Proof of Concatenation and the Sup-Convolution of a Sum in Separated Variables

Throughout, the notation is that of the statement.

Claim 1. Apply claim 2 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space with $\xi=\xi'=0_{\mathbb{R}^{m}}$ and $\eta=\eta'=0_{\mathbb{R}^{n}}$, in the form recorded there for differences:

In the real vector space $\mathbb{R}^{q}$ one has $z-z=0_{\mathbb{R}^{q}}$ for every $z\in\mathbb{R}^{q}$. The left-hand side is therefore $\iota(0_{\mathbb{R}^{m}},0_{\mathbb{R}^{n}})$ and the right-hand side is $0_{\mathbb{R}^{N}}$, which is the first assertion.

For the second assertion, put, for $k\in\mathbb{N}$,

which are nonnegative, being Euclidean norms of differences by the identity $d_{E}(x,y)=\lVert x-y\rVert$ recorded in the statement together with claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. By claim 4 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space,

Since $b_{k}^{2}$ and $a_{k}^{2}$ are nonnegative, this gives $a_{k}^{2}\le c_{k}^{2}$ and $b_{k}^{2}\le c_{k}^{2}$, hence $a_{k}\le c_{k}$ and $b_{k}\le c_{k}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, the numbers involved being nonnegative. Also $2\,a_{k}b_{k}$ is nonnegative by claim 5 of Elementary Order Arithmetic in an Ordered Field, so

and the same claim gives $c_{k}\le a_{k}+b_{k}$.

Suppose first that $\bigl(\iota(\xi_{k},\eta_{k})\bigr)$ converges to $\iota(\xi,\eta)$, and let $\varepsilon\in\mathbb{R}$ be positive. By Convergent Sequence in a Metric Space there is $K\in\mathbb{N}$ with $c_{k}<\varepsilon$ for every $k\ge K$; then $a_{k}\le c_{k}<\varepsilon$ and $b_{k}\le c_{k}<\varepsilon$ for every such $k$. Hence $(\xi_{k})$ converges to $\xi$ and $(\eta_{k})$ converges to $\eta$.

Conversely, suppose $(\xi_{k})$ converges to $\xi$ and $(\eta_{k})$ converges to $\eta$, and let $\varepsilon\in\mathbb{R}$ be positive. The number $\tfrac{\varepsilon}{2}$ is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, so there are $K_{1},K_{2}\in\mathbb{N}$ with $a_{k}<\tfrac{\varepsilon}{2}$ for $k\ge K_{1}$ and $b_{k}<\tfrac{\varepsilon}{2}$ for $k\ge K_{2}$. For $k$ at least as large as both $K_{1}$ and $K_{2}$ we get $c_{k}\le a_{k}+b_{k}<\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon$. Hence $\bigl(\iota(\xi_{k},\eta_{k})\bigr)$ converges to $\iota(\xi,\eta)$.

Claim 2. Let $x\in\mathbb{R}^{N}$. Since $\iota$ is a bijection there are $\xi\in\mathbb{R}^{m}$ and $\eta\in\mathbb{R}^{n}$ with $x=\iota(\xi,\eta)$, so that $w(x)=u_{1}(\xi)+u_{2}(\eta)$. From $u_{1}(\xi)\le C_{1}$ and $u_{2}(\eta)\le C_{2}$ and the compatibility of $\le$ with addition in an ordered field we obtain $w(x)\le C_{1}+C_{2}$. Thus $C_{1}+C_{2}$ is an upper bound for the set of values of $w$. Since the sets of values of $u_{1}$, $u_{2}$ and $w$ all have upper bounds in $\mathbb{R}$ and $0<\lambda$, the sup-convolutions $u_{1}^{\lambda}$, $u_{2}^{\lambda}$ and $w^{\lambda}$ are defined by Sup-Convolution of a Function on $\mathbb{R}^M$.

Claim 3. Fix $\xi\in\mathbb{R}^{m}$ and $\eta\in\mathbb{R}^{n}$ and put $z=\iota(\xi,\eta)$, $s_{1}=u_{1}^{\lambda}(\xi)$ and $s_{2}=u_{2}^{\lambda}(\eta)$. Write

These are the sets denoted $S_{\lambda,u_{1}}(\xi)$, $S_{\lambda,u_{2}}(\eta)$ and $S_{\lambda,w}(z)$ in Sup-Convolution of a Function on $\mathbb{R}^M$, and by that definition $s_{1}$, $s_{2}$ and $w^{\lambda}(z)$ are their least upper bounds; in particular all three sets are nonempty and bounded above.

Let $x\in\mathbb{R}^{N}$ and let $a\in\mathbb{R}^{m}$, $b\in\mathbb{R}^{n}$ be the unique pair with $x=\iota(a,b)$. By claim 2 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space we have $x-z=\iota(a-\xi,b-\eta)$, and hence by claim 3 there

Using the distributive law in the form $\frac{\lambda}{2}(s+t)=\frac{\lambda}{2}s+\frac{\lambda}{2}t$ together with the commutativity and associativity of addition, it follows that

As $x$ ranges over $\mathbb{R}^{N}$ the pair $(a,b)$ ranges over all of $\mathbb{R}^{m}\times\mathbb{R}^{n}$, because $\iota$ is a bijection. Therefore $S$ is exactly the set of sums $t_{1}+t_{2}$ with $t_{1}\in S_{1}$ and $t_{2}\in S_{2}$.

We show that $s_{1}+s_{2}$ is the least upper bound of $S$. First, if $t_{1}\in S_{1}$ and $t_{2}\in S_{2}$ then $t_{1}\le s_{1}$ and $t_{2}\le s_{2}$, so $t_{1}+t_{2}\le s_{1}+s_{2}$; hence $s_{1}+s_{2}$ is an upper bound for $S$.

Second, let $t\in\mathbb{R}$ be an upper bound for $S$ and suppose, for a contradiction, that $t<s_{1}+s_{2}$. Put

which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field. Then $s_{1}-\varepsilon<s_{1}$, so $s_{1}-\varepsilon$ is not an upper bound for $S_{1}$ and there is $t_{1}\in S_{1}$ with $s_{1}-\varepsilon<t_{1}$; likewise there is $t_{2}\in S_{2}$ with $s_{2}-\varepsilon<t_{2}$. Adding,

and $t_{1}+t_{2}\in S$, contradicting that $t$ is an upper bound for $S$. Hence $s_{1}+s_{2}\le t$ for every upper bound $t$ of $S$.

The two paragraphs together say that $s_{1}+s_{2}$ is the least upper bound of $S$, so $w^{\lambda}(z)=s_{1}+s_{2}$, which is the asserted identity.