3
$\begingroup$

I need help with the following problem:

Let $(X,\rho)$ be a compact metric space. Prove that if $K$ is a compact subset of $C(X)=C(X,\mathbb{R})$ (i.e. continuous functions with real values) whose linear span is dense in $C(X)$, then the pseudometric $d(x,y)= \sup _{f \in K} \lvert f(x) - f(y)\rvert$ on $X$ is actually a metric. Moreover, show that $d$ and $\rho$ generate the same topology.

The existence of $K$ was proved in another thread here: Dense subset of $C(X)$.

Thanks.

  • 0
    @ThomasE.: The OP is assuming that $ C(X,\mathbb{R}) $ is endowed with the sup-norm topology. I think he didn't specify it here because this problem is the sequel to another one whose thread link is provided above.2012-12-31

1 Answers 1

1

Let $ a,b,c \in X $.

  • If $ a = b $, then $ f(a) = f(b) $ for all $ f \in K $. Therefore, $ d(a,b) = 0 $.

  • If $ d(a,b) = 0 $, then $ f(a) = f(b) $ for all $ f \in K $. It follows that $ g(a) = g(b) $ for all $ g \in \text{Span}(K) $. Consider the continuous ‘distance’ function $ \phi_{a}: X \to \mathbb{R}_{\geq 0} $ defined by $ \forall x \in X: \quad {\phi_{a}}(x) \stackrel{\text{def}}{=} \rho(a,x). $ As $ \text{Span}(K) $ is assumed to be a dense subset of $ C(X,\mathbb{R}) $, there exists a sequence $ (g_{n})_{n \in \mathbb{N}} $ in $ \text{Span}(K) $ converging uniformly to $ \phi_{a} $. Hence, $ (g_{n})_{n \in \mathbb{N}} $ also converges pointwise to $ \phi_{a} $. We thus have \begin{align} 0 &= {\phi_{a}}(a) \quad (\text{As $ \rho(a,a) = 0 $.}) \\ &= \lim_{n \to \infty} {g_{n}}(a) \\ &= \lim_{n \to \infty} {g_{n}}(b) \quad (\forall g \in \text{Span}(K): ~~ g(a) = g(b).) \\ &= {\phi_{a}}(b). \end{align} We see now that $ \rho(a,b) = 0 $. However, this holds if and only if $ a = b $, as $ \rho $ is a metric. Therefore, $ a = b \iff d(a,b) = 0. $

  • It remains to show that $ d $ satisfies the Triangle Inequality. For all $ f \in K $, we have $ |f(a) - f(c)| \leq |f(a) - f(b)| + |f(b) - f(c)|. $ Hence, \begin{align} \sup_{f \in K} |f(a) - f(c)| &\leq \sup_{f \in K} (|f(a) - f(b)| + |f(b) - f(c)|) \\ &\leq \sup_{f \in K} |f(a) - f(b)| + \sup_{f \in K} |f(b) - f(c)|, \end{align} which yields $ d(a,c) \leq d(a,b) + d(b,c) $.

Conclusion: $ d $ is a metric on $ X $.


Finally, we will prove that $ \rho $ and $ d $ generate the same topology on $ X $. For all $ a \in X $ and $ r \in \mathbb{R}_{> 0} $, define \begin{align} {\mathbb{B}_{d}}(a;r) &\stackrel{\text{def}}{=} \{ x \in X ~|~ d(a,x) < r \}, \\ {\mathbb{B}_{\rho}}(a;r) &\stackrel{\text{def}}{=} \{ x \in X ~|~ \rho(a,x) < r \}. \end{align}

