[proofplan] We reduce the problem from an arbitrary finite-dimensional [Banach space](/page/Banach%20Space) $X$ to Euclidean space $\mathbb{R}^n$ via a linear isomorphism, which preserves compactness, convexity, and continuity. In $\mathbb{R}^n$, we embed the [compact](/page/Compact%20Space) convex set $K$ into a closed ball $\overline{B}_R(0)$ and retract the ball onto $K$ using the metric projection $P_K$. We then compose $f$ with the retraction to obtain a self-map of the ball, apply the [Brouwer Fixed Point Theorem](/theorems/80), and verify that the resulting fixed point lies in $K$ and is a fixed point of $f$. [/proofplan] [step:Reduce to Euclidean space via a linear isomorphism] Let $n = \dim(X)$. Since every $n$-dimensional real vector space is isomorphic to $\mathbb{R}^n$, there exists a linear isomorphism \begin{align*} L: X &\to \mathbb{R}^n. \end{align*} Since all norms on a finite-dimensional space are equivalent, $L$ is a homeomorphism. Define $\tilde{K} = L(K) \subset \mathbb{R}^n$. Since $L$ is linear and continuous, it preserves compactness and convexity, so $\tilde{K}$ is a [compact](/page/Compact%20Space) convex subset of $\mathbb{R}^n$. Define \begin{align*} \tilde{f}: \tilde{K} &\to \tilde{K}, \\ y &\mapsto L(f(L^{-1}(y))). \end{align*} Since $f$ is [continuous](/page/Continuity) and $L$, $L^{-1}$ are continuous, $\tilde{f}$ is continuous. If $\tilde{x} \in \tilde{K}$ is a fixed point of $\tilde{f}$, then $\bar{x} = L^{-1}(\tilde{x})$ satisfies \begin{align*} f(\bar{x}) = f(L^{-1}(\tilde{x})) = L^{-1}(\tilde{f}(\tilde{x})) = L^{-1}(\tilde{x}) = \bar{x}. \end{align*} It therefore suffices to prove the theorem for $\tilde{K} \subset \mathbb{R}^n$ with the Euclidean norm. We drop the tildes and assume $X = \mathbb{R}^n$. [guided] The theorem is stated for a general finite-dimensional [Banach space](/page/Banach%20Space) $X$, but the [Brouwer Fixed Point Theorem](/theorems/80) applies to continuous self-maps of the closed unit ball in $\mathbb{R}^n$. To bridge this gap, we transfer the problem to Euclidean space. Let $n = \dim(X)$. Since every $n$-dimensional real vector space is isomorphic to $\mathbb{R}^n$, there exists a linear isomorphism \begin{align*} L: X &\to \mathbb{R}^n. \end{align*} Why is $L$ a homeomorphism? Because all norms on a finite-dimensional space are equivalent, so $L$ and $L^{-1}$ are both continuous with respect to any choice of norms on $X$ and $\mathbb{R}^n$. Define $\tilde{K} = L(K)$. Since $L$ is linear and bijective: - **Compactness:** $K$ is compact and $L$ is continuous, so $\tilde{K} = L(K)$ is compact. - **Convexity:** For $\tilde{x}, \tilde{y} \in \tilde{K}$ and $t \in [0,1]$, write $\tilde{x} = L(x)$, $\tilde{y} = L(y)$. Then $t\tilde{x} + (1-t)\tilde{y} = L(tx + (1-t)y) \in L(K) = \tilde{K}$ by convexity of $K$ and linearity of $L$. Define the conjugated map \begin{align*} \tilde{f}: \tilde{K} &\to \tilde{K}, \\ y &\mapsto L(f(L^{-1}(y))). \end{align*} This is well-defined because $L^{-1}(y) \in K$, so $f(L^{-1}(y)) \in K$, so $L(f(L^{-1}(y))) \in \tilde{K}$. It is continuous as a composition of continuous maps. If $\tilde{x}$ is a fixed point of $\tilde{f}$, then setting $\bar{x} = L^{-1}(\tilde{x})$: \begin{align*} f(\bar{x}) = f(L^{-1}(\tilde{x})) = L^{-1}(L(f(L^{-1}(\tilde{x})))) = L^{-1}(\tilde{f}(\tilde{x})) = L^{-1}(\tilde{x}) = \bar{x}. \end{align*} We drop the tildes and assume $X = \mathbb{R}^n$ with the standard Euclidean norm for the remainder of the proof. [/guided] [/step] [step:Retract a closed ball onto $K$ via the metric projection] Since $K$ is compact, it is bounded. Choose $R > 0$ such that $K \subseteq \overline{B}_R(0) := \{x \in \mathbb{R}^n : \|x\|_2 \le R\}$. Write $\overline{B} = \overline{B}_R(0)$. Since $K$ is nonempty, closed, and convex in the [Hilbert space](/page/Hilbert%20Space) $\mathbb{R}^n$, the metric projection \begin{align*} P_K: \mathbb{R}^n &\to K, \\ x &\mapsto \operatorname{argmin}_{y \in K} \|x - y\|_2 \end{align*} is well-defined and unique for every $x \in \mathbb{R}^n$. By the [non-expansiveness of the projection onto a closed convex set](/theorems/647), \begin{align*} \|P_K(x) - P_K(y)\|_2 \le \|x - y\|_2 \quad \text{for all } x, y \in \mathbb{R}^n. \end{align*} In particular, $P_K$ is continuous. The restriction $r := P_K|_{\overline{B}}: \overline{B} \to K$ is a continuous retraction of $\overline{B}$ onto $K$, since $P_K(x) = x$ for all $x \in K$. [guided] We need a continuous map $r: \overline{B} \to K$ that fixes every point of $K$. The natural candidate is the metric projection (nearest-point map). Since $K$ is compact, it is bounded, so there exists $R > 0$ with $K \subseteq \overline{B}_R(0)$. Write $\overline{B} = \overline{B}_R(0)$. The set $K$ is nonempty, closed, and convex in $\mathbb{R}^n$ (a [Hilbert space](/page/Hilbert%20Space)). By the [existence and uniqueness of the projection operator](/theorems/86), for every $x \in \mathbb{R}^n$ there exists a unique point $P_K(x) \in K$ minimising $\|x - y\|_2$ over $y \in K$. This defines the metric projection \begin{align*} P_K: \mathbb{R}^n &\to K, \\ x &\mapsto \operatorname{argmin}_{y \in K} \|x - y\|_2. \end{align*} Why is $P_K$ continuous? By the [non-expansiveness of the projection onto a closed convex set](/theorems/647), $P_K$ is Lipschitz with constant $1$: \begin{align*} \|P_K(x) - P_K(y)\|_2 \le \|x - y\|_2 \quad \text{for all } x, y \in \mathbb{R}^n. \end{align*} Lipschitz continuity is stronger than continuity. The restriction $r := P_K|_{\overline{B}}$ maps $\overline{B}$ continuously into $K$, and for any $x \in K$ we have $P_K(x) = x$ (the nearest point in $K$ to a point already in $K$ is itself). So $r$ is a retraction. [/guided] [/step] [step:Compose with $f$ and apply the Brouwer Fixed Point Theorem to the ball] Define the auxiliary map \begin{align*} g: \overline{B} &\to \overline{B}, \\ x &\mapsto f(r(x)). \end{align*} We verify the hypotheses of the [Brouwer Fixed Point Theorem](/theorems/80): 1. **Domain:** $\overline{B}$ is a closed ball in $\mathbb{R}^n$, hence homeomorphic to the closed unit ball. 2. **Continuity:** $r$ is continuous and $f$ is continuous, so $g = f \circ r$ is continuous. 3. **Self-map:** For any $x \in \overline{B}$, $r(x) \in K$, so $f(r(x)) \in K \subseteq \overline{B}$. Thus $g(\overline{B}) \subseteq \overline{B}$. By the [Brouwer Fixed Point Theorem](/theorems/80), there exists $y \in \overline{B}$ with $g(y) = y$. [/step] [step:Verify that the Brouwer fixed point is a fixed point of $f$ in $K$] Since $y = g(y) = f(r(y))$ and $f(r(y)) \in K$, we have $y \in K$. Since $r$ is a retraction onto $K$, $r(y) = y$. Substituting: \begin{align*} y = g(y) = f(r(y)) = f(y). \end{align*} Setting $\bar{x} = y$ gives the desired fixed point $f(\bar{x}) = \bar{x}$ in $K$. [/step]