[proofplan]
We truncate $f$ via a radial retraction onto a large ball $\overline{B}_r(0)$ whose radius exceeds $\sup_{x \in \mathcal{F}} \|x\|_X$. The truncated map $g = \rho \circ f$ is a continuous compact self-map of the closed ball, so the [Schauder Fixed Point Theorem](/theorems/82) provides a fixed point $\bar{x}$ of $g$. We then show by contradiction, using the boundedness of $\mathcal{F}$, that the radial retraction does not activate at $\bar{x}$, so $\bar{x}$ is in fact a fixed point of $f$.
[/proofplan]
[step:Truncate $f$ via a radial retraction onto a ball of radius $r > \sup_{\mathcal{F}} \|x\|_X$]
Since $\mathcal{F}$ is bounded, let $M = \sup_{x \in \mathcal{F}} \|x\|_X$ and choose $r > M$. Define the radial retraction
\begin{align*}
\rho: X &\to \overline{B}_r(0), \\
y &\mapsto \begin{cases} y & \text{if } \|y\|_X \le r, \\ \dfrac{r}{\|y\|_X}\, y & \text{if } \|y\|_X > r. \end{cases}
\end{align*}
Define $g := \rho \circ f: X \to \overline{B}_r(0)$. Since $f$ is [continuous](/page/Continuity) and $\rho$ is continuous, $g$ is continuous.
[/step]
[step:Verify that $g$ restricts to a continuous compact self-map of $\overline{B}_r(0)$]
Let $C = \overline{B}_r(0)$. By construction, $g(C) \subseteq C$, so $g: C \to C$ is a continuous self-map. We verify compactness: since $f$ is a compact operator and $C$ is bounded, $f(C)$ is relatively compact. The radial retraction $\rho$ is continuous, so $\rho(\overline{f(C)})$ is compact (continuous image of a compact set). Since $g(C) = \rho(f(C)) \subseteq \rho(\overline{f(C)})$, the image $g(C)$ is relatively compact.
The set $C$ is nonempty, closed, bounded, and convex in the [Banach space](/page/Banach%20Space) $X$, and $g: C \to C$ is continuous and compact. The hypotheses of the [Schauder Fixed Point Theorem](/theorems/82) are satisfied.
[guided]
We need to verify three properties for the [Schauder Fixed Point Theorem](/theorems/82): the domain $C = \overline{B}_r(0)$ must be nonempty, closed, bounded, and convex; the map $g: C \to C$ must be continuous; and $g$ must be compact (i.e., $g(C)$ must be relatively compact).
- **Domain properties:** The closed ball $\overline{B}_r(0)$ is nonempty (contains $0$), closed, bounded, and convex in any normed space.
- **Self-map:** For $x \in C$, $g(x) = \rho(f(x))$, and $\rho$ maps every point of $X$ into $\overline{B}_r(0) = C$ by definition.
- **Continuity:** $f$ is continuous by hypothesis, and $\rho$ is continuous (it is the identity on $\overline{B}_r(0)$ and a continuous radial projection outside), so $g = \rho \circ f$ is continuous.
- **Compactness:** Since $C$ is bounded and $f$ is a compact operator, $f(C)$ is relatively compact, meaning $\overline{f(C)}$ is compact. The continuous image of a compact set under $\rho$ is compact, so $\rho(\overline{f(C)})$ is compact. Since $g(C) = \rho(f(C)) \subseteq \rho(\overline{f(C)})$, the closure $\overline{g(C)}$ is contained in the compact set $\rho(\overline{f(C)})$, hence $g(C)$ is relatively compact.
[/guided]
[/step]
[step:Apply the Schauder Fixed Point Theorem to obtain a fixed point of $g$]
By the [Schauder Fixed Point Theorem](/theorems/82), there exists $\bar{x} \in C$ with
\begin{align*}
g(\bar{x}) = \bar{x}.
\end{align*}
[/step]
[step:Show the retraction does not activate at $\bar{x}$, so $f(\bar{x}) = \bar{x}$]
**Case 1:** If $\|f(\bar{x})\|_X \le r$, then $g(\bar{x}) = \rho(f(\bar{x})) = f(\bar{x})$, so $\bar{x} = f(\bar{x})$ and we are done.
**Case 2:** Suppose for contradiction that $\|f(\bar{x})\|_X > r$. Then
\begin{align*}
\bar{x} = g(\bar{x}) = \frac{r}{\|f(\bar{x})\|_X}\, f(\bar{x}).
\end{align*}
Setting $\lambda = \frac{r}{\|f(\bar{x})\|_X} \in (0,1)$, we have $\bar{x} = \lambda f(\bar{x})$. By definition of $\mathcal{F}$, this means $\bar{x} \in \mathcal{F}$. Taking the norm:
\begin{align*}
\|\bar{x}\|_X = \frac{r}{\|f(\bar{x})\|_X} \|f(\bar{x})\|_X = r.
\end{align*}
This gives $\|\bar{x}\|_X = r > M = \sup_{x \in \mathcal{F}} \|x\|_X$, contradicting $\bar{x} \in \mathcal{F}$.
Therefore Case 2 is impossible, and $f(\bar{x}) = \bar{x}$.
[guided]
We must verify that the fixed point of $g$ is genuinely a fixed point of $f$, not an artifact of the truncation.
**Case 1:** If $\|f(\bar{x})\|_X \le r$, the retraction $\rho$ acts as the identity on $f(\bar{x})$, so $g(\bar{x}) = f(\bar{x}) = \bar{x}$.
**Case 2:** Suppose $\|f(\bar{x})\|_X > r$. Then the retraction rescales:
\begin{align*}
\bar{x} = g(\bar{x}) = \rho(f(\bar{x})) = \frac{r}{\|f(\bar{x})\|_X}\, f(\bar{x}).
\end{align*}
Define $\lambda = \frac{r}{\|f(\bar{x})\|_X}$. Since $\|f(\bar{x})\|_X > r > 0$, we have $\lambda \in (0,1)$. The equation $\bar{x} = \lambda f(\bar{x})$ places $\bar{x}$ in $\mathcal{F} = \{x \in X : x = \mu f(x) \text{ for some } \mu \in [0,1]\}$.
But computing the norm of $\bar{x}$:
\begin{align*}
\|\bar{x}\|_X = \left\|\frac{r}{\|f(\bar{x})\|_X}\, f(\bar{x})\right\|_X = \frac{r}{\|f(\bar{x})\|_X} \cdot \|f(\bar{x})\|_X = r.
\end{align*}
This gives $\|\bar{x}\|_X = r > M = \sup_{x \in \mathcal{F}} \|x\|_X$, which contradicts $\bar{x} \in \mathcal{F}$. The contradiction arises from the choice $r > M$: no element of $\mathcal{F}$ can have norm as large as $r$.
Therefore Case 2 is impossible, and we conclude $g(\bar{x}) = f(\bar{x}) = \bar{x}$.
[/guided]
[/step]
