Proof of Reduction of the Theorem on Sums to a Global Quadratic Bound

Throughout, the notation is that of the statement. For $s\in\mathbb{R}$ we write $s^{+}$ for the nonnegative part of $s$, as in Localisation of an Upper Semicontinuous Function by Projection onto a Closed Ball: $s^{+}=s$ if $0\le s$ and $s^{+}=0$ otherwise, so that $0\le s^{+}$ and $s\le s^{+}$. Open balls $B(x,\rho)$ are those of Open Ball in a Metric Space.

Step 1 (a quadratic bound near $\hat{x}$). By Basic Properties of Twice Differentiability at a Point (ii) the function $\varphi$ is twice differentiable at $\hat{x}$ with first-order coefficient $D\varphi(\hat{x})$ and Hessian $A$. The number $\tfrac{\theta}{2}$ is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, so by Twice Differentiability at a Point (i) there is a positive $\delta_{1}\in\mathbb{R}$ such that every $h\in\mathbb{R}^{N}$ with $\lVert h\rVert<\delta_{1}$ satisfies $\hat{x}+h\in V$ and

$$\Bigl|\varphi(\hat{x}+h)-\varphi(\hat{x})-D\varphi(\hat{x})\cdot h-\tfrac{1}{2}\,h\cdot(Ah)\Bigr|\le\tfrac{\theta}{2}\,\lVert h\rVert^{2}.$$

Since $w-\varphi$ has a local maximum at $\hat{x}$ relative to $\Omega$, there is a positive $\delta_{2}\in\mathbb{R}$ such that every $x\in\Omega$ with $d_{E}(x,\hat{x})<\delta_{2}$ satisfies $w(x)-\varphi(x)\le w(\hat{x})-\varphi(\hat{x})$. Put $\rho=\min\{\delta_{1},\delta_{2}\}$, which exists by claim 9 of Elementary Order Arithmetic in an Ordered Field and is positive.

Let $h\in\mathbb{R}^{N}$ satisfy $\lVert h\rVert<\rho$ and $\hat{x}+h\in\Omega$. Then $d_{E}(\hat{x}+h,\hat{x})=\lVert h\rVert<\delta_{2}$ by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so

$$w(\hat{x}+h)-w(\hat{x})\le\varphi(\hat{x}+h)-\varphi(\hat{x}),$$

while claim 3 of Properties of the Absolute Value in an Ordered Field turns the displayed bound into

$$\varphi(\hat{x}+h)-\varphi(\hat{x})\le D\varphi(\hat{x})\cdot h+\tfrac{1}{2}\,h\cdot(Ah)+\tfrac{\theta}{2}\,\lVert h\rVert^{2}.$$

By claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum we have $A_{\theta}h=Ah+(\theta I_{N})h$; by claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vi),

$$h\cdot(A_{\theta}h)=h\cdot(Ah)+h\cdot\bigl((\theta I_{N})h\bigr)=h\cdot(Ah)+\theta\,\lVert h\rVert^{2}.$$

Combining the three displays,

$$w(\hat{x}+h)-w(\hat{x})-D\varphi(\hat{x})\cdot h\ \le\ \tfrac{1}{2}\,h\cdot(A_{\theta}h).\tag{1}$$

Now let $\zeta_{1}\in\Omega_{1}-\hat{x}_{1}$ and $\zeta_{2}\in\Omega_{2}-\hat{x}_{2}$ satisfy $\lVert\iota(\zeta_{1},\zeta_{2})\rVert<\rho$, and put $h=\iota(\zeta_{1},\zeta_{2})$. By claim 2 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space,

$$\hat{x}+h=\iota(\hat{x}_{1},\hat{x}_{2})+\iota(\zeta_{1},\zeta_{2})=\iota(\hat{x}_{1}+\zeta_{1},\hat{x}_{2}+\zeta_{2}),$$

