**Step 1: Construction of a Retraction Map** Since $\mathcal{F}$ is bounded, let $M = \sup_{x \in \mathcal{F}} \|x\|_X$. Choose a radius $r > M$. We define a map $g: X \to X$ which retracts the image of $f$ into the closed ball $\overline{B}_r(0)$ if it falls outside: \begin{align*}

g(x) := \begin{cases} f(x) & \text{if } \|f(x)\|_X \le r, \\ \frac{r}{\|f(x)\|_X} f(x) & \text{if } \|f(x)\|_X > r. \end{cases} \end{align*} This map can be viewed as the composition $g = \rho \circ f$, where $\rho$ is the radial retraction onto the ball $\overline{B}_r(0)$. Since both $f$ and the radial retraction are continuous, $g$ is continuous. **Step 2: Compactness of the Auxiliary Map** We consider the restriction of $g$ to the closed ball $C = \overline{B}_r(0)$. The map $f$ is a compact operator, so it maps the bounded set $C$ to a relatively compact set $K = \overline{f(C)}$. The radial retraction $\rho$ is continuous, so it maps the compact set $K$ to a compact set. Thus, $g(C)$ is relatively compact. Moreover, by definition, the image of $g$ is contained in $\overline{B}_r(0) = C$. Therefore, $g: C \to C$ is a continuous, compact map on a nonempty, closed, bounded, and convex set. **Step 3: Application of Schauder's Theorem** By the [Schauder Fixed Point Theorem](/theorems/82), there exists a fixed point $\bar{x} \in C$ such that: \begin{align*} g(\bar{x}) = \bar{x}. \end{align*} **Step 4: Verification of the Solution** We show that $\bar{x}$ is a fixed point of $f$. If $\|f(\bar{x})\|_X \le r$, then $g(\bar{x}) = f(\bar{x})$, so $\bar{x} = f(\bar{x})$ and we are done. Suppose, for the sake of contradiction, that $\|f(\bar{x})\|_X > r$. Then by the definition of $g$: \begin{align*} \bar{x} = g(\bar{x}) = \frac{r}{\|f(\bar{x})\|_X} f(\bar{x}). \end{align*} Let $\lambda = \frac{r}{\|f(\bar{x})\|_X}$. Since $\|f(\bar{x})\|_X > r$, we have $\lambda \in (0, 1)$. The equation $\bar{x} = \lambda f(\bar{x})$ implies that $\bar{x} \in \mathcal{F}$. However, taking the norm of the fixed point equation: \begin{align*} \|\bar{x}\|_X = \left\| \frac{r}{\|f(\bar{x})\|_X} f(\bar{x}) \right\|_X = r. \end{align*} This implies that there exists an element $\bar{x} \in \mathcal{F}$ with $\|\bar{x}\|_X = r$. But we chose $r > \sup_{x \in \mathcal{F}} \|x\|_X$, which is a contradiction. Therefore, the case $\|f(\bar{x})\|_X > r$ is impossible. We conclude that $g(\bar{x}) = f(\bar{x}) = \bar{x}$.