Can continuous correspondence be represented via continuous functions?

Question

Let $\Theta \subset \mathbb{R}^n, \mathcal{X} \subset \mathbb{R^m}$, and suppose that $C: \Theta \rightrightarrows \mathcal{X}$ is a correspondence defined by $f: \Theta \times \mathcal{X}\to \mathbb{R}^d$ as follows:

$$ C(\theta) = \{ x\in \mathcal{X} \mid f_1(\theta, x) \geq 0, \dots, f_d(\theta, x) \geq 0 \} $$

Is it true that 1) if $f$ is continuous then $C$ also is, and 2) if $C$ is continuous then $f$ also is?

(Here continuity for correspondences is defined as upper and lower hemicontinuity.)

Answer 1

Neither implication holds.

Let $\Theta=\mathcal{X}=[-1,1]$.

First, define $f$ by $f(x,y)=xy$. Then $C$ is not lower hemicontinuous. Indeed, $$\big\{\theta\mid C(\theta)\cap (-1,0)\neq\emptyset\big\}=[-1,0]$$ is not open.

Next, let $f$ be any discontinuous function with nonnegative values. Then $C$ is constant with value $[-1,1]$ and, therefore, continuous.