Proof of Lions' Lemma on a Hilbert Space: Test Data and Form Bounds at a Sequentially Strict Maximum of a Quadratically Penalised Difference

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

Notation. Let $\Lambda$, $\Lambda^{\sharp}$, $P$, the forms $M^{\Lambda}$ for $M\in\mathcal{S}(m)$, the projection form $\Pi$ and the tail form $N$ be those determined by $e$ as in Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions, all of whose clauses are used throughout; in particular $\Lambda$ and $\Lambda^{\sharp}$ are linear, $P=\Lambda^{\sharp}\Lambda$, $\lVert\Lambda x\rVert=|Px|\le|x|$, $|\Lambda^{\sharp}\xi|=\lVert\xi\rVert$, $|x|^{2}=\Pi(x,x)+N(x,x)$ and $N(x,x)=|x-Px|^{2}$, so that $|x-Px|\le|x|$. Since $|e_{1}|=1$ we have $H\ne\{0_{H}\}$, so $\lVert I\rVert=1$ by claim 3 of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity. Write

$$M_{0}=\Phi(\bar{x},\bar{y}),\qquad \bar{z}=(\bar{x}-\bar{y})-P(\bar{x}-\bar{y}),$$

and let $g,k:H\to\mathbb{R}$ be given by

$$g(x)=-\alpha\,\langle\bar{z},x-\bar{x}\rangle-\alpha\,N(x-\bar{x},x-\bar{x}), \qquad k(y)=\alpha\,\langle\bar{z},y-\bar{y}\rangle-\alpha\,N(y-\bar{y},y-\bar{y}).$$

By clause 1 of Sequentially Strict Maxima and Minima on a Subset of a Metric Space (i), $\Phi(x,y)\le M_{0}$ for all $(x,y)\in A\times A$. Every component of $e$ lies in $A$, so $A$ is nonempty.

Claim 1 (splitting off the tail of the penalty). For all $x,y\in H$,

$$-\tfrac{\alpha}{2}\,N(x-y,x-y)\ \ge\ -\tfrac{\alpha}{2}\,|\bar{z}|^{2}+g(x)+k(y),$$

with equality when $x=\bar{x}$ and $y=\bar{y}$; moreover $g(x)\le\tfrac{\alpha}{4}|\bar{z}|^{2}$ and $k(y)\le\tfrac{\alpha}{4}|\bar{z}|^{2}$ for all $x,y\in H$.

Proof. Put $a=(x-\bar{x})-P(x-\bar{x})$ and $b=(y-\bar{y})-P(y-\bar{y})$. Since $P$ is linear,

$$(x-y)-P(x-y)=a-b+\bar{z},$$

because $(x-y)=(x-\bar{x})-(y-\bar{y})+(\bar{x}-\bar{y})$. By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (iii),

$$N(x-y,x-y)=|a-b+\bar{z}|^{2},\qquad N(x-\bar{x},x-\bar{x})=|a|^{2},\qquad N(y-\bar{y},y-\bar{y})=|b|^{2},$$

and, using the second expression $N(z,w)=\langle z-Pz,w\rangle$ there together with the first,

$$\langle\bar{z},x-\bar{x}\rangle=N(\bar{x}-\bar{y},x-\bar{x})=\langle\bar{z},a\rangle, \qquad \langle\bar{z},y-\bar{y}\rangle=\langle\bar{z},b\rangle .$$

By Elementary Identities in a Real Inner Product Space (v) applied twice, and by Elementary Identities in a Real Inner Product Space (i),

$$|a-b+\bar{z}|^{2}=|a-b|^{2}+2\langle a-b,\bar{z}\rangle+|\bar{z}|^{2} =|a-b|^{2}+2\langle\bar{z},a\rangle-2\langle\bar{z},b\rangle+|\bar{z}|^{2},$$

and $|a-b|^{2}=|a|^{2}-2\langle a,b\rangle+|b|^{2}$ while $0\le|a+b|^{2}=|a|^{2}+2\langle a,b\rangle+|b|^{2}$, so $-2\langle a,b\rangle\le|a|^{2}+|b|^{2}$ and therefore $|a-b|^{2}\le2|a|^{2}+2|b|^{2}$. Hence

$$N(x-y,x-y)\le|\bar{z}|^{2}+2\langle\bar{z},a\rangle-2\langle\bar{z},b\rangle+2|a|^{2}+2|b|^{2}.$$

Multiplying by the positive number $\tfrac{\alpha}{2}$ (claim 5 of Elementary Arithmetic in an Ordered Field) and reversing the sign (claim 4 of Elementary Order Arithmetic in an Ordered Field) gives

$$-\tfrac{\alpha}{2}N(x-y,x-y)\ \ge\ -\tfrac{\alpha}{2}|\bar{z}|^{2}-\alpha\langle\bar{z},a\rangle+\alpha\langle\bar{z},b\rangle-\alpha|a|^{2}-\alpha|b|^{2} =-\tfrac{\alpha}{2}|\bar{z}|^{2}+g(x)+k(y),$$

by the identities displayed above. If $x=\bar{x}$ and $y=\bar{y}$ then $a=b=0_{H}$, so $g(\bar{x})=0$, $k(\bar{y})=0$ and $N(\bar{x}-\bar{y},\bar{x}-\bar{y})=|\bar{z}|^{2}$, and both sides equal $-\tfrac{\alpha}{2}|\bar{z}|^{2}$.

For the upper bounds, The Cauchy–Schwarz Inequality in a Real Inner Product Space and claims 1 and 3 of Properties of the Absolute Value in an Ordered Field give $-\langle\bar{z},a\rangle\le|\langle\bar{z},a\rangle|\le|\bar{z}|\,|a|$, so $g(x)\le\alpha|\bar{z}||a|-\alpha|a|^{2}$ by claim 5 of Elementary Arithmetic in an Ordered Field. Since $0\le\alpha\bigl(|a|-\tfrac{|\bar{z}|}{2}\bigr)^{2}=\alpha|a|^{2}-\alpha|\bar{z}||a|+\tfrac{\alpha}{4}|\bar{z}|^{2}$, we get $\alpha|\bar{z}||a|-\alpha|a|^{2}\le\tfrac{\alpha}{4}|\bar{z}|^{2}$, whence $g(x)\le\tfrac{\alpha}{4}|\bar{z}|^{2}$. The same computation with $b$ in place of $a$ bounds $k$. This proves Claim 1.