  • Let $ x_{0} \in {\mathbb{B}_{d}}(a;r) $. The compactness of $ K $ implies the equicontinuity of $ K $. As $ \dfrac{1}{2} [r - d(a,x_{0})] $ is a positive number, we can thus find a $ \delta > 0 $ such that for all $ x \in X $ satisfying $ \rho(x_{0},x) < \delta $, the inequality $ |f(x_{0}) - f(x)| < \dfrac{1}{2} [r - d(a,x_{0})] $ holds for all $ f \in K $. Hence, \begin{align} x_{0} &\in {\mathbb{B}_{\rho}}(x_{0};\delta) \\ &\subseteq {\mathbb{B}_{d}} \left( x_{0};\frac{1}{2} [r - d(a,x_{0})] \right) \\ &\subseteq {\mathbb{B}_{d}}(a;r). \end{align}

  • Let $ x_{0} \in {\mathbb{B}_{\rho}}(a;r) $. Consider the ‘distance’ function $ \phi_{x_{0}}: X \to \mathbb{R}_{\geq 0} $ defined by $ \forall x \in X: \quad {\phi_{x_{0}}}(x) \stackrel{\text{def}}{=} \rho(x_{0},x). $ Fix $ \epsilon := \dfrac{1}{2} [r - \rho(a,x_{0})] $, which is a positive number. By the denseness of $ \text{Span}(K) $ in $ C(X,\mathbb{R}) $, we can find $ \lambda_{1},\ldots,\lambda_{n} \in \mathbb{R} \setminus \{ 0 \} $ and $ f_{1},\ldots,f_{n} \in K $ such that $ \left\| \phi_{x_{0}} - \sum_{i=1}^{n} \lambda_{i} f_{i} \right\|_{\infty} < \frac{\epsilon}{3}. $ Let $ \delta := \dfrac{\epsilon}{3n \cdot \max(|\lambda_{1}|,\ldots,|\lambda_{n}|)} $. Then \begin{align} \forall x \in {\mathbb{B}_{d}}(x_{0};\delta): \quad \left| \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x_{0}) - \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x) \right| &\leq \sum_{i=1}^{n} |\lambda_{i}||{f_{i}}(x_{0}) - {f_{i}}(x)| \\ &< \sum_{i=1}^{n} \frac{\epsilon}{3n} \\ &= \frac{\epsilon}{3}. \end{align} Hence, for all $ x \in {\mathbb{B}_{d}}(x_{0};\delta) $, we have \begin{align} &\left| {\phi_{x_{0}}}(x) \right| \\ = &\left| {\phi_{x_{0}}}(x_{0}) - {\phi_{x_{0}}}(x) \right| \quad (\text{As $ \rho(x_{0},x_{0}) = 0 $.}) \\ \leq &\left| {\phi_{x_{0}}}(x_{0}) - \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x_{0}) \right| + \left| \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x_{0}) - \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x) \right| + \left| \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x) - {\phi_{x_{0}}}(x) \right| \\ = &\left| {\phi_{x_{0}}}(x_{0}) - \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x_{0}) \right| + \left| \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x_{0}) - \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x) \right| + \left| {\phi_{x_{0}}}(x) - \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x) \right| \\ \leq &\left\| \phi_{x_{0}} - \sum_{i=1}^{n} \lambda_{i} f_{i} \right\|_{\infty} + \left| \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x_{0}) - \sum_{i=1}^{n} \lambda_{i} {f_{i}}(x) \right| + \left\| \phi_{x_{0}} - \sum_{i=1}^{n} \lambda_{i} f_{i} \right\|_{\infty} \\ < &\frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\ = &\epsilon \\ := &\frac{1}{2} [r - \rho(a,x_{0})]. \end{align} Therefore, \begin{align} x_{0} &\in {\mathbb{B}_{d}}(x_{0};\delta) \\ &\subseteq {\mathbb{B}_{\rho}} \left( x_{0};\frac{1}{2} [r - \rho(a,x_{0})] \right) \\ &\subseteq {\mathbb{B}_{\rho}}(a;r). \end{align}

Conclusion: The foregoing argument shows that open $ d $-balls are a union of open $ \rho $-balls and vice-versa. Therefore, $ \rho $ and $ d $ generate the same topology.

  • 0
    @Dan: You can use the equicontinuity of $ K $ to prove that they generate the same topology.2012-12-31