which lies in $\Omega$ because $\zeta_{i}+\hat{x}_{i}\in\Omega_{i}$ for $i\in\{1,2\}$. Hence

$$w(\hat{x}+h)=u_{1}(\zeta_{1}+\hat{x}_{1})+u_{2}(\zeta_{2}+\hat{x}_{2}),\qquad w(\hat{x})=u_{1}(\hat{x}_{1})+u_{2}(\hat{x}_{2}),$$

and by claim 3 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space,

$$D\varphi(\hat{x})\cdot h=\iota(p_{1},p_{2})\cdot\iota(\zeta_{1},\zeta_{2})=p_{1}\cdot\zeta_{1}+p_{2}\cdot\zeta_{2}.$$

Substituting these three identities into (1) and regrouping gives

$$\tilde{u}_{1}(\zeta_{1})+\tilde{u}_{2}(\zeta_{2})\ \le\ \tfrac{1}{2}\,\iota(\zeta_{1},\zeta_{2})\cdot\bigl(A_{\theta}\,\iota(\zeta_{1},\zeta_{2})\bigr)\tag{2}$$

whenever $\zeta_{i}\in\Omega_{i}-\hat{x}_{i}$ for $i\in\{1,2\}$ and $\lVert\iota(\zeta_{1},\zeta_{2})\rVert<\rho$.

Step 2 (choice of the radius). For $i\in\{1,2\}$ the set $\Omega_{i}-\hat{x}_{i}$ is open and contains $0_{\mathbb{R}^{n_{i}}}$, so by Open Subset of a Metric Space there is a positive $r_{i}\in\mathbb{R}$ with $B(0_{\mathbb{R}^{n_{i}}},r_{i})\subseteq\Omega_{i}-\hat{x}_{i}$. Put

$$r=\min\bigl\{\,r_{1}\cdot2^{-1},\ r_{2}\cdot2^{-1},\ \rho\cdot2^{-1}\,\bigr\},$$

which exists by claim 9 of Elementary Order Arithmetic in an Ordered Field applied twice and is positive by claim 8 there.

If $\zeta\in\mathbb{R}^{n_{i}}$ satisfies $\lVert\zeta\rVert\le r$ then $\lVert\zeta\rVert\le r_{i}\cdot2^{-1}<r_{i}$ by claim 8, and $d_{E}(0_{\mathbb{R}^{n_{i}}},\zeta)=\lVert\zeta\rVert$ by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so $\zeta\in B(0_{\mathbb{R}^{n_{i}}},r_{i})\subseteq\Omega_{i}-\hat{x}_{i}$. As $\bar{B}(0_{\mathbb{R}^{n_{i}}},r)$ is exactly the set of such $\zeta$, we obtain

$$\bar{B}\bigl(0_{\mathbb{R}^{n_{i}}},r\bigr)\subseteq\Omega_{i}-\hat{x}_{i}\qquad(i\in\{1,2\}).\tag{3}$$

Moreover, if $\lVert\zeta_{1}\rVert\le r$ and $\lVert\zeta_{2}\rVert\le r$ then claim 3 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space and claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field give

$$\lVert\iota(\zeta_{1},\zeta_{2})\rVert^{2}=\lVert\zeta_{1}\rVert^{2}+\lVert\zeta_{2}\rVert^{2}\le\bigl(\rho\cdot2^{-1}\bigr)^{2}+\bigl(\rho\cdot2^{-1}\bigr)^{2}=\rho^{2}\cdot2^{-1}<\rho^{2},$$

the last step by claim 8 of Elementary Order Arithmetic in an Ordered Field applied to the number $\rho^{2}$, which is positive by claim 5 there. Hence

$$\lVert\iota(\zeta_{1},\zeta_{2})\rVert<\rho\tag{4}$$

by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, both numbers being nonnegative.

Step 3 (the localised summands; proof of claim 1). We first check that $\tilde{u}_{i}$ is upper semicontinuous on $\Omega_{i}-\hat{x}_{i}$.