Claim 2 (the decoupled functional). Let $\hat{u},\hat{v}:A\to\mathbb{R}$ be given by $\hat{u}(x)=u(x)+g(x)$ and $\hat{v}(y)=v(y)-k(y)$. Then the set of values of $\hat{u}$ is bounded above and that of $\hat{v}$ is bounded below;

$$\hat{u}(x)-\hat{v}(y)-\tfrac{\alpha}{2}\lVert\Lambda x-\Lambda y\rVert^{2}-\tfrac{\alpha}{2}|\bar{z}|^{2}\ \le\ \Phi(x,y)\qquad\text{for all }x,y\in A,$$

with equality when $x=\bar{x}$, $y=\bar{y}$; and

$$\hat{u}(x)-\hat{v}(y)-\tfrac{\alpha}{2}\lVert\Lambda x-\Lambda y\rVert^{2}\ \le\ \hat{u}(\bar{x})-\hat{v}(\bar{y})-\tfrac{\alpha}{2}\lVert\Lambda\bar{x}-\Lambda\bar{y}\rVert^{2}\qquad\text{for all }x,y\in A .$$

Proof. By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (i) and Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (iii), $\lVert\Lambda x-\Lambda y\rVert=\lVert\Lambda(x-y)\rVert$ and $|x-y|^{2}=\lVert\Lambda(x-y)\rVert^{2}+N(x-y,x-y)$. Hence, by Claim 1,

$$\Phi(x,y)=u(x)-v(y)-\tfrac{\alpha}{2}\lVert\Lambda x-\Lambda y\rVert^{2}-\tfrac{\alpha}{2}N(x-y,x-y) \ \ge\ u(x)+g(x)-\bigl(v(y)-k(y)\bigr)-\tfrac{\alpha}{2}\lVert\Lambda x-\Lambda y\rVert^{2}-\tfrac{\alpha}{2}|\bar{z}|^{2},$$

which is the first display; equality at $(\bar{x},\bar{y})$ follows from the equality case of Claim 1. For the second display, the first one and $\Phi(x,y)\le M_{0}$ give

$$\hat{u}(x)-\hat{v}(y)-\tfrac{\alpha}{2}\lVert\Lambda x-\Lambda y\rVert^{2}\le\Phi(x,y)+\tfrac{\alpha}{2}|\bar{z}|^{2}\le M_{0}+\tfrac{\alpha}{2}|\bar{z}|^{2},$$

and by the equality case the right-hand side equals $\hat{u}(\bar{x})-\hat{v}(\bar{y})-\tfrac{\alpha}{2}\lVert\Lambda\bar{x}-\Lambda\bar{y}\rVert^{2}$. Finally, let $C_{u}$ be an upper bound for the values of $u$ and $C_{v}$ one for the values of $-v$; by Claim 1, $\hat{u}(x)\le C_{u}+\tfrac{\alpha}{4}|\bar{z}|^{2}$ and $-\hat{v}(y)=-v(y)+k(y)\le C_{v}+\tfrac{\alpha}{4}|\bar{z}|^{2}$, so the values of $\hat{v}$ are bounded below. This proves Claim 2.

Claim 3 (the two forms, and clauses 2 to 5). Claim 2 supplies the hypotheses of Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes for the data $A$, $e$, $\hat{u}$, $\hat{v}$, $\alpha$, $\bar{x}$, $\bar{y}$. Let $U$, $\mathcal{V}$, $U^{*}$, $\mathcal{V}_{*}$, $\bar{\zeta}=\Lambda\bar{x}$ and $\bar{\omega}=\Lambda\bar{y}$ be as there. By Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes (ii), $U^{*}$ is upper semicontinuous on $\mathbb{R}^{m}$, $\mathcal{V}_{*}$ is lower semicontinuous on $\mathbb{R}^{m}$, and

$$U^{*}(\zeta)-\mathcal{V}_{*}(\omega)-\tfrac{\alpha}{2}\lVert\zeta-\omega\rVert^{2}\ \le\ U^{*}(\bar{\zeta})-\mathcal{V}_{*}(\bar{\omega})-\tfrac{\alpha}{2}\lVert\bar{\zeta}-\bar{\omega}\rVert^{2} \qquad\text{for all }\zeta,\omega\in\mathbb{R}^{m};$$

and by Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes (iii), $U^{*}(\bar{\zeta})=\hat{u}(\bar{x})$ and $\mathcal{V}_{*}(\bar{\omega})=\hat{v}(\bar{y})$.

The set $\mathbb{R}^{m}$ is open in $\mathbb{R}^{m}$ by claim 1 of Euclidean Space Is Open in Itself, so Ishii's Lemma: Test Data and Matrix Bounds at a Maximum of a Quadratically Penalised Difference applies with $m$ in the role of $n$, with $\Omega=\mathbb{R}^{m}$, with $U^{*}$ and $\mathcal{V}_{*}$ in the roles of $u$ and $v$, with $\alpha$, and with $\bar{\zeta}$ and $\bar{\omega}$ in the roles of the two points; its local hypothesis holds with $\delta=1$ because the displayed inequality holds for all $\zeta,\omega$. Writing $p_{0}=\alpha(\bar{\zeta}-\bar{\omega})$, we obtain $X_{0},Y_{0}\in\mathcal{S}(m)$ such that (i) the quadruple $(\bar{\zeta},U^{*}(\bar{\zeta}),p_{0},X_{0})$ is approximable by test data from above for $U^{*}$ and $(\bar{\omega},\mathcal{V}_{*}(\bar{\omega}),p_{0},Y_{0})$ is approximable by test data from below for $\mathcal{V}_{*}$, the open set being $\mathbb{R}^{m}$ in both cases (clause 1 there); (ii) for all $\xi,\eta\in\mathbb{R}^{m}$,

$$-3\alpha\bigl(\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}\bigr)\le\xi\cdot(X_{0}\xi)-\eta\cdot(Y_{0}\eta)\le3\alpha\lVert\xi-\eta\rVert^{2}$$

(clause 3 there); (iii) $X_{0}\preceq Y_{0}$ (clause 4 there); and (iv) $\lVert X_{0}\rVert\le6\alpha$ and $\lVert Y_{0}\rVert\le6\alpha$ (clause 5 there).

