Proof of A Smooth Variational Principle of Borwein-Preiss Type with a Gauge on a Complete Metric Space

Each result cited is universally quantified over the data in its own statement. Natural numbers are read in $\mathbb{R}$ through the canonical map as in The Real Numbers: Standing Notation and Background §numbers, so $1/n$ is a positive real for every $n\in\mathbb{N}$; the successor of $n$ is written $n+1$, as in claim 1 of Natural Numbers. Since $X\ne\varnothing$ and $f$ is bounded above, $\sigma=\sup_{x\in X}f(x)$ exists by The Real Numbers: Standing Notation and Background §bounds, and $f(x)\le\sigma$ for every $x\in X$. Since each $c_{k}$ is positive and $0\le g(x,y)$, claim 5 of Elementary Arithmetic in an Ordered Field gives

$$0\le c_{k}\,g(x,y)\qquad\text{for all }k\in\mathbb{N}\text{ and }x,y\in X.\tag{0}$$

Step 1 (Gauge radii). For $n\in\mathbb{N}$ let $B_{n}$ be the set of reals $\beta>0$ such that $d(x,y)<1/n$ for all $x,y\in X$ with $g(x,y)\le\beta$. The hypothesis on $g$, applied with $\eta=1/n$, shows that $B_{n}$ is nonempty. Thus $(B_{n})_{n\in\mathbb{N}}$ is a family of nonempty subsets of $\mathbb{R}$, and Axiom of Countable Choice gives a sequence $(\beta_{n})_{n\in\mathbb{N}}$ with $\beta_{n}\in B_{n}$ for every $n$, that is,

$$0<\beta_{n},\qquad\text{and}\qquad g(x,y)\le\beta_{n}\ \Longrightarrow\ d(x,y)<\tfrac{1}{n}\quad(x,y\in X).\tag{1}$$

Step 2 (Recursive construction). Let $\mathcal{T}$ be the set of quadruples $(n,z,H,A)$ with $n\in\mathbb{N}$, $A\subseteq X$, $z\in A$, and $H:X\to\mathbb{R}$ bounded above. For $(n,z,H,A)\in\mathcal{T}$ put $A^{+}=\{x\in A:\ H(z)\le H(x)\}$. Then $z\in A^{+}$, so $H(A^{+})$ is nonempty, and it is bounded above because $H$ is; let $M(n,z,H,A)=\sup_{x\in A^{+}}H(x)$ (The Real Numbers: Standing Notation and Background §bounds).

Let $R$ be the relation on $\mathcal{T}$ consisting of the pairs $\bigl((n,z,H,A),(n+1,y,H',A^{+})\bigr)$ with $(n,z,H,A)\in\mathcal{T}$, $y\in A^{+}$, $H'(x)=H(x)-c_{n+1}\,g(x,y)$ for every $x\in X$, and

$$H(y)>M(n,z,H,A)-c_{n+1}\beta_{n}.$$

Every $(n,z,H,A)\in\mathcal{T}$ has an $R$-successor in $\mathcal{T}$: $c_{n+1}\beta_{n}$ is positive by (1) and claim 5 of Elementary Order Arithmetic in an Ordered Field, so claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to $H(A^{+})$ with $\varepsilon=c_{n+1}\beta_{n}$, gives $y\in A^{+}$ with $H(y)>M(n,z,H,A)-c_{n+1}\beta_{n}$; the function $H'$ so defined satisfies $H'(x)\le H(x)$ for every $x$ by (0) and claim 3 of Elementary Arithmetic in an Ordered Field, so it is bounded above, and $y\in A^{+}$, whence $(n+1,y,H',A^{+})\in\mathcal{T}$.

Let $F_{1}:X\to\mathbb{R}$ be given by $F_{1}(x)=f(x)-c_{1}\,g(x,x_{1})$; as above $F_{1}(x)\le f(x)\le\sigma$, so $(1,x_{1},F_{1},X)\in\mathcal{T}$. By Axiom of Dependent Choice there is a sequence $(a_{m})_{m\in\mathbb{N}}$ in $\mathcal{T}$ with $a_{1}=(1,x_{1},F_{1},X)$ and $(a_{m},a_{m+1})\in R$ for every $m$. By the form of $R$ and induction on $m$, $a_{m}=(m,x_{m},F_{m},T_{m})$, where $(x_{m})_{m\in\mathbb{N}}$ is a sequence in $X$ whose first term is the given point $x_{1}$, each $F_{m}:X\to\mathbb{R}$ is a function and each $T_{m}\subseteq X$, such that $T_{1}=X$ and, for every $m\in\mathbb{N}$,

