9

Background

I'm reading Cedric Villani's Optimal Transport: Old and New [1] and came across a result (below) I'm not quite sure how to prove. It is used to prove Lemma 4.3 and through my research, I've found it to be known as "Baire's Theorem for Lower Semi-continuous Functions" with topological approaches found in other StackExchange posts like [3] and [4] but never formally worked out.

Question

If $(X, d)$ is a metric space and $F$ is a nonnegative lower semi-continuous function on $X$, then it can be written as the supremum of an increasing sequence of (uniformly?) continuous nonnegative functions. To see this, choose $$ F_{n}(x) = \inf\limits_{y~\in~X}\{~ F(y) + n\cdot d(x,y) ~\} $$ and show it is: (i) increasing; (ii) (uniformly?) continuous; (iii) convergent to $F$ [1, pg. 26; 2, pg. 55].

References:

  1. C. Villani, Optimal Transport, vol. 338. Berlin, Heidelberg: Springer Berlin Heidelberg, 2009. Available: https://ljk.imag.fr/membres/Emmanuel.Maitre/lib/exe/fetch.php?media=b07.stflour.pdf
  2. C. Villani, Topics in Optimal Transportation, 1st ed. American Mathematical Society, 2003.
  3. “Prove by definition that every upper semi-continuous function can be expressed as infimum of a sequence of continuous functions.,” Stack Exchange, 2017. [Online]. Available: https://math.stackexchange.com/questions/2227074/prove-by-definition-that-every-upper-semi-continuous-function-can-be-expressed-a?noredirect=1&lq=1. [Accessed: 28-Dec-2019].
  4. “Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions,” Stack Exchange, 2015. [Online]. Available: https://math.stackexchange.com/questions/1279763/show-that-lower-semicontinuous-function-is-the-supremum-of-an-increasing-sequenc/1284586. [Accessed: 28-Dec-2019].
  5. “What's behind the function g(x)=inf{f(p)+d(x,p):p∈X}?,” Stack Exchange, 2013. [Online]. Available: https://math.stackexchange.com/questions/616071/whats-behind-the-function-gx-operatornameinf-fpdx-pp-in-x?rq=1. [Accessed: 28-Dec-2019]
Community
  • 1
Grant
  • 398
  • Thank you very much for linking [5] https://math.stackexchange.com/questions/616071/whats-behind-the-function-gx-operatornameinf-fpdx-pp-in-x ! The intuition behind the construction of $F_n$ was the first question I had in mind when I read you question, and that link answered it beautifully: the greatest $n$-Lipschitz minorant (or $n$-Lipschitz lower envelope) of $f$! Wow! – D.R. Nov 17 '24 at 07:06

2 Answers2

5

Hints:

$1).\ F(y) + n\cdot d(x,y)\le F(y) + (n+1)\cdot d(x,y)\Rightarrow F_n(x)\le F_{n+1}(x).$

$2).$ Since $F\ge 0$ we have by the triangle inequality $F(y)+nd(y,z)\le F(y)+nd(w,y)+nd(w,z)$ so $F_n(z)-F_n(w)\le nd(w,z)$. Now interchange $z$ and $w.$

$3).\ $ Fix $x_0\in X,\ \epsilon>0.$ Lower semicontinuity of $F$ implies that there is $\delta>0$ such that $d(x_0,y)<\delta\Rightarrow F(y)>F(x_0)-\epsilon.$ Now suppose $d(y,x_0)>\delta>0.$ Then, because $F\ge 0,$ there is a positive integer $n$ such that $F(y)+nd(x_0,y)>F(x_0).$ Now take the inf$:=\alpha$, over all such $y.$ Then, $\alpha\ge F(x_0)$. This shows that the inf in the definition of $F_n$ must lie in $B_{\delta}(x_0)$. But $y\in B_{\delta}(x_0)\Rightarrow F(y)+nd(x_0,y)\ge F(y)>F(x_0)-\epsilon\Rightarrow F_n(x_0)\ge F(x_0)-\epsilon.$

Remark: the proof goes through virtually unchanged if $F$ is only bounded below.