Let $T_{i}:\Omega_{i}-\hat{x}_{i}\to\mathbb{R}^{n_{i}}$ be given by $T_{i}(\zeta)=\zeta+\hat{x}_{i}$; by the definition of $\Omega_{i}-\hat{x}_{i}$ its values lie in $\Omega_{i}$. For $\zeta,\zeta'$ in its domain the vector space identities of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space give $T_{i}(\zeta)-T_{i}(\zeta')=\zeta-\zeta'$, so $d_{E}(T_{i}(\zeta),T_{i}(\zeta'))=d_{E}(\zeta,\zeta')$ and $T_{i}$ is continuous on $\Omega_{i}-\hat{x}_{i}$ relative to that set, one may take $\delta=\varepsilon$ in that definition. Hence $u_{i}\circ T_{i}$ is upper semicontinuous on $\Omega_{i}-\hat{x}_{i}$ by claim 1 of Semicontinuity and Continuity Under Composition with a Continuous Map.

Let $\ell_{i}:\Omega_{i}-\hat{x}_{i}\to\mathbb{R}$ be given by $\ell_{i}(\zeta)=-p_{i}\cdot\zeta-u_{i}(\hat{x}_{i})$. For $\zeta,\zeta'$ in its domain, claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, which records the behaviour of the dot product in its second argument, gives $\ell_{i}(\zeta')-\ell_{i}(\zeta)=p_{i}\cdot\zeta-p_{i}\cdot\zeta'=p_{i}\cdot(\zeta-\zeta')$, so by The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ and claim 3 of Properties of the Absolute Value in an Ordered Field,

$$\ell_{i}(\zeta')-\ell_{i}(\zeta)\le\bigl|p_{i}\cdot(\zeta-\zeta')\bigr|\le\lVert p_{i}\rVert\,\lVert\zeta-\zeta'\rVert\le\bigl(1+\lVert p_{i}\rVert\bigr)\lVert\zeta-\zeta'\rVert .$$

Given a positive $\varepsilon\in\mathbb{R}$, the number $1+\lVert p_{i}\rVert$ is positive by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and claim 6 of Elementary Order Arithmetic in an Ordered Field, so $\delta=\varepsilon\bigl(1+\lVert p_{i}\rVert\bigr)^{-1}$ is positive by claims 5 and 7 there; and if $d_{E}(\zeta,\zeta')=\lVert\zeta-\zeta'\rVert<\delta$ then $\bigl(1+\lVert p_{i}\rVert\bigr)\lVert\zeta-\zeta'\rVert<\bigl(1+\lVert p_{i}\rVert\bigr)\delta=\varepsilon$ by claim 10 there, whence $\ell_{i}(\zeta')-\ell_{i}(\zeta)<\varepsilon$ by claim 2 there and $\ell_{i}(\zeta')<\ell_{i}(\zeta)+\varepsilon$ by claim 1 there. Thus $\ell_{i}$ is upper semicontinuous on $\Omega_{i}-\hat{x}_{i}$.

Since $\tilde{u}_{i}(\zeta)=(u_{i}\circ T_{i})(\zeta)+\ell_{i}(\zeta)$ for every $\zeta$, claim 1 of Sums and Nonnegative Multiples of Semicontinuous Functions shows that $\tilde{u}_{i}$ is upper semicontinuous on $\Omega_{i}-\hat{x}_{i}$.