Put $X=X_{0}^{\Lambda}$ and $Y=Y_{0}^{\Lambda}$, elements of $\mathrm{Sym}(H)$ by Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (ii). Clause 5 is then the identity $M^{\Lambda}(z,w)=M^{\Lambda}(Pz,Pw)$ of that clause, clause 4 follows from (iv) and the bound $\lVert M^{\Lambda}\rVert\le\lVert M\rVert$ there, and clause 3 follows from (iii) and the order transfer there. For clause 2, apply (ii) with $\xi=\Lambda z$ and $\eta=\Lambda w$, so that $\xi\cdot(X_{0}\xi)=X(z,z)$ and $\eta\cdot(Y_{0}\eta)=Y(w,w)$. Since $0\le\lVert\Lambda(z-w)\rVert\le|z-w|$, two applications of claim 5 of Elementary Arithmetic in an Ordered Field give $\lVert\Lambda z-\Lambda w\rVert^{2}=\lVert\Lambda(z-w)\rVert^{2}\le|z-w|^{2}$, and multiplying by the nonnegative $3\alpha$ gives the upper bound of clause 2. Likewise $\lVert\Lambda z\rVert^{2}+\lVert\Lambda w\rVert^{2}\le|z|^{2}+|w|^{2}$, so multiplying by $3\alpha$ and reversing the sign gives $-3\alpha(|z|^{2}+|w|^{2})\le-3\alpha(\lVert\Lambda z\rVert^{2}+\lVert\Lambda w\rVert^{2})$, which with (ii) gives the lower bound.

We record for later use that $\bar{\zeta}-\bar{\omega}=\Lambda(\bar{x}-\bar{y})$, so that by the linearity of $\Lambda^{\sharp}$ and $P=\Lambda^{\sharp}\Lambda$,

$$\Lambda^{\sharp}p_{0}=\alpha\,P(\bar{x}-\bar{y}), \qquad\text{whence}\qquad p=\alpha(\bar{x}-\bar{y})=\alpha P(\bar{x}-\bar{y})+\alpha\bar{z}=\Lambda^{\sharp}p_{0}+\alpha\bar{z}.$$

This proves Claim 3.

Claim 4 (clause 1). Let $\varepsilon\in\mathbb{R}$ be positive. We produce $x_{1},y_{1}\in A$ and $\varphi,\psi\in C^{2}(H)$ such that the function on $A$ with value $u(x)-\varphi(x)$ at $x$ has a local maximum at $x_{1}$ relative to $A$, the function on $A$ with value $v(y)-\psi(y)$ at $y$ has a local minimum at $y_{1}$ relative to $A$, and

$$|x_{1}-\bar{x}|<\varepsilon,\quad |u(x_{1})-u(\bar{x})|<\varepsilon,\quad \bigl|D\varphi(x_{1})-p\bigr|<\varepsilon,\quad \bigl\lVert D^{2}\varphi(x_{1})-(X+2\alpha N)\bigr\rVert<\varepsilon,$$ $$|y_{1}-\bar{y}|<\varepsilon,\quad |v(y_{1})-v(\bar{y})|<\varepsilon,\quad \bigl|D\psi(y_{1})-p\bigr|<\varepsilon,\quad \bigl\lVert D^{2}\psi(y_{1})-(Y-2\alpha N)\bigr\rVert<\varepsilon .$$

Since $\varepsilon$ is arbitrary, this is clause 1, by Quadruple Approximable by Test Data on a Subset of a Real Inner Product Space (i) and Quadruple Approximable by Test Data on a Subset of a Real Inner Product Space (ii).

Step 1 (the localisation radius). Let $\varepsilon_{1}$ be the lesser of $\varepsilon$ and $\tfrac{\varepsilon}{8\alpha}$, positive because it is one of them. The function $u$ and the function $-v$ have closed superlevel sets in $H$, $A$ is nonempty, and $\Phi$ attains a sequentially strict maximum on $A\times A$ at $(\bar{x},\bar{y})$, so Localisation at a Sequentially Strict Maximum of a Quadratically Penalised Difference (i) provides a positive $\eta_{0}\in\mathbb{R}$ such that every $(x,y)\in A\times A$ with $M_{0}-\eta_{0}<\Phi(x,y)$ satisfies

$$|x-\bar{x}|<\varepsilon_{1},\quad |y-\bar{y}|<\varepsilon_{1},\quad |u(x)-u(\bar{x})|<\varepsilon_{1},\quad |v(y)-v(\bar{y})|<\varepsilon_{1}.$$

Step 2 (accuracy of the Euclidean data). Put $D_{0}=\lVert\bar{\zeta}-\bar{\omega}\rVert$. Let $\varepsilon'$ be a positive real number with

$$\varepsilon'\le1,\qquad \varepsilon'\le\tfrac{\varepsilon}{16},\qquad \varepsilon'\le\tfrac{\eta_{0}}{16},\qquad 2|\alpha|\,\varepsilon'\,(D_{0}+2)\le\tfrac{\eta_{0}}{16},$$

which exists: the first three are satisfied by the least of $1$, $\tfrac{\varepsilon}{16}$ and $\tfrac{\eta_{0}}{16}$, and the fourth by any positive number at most $\tfrac{\eta_{0}}{16\,(2|\alpha|(D_{0}+2)+1)}$, so the least of these four positive numbers serves. Let $\eta_{1}$ be a positive real with $2\eta_{1}\le\tfrac{\varepsilon}{16}$.

