Let $X, Y$ be topological spaces, and $Y$ is compact, and let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous function. Define $g: X \rightarrow \mathbb{R}$ as $g(x) = \inf_{y \in Y} f(x,y)$. Is $g$ continuous? I think the answer is yes, but can someone provide a simple proof?
Does it hold if $Y$ is non-compact?
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A. Prove that $g$ is continuous.
B. Prove that if $f$ is bounded below, then $g$ is bounded below.
C. Prove that if $Y$ is Hausdorff, and $f(x,y) = 0$ for all but finitely many $y \in Y$ for each $x \in X$, then $g$ is continuous.
D. Prove that if $Y$ is Hausdorff, $\inf_{y \in Y} f(x,y) = 0$ for all $x \in X$, and $\inf_{x \in X} \inf_{y \in Y} f(x,y) = 0$, then $g$ is continuous.
E. Prove that if $Y$ is Hausdorff, and $f(x,y) \geq 0$ for all $(x,y) \in X \times Y$, and $\inf_{y \in Y} f(x,y) = 0$ for all $x \in X$, and $\inf_{x \in X} \inf_{y \in Y} f(x,y) = 0$, then $g$ is continuous.
F. Prove that if $Y$ is Hausdorff, and $f(x,y) = 0$ for all $(x,y) \in X \times Y$, then $g$ is continuous.
A. Let $x_0 \in X$ and $\varepsilon>0$ be given. We need to show that there exists a $\delta>0$ such that $|g(x)-g(x_0)| < \varepsilon$ whenever $|x-x_0|<\delta$.
Since $f$ is continuous, for each $y \in Y$, the function $f_y: X \rightarrow \mathbb{R}$ given by $f_y(x) = f(x,y)$ is continuous. Therefore, for each $y \in Y$, there exists $\delta_y > 0$ such that $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$ whenever $|x-x_0| < \delta_y$.
Since $Y$ is compact, there exists a finite subset $Y_0 \subseteq Y$ such that $Y \subseteq \bigcup_{y \in Y_0} B(y, \frac{\varepsilon}{2})$.
Let $\delta = \min \{ \delta_y : y \in Y_0 \}$. Then for all $y \in Y$, we have $y \in B(y, \frac{\varepsilon}{2}) \subseteq B(y_0, \delta)$ for some $y_0 \in Y_0$. Therefore, $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$ whenever $|x-x_0| < \delta$.
Now let $x \in X$ be such that $|x-x_0| < \delta$. Then for all $y \in Y$, we have $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$. Therefore, \begin{align*} |g(x)-g(x_0)| &= \left| \inf_{y \in Y} f_y(x) - \inf_{y \in Y} f_y(x_0) \right| \\ &\leq \left| \inf_{y \in Y} f_y(x) - f_y(x_0) \right| + \left| f_y(x_0) - \inf_{y \in Y} f_y(x_0) \right| \\ &\leq \sup_{y \in Y} |f_y(x)-f_y(x_0)| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align*} Therefore, $g$ is continuous.
B. Let $x_0 \in X$ and $\varepsilon>0$ be given. We need to show that there exists a $\delta>0$ such that $|g(x)-g(x_0)| < \varepsilon$ whenever $|x-x_0|<\delta$.
Since $f$ is bounded below, there exists $M \in \mathbb{R}$ such that $f(x,y) \geq M$ for all $(x,y) \in X \times Y$.
Since $f$ is continuous, for each $y \in Y$, the function $f_y: X \rightarrow \mathbb{R}$ given by $f_y(x) = f(x,y)$ is continuous. Therefore, for each $y \in Y$, there exists $\delta_y > 0$ such that $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$ whenever $|x-x_0| < \delta_y$.
Since $Y$ is compact, there exists a finite subset $Y_0 \subseteq Y$ such that $Y \subseteq \bigcup_{y \in Y_0} B(y, \frac{\varepsilon}{2})$.
Let $\delta = \min \{ \delta_y : y \in Y_0 \}$. Then for all $y \in Y$, we have $y \in B(y, \frac{\varepsilon}{2}) \subseteq B(y_0, \delta)$ for some $y_0 \in Y_0$. Therefore, $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$ whenever $|x-x_0| < \delta$.
Now let $x \in X$ be such that $|x-x_0| < \delta$. Then for all $y \in Y$, we have $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$. Therefore, \begin{align*} |g(x)-g(x_0)| &= \left| \inf_{y \in Y} f_y(x) - \inf_{y \in Y} f_y(x_0) \right| \\ &\leq \left| \inf_{y \in Y} f_y(x) - f_y(x_0) \right| + \left| f_y(x_0) - \inf_{y \in Y} f_y(x_0) \right| \\ &\leq \sup_{y \in Y} |f_y(x)-f_y(x_0)| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align*} Therefore, $g$ is continuous.
C. Let $x_0 \in X$ and $\varepsilon>0$ be given. We need to show that there exists a $\delta>0$ such that $|g(x)-g(x_0)| < \varepsilon$ whenever $|x-x_0|<\delta$.
Since $f(x,y) = 0$ for all but finitely many $y \in Y$ for each $x \in X$, there exists a finite subset $Y_0 \subseteq Y$ such that $f(x,y) = 0$ for all $(x,y) \in X \times (Y \setminus Y_0)$.
Since $f$ is continuous, for each $y \in Y_0$, the function $f_y: X \rightarrow \mathbb{R}$ given by $f_y(x) = f(x,y)$ is continuous. Therefore, for each $y \in Y_0$, there exists $\delta_y > 0$ such that $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$ whenever $|x-x_0| < \delta_y$.
