{H}-g(y_{1}).$$

By Elementary Properties of the Trace of a Form along a Square-Summable Sequence §linear and Elementary Properties of the Trace of a Form along a Square-Summable Sequence §identity, $\mathrm{Tr}_{f}(X|_{V}+\delta I_{V})=\mathrm{Tr}_{f}(X|_{V})+\delta\sigma(f)$ and $\mathrm{Tr}_{f}(Y|_{V}-\delta I_{V})=\mathrm{Tr}_{f}(Y|_{V})-\delta\sigma(f)$. Subtracting the second inequality of (6.1) from the first and solving for $\lambda_{0}(s-t)$,

$$\lambda_{0}(s-t)\le2\varepsilon+\omega(\tau_{1})+\omega(\tau_{2})+T_{1}+T_{2}+\nu\delta\sigma(f)+T_{3}+T_{4}+T_{5}, \tag{7.1}$$

where

$$T_{1}=-\lambda_{0}\delta\bigl(h(x_{1})+h(y_{1})\bigr),\qquad T_{2}=\tfrac{\nu}{2}\bigl(\mathrm{Tr}_{f}(X|_{V})-\mathrm{Tr}_{f}(Y|_{V})\bigr),$$ $$T_{3}=\tfrac12\Gamma(P_{1}-\delta b_{1},P_{1}-\delta b_{1})-\tfrac12\Gamma(P_{1}+\delta a_{1},P_{1}+\delta a_{1}),$$ $$T_{4}=-\langle a_{1}+B(x_{1}),P_{1}+\delta a_{1}\rangle_{H}+\langle b_{1}+B(y_{1}),P_{1}-\delta b_{1}\rangle_{H},\qquad T_{5}=g(x_{1})-g(y_{1}).$$

We estimate these terms in turn.

$T_{1}\le0$, because $0<\lambda_{0}$, $0<\delta$ and $h(x_{1}),h(y_{1})\ge0$ by The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds §nonneg.

$T_{2}\le0$: from $X\preceq Y$, Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity §restriction gives $X|_{V}\preceq Y|_{V}$ in $\mathrm{Sym}(V)$, hence $\mathrm{Tr}_{f}(X|_{V})\le\mathrm{Tr}_{f}(Y|_{V})$ by Elementary Properties of the Trace of a Form along a Square-Summable Sequence §monotone, and $0\le\nu$.

For $T_{3}$, bilinearity and symmetry of $\Gamma$ (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §form) give

$$T_{3}=-\delta\,\Gamma(P_{1},a_{1}+b_{1})+\tfrac{\delta^{2}}{2}\bigl(\Gamma(b_{1},b_{1})-\Gamma(a_{1},a_{1})\bigr).$$

By the definition of the order in Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §order, $0_{\mathrm{Sym}}\preceq\Gamma\preceq I_{H}$ means $0\le\Gamma(w,w)\le|w|_{H}^{2}$ for every $w\in H$. Taking $w=2P_{1}+a_{1}+b_{1}$ and expanding, $0\le4\Gamma(P_{1},P_{1})+4\Gamma(P_{1},a_{1}+b_{1})+\Gamma(a_{1}+b_{1},a_{1}+b_{1})$, so, with the parallelogram law Elementary Identities in a Real Inner Product Space §parallelogram,

$$-4\Gamma(P_{1},a_{1}+b_{1})\le4|P_{1}|_{H}^{2}+|a_{1}+b_{1}|_{H}^{2}\le4|P_{1}|_{H}^{2}+2|a_{1}|_{H}^{2}+2|b_{1}|_{H}^{2}.$$

Using also $\Gamma(b_{1},b_{1})\le|b_{1}|_{H}^{2}$, $0\le\Gamma(a_{1},a_{1})$ and $\delta^{2}\le\delta$ (as $0<\delta<1$),

$$T_{3}\le\delta|P_{1}|_{H}^{2}+\tfrac{\delta}{2}|a_{1}|_{H}^{2}+\tfrac{\delta}{2}|b_{1}|_{H}^{2}+\tfrac{\delta}{2}|b_{1}|_{H}^{2}.$$

For $T_{4}$, expanding by Elementary Identities in a Real Inner Product Space §bilinear,

$$T_{4}=-\langle a_{1}-b_{1},P_{1}\rangle_{H}-\langle B(x_{1})-B(y_{1}),P_{1}\rangle_{H}-\delta|a_{1}|_{H}^{2}-\delta|b_{1}|_{H}^{2}-\delta\langle B(x_{1}),Ax_{1}\rangle_{H}-\delta\langle B(y_{1}),Ay_{1}\rangle_{H}.$$