Step 3 (the Euclidean test data). By (i) of Claim 3 and Quadruple Approximable by Test-Function Data (i), applied with the accuracy $\varepsilon'$, there are $\zeta_{1}\in\mathbb{R}^{m}$ and a function $\chi:\mathbb{R}^{m}\to\mathbb{R}$ of class $C^{2}$ on $\mathbb{R}^{m}$ such that the function with value $U^{*}(\zeta)-\chi(\zeta)$ at $\zeta$ has a local maximum at $\zeta_{1}$ relative to $\mathbb{R}^{m}$ and

$$d_{E}(\zeta_{1},\bar{\zeta})<\varepsilon',\quad |U^{*}(\zeta_{1})-U^{*}(\bar{\zeta})|<\varepsilon',\quad \lVert D\chi(\zeta_{1})-p_{0}\rVert<\varepsilon',\quad d_{\mathcal{S}(m)}\bigl(D^{2}\chi(\zeta_{1}),X_{0}\bigr)<\varepsilon' .$$

Likewise, by Quadruple Approximable by Test-Function Data (ii), there are $\omega_{1}\in\mathbb{R}^{m}$ and $\theta:\mathbb{R}^{m}\to\mathbb{R}$ of class $C^{2}$ on $\mathbb{R}^{m}$ such that the function with value $\mathcal{V}_{*}(\omega)-\theta(\omega)$ at $\omega$ has a local minimum at $\omega_{1}$ relative to $\mathbb{R}^{m}$ and the four analogous inequalities hold with $\omega_{1},\theta,\mathcal{V}_{*},\bar{\omega},Y_{0}$ in place of $\zeta_{1},\chi,U^{*},\bar{\zeta},X_{0}$.

Step 4 (replacing the test functions by coordinate quadratics). Apply Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (v) with $\Omega=\mathbb{R}^{m}$, with $\chi$, $\zeta_{1}$ and the positive $\eta_{1}$: the associated quadratic

$$T_{0}(\zeta)=\chi(\zeta_{1})+D\chi(\zeta_{1})\cdot(\zeta-\zeta_{1})+\tfrac{1}{2}(\zeta-\zeta_{1})\cdot\bigl(D^{2}\chi(\zeta_{1})(\zeta-\zeta_{1})\bigr)+\eta_{1}\lVert\zeta-\zeta_{1}\rVert^{2}$$

satisfies $T_{0}(\zeta_{1})=\chi(\zeta_{1})$, and there is a positive $\rho_{1}$ with $\chi(\zeta)\le T_{0}(\zeta)$ whenever $\lVert\zeta-\zeta_{1}\rVert<\rho_{1}$. Let $\rho_{0}$ be positive and such that $U^{*}(\zeta)-\chi(\zeta)\le U^{*}(\zeta_{1})-\chi(\zeta_{1})$ for every $\zeta$ with $d_{E}(\zeta,\zeta_{1})<\rho_{0}$, as the local maximum property provides. Then for every $\zeta$ with $\lVert\zeta-\zeta_{1}\rVert$ less than both $\rho_{0}$ and $\rho_{1}$,

$$U^{*}(\zeta)-T_{0}(\zeta)\ \le\ U^{*}(\zeta)-\chi(\zeta)\ \le\ U^{*}(\zeta_{1})-\chi(\zeta_{1})=U^{*}(\zeta_{1})-T_{0}(\zeta_{1}).$$

Let $T=T_{0}\circ\Lambda$. By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (iv), $T\in C^{2}(H)$, $T(x)$ depends only on $\Lambda x$, the Hessian $D^{2}T(x)$ equals the constant form

$$b_{u}=\bigl(D^{2}\chi(\zeta_{1})\bigr)^{\Lambda}+2\eta_{1}\,\Pi ,$$

and $\bigl|DT(x)-\Lambda^{\sharp}D\chi(\zeta_{1})\bigr|\le\bigl(\lVert D^{2}\chi(\zeta_{1})\rVert+2\eta_{1}\bigr)\lVert\Lambda x-\zeta_{1}\rVert$ for every $x\in H$. By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (ii) and Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (iii), together with claims 1 and 3 of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity,

$$\lVert b_{u}-X\rVert\le\bigl\lVert\bigl(D^{2}\chi(\zeta_{1})-X_{0}\bigr)^{\Lambda}\bigr\rVert+2\eta_{1}\lVert\Pi\rVert\le\varepsilon'+2\eta_{1}.$$

Applying Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (v) instead with $\theta$, $\omega_{1}$ and the negative number $-\eta_{1}$ gives a quadratic $S_{0}$ with $S_{0}(\omega_{1})=\theta(\omega_{1})$ and a positive $\rho_{1}'$ with $S_{0}(\omega)\le\theta(\omega)$ for $\lVert\omega-\omega_{1}\rVert<\rho_{1}'$; with $\rho_{0}'$ a radius for the local minimum of $\mathcal{V}_{*}-\theta$ at $\omega_{1}$, we get

$$\mathcal{V}_{*}(\omega_{1})-S_{0}(\omega_{1})\ \le\ \mathcal{V}_{*}(\omega)-S_{0}(\omega)$$

whenever $\lVert\omega-\omega_{1}\rVert$ is less than both $\rho_{0}'$ and $\rho_{1}'$. Setting $S=S_{0}\circ\Lambda$, we have $S\in C^{2}(H)$, $S(y)$ depends only on $\Lambda y$, $D^{2}S(y)$ is the constant form $b_{v}=(D^{2}\theta(\omega_{1}))^{\Lambda}-2\eta_{1}\Pi$ with $\lVert b_{v}-Y\rVert\le\varepsilon'+2\eta_{1}$, and $\bigl|DS(y)-\Lambda^{\sharp}D\theta(\omega_{1})\bigr|\le(\lVert D^{2}\theta(\omega_{1})\rVert+2\eta_{1})\lVert\Lambda y-\omega_{1}\rVert$.