Put $c=3\,\lVert A_{\theta}\rVert\cdot2^{-1}$, which is nonnegative by claim 1 of Properties of the Norm of a Symmetric Real Matrix, and for $i\in\{1,2\}$ let $\pi^{(i)}_{r}:\mathbb{R}^{n_{i}}\to\mathbb{R}^{n_{i}}$ be the projection onto $\bar{B}(0_{\mathbb{R}^{n_{i}}},r)$ of Localisation of an Upper Semicontinuous Function by Projection onto a Closed Ball. By Localisation of an Upper Semicontinuous Function by Projection onto a Closed Ball (iii), applied in dimension $n_{i}$ with the radius $r$, the open set $\Omega_{i}-\hat{x}_{i}$, which contains $\bar{B}(0_{\mathbb{R}^{n_{i}}},r)$ by (3), the upper semicontinuous function $\tilde{u}_{i}$ and the constant $c$, the function $v_{i}:\mathbb{R}^{n_{i}}\to\mathbb{R}$ given by

$$v_{i}(\xi)=\tilde{u}_{i}\bigl(\pi^{(i)}_{r}(\xi)\bigr)-c\,\bigl(\lVert\xi\rVert^{2}-r^{2}\bigr)^{+}$$

is upper semicontinuous on $\mathbb{R}^{n_{i}}$, its set of values has an upper bound in $\mathbb{R}$, and $v_{i}(\zeta)=\tilde{u}_{i}(\zeta)$ for every $\zeta$ with $\lVert\zeta\rVert\le r$. Since $\lVert 0_{\mathbb{R}^{n_{i}}}\rVert=0\le r$ by claim 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, we get $v_{i}(0_{\mathbb{R}^{n_{i}}})=\tilde{u}_{i}(0_{\mathbb{R}^{n_{i}}})=0$. Together with (3) this proves claim 1.

Step 4 (proof of claim 2). Let $\xi\in\mathbb{R}^{n_{1}}$ and $\eta\in\mathbb{R}^{n_{2}}$. Put

$$\bar{\xi}=\pi^{(1)}_{r}(\xi),\quad \bar{\eta}=\pi^{(2)}_{r}(\eta),\quad z=\iota(\xi,\eta),\quad \bar{z}=\iota(\bar{\xi},\bar{\eta}),$$

and

$$t_{1}=\lVert\xi\rVert,\quad t_{2}=\lVert\eta\rVert,\quad a_{i}=(t_{i}-r)^{+},\quad b_{i}=t_{i}+r,\quad S=\bigl(t_{1}^{2}-r^{2}\bigr)^{+}+\bigl(t_{2}^{2}-r^{2}\bigr)^{+}.$$

By Localisation of an Upper Semicontinuous Function by Projection onto a Closed Ball (i) we have $\lVert\bar{\xi}\rVert\le r$, $\lVert\bar{\eta}\rVert\le r$,

$$\lVert\bar{\xi}-\xi\rVert=a_{1},\quad \lVert\bar{\eta}-\eta\rVert=a_{2},\quad \lVert\bar{\xi}+\xi\rVert\le b_{1},\quad \lVert\bar{\eta}+\eta\rVert\le b_{2},$$

and by Localisation of an Upper Semicontinuous Function by Projection onto a Closed Ball (ii) we have $a_{i}b_{i}=(t_{i}^{2}-r^{2})^{+}$, so that $S=a_{1}b_{1}+a_{2}b_{2}$. All of $a_{1},a_{2},b_{1},b_{2}$ are nonnegative, the $a_{i}$ because $0\le s^{+}$ and the $b_{i}$ by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square.

From $\lVert\bar{\xi}\rVert\le r$ and $\lVert\bar{\eta}\rVert\le r$, together with (3) and (4), we get $\bar{\xi}\in\Omega_{1}-\hat{x}_{1}$, $\bar{\eta}\in\Omega_{2}-\hat{x}_{2}$ and $\lVert\bar{z}\rVert<\rho$, so (2) applies to the pair $(\bar{\xi},\bar{\eta})$ and gives

$$\tilde{u}_{1}(\bar{\xi})+\tilde{u}_{2}(\bar{\eta})\ \le\ \tfrac{1}{2}\,\bar{z}\cdot(A_{\theta}\bar{z}).\tag{5}$$