Matematleta
  • 30,081
  • I do not see how the inf in (iii) lies in a ball of radius $\delta$ about $x_{0}$. – Grant Apr 26 '20 at 19:13
  • 1
    Because the sentence before shows that $ \inf\limits_{y~\in~X}{~ F(y) + n\cdot d(x,y) ~}=\inf\limits_{y~\in B_{\delta}(x_0)}{~ F(y) + n\cdot d(x,y) ~}$ so to compute the inf, it is enough to consider values of $y$ in that ball. – Matematleta Apr 26 '20 at 20:40
  • But we suppose the distance between $d(y,x_{0}) > \delta$, so how do we get to inf over $d(y,x_{0}) < \delta$? – Grant Apr 26 '20 at 20:49
  • 1
    I think my wording may be confusing. What I showed was just that if $y$ is outside the ball of radius $\delta$ then the inf in the defnition of $F_n$ is always greater than or equal to the values $F(y) + n\cdot d(x,y)$ and this means that the only values of $y$ that we need to consider when computing the inf, are those inside the ball. And this is all you need to get the desired inequality which is the next sentence. – Matematleta Apr 26 '20 at 20:56
  • So I have that $\inf\limits_{y\in X} (...) \leqq \inf\limits_{y\not\in B_{\delta}(x_{0})}(...)$ since the domain of the lower bound $\inf$ is bigger, but how does that lead to $\inf\limits_{y\in B_{\delta}(x_{0})}(...) \leqq \inf\limits_{y\not\in B_{\delta}(x_{0})}(...)$. I'm sorry, I feel like I'm close, just missing a small detail. – Grant Apr 26 '20 at 21:16
  • 1
    It's confusing. How about this example: suppose you want to inf over all $y$ such that $y>2$ Obviously, the inf is $2$, but you do not need to consider $all$ values of $y$ that are greater than $2$. Because of the nature of the set over which you are infing, you need only consider for example those $y$ such that $2<y<3$. For another example, suppose you want the distance from a point $x$ to a set $A$. You do not have to inf over $all$ values of $a\in A$, you only need to pick one $a$ say $a_0$ and inf over all $a$ such that $d(x,a)\le d(x,a_0).$ Cont. below.. – Matematleta Apr 26 '20 at 21:35
  • 1
    ....In the case of my answer, all we did was prove that the inf can be taken over just those values of $y$ in the ball. The others do not matter. – Matematleta Apr 26 '20 at 21:36
0

Since there are the good proofs for (1) and (2), I want to give a more detailed proof of (3).

Let $x\in X$ and $\varepsilon>0$. As $F$ is l.s.c., there is $\delta>0$ such that $$F(y)>F(x)-\varepsilon, \text{ for all } y\in B(x,\delta).$$ Let $\inf_{x\in X}F(x)=M$. Hence, for any sufficiently large $n\in\mathbb N$, we have $F(y)+nd(x,y)\geq M+nd(x,y)\geq F(x)$ for any $y\not\in B(x,\delta)$. This implies $$\inf_{y\in X\setminus B(x,\delta)}F(y)+nd(x,y)\geq F(x)\geq\inf_{y\in B(x,\delta)}F(y)+nd(x,y).$$ Then $$F_n(x)=\inf_{y\in B(x,\delta)}F(y)+nd(x,y).$$ By the definition of infimum, there is $y_\delta\in B(x,\delta)$ such that $$F_n(x)\geq F(y_\delta)+nd(x,y_\delta)-\varepsilon.$$ Therefore, it follows that $$|F(x)-F_n(x)|=F(x)-F_n(x)\leq F(y_\delta)+\varepsilon-F(y_\delta)-nd(x,y_\delta)+\varepsilon\leq 2\varepsilon$$ for any sufficiently large $n\in\mathbb N$.

Remark: The proof only requires that $F$ has a lower bound, rather than being non-negative. For example, this condition is satisfied if $X$ is compact.

Dingxuan Tang
  • 9