$$x_{m}\in T_{m},\qquad T_{m+1}=\{x\in T_{m}:\ F_{m}(x_{m})\le F_{m}(x)\},\qquad x_{m+1}\in T_{m+1},\tag{2}$$ $$F_{m+1}(x)=F_{m}(x)-c_{m+1}\,g(x,x_{m+1})\quad(x\in X),\tag{3}$$ $$F_{m}(x_{m+1})>\sup_{x\in T_{m+1}}F_{m}(x)-c_{m+1}\beta_{m}.\tag{4}$$

Step 3 (Partial sums, the series, and semicontinuity). For $x\in X$ and $n\in\mathbb{N}$ let $p_{n}(x)=\sum_{k=1}^{n}c_{k}\,g(x,x_{k})$, the $n$-th partial sum of $(c_{k}\,g(x,x_{k}))_{k\in\mathbb{N}}$. Then

$$F_{n}(x)=f(x)-p_{n}(x)\qquad(x\in X,\ n\in\mathbb{N}),\tag{5}$$

by induction on $n$: for $n=1$ this is the definition of $F_{1}$ together with $\sum_{k=1}^{1}a_{k}=a_{1}$ from claim 1 of Properties of Finite Sums, and the step from $n$ to $n+1$ is (3) together with the recursion $p_{n+1}(x)=p_{n}(x)+c_{n+1}\,g(x,x_{n+1})$ of the same claim. In particular $F_{1}(x_{1})=f(x_{1})-c_{1}\,g(x_{1},x_{1})=f(x_{1})$, since $g(x_{1},x_{1})=0$.

Fix $x\in X$. By (0) and the hypotheses, $0\le c_{k}$, $0\le g(x,x_{k})\le G$ for every $k$, and $\sum_{k=1}^{\infty}c_{k}$ converges, so the tail bound for dominated series (with $\mu_{k}=c_{k}$, $w_{k}=g(x,x_{k})$ and $M=G$) shows that the series $P(x)=\sum_{k=1}^{\infty}c_{k}\,g(x,x_{k})$ converges and $0\le P(x)-p_{n}(x)$ for every $n$. Hence $\Phi(x)=f(x)-P(x)$ is defined and, by claim 3 of Elementary Arithmetic in an Ordered Field and (5), since $F_{n}(x)-\Phi(x)=P(x)-p_{n}(x)$,

$$\Phi(x)\le F_{n}(x)\qquad(x\in X,\ n\in\mathbb{N}).\tag{6}$$

Each $F_{n}$ is upper semicontinuous on $X$. Indeed, for every $k$ the function $c_{k}\,g(\cdot,x_{k})$ is lower semicontinuous on $X$ by claim 3 of Sums and Nonnegative Multiples of Semicontinuous Functions, since $g(\cdot,x_{k})$ is and $0\le c_{k}$. As $f$ is upper semicontinuous on $X$, claim 3 of Negation, Restriction, and Separated Differences of Semicontinuous Functions shows that $F_{1}=f-c_{1}\,g(\cdot,x_{1})$ is upper semicontinuous on $X$; and if $F_{n}$ is, then so is $F_{n+1}=F_{n}-c_{n+1}\,g(\cdot,x_{n+1})$, by (3) and the same claim.

Step 4 (Nesting and monotone values). We claim that, whenever $n\le m$,

$$T_{m}\subseteq T_{n},\qquad x_{m}\in T_{n},\qquad F_{n}(x_{n})\le F_{m}(x_{m}).\tag{7}$$

First, for every $m$ we have $T_{m+1}\subseteq T_{m}$ by (2), and

$$F_{m+1}(x_{m+1})=F_{m}(x_{m+1})-c_{m+1}\,g(x_{m+1},x_{m+1})=F_{m}(x_{m+1})\ge F_{m}(x_{m})$$

by (3), $g(x_{m+1},x_{m+1})=0$, and $x_{m+1}\in T_{m+1}$ in (2). Now argue by induction on $m$. For $m=1$, $n\le1$ and $1\le n$ (claim 4 of Properties of the Order on the Natural Numbers) force $n=1$ by claim 2 there, and the first and third assertions are trivial. If they hold for $m$ and $n\le m+1$, then either $n=m+1$, where they are trivial, or $n\le m$ by claim 5 of Properties of the Order on the Natural Numbers, in which case $T_{m+1}\subseteq T_{m}\subseteq T_{n}$ and $F_{n}(x_{n})\le F_{m}(x_{m})\le F_{m+1}(x_{m+1})$. The second assertion follows from the first because $x_{m}\in T_{m}$ by (2).