We next bound $\lVert\bar{z}-z\rVert\,\lVert\bar{z}+z\rVert$. By claim 2 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space we have $\bar{z}-z=\iota(\bar{\xi}-\xi,\bar{\eta}-\eta)$ and $\bar{z}+z=\iota(\bar{\xi}+\xi,\bar{\eta}+\eta)$, so claim 3 there gives

$$\lVert\bar{z}-z\rVert^{2}=a_{1}^{2}+a_{2}^{2}\le a_{1}^{2}+2\,a_{1}a_{2}+a_{2}^{2}=(a_{1}+a_{2})^{2},$$ $$\lVert\bar{z}+z\rVert^{2}=\lVert\bar{\xi}+\xi\rVert^{2}+\lVert\bar{\eta}+\eta\rVert^{2}\le b_{1}^{2}+b_{2}^{2}\le(b_{1}+b_{2})^{2},$$

where the products $a_{1}a_{2}$ and $b_{1}b_{2}$ are nonnegative (they vanish if a factor is $0$ and are positive otherwise, by claim 5 of Elementary Order Arithmetic in an Ordered Field) and the inequalities $\lVert\bar{\xi}+\xi\rVert^{2}\le b_{1}^{2}$ and $\lVert\bar{\eta}+\eta\rVert^{2}\le b_{2}^{2}$ come from claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. That same claim, applied to the nonnegative numbers involved, yields

$$\lVert\bar{z}-z\rVert\le a_{1}+a_{2},\qquad \lVert\bar{z}+z\rVert\le b_{1}+b_{2}.$$

The map $t\mapsto(t-r)^{+}$ is nondecreasing: if $t\le t'$ and $(t-r)^{+}=0$ then $(t-r)^{+}\le(t'-r)^{+}$ because the latter is nonnegative, while if $(t-r)^{+}=t-r$ then $t-r\le t'-r\le(t'-r)^{+}$. The map $t\mapsto t+r$ is nondecreasing as well. Since the order of $\mathbb{R}$ is total, either $t_{1}\le t_{2}$, in which case $a_{1}\le a_{2}$ and hence $a_{1}b_{2}\le a_{2}b_{2}$, or $t_{2}\le t_{1}$, in which case $b_{2}\le b_{1}$ and hence $a_{1}b_{2}\le a_{1}b_{1}$; in either case, the two numbers $a_{1}b_{1}$ and $a_{2}b_{2}$ being nonnegative, $a_{1}b_{2}\le a_{1}b_{1}+a_{2}b_{2}=S$. Exchanging the roles of the indices $1$ and $2$ gives $a_{2}b_{1}\le S$. Multiplying the two bounds of the previous display, which is legitimate because all four numbers are nonnegative, we obtain

$$\lVert\bar{z}-z\rVert\,\lVert\bar{z}+z\rVert\le(a_{1}+a_{2})(b_{1}+b_{2})=a_{1}b_{1}+a_{2}b_{2}+a_{1}b_{2}+a_{2}b_{1}\le 3S.\tag{6}$$

Now Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vii), applied to $A_{\theta}$ with the points $\bar{z}$ and $z$, together with claim 3 of Properties of the Absolute Value in an Ordered Field and (6), gives

$$\bar{z}\cdot(A_{\theta}\bar{z})-z\cdot(A_{\theta}z)\ \le\ \bigl|\bar{z}\cdot(A_{\theta}\bar{z})-z\cdot(A_{\theta}z)\bigr|\ \le\ \lVert A_{\theta}\rVert\,\lVert\bar{z}-z\rVert\,\lVert\bar{z}+z\rVert\ \le\ 3\,\lVert A_{\theta}\rVert\,S,$$

the last step because $\lVert A_{\theta}\rVert$ is nonnegative. Multiplying by the positive number $2^{-1}$ and recalling $c=3\lVert A_{\theta}\rVert\cdot2^{-1}$,

$$\tfrac{1}{2}\,\bar{z}\cdot(A_{\theta}\bar{z})\ \le\ \tfrac{1}{2}\,z\cdot(A_{\theta}z)+c\,S.$$