Since $x_{1}-y_{1}\in V$, Hilbert Triples: Standing Notation and Background §operator gives $\langle Ax_{1},x_{1}-y_{1}\rangle_{H}=\langle x_{1},x_{1}-y_{1}\rangle_{V}$ and $\langle Ay_{1},x_{1}-y_{1}\rangle_{H}=\langle y_{1},x_{1}-y_{1}\rangle_{V}$, so $\langle a_{1}-b_{1},P_{1}\rangle_{H}=\alpha|x_{1}-y_{1}|_{V}^{2}\ge0$. Since $B$ is a monotone nonlinearity, Monotone, $A$-Monotone and Locally Bounded Nonlinearities on a Hilbert Triple §monotone gives $\langle B(x_{1})-B(y_{1}),P_{1}\rangle_{H}=\alpha\langle B(x_{1})-B(y_{1}),x_{1}-y_{1}\rangle_{H}\ge0$, and Monotone, $A$-Monotone and Locally Bounded Nonlinearities on a Hilbert Triple §a-monotone gives $\langle B(x_{1}),Ax_{1}\rangle_{H}\ge0$ and $\langle B(y_{1}),Ay_{1}\rangle_{H}\ge0$. Hence $T_{4}\le-\delta|a_{1}|_{H}^{2}-\delta|b_{1}|_{H}^{2}$, and

$$T_{3}+T_{4}\le\delta|P_{1}|_{H}^{2}-\tfrac{\delta}{2}|a_{1}|_{H}^{2}\le\delta|P_{1}|_{H}^{2}.$$

Finally $T_{5}\le|g(x_{1})-g(y_{1})|\le\ell d_{1}\le\ell d_{0}+2\ell\varepsilon$, by claim 3 of Properties of the Absolute Value in an Ordered Field, the hypothesis on $g$ and Step 6.

Inserting these bounds, (2.1) and (6.2) into (7.1),

$$\lambda_{0}(s-t)\le2\varepsilon+\omega(\tau_{1})+\omega(\tau_{2})+\nu\delta\sigma(f)+\delta|P_{1}|_{H}^{2}+\ell d_{0}+2\ell\varepsilon\le(2+2\ell)\varepsilon+4\eta+\ell d_{0}. \tag{7.2}$$

Step 8 (the lower bound and the completion of the square). By claim 9 of Properties of the Absolute Value in an Ordered Field, $s>u^{-}_{\delta}(\hat{x})-\varepsilon$ and $t<u^{+}_{\delta}(\hat{y})+\varepsilon$, so

$$s-t>u^{-}_{\delta}(\hat{x})-u^{+}_{\delta}(\hat{y})-2\varepsilon=\Phi(\hat{x},\hat{y})+\tfrac{\alpha}{2}d_{0}^{2}-2\varepsilon .$$

Multiplying by $\lambda_{0}>0$ and combining with (7.2),

$$\lambda_{0}\Phi(\hat{x},\hat{y})\le(2+2\lambda_{0}+2\ell)\varepsilon+4\eta+\Bigl(\ell d_{0}-\tfrac{\lambda_{0}\alpha}{2}d_{0}^{2}\Bigr)\le5\eta+\frac{\ell^{2}}{2\lambda_{0}\alpha},$$

where we used the choice of $\varepsilon$ and $\ell d_{0}-\tfrac{\lambda_{0}\alpha}{2}d_{0}^{2}\le\tfrac{\ell^{2}}{2\lambda_{0}\alpha}$, which holds because the difference of the two sides is $\tfrac{\lambda_{0}\alpha}{2}\bigl(d_{0}-\tfrac{\ell}{\lambda_{0}\alpha}\bigr)^{2}\ge0$. With (4.2) and $\lambda_{0}>0$ this gives $\lambda_{0}(L_{0}-2\eta)\le5\eta+\tfrac{\ell^{2}}{2\lambda_{0}\alpha}$, which is (1.1).

Step 9 (conclusion of claim 1). Inequality (1.1) holds for every $\eta$ with $0<\eta\le1$. Given a positive real $\mu$, apply it with $\eta$ the smaller of $1$ and $\tfrac{\mu}{5+2\lambda_{0}}$ to obtain $\lambda_{0}L_{0}\le\tfrac{\ell^{2}}{2\lambda_{0}\alpha}+\mu$; so Comparison of Real Numbers with Arbitrary Positive Slack §slack-above gives $\lambda_{0}L_{0}\le\tfrac{\ell^{2}}{2\lambda_{0}\alpha}$, and dividing by $\lambda_{0}>0$, $L_{0}\le\tfrac{\ell^{2}}{2\lambda_{0}^{2}\alpha}$. As $\alpha>1$ and $x_{0},y_{0}\in V$ were arbitrary, claim 1 is proved.

Proof of claims 2 and 3.

Step 10 (two consequences of claim 1 on $V$). Let $x,y\in V$ and $d=|x-y|_{H}$.

(a) $u(x)-u(y)\le\tfrac12d^{2}+\tfrac12c^{2}=cd+\tfrac12(d-c)^{2}$. Indeed, let $\mu$ be a positive real and put $\alpha=1+\tfrac{2\mu}{d^{2}+1}$, so $\alpha>1$. By claim 1, $u(x)-u(y)\le\tfrac{\alpha}{2}d^{2}+\tfrac{c^{2}}{2\alpha}$; here $\tfrac{c^{2}}{2\alpha}\le\tfrac{c^{2}}{2}$ since $\tfrac{1}{\alpha}<1$, and $\tfrac{\alpha}{2}d^{2}=\tfrac12d^{2}+\tfrac{\mu d^{2}}{d^{2}+1}\le\tfrac12d^{2}+\mu$. Hence $u(x)-u(y)\le\tfrac12d^{2}+\tfrac12c^{2}+\mu$ for every positive $\mu$, and Comparison of Real Numbers with Arbitrary Positive Slack §slack-above gives the claim; the identity $\tfrac12d^{2}+\tfrac12c^{2}=cd+\tfrac12(d-c)^{2}$ is elementary.