Step 5 (The sets $T_{n}$ are closed). We show by induction on $n$: if $y\in X$ is such that for every real $\delta>0$ some $w\in T_{n}$ satisfies $d(y,w)<\delta$, then $y\in T_{n}$. For $n=1$ this is clear since $T_{1}=X$. Assume it for $n$, and let $y$ have this property relative to $T_{n+1}$. Since $T_{n+1}\subseteq T_{n}$, $y$ has it relative to $T_{n}$, so $y\in T_{n}$. Suppose $F_{n}(x_{n})\le F_{n}(y)$ fails; then $F_{n}(y)<F_{n}(x_{n})$ since the order is total, and $\gamma=F_{n}(x_{n})-F_{n}(y)$ is positive by claim 1 of Elementary Order Arithmetic in an Ordered Field. As $F_{n}$ is upper semicontinuous at $y$ relative to $X$ (Step 3), there is $\delta>0$ with $F_{n}(w)<F_{n}(y)+\gamma=F_{n}(x_{n})$ for every $w\in X$ with $d(y,w)<\delta$. Taking such a $w$ in $T_{n+1}$ contradicts $F_{n}(x_{n})\le F_{n}(w)$, which holds by (2). Hence $F_{n}(x_{n})\le F_{n}(y)$, and $y\in T_{n+1}$ by (2).

Step 6 (Small sets). Let $n\in\mathbb{N}$ and $x\in T_{n+2}$. Then $x\in T_{n+1}$ by (7), so $F_{n}(x)\le\sup_{w\in T_{n+1}}F_{n}(w)$, and by (4) and claims 1 and 2 of Elementary Order Arithmetic in an Ordered Field, $F_{n}(x)<F_{n}(x_{n+1})+c_{n+1}\beta_{n}$. On the other hand $F_{n+1}(x_{n+1})\le F_{n+1}(x)$ by (2) applied with $m=n+1$, which by (3) and $g(x_{n+1},x_{n+1})=0$ reads $F_{n}(x_{n+1})\le F_{n}(x)-c_{n+1}\,g(x,x_{n+1})$. Combining the two,

$$c_{n+1}\,g(x,x_{n+1})\le F_{n}(x)-F_{n}(x_{n+1})<c_{n+1}\beta_{n},$$

and multiplying by $c_{n+1}^{-1}>0$ (claims 7 and 10 of Elementary Order Arithmetic in an Ordered Field) gives $g(x,x_{n+1})<\beta_{n}$. By (1),

$$d(x,x_{n+1})<\tfrac{1}{n}\qquad\text{for every }n\in\mathbb{N}\text{ and every }x\in T_{n+2}.\tag{8}$$

Step 7 (Centres). Let a real $\varepsilon>0$ be given. By claim 8 of Elementary Order Arithmetic in an Ordered Field and claim 3 of The Archimedean Property of the Real Numbers there is $n\in\mathbb{N}$ with $1/n<\varepsilon/2$. For $m,\ell\in\mathbb{N}$ with $m\ge n+2$ and $\ell\ge n+2$, (7) gives $x_{m},x_{\ell}\in T_{n+2}$, so (8) and the symmetry and triangle inequality of Metric Space give

$$d(x_{m},x_{\ell})\le d(x_{m},x_{n+1})+d(x_{\ell},x_{n+1})<\tfrac{1}{n}+\tfrac{1}{n}<\varepsilon,$$

using claims 3 and 8 of Elementary Order Arithmetic in an Ordered Field. Thus $(x_{m})$ is a Cauchy sequence, and since $(X,d)$ is complete it converges to some $\bar{x}\in X$. This is the centres assertion.

Moreover $\bar{x}\in T_{n}$ for every $n\in\mathbb{N}$. Given a real $\delta>0$, convergence gives $N\in\mathbb{N}$ with $d(x_{m},\bar{x})<\delta$ for all $m\ge N$; let $m$ be the larger of $N$ and $n$ (trichotomy, claim 3 of Properties of the Order on the Natural Numbers). Then $x_{m}\in T_{n}$ by (7) and $d(\bar{x},x_{m})<\delta$, so Step 5 gives $\bar{x}\in T_{n}$.