Step 5 (the fibre radius). By (iv) of Claim 3 and Euclidean Space and Real Matrices: Standing Notation (v), $\lVert D^{2}\chi(\zeta_{1})\rVert\le\lVert X_{0}\rVert+\varepsilon'\le6\alpha+\varepsilon'$, and likewise for $\theta$; put $K=6\alpha+\varepsilon'+2\eta_{1}$, so that the two gradient estimates of Step 4 hold with the factor $K$. Let $r$ be a positive real number with

$$2r\le\rho_{0},\quad 2r\le\rho_{1},\quad 2r\le\rho_{0}',\quad 2r\le\rho_{1}',\quad r\le1,\quad Kr\le\tfrac{\varepsilon}{16},\quad 2|\alpha|\,r\,(D_{0}+2)\le\tfrac{\eta_{0}}{16},$$

and such that

$$\bigl|T_{0}(\zeta)-T_{0}(\zeta_{1})\bigr|\le\tfrac{\eta_{0}}{16}\ \text{ whenever }\lVert\zeta-\zeta_{1}\rVert\le r, \qquad \bigl|S_{0}(\omega)-S_{0}(\omega_{1})\bigr|\le\tfrac{\eta_{0}}{16}\ \text{ whenever }\lVert\omega-\omega_{1}\rVert\le r .$$

Such an $r$ exists: the first seven conditions hold for every positive $r$ at most the least of the seven positive numbers

$$\tfrac{\rho_{0}}{2},\qquad\tfrac{\rho_{1}}{2},\qquad\tfrac{\rho_{0}'}{2},\qquad\tfrac{\rho_{1}'}{2},\qquad 1,\qquad\tfrac{\varepsilon}{16K},\qquad\tfrac{\eta_{0}}{16\,(2|\alpha|(D_{0}+2)+1)},$$

the sixth of which is defined because $K$ is positive; and for the last two it suffices, by the final estimate of Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (iv), that $r\le1$ and $(\lVert q\rVert+\tfrac12\lVert M\rVert+|\eta|)\,r\le\tfrac{\eta_{0}}{16}$ for each of the two quadratics, since then $\lVert\zeta-\zeta_{1}\rVert\le r\le1$ gives $\lVert\zeta-\zeta_{1}\rVert^{2}\le\lVert\zeta-\zeta_{1}\rVert\le r$. Put

$$A_{u}=\{\,x\in A:\lVert\Lambda x-\zeta_{1}\rVert\le r\,\}, \qquad A_{v}=\{\,y\in A:\lVert\Lambda y-\omega_{1}\rVert\le r\,\}.$$

Step 6 (the auxiliary functions). Let $\Theta_{u},\Theta_{v}:A\to\mathbb{R}$ be given by $\Theta_{u}(x)=\hat{u}(x)-T(x)$ and $\Theta_{v}(y)=-\hat{v}(y)+S(y)$, and put

$$M_{u}=U^{*}(\zeta_{1})-T_{0}(\zeta_{1}),\qquad M_{v}=-\mathcal{V}_{*}(\omega_{1})+S_{0}(\omega_{1}).$$

(a) Upper bounds. Let $x\in A_{u}$. Then $\hat{u}(x)\le U(\Lambda x)\le U^{*}(\Lambda x)$, by the definition of $U$ in Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes (i) and claim 1 of Properties of the Upper Semicontinuous Envelope; and $T(x)=T_{0}(\Lambda x)$ with $\lVert\Lambda x-\zeta_{1}\rVert\le r$, which is less than both $\rho_{0}$ and $\rho_{1}$. So Step 4 gives $\Theta_{u}(x)\le U^{*}(\Lambda x)-T_{0}(\Lambda x)\le M_{u}$. Symmetrically $\Theta_{v}(y)\le M_{v}$ for every $y\in A_{v}$, using $-\hat{v}(y)\le-\mathcal{V}(\Lambda y)\le-\mathcal{V}_{*}(\Lambda y)$ (claim 2 of Properties of the Lower Semicontinuous Envelope, by Duality).

(b) Near-maximisers on a prescribed fibre neighbourhood. Let $\tau$ and $s$ be positive with $s\le r$. Let $s'$ be a positive number, at most $s$, with $\bigl|T_{0}(\zeta)-T_{0}(\zeta_{1})\bigr|\le\tfrac{\tau}{2}$ whenever $\lVert\zeta-\zeta_{1}\rVert\le s'$; such an $s'$ exists by the argument used for $r$ in Step 5. Applying Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes (iv) with $\zeta_{1}$, the radius $s'$ and the accuracy $\tfrac{\tau}{2}$ gives $x\in A$ with $\lVert\Lambda x-\zeta_{1}\rVert\le s'\le s\le r$, hence $x\in A_{u}$, and $U^{*}(\zeta_{1})-\tfrac{\tau}{2}<\hat{u}(x)$. Since $T(x)=T_{0}(\Lambda x)\le T_{0}(\zeta_{1})+\tfrac{\tau}{2}$,

$$\Theta_{u}(x)=\hat{u}(x)-T_{0}(\Lambda x)>U^{*}(\zeta_{1})-\tfrac{\tau}{2}-T_{0}(\zeta_{1})-\tfrac{\tau}{2}=M_{u}-\tau .$$

Applying the same clause of Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes again, this time with $\omega_{1}$ in the role of $\zeta_{1}$, and using the second half of its conclusion (the one concerning $\hat{v}$ and $\mathcal{V}_{*}$) together with $S_{0}$ in place of $T_{0}$, gives in the same way a $y\in A_{v}$ with $\lVert\Lambda y-\omega_{1}\rVert\le s$ and $M_{v}-\tau<\Theta_{v}(y)$. In particular $A_{u}$ and $A_{v}$ are nonempty.

