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Let $f:(0,\infty) \to [0,\infty)$ be a continuous function satisfying $f(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$. Define $$ F(s)=\min_{xy=s,x,y>0} f(x)+ f(y), \, \, \, \, \text{for } \, \, s \in (0,\infty). $$

Claim: $F$ is continuous.

I am looking for a reference for such a claim. (not that claim exactly, but perhaps a slightly more general claim which implies it , or is similar to it).

I think that it follows from a result in the book "Perturbation Analysis of Optimization Problems" by Bonnans and Shapiro, but that book phrases things in much more abstract setting than I find necessary.

BTW, here is my proof:

Suppose that $s \le 1$. Define $g(x,y)=f(x)+f(y)$. Suppose that $s_n \to s$. Write $F(s_n)=g(x_n,y_n)$ for some $x_n,y_n \in [s_n,1]^2, x_ny_n=s_n$. By compactness we may assume that $x_{n_{k}} \to x, y_{n_{k}} \to y$. Thus $$ F(s) \le g(x,y)=\lim_{k \to \infty} g(x_{n_k},x_{y_k})=\lim_{k \to \infty}F(s_{n_k}) \le \liminf F(s_n). $$

On the other hand, take $(x,y) \in (0,\infty)^2$ such that $xy=s$ and $F(s)=g(x,y)$. Now take $x_n,y_n$ such that $x_ny_n=s_n$, and $(x_n,y_n) \to (x,y)$. Then $$ F(s_n) \le g(x_n,y_n) \Rightarrow \limsup F(s_n) \le \lim_{n \to \infty}g(x_n,y_n)=g(x,y)=F(s). $$

Asaf Shachar
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  • Is the book result abstract to the point of ambiguity (i.e.unclear if it entirely proves the result)? – Zim Oct 05 '20 at 20:36
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    I am not certain. I think that it implies the result, but I am not sure since there are quite a few definitions to unpack. This makes the abstract statement less appealing to me, since I thought there should be a rather simple, concrete and easy to state claim which implies the result. (and indeed Martin R's answer provides such a thing.) – Asaf Shachar Oct 07 '20 at 11:56

1 Answers1

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That is a consequence of the following theorem:

If $X,Y$ are topological spaces and $Y$ is compact. Then for any continuous $f:X\times Y\to\mathbb{R}$, the function $g(x) := \inf_{y\in Y} f(x,y)$ is well-defined and continuous.

Source: Continuity of the infimum, by Willie Wong. See also Infimum is a continuous function, compact set.

In our case we have $$ F(s) = \inf_{x > 0} f(x) + f(s/x) \, . $$ In order to apply the above theorem we fix $0 < s_1 < s_2$ and show that one can replace the $\inf_{x > 0}$ by an infimum over a compact set if $s$ is restricted to $[s_1, s_2]$: Let $$ x_1 = \min(1, s_1) \, , \, x_2 = \max(1, s_2) \, . $$ Then, due to the given monotony conditions on $f$, $$ x < x_1 \implies f(x)+f(s/x) \ge f(x_1) + f(s/x_1) \\ x > x_2 \implies f(x)+f(s/x) \ge f(x_2) + f(s/x_2) $$ and therefore $$ F(s) = \inf_{x_1 \le x \le x_2} f(x) + f(s/x) \, . $$

Martin R
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