Problem 1518

### Part (a)
1. Define the function $f: \mathbb{R} \rightarrow \mathbb{R}$ as follows:
$$f(x) = \chi_{\mathbb{R}^+} (x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x \leq 0 \end{cases}$$
2. Verify the hypothesis for different cases:
- If both $a, b > 0$ or $a, b \leq 0$, then $f(a) = f(b)$ and the hypothesis $f\left(\frac{a+b}{2}\right) \in \{f(a), f(b)\}$ is trivially satisfied.
- If $a > 0$ and $b \leq 0$, then $\{ f(a), f(b) \} = \{1, 0\}$. The midpoint $\frac{a+b}{2}$ will be either positive or non-positive, so $f\left(\frac{a+b}{2}\right)$ will be either 1 or 0, which is in $\{f(a), f(b)\}$.

Thus, the function $f(x) = \chi_{\mathbb{R}^+} (x)$ satisfies the hypothesis and is a non-constant function.

### Part (b)
1. Assume $f$ is continuous and suppose, for contradiction, that there exist $a_0, b_0 \in \mathbb{R}$ such that $f(a_0) \neq f(b_0)$.
2. Define $c_1 = \frac{a_0 + b_0}{2}$. By the hypothesis, $f(c_1)$ must be either $f(a_0)$ or $f(b_0)$.
3. Depending on the value of $f(c_1)$:
- If $f(c_1) = f(a_0)$, set $a_1 = c_1$ and $b_1 = b_0$.
- If $f(c_1) = f(b_0)$, set $a_1 = a_0$ and $b_1 = c_1$.
4. Note that $f(a_1) \neq f(b_1)$ because $f(a_0) \neq f(b_0)$.
5. Repeat this process to construct sequences $(a_k)$ and $(b_k)$ such that:
- $a_{k+1} = \frac{a_k + b_k}{2}$ or $b_{k+1} = \frac{a_k + b_k}{2}$.
- $f(a_k) \neq f(b_k)$ for all $k$.
6. Since $a_k$ and $b_k$ are bounded and the interval $[a_k, b_k]$ is halved at each step, both sequences $(a_k)$ and $(b_k)$ converge to the same limit $\alpha \in \mathbb{R}$.
7. By the continuity of $f$, we have:
$$\lim_{k \to \infty} f(a_k) = f(\alpha) \quad \text{and} \quad \lim_{k \to \infty} f(b_k) = f(\alpha)$$
8. However, since $f(a_k) = f(a_0)$ and $f(b_k) = f(b_0)$ for all $k$, we have:
$$f(a_0) = f(\alpha) \quad \text{and} \quad f(b_0) = f(\alpha)$$
9. This leads to a contradiction because $f(a_0) \neq f(b_0)$.

Therefore, the assumption that $f(a_0) \neq f(b_0)$ must be false, implying that $f$ is constant.

$\blacksquare$