(c) Closed superlevel sets. The functions $g$, $k$, $T$ and $S$ belong to $C^{2}(H)$: for $T$ and $S$ this is Step 4, and for $g$ note that

$$g(x)=-\alpha\langle\bar{z},x\rangle+\alpha\langle\bar{z},\bar{x}\rangle-\alpha\Bigl(N(x,x)-2\langle T_{N}\bar{x},x\rangle+N(\bar{x},\bar{x})\Bigr),$$

by the bilinearity and symmetry of $N$ and claim 1 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space, which exhibits $g$ as a sum of an affine function and a scalar multiple of $x\mapsto\tfrac12N(x,x)$; so claims 1 and 2 of Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$ and claims 2 and 3 of Constants, Sums, Scalar Multiples and Differences of Differentiable Functions on an Open Subset of a Real Inner Product Space give $g\in C^{2}(H)$ with

$$Dg(x)=-\alpha\bar{z}-2\alpha\,T_{N}(x-\bar{x}),\qquad D^{2}g(x)=-2\alpha N ,$$

and likewise $k\in C^{2}(H)$ with $Dk(y)=\alpha\bar{z}-2\alpha T_{N}(y-\bar{y})$ and $D^{2}k(y)=-2\alpha N$. By claim 4 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space and Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (iii), $T_{N}z=z-Pz$, so $|T_{N}z|\le|z|$.

Members of $C^{2}(H)$ are differentiable on $H$ and hence continuous on $H$, by The Classes $C^1$ and $C^2$ on an Open Subset of a Real Inner Product Space (ii) and Basic Properties of Differentiability on an Open Subset of a Real Inner Product Space (ii). Thus $T-g$ is continuous on $H$, and $\Theta_{u}(x)=u(x)-\bigl(T(x)-g(x)\bigr)$, so $\Theta_{u}$ has closed superlevel sets in $H$ as a function on $A$, by claim 3 of Functions with Closed Superlevel Sets: Sequential Characterisation, Semicontinuity, Perturbation and Limits.

The restriction of $\Theta_{u}$ to $A_{u}$ also has closed superlevel sets in $H$. Indeed, let $(x_{j})_{j\in\mathbb{N}}$ be a sequence in $A_{u}$ converging to $x\in H$ and let $t\in\mathbb{R}$ satisfy $t\le\Theta_{u}(x_{j})$ for every $j$. By claim 1 of Functions with Closed Superlevel Sets: Sequential Characterisation, Semicontinuity, Perturbation and Limits applied to $\Theta_{u}$ on $A$ we get $x\in A$ and $t\le\Theta_{u}(x)$. Moreover, for each $j$,

$$\lVert\Lambda x-\zeta_{1}\rVert\le\lVert\Lambda x-\Lambda x_{j}\rVert+\lVert\Lambda x_{j}-\zeta_{1}\rVert\le|x-x_{j}|+r ,$$

by $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence and Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (i); since $|x-x_{j}|$ is smaller than any prescribed positive number for suitable $j$, Comparison of Real Numbers with Arbitrary Positive Slack gives $\lVert\Lambda x-\zeta_{1}\rVert\le r$, so $x\in A_{u}$. By claim 1 of Functions with Closed Superlevel Sets: Sequential Characterisation, Semicontinuity, Perturbation and Limits again, in the converse direction, the restriction has closed superlevel sets in $H$. The same argument applies to $\Theta_{v}$ on $A_{v}$.

Step 7 (the perturbed maxima). Let $\lambda$ be the lesser of $\tfrac{r}{8}$ and $1$, and let $\mu$ be a positive real with

$$2\mu\le\tfrac{\varepsilon}{16},\qquad 16\mu\lambda\le\tfrac{\varepsilon}{16},\qquad 8\mu\lambda^{2}\le\tfrac{\eta_{0}}{16},$$

which exists as the least of three positive numbers of the required form. Then $4\lambda\le\tfrac{r}{2}$.

By (b) with $\tau=\mu\lambda^{2}$ and $s=\tfrac{r}{4}$ there is $x_{0}\in A_{u}$ with $\lVert\Lambda x_{0}-\zeta_{1}\rVert\le\tfrac{r}{4}$ and $M_{u}-\mu\lambda^{2}<\Theta_{u}(x_{0})$. By (a), $M_{u}$ is an upper bound for the values of $\Theta_{u}$ on $A_{u}$, so the least upper bound of those values is at most $\Theta_{u}(x_{0})+\mu\lambda^{2}$. Hence A Perturbed Maximum Principle of Borwein-Preiss Type in a Real Hilbert Space applies to $A_{u}$, $\Theta_{u}$, $\mu$, $\lambda$ and $x_{0}$, and provides two points, written here $c_{1}\in H$ and $x_{1}\in A_{u}$ — they are the points called $\bar{y}$ and $\bar{x}$ in that theorem, and are unrelated to the point $\bar{y}$ of the present lemma — such that

$$|x_{1}-x_{0}|\le4\lambda,\qquad |x_{1}-c_{1}|\le8\lambda,$$

the function on $A_{u}$ with value $\Theta_{u}(x)-\mu|x-c_{1}|^{2}$ at $x$ attains a sequentially strict maximum on $A_{u}$ at $x_{1}$ (claim 2 there), and the least upper bound of the values of $\Theta_{u}$ on $A_{u}$ is at most $\Theta_{u}(x_{1})+2\mu\lambda^{2}$ (claim 3 there). Combining the last bound with $M_{u}-\mu\lambda^{2}<\Theta_{u}(x_{0})$ gives

$$M_{u}-3\mu\lambda^{2}<\Theta_{u}(x_{1}).$$

Also

$$\lVert\Lambda x_{1}-\zeta_{1}\rVert\le\lVert\Lambda(x_{1}-x_{0})\rVert+\lVert\Lambda x_{0}-\zeta_{1}\rVert\le|x_{1}-x_{0}|+\tfrac{r}{4}\le4\lambda+\tfrac{r}{4}\le\tfrac{r}{2}+\tfrac{r}{4}<r .$$

The same construction on the $v$ side provides $c_{2}\in H$ and $y_{1}\in A_{v}$ with $|y_{1}-c_{2}|\le8\lambda$, with $\Theta_{v}-\mu|\cdot-c_{2}|^{2}$ attaining a sequentially strict maximum on $A_{v}$ at $y_{1}$, with $M_{v}-3\mu\lambda^{2}<\Theta_{v}(y_{1})$ and with $\lVert\Lambda y_{1}-\omega_{1}\rVert<r$.

