# Equivalent metric generated by a compact subset of $C(X)$ where $X$ is a compact metric space.

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.

-
Usually when you refer to another thread it makes sense to link to it. In the case of other threads on this site, you can just put the URL directly into the text and it will be replaced by the title of the post; for external URLs you can use the chain icon above the editing area to create links. – joriki Dec 31 '12 at 1:33
@Joriki: Sorry, I am just new here. Someone has done it already. – Dan Dec 31 '12 at 2:00
What topology do you assume on $C(X)$? – T. Eskin Dec 31 '12 at 18:35
@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. – Haskell Curry Dec 31 '12 at 21:32

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.

-
Your second bullet can be summarized/simplified by saying: "The function $g \mapsto g(a) - g(b)$ is linear, uniformly continuous and zero on $K$, hence on the dense set $\operatorname{Span}(K)$, so it is zero everywhere. Thus, $\rho = 0$" – Martin Dec 31 '12 at 1:45
Thanks. Any hints about the question that generate the same topology? – Dan Dec 31 '12 at 2:02
@Dan: You can use the equicontinuity of $K$ to prove that they generate the same topology. – Haskell Curry Dec 31 '12 at 2:21