Finally, by the definition of $v_{1}$ and $v_{2}$ in Step 3 and by (5),

$$v_{1}(\xi)+v_{2}(\eta)=\tilde{u}_{1}(\bar{\xi})+\tilde{u}_{2}(\bar{\eta})-c\,S\ \le\ \tfrac{1}{2}\,\bar{z}\cdot(A_{\theta}\bar{z})-c\,S\ \le\ \tfrac{1}{2}\,z\cdot(A_{\theta}z),$$

which is claim 2.

Step 5 (proof of claim 3). Let $i\in\{1,2\}$ and $X\in\mathcal{S}(n_{i})$, and suppose that $\bigl(0_{\mathbb{R}^{n_{i}}},v_{i}(0_{\mathbb{R}^{n_{i}}}),0_{\mathbb{R}^{n_{i}}},X\bigr)$ is approximable by test data from above for $v_{i}$, the domain being $\mathbb{R}^{n_{i}}$.

Put $W=B(0_{\mathbb{R}^{n_{i}}},r)$, which is open by Open Ball in a Metric Space is Open and contains $0_{\mathbb{R}^{n_{i}}}$. Every $\zeta\in W$ satisfies $\lVert\zeta\rVert<r$, so $W\subseteq\bar{B}(0_{\mathbb{R}^{n_{i}}},r)\subseteq\Omega_{i}-\hat{x}_{i}$ by (3), and $v_{i}(\zeta)=\tilde{u}_{i}(\zeta)$ for every $\zeta\in W$ by claim 1. Therefore Quadratic Test Functions, Limits, Translation and Locality for Approximability by Test Data (v), applied with $U=\mathbb{R}^{n_{i}}$, $u=v_{i}$, $U'=\Omega_{i}-\hat{x}_{i}$, $u'=\tilde{u}_{i}$, the point $x_{0}=0_{\mathbb{R}^{n_{i}}}$, the vector $p=0_{\mathbb{R}^{n_{i}}}$, the matrix $X$ and the open set $W$, shows that

$$\bigl(0_{\mathbb{R}^{n_{i}}},\tilde{u}_{i}(0_{\mathbb{R}^{n_{i}}}),0_{\mathbb{R}^{n_{i}}},X\bigr)$$

is approximable by test data from above for $\tilde{u}_{i}$, the domain being $\Omega_{i}-\hat{x}_{i}$.

Finally apply Quadratic Test Functions, Limits, Translation and Locality for Approximability by Test Data (iv) with $U=\Omega_{i}$, $u=u_{i}$, $b=\hat{x}_{i}$, $q=-p_{i}$ and $c=-u_{i}(\hat{x}_{i})$. The set $U-b$ of that clause is $\Omega_{i}-\hat{x}_{i}$ and the function it produces is $\zeta\mapsto u_{i}(\zeta+\hat{x}_{i})-p_{i}\cdot\zeta-u_{i}(\hat{x}_{i})$, that is $\tilde{u}_{i}$. Taking $x_{0}=\hat{x}_{i}$ and $p=p_{i}$, and using $\hat{x}_{i}-\hat{x}_{i}=0_{\mathbb{R}^{n_{i}}}$ and $p_{i}+(-p_{i})=0_{\mathbb{R}^{n_{i}}}$ from Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, that clause states that $\bigl(\hat{x}_{i},u_{i}(\hat{x}_{i}),p_{i},X\bigr)$ is approximable by test data from above for $u_{i}$ if and only if $\bigl(0_{\mathbb{R}^{n_{i}}},\tilde{u}_{i}(0_{\mathbb{R}^{n_{i}}}),0_{\mathbb{R}^{n_{i}}},X\bigr)$ is approximable by test data from above for $\tilde{u}_{i}$. The latter has just been established, so the former holds. This is claim 3.