Step 8 (the test functions and the local extrema). Define $\varphi,\psi:H\to\mathbb{R}$ by

$$\varphi(x)=T(x)-g(x)+\mu\,|x-c_{1}|^{2}, \qquad \psi(y)=S(y)+k(y)-\mu\,|y-c_{2}|^{2},$$

so that, for $x,y\in A$,

$$u(x)-\varphi(x)=\Theta_{u}(x)-\mu|x-c_{1}|^{2}, \qquad v(y)-\psi(y)=-\Bigl(\Theta_{v}(y)-\mu|y-c_{2}|^{2}\Bigr).$$

Put $\rho_{2}=r-\lVert\Lambda x_{1}-\zeta_{1}\rVert$, positive by Step 7. If $x\in A$ and $|x-x_{1}|<\rho_{2}$ then

$$\lVert\Lambda x-\zeta_{1}\rVert\le\lVert\Lambda(x-x_{1})\rVert+\lVert\Lambda x_{1}-\zeta_{1}\rVert\le|x-x_{1}|+\lVert\Lambda x_{1}-\zeta_{1}\rVert<r,$$

so $x\in A_{u}$ and hence, the sequentially strict maximum of Step 7 being in particular a maximum on $A_{u}$ (clause 1 of Sequentially Strict Maxima and Minima on a Subset of a Metric Space (i)),

$$u(x)-\varphi(x)=\Theta_{u}(x)-\mu|x-c_{1}|^{2}\le\Theta_{u}(x_{1})-\mu|x_{1}-c_{1}|^{2}=u(x_{1})-\varphi(x_{1}).$$

Thus $u-\varphi$ has a local maximum at $x_{1}$ relative to $A$. The same argument on the $v$ side shows that $\Theta_{v}-\mu|\cdot-c_{2}|^{2}$ has a local maximum at $y_{1}$ relative to $A$, hence that $v-\psi$, being its additive inverse there, has a local minimum at $y_{1}$ relative to $A$.

By Step 4, Step 6(c) and claim 3 of Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$ (applied with $2\mu$ in the role of the coefficient), each of $T$, $g$, $S$, $k$ and the two squared-distance terms belongs to $C^{2}(H)$; so $\varphi,\psi\in C^{2}(H)$ by claims 2 and 4 of Constants, Sums, Scalar Multiples and Differences of Differentiable Functions on an Open Subset of a Real Inner Product Space, with

$$D\varphi(x_{1})=DT(x_{1})-Dg(x_{1})+2\mu(x_{1}-c_{1}), \qquad D^{2}\varphi(x_{1})=b_{u}+2\alpha N+2\mu I,$$ $$D\psi(y_{1})=DS(y_{1})+Dk(y_{1})-2\mu(y_{1}-c_{2}), \qquad D^{2}\psi(y_{1})=b_{v}-2\alpha N-2\mu I .$$

Step 9 (the localisation). Write $\Xi(x,y)=\hat{u}(x)-\hat{v}(y)-\tfrac{\alpha}{2}\lVert\Lambda x-\Lambda y\rVert^{2}$ for $x,y\in A$, and $M_{1}=\Xi(\bar{x},\bar{y})$; by Claim 2, $\Phi(x,y)\ge\Xi(x,y)-\tfrac{\alpha}{2}|\bar{z}|^{2}$ and $M_{1}=M_{0}+\tfrac{\alpha}{2}|\bar{z}|^{2}$. Since $T(x_{1})=T_{0}(\Lambda x_{1})$ and $S(y_{1})=S_{0}(\Lambda y_{1})$,

$$\Xi(x_{1},y_{1})=\Theta_{u}(x_{1})+\Theta_{v}(y_{1})+T_{0}(\Lambda x_{1})-S_{0}(\Lambda y_{1})-\tfrac{\alpha}{2}\lVert\Lambda x_{1}-\Lambda y_{1}\rVert^{2}.$$

By Step 3 and Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes (iii) (through Claim 3),

$$M_{u}+M_{v}=U^{*}(\zeta_{1})-\mathcal{V}_{*}(\omega_{1})-T_{0}(\zeta_{1})+S_{0}(\omega_{1}) >U^{*}(\bar{\zeta})-\mathcal{V}_{*}(\bar{\omega})-2\varepsilon'-T_{0}(\zeta_{1})+S_{0}(\omega_{1}),$$

and $U^{*}(\bar{\zeta})-\mathcal{V}_{*}(\bar{\omega})=\hat{u}(\bar{x})-\hat{v}(\bar{y})=M_{1}+\tfrac{\alpha}{2}D_{0}^{2}$. With $M_{u}-3\mu\lambda^{2}<\Theta_{u}(x_{1})$ and $M_{v}-3\mu\lambda^{2}<\Theta_{v}(y_{1})$ from Step 7, this gives

$$\Xi(x_{1},y_{1})>M_{1}+\tfrac{\alpha}{2}D_{0}^{2}-2\varepsilon'-6\mu\lambda^{2} +\bigl[T_{0}(\Lambda x_{1})-T_{0}(\zeta_{1})\bigr]-\bigl[S_{0}(\Lambda y_{1})-S_{0}(\omega_{1})\bigr]-\tfrac{\alpha}{2}\lVert\Lambda x_{1}-\Lambda y_{1}\rVert^{2}.$$

