Infimum is a continuous function, compact set

Solution 1:

Following the suggestion of Joe Johnson (I was going using intervals $(a,b)$, but this is simpler) (do you see why Joe's suggestion is enough? The sets $(a,\infty)$ and $(-\infty,a)$ form a subbasis for the topology of $\mathbb{R}$, and for a function to be continuous, it suffices to show that the inverse image of every set in a given subbasis is open).

Let $a\in\mathbb{R}$. What is $g^{-1}(-\infty,a)$? It consists of all $x\in X$ such that $g(x)\lt a$. If $g(x)\lt a$, there exists $y_x$ such that $f(x,y_x)\lt a$. Since $f$ is continuous, $f^{-1}(-\infty,a)$ is open, and $(x,y_x)\in f^{-1}(-\infty,a)$, so there exist open sets $U_x$ and $V_y$ of $X$ and $Y$, respectively, such that $U_x\times V_y\subseteq f^{-1}(-\infty,a)$.

Now, suppose $x'\in U_x$. Then for all $y\in V_y$ (in particular, for $y_x$) you have $f(x',y_x)\in (-\infty,a)$. What does that tell you about $g(x')$? What does that tell you about $g^{-1}(-\infty,a)$?

Now try to do something along those lines with $g^{-1}(a,\infty)$.

Solution 2:

Perhaps, we can suppose that $X$ and $Y$ are metric spaces (in particular subset of real line $\mathbb R$) and that $f(x,y)$ is real-valued function in $x ,y$. For a simple model we can take $X=\mathbb R$ and $Y=[a,b]$ closed interval in $\mathbb R$.

Set $I(x) = \inf_{y\in Y} f(x,y)$. Fix a point $x_0$ in $X$ .

Let us prove that $I(x)$ is continuous at $x_0$.

Consider a sequence $x_n$ in $X$ such that $x_n$ converges to $x_0$.

Let us prove that $I(x_n)$ converges to $I(x_0)$.

Since $Y$ is compact there is a sequence $(y_n)$ and $y_0$ in $Y$ such that $I(x_n) = f(x_n,y_n)$ and $a_0:= I(x_0)= f(x_0,y_0)$. Since $Y$ is compact there is a subsequence of $y_n$ which converges to $y^0$ in $Y$. By abusing notation we can suppose that $(x_n,y_n)$ converges to $(x_0, y^0)$. By continuity of $f$ in $x ,y$ we conclude that $f(x_n,y_n)$ converges to $f(x_0, y^0):=b_0$. By definition of $I(x_n)$, $f(x_n,y_n) \leq f(x_n,y_0)$ and therefore since $(x_n,y_n)$ converges to $(x_0, y^0)$ and $(x_n,y_0)$ converges to $(x_0, y_0)$, by continuity of $f$, we find

$b_0=f(x_0,y^0)\leq a_0$. On the other hand, by definition $a_0\leq b_0$. Hence $a_0=b_0$.

This shows that the set of cluster points (or accumulation points) of sequence $I(x_n)$ is a single point $a_0$ and therefore $I(x_n)$ converges to $a_0=I(x_0)$.

Therefore $I$ is continuous in $x_0$ .

If necessary, we can give more details.

You can also see Mathematical Analyis I/II. by Vladimir A. Zorich.

Can I get a text book reference for the proof that $g(x) = \inf_{y\in Y} f(x,y)$ is continuous in $x$ given compactness of $Y$ and continuity of $f$ in $x ,y$? - ResearchGate. Available from: https://www.researchgate.net/post/Can_I_get_a_text_book_reference_for_the_proof_that_gx_inf_yin_Y_fx_y_is_continuous_in_x_given_compactness_of_Y_and_continuity_of_f_in_x_y [accessed Jul 29, 2015].

Solution 3:

Old question, I know, but there’s a simple synthetic approach:

If $f: X × Y → ℝ$ is continuous, then so is the curried version $f^\mathrm{curr}\colon X → C(Y,ℝ)$, where $C(Y,ℝ)$ is the space of continuous functions $Y → ℝ$ with the compact-open topology. Since $Y$ is compact, taking infima is continuous as a map $\inf \colon C(Y,ℝ) → ℝ, f ↦ \inf f$. Thus $\inf ∘ f^\mathrm{curr} \colon X → ℝ$ is continuous as well, which is exactly the function in question.