Since $Y$ is Hausdorff, there exists a $\delta_0 > 0$ such that $B(y, \delta_0) \cap B(y', \delta_0) = \emptyset$ for all $y, y' \in Y$ with $y \neq y'$.
Let $\delta = \min \{ \delta_y, \delta_0 : y \in Y_0 \}$. Then for all $y \in Y_0$, we have $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$ whenever $|x-x_0| < \delta$.
Now let $x \in X$ be such that $|x-x_0| < \delta$. Then for all $y \in Y$, we have $|f_y(x)-f_y(x_0)| < \frac{\varepsilon}{2}$ whenever $|x-x_0| < \delta$. Therefore, \begin{align*} |g(x)-g(x_0)| &= \left| \inf_{y \in Y} f_y(x) - \inf_{y \in Y} f_y(x_0) \right| \\ &\leq \left| \inf_{y \in Y} f_y(x) - f_y(x_0) \right| + \left| f_y(x_0) - \inf_{y \in Y} f_y(x_0) \right| \\ &\leq \sup_{y \in Y} |f_y(x)-f_y(x_0)| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align*} Therefore, $g$ is continuous.
On
For the compact-open topology on $C(Y,\mathbb R)$, the map $X\to C(Y,\mathbb R),\;x\mapsto f(x,\cdot)$ is continuous.
Since $Y$ is compact and $\mathbb R$ metric, this topology coincides with that of uniform convergence, for which $C(Y,\mathbb R)\to\mathbb R,\;\varphi\mapsto\min(\varphi(Y))$ is continuous.
By composition, $g$ is continuous.
A counterexample in the non-compact case (similar to the one Daniel Schepler gave a few minutes after your post) is $$f:\mathbb R^2\to\mathbb R,(x,y)\mapsto\begin{cases}0&\text{if }x=0\\-\Psi(y-\frac1x)&\text{else,}\end{cases}$$ where $\Psi$ is any non-negative bump function. The key point in this construction is that as $t\to\infty$, $\Psi(\cdot-t)\to0$ uniformly on every compact but not uniformly on $\mathbb R.$
For general, not necessarily compact $Y$, this map is upper-semicontinuous, but not necessarily lower-semicontinuous. When $Y$ is compact, then we get full continuity.
Fix $x_0 \in X$, let $\alpha = g(x_0)$, and suppose $\varepsilon > 0$. The set $f^{-1}(-\infty, \alpha + \varepsilon)$ must contain points in $\{x_0\} \times Y$, otherwise $\alpha = \inf_{y \in Y} f(x_0, y) \ge \alpha + \varepsilon$, which is impossible. Pick any $(x_0, y_0) \in f^{-1}(-\infty, \alpha + \varepsilon)$, and let $\mathcal{U} \times \mathcal{V}$ be a basic open neighbourhood of $(x_0, y_0)$ contained in $f^{-1}(-\infty, \alpha + \varepsilon)$. Then, $$x \in \mathcal{U} \implies g(x) = \inf_{y \in Y} f(x, y) \le \inf_{y \in \mathcal{V}} f(x, y) < \alpha + \varepsilon.$$ That is, $g$ is upper-semicontinuous.
Suppose now that $Y$ is compact. Again, fix $x_0 \in X$, let $\alpha = g(x_0)$, and suppose $\varepsilon > 0$. The set $f^{-1}(\alpha - \varepsilon, \infty)$ is open in $X \times Y$ by the continuity of $f$, and this set must contain $\{x_0\} \times Y$.
For each $(x_0, y) \in \{x_0\} \times Y \subseteq f^{-1}(\alpha - \varepsilon, \infty)$, we can find a basic open neighbourhood $\mathcal{U}_y \times \mathcal{V}_y$ of $(x_0, y_0)$ that is contained in $f^{-1}(\alpha - \varepsilon, \infty)$, and these form an open cover of $\{x_0\} \times Y$.
By the compactness of $Y$ there exist $y_1, \ldots, y_n \in Y$ such that $$\{x_0\} \times Y \subseteq \bigcup_{i=1}^n \left(\mathcal{U}_{y_1} \times \mathcal{V}_{y_n}\right) \subseteq f^{-1}(\alpha - \varepsilon, \infty).$$ Let $\mathcal{U} = \bigcap_i \mathcal{U}_{y_i}$. Then $$\{x_0\} \times Y \subseteq \mathcal{U} \times Y \subseteq f^{-1}(\alpha - \varepsilon, \infty).$$ That is, for every $y \in Y$ and $x \in \mathcal{U}$, $f(x, y) > \alpha - \varepsilon$, hence taking the infimum over $y \in Y$, $$x \in \mathcal{U} \implies g(x) = \inf_{y \in Y} f(x, y) \ge \alpha - \varepsilon.$$ Thus, when $Y$ is compact, we have lower-semicontinuity as well.
Finally, let's finish with an example to show sharpness. Due primarily to laziness, I'm going to let $X = Y = \Bbb{R}$, and use Urysohn's lemma on $X \times Y = \Bbb{R}^2$. Consider the sets $A = \{(x, y) \in \Bbb{R}^2 : x \ge e^y\}$ and $B = \{(x, y) : x \le 0\}$. These are closed, disjoint sets in a normal toplogical space $\Bbb{R}^2$, hence by Urysohn, there must exist a function $f : \Bbb{R}^2 \to [0, 1]$ such that $f(A) = 0$ and $f(B) = 1$.
Since the entirety of $\{0\} \times Y$ lies in $B$, we have $g(0) = \inf_{y \in \Bbb{R}} f(0, y) = 1$. But, if we take $x > 0$, then $(x, \ln x) \in A$, so $g(x) = \inf_{y \in \Bbb{R}} f(x, y) \le f(x, \ln x) = 0$. Thus, $g$ is not continuous.