By Step 7 we have $\lVert\Lambda x_{1}-\zeta_{1}\rVert\le r$ and $\lVert\Lambda y_{1}-\omega_{1}\rVert\le r$, so each of the two bracketed terms has absolute value at most $\tfrac{\eta_{0}}{16}$ by Step 5, and therefore each is at least $-\tfrac{\eta_{0}}{16}$ by claim 3 of Properties of the Absolute Value in an Ordered Field. Next, $\lVert\bar{\zeta}-\Lambda x_{1}\rVert\le\lVert\bar{\zeta}-\zeta_{1}\rVert+\lVert\zeta_{1}-\Lambda x_{1}\rVert<\varepsilon'+r$ and similarly $\lVert\bar{\omega}-\Lambda y_{1}\rVert<\varepsilon'+r$; so Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (vi), applied with $(\bar{\zeta},\bar{\omega})$ and $(\Lambda x_{1},\Lambda y_{1})$, gives $\lVert\Lambda x_{1}-\Lambda y_{1}\rVert\le D_{0}+2(\varepsilon'+r)\le D_{0}+4$ (as $\varepsilon'\le1$ and $r\le1$) and hence

$$\Bigl|\tfrac{\alpha}{2}D_{0}^{2}-\tfrac{\alpha}{2}\lVert\Lambda x_{1}-\Lambda y_{1}\rVert^{2}\Bigr| \le\tfrac{|\alpha|}{2}\,(2D_{0}+4)\cdot2(\varepsilon'+r) =2|\alpha|\,(\varepsilon'+r)\,(D_{0}+2)\le\tfrac{\eta_{0}}{16}+\tfrac{\eta_{0}}{16},$$

by Steps 2 and 5. Since also $2\varepsilon'\le\tfrac{\eta_{0}}{8}$ and $6\mu\lambda^{2}\le\tfrac{\eta_{0}}{16}$, we conclude

$$\Xi(x_{1},y_{1})>M_{1}-\tfrac{\eta_{0}}{8}-\tfrac{\eta_{0}}{16}-\tfrac{\eta_{0}}{16}-\tfrac{\eta_{0}}{16}-\tfrac{\eta_{0}}{16}-\tfrac{\eta_{0}}{16}\ \ge\ M_{1}-\eta_{0},$$

and therefore $\Phi(x_{1},y_{1})\ge\Xi(x_{1},y_{1})-\tfrac{\alpha}{2}|\bar{z}|^{2}>M_{1}-\eta_{0}-\tfrac{\alpha}{2}|\bar{z}|^{2}=M_{0}-\eta_{0}$. By Step 1,

$$|x_{1}-\bar{x}|<\varepsilon_{1}\le\varepsilon,\quad |y_{1}-\bar{y}|<\varepsilon_{1}\le\varepsilon,\quad |u(x_{1})-u(\bar{x})|<\varepsilon_{1}\le\varepsilon,\quad |v(y_{1})-v(\bar{y})|<\varepsilon_{1}\le\varepsilon,$$

which are four of the eight required inequalities.

Step 10 (the gradients and Hessians). By Step 8, Step 6(c) and Claim 3,

$$D\varphi(x_{1})-p=\Bigl(DT(x_{1})-\Lambda^{\sharp}D\chi(\zeta_{1})\Bigr)+\Lambda^{\sharp}\bigl(D\chi(\zeta_{1})-p_{0}\bigr)+2\alpha\,T_{N}(x_{1}-\bar{x})+2\mu(x_{1}-c_{1}),$$

because $-Dg(x_{1})=\alpha\bar{z}+2\alpha T_{N}(x_{1}-\bar{x})$, because $\Lambda^{\sharp}$ is linear, and because $p=\Lambda^{\sharp}p_{0}+\alpha\bar{z}$. The four terms are estimated in turn: the first has norm at most $K\lVert\Lambda x_{1}-\zeta_{1}\rVert\le Kr\le\tfrac{\varepsilon}{16}$ by Steps 4, 5 and 7; the second has norm $\lVert D\chi(\zeta_{1})-p_{0}\rVert<\varepsilon'\le\tfrac{\varepsilon}{16}$, since $|\Lambda^{\sharp}\xi|=\lVert\xi\rVert$; the third has norm at most $2\alpha|x_{1}-\bar{x}|<2\alpha\varepsilon_{1}\le\tfrac{\varepsilon}{4}$ by $|T_{N}z|\le|z|$, Step 9 and $\varepsilon_{1}\le\tfrac{\varepsilon}{8\alpha}$; and the fourth has norm $2\mu|x_{1}-c_{1}|\le16\mu\lambda\le\tfrac{\varepsilon}{16}$ by Step 7. By the triangle inequality (claim 1 of The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity),

$$\bigl|D\varphi(x_{1})-p\bigr|\le\tfrac{\varepsilon}{16}+\tfrac{\varepsilon}{16}+\tfrac{\varepsilon}{4}+\tfrac{\varepsilon}{16}<\varepsilon .$$

For the Hessian, $D^{2}\varphi(x_{1})-(X+2\alpha N)=(b_{u}-X)+2\mu I$ by Step 8, so by claims 1 and 3 of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity, Step 4 and Step 7,

$$\bigl\lVert D^{2}\varphi(x_{1})-(X+2\alpha N)\bigr\rVert\le\lVert b_{u}-X\rVert+2\mu\lVert I\rVert\le(\varepsilon'+2\eta_{1})+2\mu\le\tfrac{\varepsilon}{16}+\tfrac{\varepsilon}{16}+\tfrac{\varepsilon}{16}<\varepsilon .$$

On the $v$ side, $Dk(y_{1})=\alpha\bar{z}-2\alpha T_{N}(y_{1}-\bar{y})$, so

$$D\psi(y_{1})-p=\Bigl(DS(y_{1})-\Lambda^{\sharp}D\theta(\omega_{1})\Bigr)+\Lambda^{\sharp}\bigl(D\theta(\omega_{1})-p_{0}\bigr)-2\alpha\,T_{N}(y_{1}-\bar{y})-2\mu(y_{1}-c_{2}),$$

and the same four estimates, with $\lVert\Lambda y_{1}-\omega_{1}\rVert\le r$ and $|y_{1}-\bar{y}|<\varepsilon_{1}$ in place of their counterparts, give $|D\psi(y_{1})-p|<\varepsilon$. Finally $D^{2}\psi(y_{1})-(Y-2\alpha N)=(b_{v}-Y)-2\mu I$, whose norm is at most $(\varepsilon'+2\eta_{1})+2\mu<\varepsilon$ as before.

This establishes all eight inequalities and completes the proof of Claim 4, hence of clause 1 and of the lemma.