Proof of Localisation of an Upper Semicontinuous Function by Projection onto a Closed Ball

Throughout we use the notation of the statement, and we write $\bar{B}_{r}=\{\xi\in\mathbb{R}^{n}:\lVert\xi\rVert\le r\}$ as recorded there.

Claim 1. By claim 1 of Nearest-Point Projection onto a Nonempty Closed Convex Subset of Euclidean Space the point $\pi_{r}(\xi)$ lies in $\bar{B}_{r}$, so $\lVert\pi_{r}(\xi)\rVert\le r$. If $\lVert\xi\rVert\le r$ then $\xi\in\bar{B}_{r}$ and claim 3 of Nearest-Point Projection onto a Nonempty Closed Convex Subset of Euclidean Space gives $\pi_{r}(\xi)=\xi$; conversely if $\pi_{r}(\xi)=\xi$ then $\xi\in\bar{B}_{r}$, that is $\lVert\xi\rVert\le r$.

Suppose now $r<\lVert\xi\rVert$. Then $\lVert\xi\rVert$ is positive, since $0<r$, so it has a positive multiplicative inverse by claim 7 of Elementary Order Arithmetic in an Ordered Field. Put $\beta=r\,\lVert\xi\rVert^{-1}$, which is positive, and $p=\beta\,\xi$. Multiplying $r<\lVert\xi\rVert$ by $\lVert\xi\rVert^{-1}$, using claim 10 of Elementary Order Arithmetic in an Ordered Field, gives $\beta<1$; put $\alpha=1-\beta$, which is positive. By claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, $\lVert p\rVert=|\beta|\,\lVert\xi\rVert=\beta\lVert\xi\rVert=r$, so $p\in\bar{B}_{r}$. By the vector space identities of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, $\xi-p=\alpha\,\xi$. Moreover, by claims 1 and 4 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and 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,

$$\xi\cdot p=\beta\,(\xi\cdot\xi)=\beta\,\lVert\xi\rVert^{2}=r\,\lVert\xi\rVert .$$

Let $y\in\bar{B}_{r}$. By claim 3 of Properties of the Absolute Value in an Ordered Field and The Cauchy-Schwarz Inequality on $\mathbb{R}^n$,

$$\xi\cdot y\le|\xi\cdot y|\le\lVert\xi\rVert\,\lVert y\rVert\le r\,\lVert\xi\rVert,$$

the last step because $\lVert y\rVert\le r$ and $\lVert\xi\rVert$ is nonnegative. Hence, by claims 4 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$(\xi-p)\cdot(y-p)=\alpha\,\bigl(\xi\cdot y-\xi\cdot p\bigr)=\alpha\,\bigl(\xi\cdot y-r\lVert\xi\rVert\bigr)\le0 .$$

By claim 2 of Nearest-Point Projection onto a Nonempty Closed Convex Subset of Euclidean Space this characterises $p$ as $\pi_{r}(\xi)$, so $\pi_{r}(\xi)=\bigl(r\lVert\xi\rVert^{-1}\bigr)\xi$ as asserted.

For the identity $\lVert\pi_{r}(\xi)-\xi\rVert=(\lVert\xi\rVert-r)^{+}$: if $\lVert\xi\rVert\le r$ then $\pi_{r}(\xi)-\xi$ is the origin, so the left-hand side is $0$ 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, while $\lVert\xi\rVert-r\le0$ makes the right-hand side $0$. If $r<\lVert\xi\rVert$ then $\pi_{r}(\xi)-\xi=-\alpha\,\xi$ and claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square gives $\lVert\pi_{r}(\xi)-\xi\rVert=\alpha\,\lVert\xi\rVert=\lVert\xi\rVert-r$, which is positive and hence equals $(\lVert\xi\rVert-r)^{+}$. The inequality $\lVert\pi_{r}(\xi)+\xi\rVert\le\lVert\pi_{r}(\xi)\rVert+\lVert\xi\rVert\le r+\lVert\xi\rVert$ is claim 6 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square together with the first assertion of this claim.

Finally $\pi_{r}$ is Lipschitz with constant $1$ by claim 4 of Nearest-Point Projection onto a Nonempty Closed Convex Subset of Euclidean Space, hence continuous by A Lipschitz Map is Uniformly Continuous and Continuous.

Claim 2. Suppose first $\lVert\xi\rVert\le r$. Then $(\lVert\xi\rVert-r)^{+}=0$, so the left-hand side of the first identity is $0$; and $\lVert\xi\rVert^{2}\le r^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, both numbers being nonnegative, so $(\lVert\xi\rVert^{2}-r^{2})^{+}=0$ as well. For the second identity, $\lVert\xi\rVert+r\le r+r=2r=(\lVert\xi\rVert-r)^{+}+2r$.

Suppose now $r<\lVert\xi\rVert$. Then $(\lVert\xi\rVert-r)^{+}=\lVert\xi\rVert-r$ and, by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, $r^{2}<\lVert\xi\rVert^{2}$, so $(\lVert\xi\rVert^{2}-r^{2})^{+}=\lVert\xi\rVert^{2}-r^{2}$. Distributivity in $\mathbb{R}$ gives $(\lVert\xi\rVert-r)(\lVert\xi\rVert+r)=\lVert\xi\rVert^{2}-r^{2}$, which is the first identity. The second is the equality $(\lVert\xi\rVert-r)+2r=\lVert\xi\rVert+r$.