(b) If $d\le c$, then $u(x)-u(y)\le cd$. If $d=c$ this is (a). If $d<c$ and $d=0$, then $x=y$ by Elementary Identities in a Real Inner Product Space §vanishing, and $u(x)-u(y)=0=cd$. If $0<d<c$, then $\alpha=\tfrac{c}{d}>1$, and claim 1 gives $u(x)-u(y)\le\tfrac{c}{2d}d^{2}+\tfrac{c^{2}d}{2c}=cd$.

(c) Consequently, for every real $e\ge0$ with $d\le c+e$ we have $u(x)-u(y)\le cd+\tfrac12e^{2}$: if $d\le c$ this follows from (b) as $0\le\tfrac12e^{2}$; otherwise $0<d-c\le e$, so $(d-c)^{2}\le e^{2}$ by Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, and (a) applies.

Step 11 (Lipschitz bound on $V$). Let $x,y\in V$ and $d=|x-y|_{H}$; we show $u(x)-u(y)\le Kd$. If $d\le c$, then $u(x)-u(y)\le cd\le Kd$ by Step 10(b). If $c<d<1$, then $c^{2}\le d^{2}$ by Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, so Step 10(a) gives $u(x)-u(y)\le\tfrac12d^{2}+\tfrac12c^{2}\le d^{2}=d\cdot d\le d\le Kd$, as $0\le d<1\le K$. If $1\le d$, then $u(x)-u(y)\le|u(x)|+|u(y)|\le2C'\le2C'd\le Kd$.

Step 12 (passage from $V$ to $H$). Let $x,y\in H$ and let $\mu$ be a real with $0<\mu\le1$. Since $u$ is continuous on $H$ (Real Hilbert Spaces: Standing Notation and Background §topology), there is a positive real $\theta$ such that every $z\in H$ with $|z-x|_{H}<\theta$ satisfies $|u(z)-u(x)|<\mu$ and every $z\in H$ with $|z-y|_{H}<\theta$ satisfies $|u(z)-u(y)|<\mu$. By The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds §dense, applied with $U=H$ and the smaller of $\theta$ and $\mu$, there are $x',y'\in D(A)\subseteq V$ with $|x'-x|_{H}<\min\{\theta,\mu\}$ and $|y'-y|_{H}<\min\{\theta,\mu\}$. Put $d'=|x'-y'|_{H}$. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle, $d'\le|x-y|_{H}+2\mu$, and by claim 9 of Properties of the Absolute Value in an Ordered Field,

$$u(x)-u(y)<u(x')-u(y')+2\mu . \tag{12.1}$$

Claim 2. By Step 11 and (12.1), $u(x)-u(y)<Kd'+2\mu\le K|x-y|_{H}+(2K+2)\mu$. Given a positive real $\mu'$, taking $\mu$ the smaller of $1$ and $\tfrac{\mu'}{2K+2}$ gives $u(x)-u(y)\le K|x-y|_{H}+\mu'$, so $u(x)-u(y)\le K|x-y|_{H}$ by Comparison of Real Numbers with Arbitrary Positive Slack §slack-above. Exchanging the roles of $x$ and $y$, and using $|y-x|_{H}=|x-y|_{H}$ (Elementary Identities in a Real Inner Product Space §homogeneity), gives $u(y)-u(x)\le K|x-y|_{H}$, and claim 6 of Properties of the Absolute Value in an Ordered Field yields $|u(x)-u(y)|\le\bigl(\tfrac{\ell_{g}}{\lambda_{0}}+2C'+1\bigr)|x-y|_{H}$.

Claim 3. Assume $|x-y|_{H}\le c$. Then $d'\le c+2\mu$, so Step 10(c) with $e=2\mu$ gives $u(x')-u(y')\le cd'+2\mu^{2}\le c|x-y|_{H}+2c\mu+2\mu$, using $\mu^{2}\le\mu$. By (12.1), $u(x)-u(y)<c|x-y|_{H}+(2c+4)\mu$. Given a positive real $\mu'$, taking $\mu$ the smaller of $1$ and $\tfrac{\mu'}{2c+4}$ and applying Comparison of Real Numbers with Arbitrary Positive Slack §slack-above gives $u(x)-u(y)\le c|x-y|_{H}$. The hypothesis $|y-x|_{H}\le c$ is symmetric, so likewise $u(y)-u(x)\le c|x-y|_{H}$, and claim 6 of Properties of the Absolute Value in an Ordered Field yields $|u(x)-u(y)|\le\tfrac{\ell_{g}}{\lambda_{0}}|x-y|_{H}$. This holds also when $\ell_{g}=0$, in which case the hypothesis forces $x=y$.