Step 8 (The sets $T_{n}$ shrink to $\bar{x}$). Let $y\in X$ lie in $T_{n}$ for every $n\in\mathbb{N}$. For each $n$, both $y$ and $\bar{x}$ lie in $T_{n+2}$, so as in Step 7, (8) gives $d(y,\bar{x})\le d(y,x_{n+1})+d(\bar{x},x_{n+1})<2/n$. Given a real $\varepsilon>0$, choosing $n$ with $1/n<\varepsilon/2$ as in Step 7 yields $d(y,\bar{x})<\varepsilon$; since $0\le d(y,\bar{x})$, vanishing gives $d(y,\bar{x})=0$, so $y=\bar{x}$ by Metric Space.

Step 9 (Value). We show

$$F_{n}(x_{n})\le\Phi(\bar{x})\qquad\text{for every }n\in\mathbb{N}.\tag{9}$$

Fix $n$ and suppose instead that $\Phi(\bar{x})<F_{n}(x_{n})$, so that $\gamma=F_{n}(x_{n})-\Phi(\bar{x})$ is positive by claim 1 of Elementary Order Arithmetic in an Ordered Field. By Step 3 and the definition of the sum, $(p_{m}(\bar{x}))_{m\in\mathbb{N}}$ converges to $P(\bar{x})$, so there is $N\in\mathbb{N}$ with $|p_{m}(\bar{x})-P(\bar{x})|<\gamma$, hence $P(\bar{x})-p_{m}(\bar{x})<\gamma$ by claim 9 of Properties of the Absolute Value in an Ordered Field, for all $m\ge N$. Let $m$ be the larger of $N$ and $n$. Since $\bar{x}\in T_{m+1}$ by Step 7, (2) gives $F_{m}(x_{m})\le F_{m}(\bar{x})$, and $F_{n}(x_{n})\le F_{m}(x_{m})$ by (7). Using (5) and claim 1 of Elementary Order Arithmetic in an Ordered Field,

$$F_{n}(x_{n})\le F_{m}(\bar{x})=f(\bar{x})-p_{m}(\bar{x})<f(\bar{x})-P(\bar{x})+\gamma=\Phi(\bar{x})+\gamma=F_{n}(x_{n}),$$

so claim 2 of Elementary Order Arithmetic in an Ordered Field would give $F_{n}(x_{n})<F_{n}(x_{n})$, which is impossible because $a<b$ requires $a\ne b$. Hence (9) holds, since the order is total. Taking $n=1$ and recalling $F_{1}(x_{1})=f(x_{1})$ from Step 3 gives $\Phi(\bar{x})\ge f(x_{1})$, the value assertion.

Step 10 (Strict maximum). Let $x\in X$ with $x\ne\bar{x}$. By Step 8, $x$ does not lie in every $T_{j}$. Hence there is $n\in\mathbb{N}$ with $x\in T_{n}$ and $x\notin T_{n+1}$: otherwise, since $x\in T_{1}=X$, induction on $n$ would give $x\in T_{n}$ for every $n$. By (2), $x\notin T_{n+1}$ with $x\in T_{n}$ means that $F_{n}(x_{n})\le F_{n}(x)$ fails, so $F_{n}(x)<F_{n}(x_{n})$. By (6) and (9),

$$\Phi(x)\le F_{n}(x)<F_{n}(x_{n})\le\Phi(\bar{x}),$$

and claim 2 of Elementary Order Arithmetic in an Ordered Field gives $\Phi(x)<\Phi(\bar{x})$, the strict maximum assertion.

Step 11 (Localisation). By (6) with $n=1$, the value assertion, the hypothesis on $x_{1}$ and $f(\bar{x})\le\sigma$,

$$f(\bar{x})-c_{1}\,g(\bar{x},x_{1})=F_{1}(\bar{x})\ge\Phi(\bar{x})\ge f(x_{1})\ge\sigma-\varepsilon\ge f(\bar{x})-\varepsilon .$$

By claim 3 of Elementary Arithmetic in an Ordered Field, $0\le\bigl(f(\bar{x})-c_{1}\,g(\bar{x},x_{1})\bigr)-\bigl(f(\bar{x})-\varepsilon\bigr)=\varepsilon-c_{1}\,g(\bar{x},x_{1})$, and by the same claim $c_{1}\,g(\bar{x},x_{1})\le\varepsilon$, the localisation assertion.