Claim 3. Since $\pi_{r}(\xi)\in\bar{B}_{r}\subseteq\Omega$ for every $\xi$, the function $v$ is well defined. If $\lVert\xi\rVert\le r$ then $\pi_{r}(\xi)=\xi$ by claim 1 and $(\lVert\xi\rVert^{2}-r^{2})^{+}=0$ as computed in claim 2, so $v(\xi)=u(\xi)$.

Let $u_{0}$ denote the restriction of $u$ to $\bar{B}_{r}$, which is upper semicontinuous on $\bar{B}_{r}$ by claim 4 of Semicontinuity and Continuity Under Composition with a Continuous Map. The set $\bar{B}_{r}$ is nonempty and compact, so by claim 1 of Semicontinuous Functions Attain Their Extrema on a Compact Set there is $x_{\max}\in\bar{B}_{r}$ with $u(x)\le u(x_{\max})$ for every $x\in\bar{B}_{r}$. Since $c$ and $(\lVert\xi\rVert^{2}-r^{2})^{+}$ are nonnegative, their product is nonnegative by claim 5 of Elementary Order Arithmetic in an Ordered Field in the strict case and trivially otherwise, so

$$v(\xi)\le u\bigl(\pi_{r}(\xi)\bigr)\le u(x_{\max})\qquad\text{for every }\xi\in\mathbb{R}^{n},$$

and $u(x_{\max})$ is an upper bound for the set of values of $v$.

It remains to prove upper semicontinuity. The map $\pi_{r}:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is continuous by claim 1 and takes its values in $\bar{B}_{r}$, so by claim 1 of Semicontinuity and Continuity Under Composition with a Continuous Map the function $\xi\mapsto u_{0}(\pi_{r}(\xi))=u(\pi_{r}(\xi))$ is upper semicontinuous on $\mathbb{R}^{n}$.

Next, let $G:\mathbb{R}^{n}\to\mathbb{R}$ be given by $G(\xi)=\lVert\xi\rVert^{2}-r^{2}$. Taking $M=2I_{n}$, which lies in $\mathcal{S}(n)$ by Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vi), together with the origin of $\mathbb{R}^{n}$ and the constant $-r^{2}$ in Quadratic and Affine Functions of Class $C^2$, Translation, and Quadratic Perturbation of Semiconvexity (i), and noting that $\tfrac{1}{2}\xi\cdot\bigl((2I_{n})\xi\bigr)=\xi\cdot\xi=\lVert\xi\rVert^{2}$ by claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and 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, we conclude that $G$ is of class $C^{2}$ on $\mathbb{R}^{n}$; in particular $G$ is of class $C^{1}$, hence continuous at every point of $\mathbb{R}^{n}$ by clauses 1 and 2 of Functions of Class Cᵏ on an Open Subset of ℝⁿ.

Let $\mu:\mathbb{R}\to\mathbb{R}$ be given by $\mu(s)=s^{+}$. Then $|\mu(s)-\mu(t)|\le|s-t|$ for all $s,t\in\mathbb{R}$: if $0\le s$ and $0\le t$ both sides are equal; if $s<0$ and $t<0$ the left side is $0$ and the right side is nonnegative; and if $0\le s$ and $t<0$ then $|\mu(s)-\mu(t)|=s$ while $s\le s-t=|s-t|$ because $-t$ is positive, the remaining case following by symmetry using claim 2 of Properties of the Absolute Value in an Ordered Field. So $\mu$ is Lipschitz with constant $1$ with respect to the metric of The Absolute Value Metric on the Real Numbers and hence continuous by A Lipschitz Map is Uniformly Continuous and Continuous. By claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map the composite $\mu\circ G$, whose value at $\xi$ is $(\lVert\xi\rVert^{2}-r^{2})^{+}$, is continuous on $\mathbb{R}^{n}$, and by claim 4 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space so is the function $\xi\mapsto(-c)\,(\lVert\xi\rVert^{2}-r^{2})^{+}$.

A continuous real-valued function $\theta$ on $\mathbb{R}^{n}$ is upper semicontinuous: given $\xi$ and a positive $\varepsilon$, continuity provides a positive $\delta$ with $|\theta(\zeta)-\theta(\xi)|<\varepsilon$ whenever $d_{E}(\zeta,\xi)<\delta$, and claim 9 of Properties of the Absolute Value in an Ordered Field then gives $\theta(\zeta)<\theta(\xi)+\varepsilon$, which is the condition of Upper Semicontinuous Function on a Subset of a Metric Space.

Finally $v$ is the sum of the upper semicontinuous function $\xi\mapsto u(\pi_{r}(\xi))$ and the upper semicontinuous function $\xi\mapsto(-c)(\lVert\xi\rVert^{2}-r^{2})^{+}$, so $v$ is upper semicontinuous on $\mathbb{R}^{n}$ by claim 1 of Sums and Nonnegative Multiples of Semicontinuous Functions.