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    Continuity of the evaluation map

    E-4P6A6

    Area
    Topology
    Topics
    Continuous Functions
    Status
    draft

    If \(X\) and \(Y\) are spaces, define

    \[ e: X \times \mathcal{C}(X, Y) \to Y \]

    by the equation \(e(x, f) = f(x)\); the map \(e\) is called the evaluation map. Show that if \(d\) is a metric for \(Y\) and \(\mathcal{C}(X, Y)\) has the corresponding uniform topology, then \(e\) is continuous. We shall generalize this result in §46.

    Solution

    The uniform topology on \(\mathcal{C}(X, Y)\) is induced by the uniform metric \(\bar{\rho}(f, g) = \sup_{x \in X} \bar{d}(f(x), g(x))\), where \(\bar{d} = \min(d, 1)\).

    1. Reduction:Let \((x_0, f_0) \in X \times \mathcal{C}(X, Y)\) and let \(\varepsilon > 0\). We may assume without loss of generality that \(\varepsilon < 1\).It suffices to find an open neighborhood \(U \subseteq X\) of \(x_0\) and an open neighborhood \(V \subseteq \mathcal{C}(X, Y)\) of \(f_0\) such that \(e(U \times V) \subseteq B_d(f_0(x_0), \varepsilon)\).

    2. Construction of neighborhoods:Proof:

    2.1. Since \(f_0: X \to Y\) is continuous at \(x_0\), there exists an open neighborhood \(U \subseteq X\) of \(x_0\) such that:\[d(f_0(x), f_0(x_0)) < \frac{\varepsilon}{2} \quad \text{for all } x \in U.\]

    2.2. In the uniform metric space \(\mathcal{C}(X, Y)\), define the open ball \(V = B_{\bar{\rho}}(f_0, \frac{\varepsilon}{2})\).

    2.3. For any \(f \in V\), the uniform bound holds:\[\bar{\rho}(f, f_0) = \sup_{x \in X} \bar{d}(f(x), f_0(x)) < \frac{\varepsilon}{2}.\]

    2.4. Since \(\frac{\varepsilon}{2} < 1\), \(\bar{d}(f(x), f_0(x)) = d(f(x), f_0(x))\), which implies:\[d(f(x), f_0(x)) < \frac{\varepsilon}{2} \quad \text{for all } x \in X.\]

    3. Verification of continuity:Proof:

    3.1. Let \((x, f) \in U \times V\).

    3.2. By the triangle inequality in the metric space \((Y, d)\):\[d(e(x, f), e(x_0, f_0)) = d(f(x), f_0(x_0)) \le d(f(x), f_0(x)) + d(f_0(x), f_0(x_0)).\]

    3.3. Since \(f \in V\), \(d(f(x), f_0(x)) < \frac{\varepsilon}{2}\).

    3.4. Since \(x \in U\), \(d(f_0(x), f_0(x_0)) < \frac{\varepsilon}{2}\).

    3.5. Summing these bounds gives:\[d(e(x, f), e(x_0, f_0)) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\]

    3.6. Thus \(e(U \times V) \subseteq B_d(f_0(x_0), \varepsilon)\).

    4. Q.E.D.By steps 1 and 3, \(e\) is continuous at every point \((x_0, f_0) \in X \times \mathcal{C}(X, Y)\).