[proofplan] We first reduce to a [compact](/page/Compact%20Space) convex set by passing from the closed convex hull of the image $f(C)$ to a compact convex set $K$ via Mazur's theorem. We then approximate the identity on $K$ by Schauder projections onto finite-dimensional convex polytopes, apply the [Brouwer Fixed Point Theorem](/theorems/80) in each finite-dimensional approximation, and extract a convergent subsequence whose limit is a fixed point of $f$ by compactness of $K$. [/proofplan] [step:Reduce to a compact convex domain via Mazur's theorem] Since $f$ is compact, $f(C)$ is relatively compact in $X$. Let $K = \overline{\operatorname{co}}(f(C))$ denote the closed convex hull of $f(C)$. By Mazur's compactness theorem, the closed convex hull of a relatively compact subset of a [Banach space](/page/Banach%20Space) is compact. Therefore $K$ is a compact convex set. Since $C$ is closed and convex and $f(C) \subset C$, every convex combination of points in $f(C)$ lies in $C$, and every limit of such combinations lies in $C$ by closedness. Hence $K = \overline{\operatorname{co}}(f(C)) \subset C$. The map $f$ sends $C$ into $f(C) \subseteq K$, so $f$ restricts to a continuous map $f: K \to K$. It suffices to find a fixed point of this restriction. [guided] The set $C$ is closed, bounded, and convex, but it may be infinite-dimensional, so we cannot apply Brouwer directly. The key idea is to exploit the compactness of $f$ to find a compact convex set that $f$ maps into itself. Since $f$ is compact, $\overline{f(C)}$ is compact. Define $K = \overline{\operatorname{co}}(f(C))$, the closed convex hull of $f(C)$. Why is $K$ compact? By Mazur's compactness theorem: in a [Banach space](/page/Banach%20Space), the closed convex hull of a relatively compact set is itself compact. The hypothesis that $\overline{f(C)}$ is compact provides the relative compactness of $f(C)$. Why does $K \subset C$? Every point in $\operatorname{co}(f(C))$ is a finite convex combination of points in $f(C) \subset C$. Since $C$ is convex, every such combination lies in $C$. Since $C$ is closed, taking the closure preserves containment: $K = \overline{\operatorname{co}}(f(C)) \subset C$. Since $f$ maps $K \subset C$ into $f(C) \subset K$, the restriction $f: K \to K$ is a well-defined continuous self-map of a [compact](/page/Compact%20Space) convex set. [/guided] [/step] [step:Construct Schauder projections approximating the identity on $K$] Let $\varepsilon > 0$. Since $K$ is compact, there exists a finite set $\{x_1, \ldots, x_n\} \subset K$ such that the open balls $B_\varepsilon(x_1), \ldots, B_\varepsilon(x_n)$ cover $K$. Let $K_\varepsilon = \operatorname{co}(\{x_1, \ldots, x_n\})$, a finite-dimensional compact convex polytope contained in $K$. Define the Schauder projection $p_\varepsilon: K \to K_\varepsilon$ by \begin{align*} p_\varepsilon(x) = \frac{\sum_{k=1}^n \operatorname{dist}(x, K \setminus B_\varepsilon(x_k)) \, x_k}{\sum_{k=1}^n \operatorname{dist}(x, K \setminus B_\varepsilon(x_k))}. \end{align*} Since the balls cover $K$, every $x \in K$ belongs to at least one $B_\varepsilon(x_k)$, so the corresponding distance term is strictly positive. The denominator is therefore nonzero, and $p_\varepsilon$ is well-defined and [continuous](/page/Continuity). For any $x \in K$, $p_\varepsilon(x)$ is a convex combination of those $x_k$ satisfying $\|x - x_k\| < \varepsilon$. Hence \begin{align*} \|p_\varepsilon(x) - x\| < \varepsilon \quad \text{for all } x \in K. \end{align*} [guided] We need to approximate the infinite-dimensional problem by finite-dimensional ones. The Schauder projection is a partition-of-unity construction that maps $K$ into a finite-dimensional convex polytope while staying uniformly close to the identity. Let $\varepsilon > 0$. Compactness of $K$ provides a finite $\varepsilon$-net $\{x_1, \ldots, x_n\} \subset K$ with $K \subset \bigcup_{k=1}^n B_\varepsilon(x_k)$. Define $K_\varepsilon = \operatorname{co}(\{x_1, \ldots, x_n\})$, which is a compact convex subset of the finite-dimensional subspace $\operatorname{span}\{x_1, \ldots, x_n\}$, and $K_\varepsilon \subset K$ since $K$ is convex and contains all the $x_k$. Define the weight functions $\mu_k: K \to [0, \infty)$ by $\mu_k(x) = \operatorname{dist}(x, K \setminus B_\varepsilon(x_k))$. Each $\mu_k$ is continuous and nonneg. If $x \in B_\varepsilon(x_k)$, then $x \notin K \setminus B_\varepsilon(x_k)$, so $\mu_k(x) > 0$. Since the balls cover $K$, at least one $\mu_k(x) > 0$ for every $x \in K$. The Schauder projection is \begin{align*} p_\varepsilon(x) = \frac{\sum_{k=1}^n \mu_k(x) \, x_k}{\sum_{k=1}^n \mu_k(x)}. \end{align*} This is a convex combination of the $x_k$'s (since the normalised weights sum to $1$ and are nonneg), so $p_\varepsilon(x) \in K_\varepsilon$. Why is $\|p_\varepsilon(x) - x\| < \varepsilon$? Only those $x_k$ with $\mu_k(x) > 0$ contribute to the sum, and $\mu_k(x) > 0$ implies $x \in B_\varepsilon(x_k)$, i.e., $\|x - x_k\| < \varepsilon$. Since $p_\varepsilon(x)$ is a convex combination of such $x_k$'s, and the ball $B_\varepsilon(x)$ is convex: \begin{align*} \|p_\varepsilon(x) - x\| = \left\|\sum_k \frac{\mu_k(x)}{\sum_j \mu_j(x)}(x_k - x)\right\| \le \sum_k \frac{\mu_k(x)}{\sum_j \mu_j(x)} \|x_k - x\| < \varepsilon. \end{align*} [/guided] [/step] [step:Apply the Brouwer Fixed Point Theorem in each finite-dimensional approximation] Define \begin{align*} g_\varepsilon: K_\varepsilon &\to K_\varepsilon, \\ x &\mapsto p_\varepsilon(f(x)). \end{align*} This is well-defined: $f$ maps $K_\varepsilon \subset K$ into $K$, and $p_\varepsilon$ maps $K$ into $K_\varepsilon$. Since $f$ and $p_\varepsilon$ are continuous, $g_\varepsilon$ is continuous. The set $K_\varepsilon$ is a compact convex subset of the finite-dimensional subspace $\operatorname{span}\{x_1, \ldots, x_n\}$. By the [Generalized Brouwer Fixed Point Theorem](/theorems/81) (or equivalently the [Brouwer Fixed Point Theorem](/theorems/80)), there exists $x_\varepsilon \in K_\varepsilon$ with \begin{align*} g_\varepsilon(x_\varepsilon) = p_\varepsilon(f(x_\varepsilon)) = x_\varepsilon. \end{align*} [/step] [step:Extract a convergent subsequence and pass to the limit] Choose a sequence $\varepsilon_j \to 0$ and let $x_j = x_{\varepsilon_j}$ denote the corresponding fixed points. By the approximation property of $p_{\varepsilon_j}$: \begin{align*} \|x_j - f(x_j)\| = \|p_{\varepsilon_j}(f(x_j)) - f(x_j)\| < \varepsilon_j. \end{align*} Since $\{x_j\} \subset K$ and $K$ is compact, there exists a subsequence (still denoted $x_j$) converging to some $\bar{x} \in K$. By continuity of $f$, $f(x_j) \to f(\bar{x})$. Taking the limit in $\|x_j - f(x_j)\| < \varepsilon_j$: \begin{align*} \|\bar{x} - f(\bar{x})\| = \lim_{j \to \infty} \|x_j - f(x_j)\| = 0. \end{align*} Therefore $f(\bar{x}) = \bar{x}$. [/step]