Condition for showing families of seminorms generate same topology

Question

There is a statement about locally convex spaces in Reed & Simon, Methods of Modern Mathematical Physics (Vol I, Section V.1) that is given without a proof.

The statement is:

Given two families of seminorms $\{\rho_\alpha\}_{\alpha \in A}$ and $\{d_\beta\}_{\beta \in B}$ over a locally convex space X, if the families generate the same natural topologies, then for each $\alpha \in A$, there are $\beta_1, \ldots, \beta_n \in B$ and $C > 0$ such that for all $x \in X$:

$$\rho_\alpha(x) \leq C(d_{\beta_1}(x) + \ldots + d_{\beta_n}(x))$$

And correspondingly, for each $\beta \in B$, there are $\alpha_1, \ldots, \alpha_n \in A$ and $D > 0$ such that for all $x \in X$:

$$d_\beta(x) \leq D(\rho_{\alpha_1}(x) + \ldots + \rho_{\alpha_n}(x))$$

The natural topology generated by a family of seminorms $\{\rho_\alpha\}_{\alpha \in A}$ is defined to be the weakest topology such that each of the seminorms and vector addition are continuous.

How would you go about proving the existence of C and $\beta_1, \ldots, \beta_n \in B$, given that the natural topologies of the two families are equivalent? The previous part of the proposition also states that we know that each $\rho_\alpha$ is continuous in the d-natural topology (and vice versa).

Answer 1

This follows from the following result: Let $X$ be a locally convex space where the topology is generated by the family of seminorms $\{p_{\alpha} : \alpha \in \mathcal{A} \}$. If $p$ is a continuous seminorm on $X$, then there exist $\alpha_{1}, \ldots , \alpha_{n}\in \mathcal{A}$ and $c_{1}, \ldots , c_{n} \in (0, \infty )$ such that for all $x\in X$,

$$p(x) \leq \sum_{i=1}^{n}c_{i}p_{\alpha_{i}}(x).$$

To show this, note that $\{x\in X:p(x) < 1\}$ is open in $X$, and as the collection

$$\left\{ \bigcap_{\alpha \in F}\{x\in X: p_{\alpha}(x) < \varepsilon_{\alpha} \}: \varepsilon_{\alpha} \in (0, \infty ), F\subseteq \mathcal{A} \text{ finite} \right\}$$

forms a neighbourhood base of $0$, there exist $\alpha_{1}, \ldots , \alpha_{n} \in \mathcal{A}$ and $\varepsilon_{1}, \ldots , \varepsilon_{n} \in (0, \infty )$ such that

$$\bigcap_{i=1}^{n}\{x\in X: p_{\alpha_{i}}(x) < \varepsilon_{i} \} \subseteq \{x\in X:p(x) < 1\}.$$

If $x\in X$ and $\sum_{i=1}^{n}\varepsilon_{i}^{-1}p_{\alpha_{i}}(x) < 1$, then $p_{\alpha_{i}}(x) < \varepsilon_{i}$ for each $i \in \{1, \ldots , n\}$, so it follows that $p(x) < 1$. So it follows that

$$\{x\in X: \sum_{i=1}^{n}\varepsilon_{i}^{-1}p_{\alpha_{i}}(x) < 1 \} \subseteq \{x\in X: p(x) < 1\}.$$

For $x\in X$, let $\alpha := \sum_{i=1}^{n}\varepsilon_{i}^{-1}p_{\alpha_{i}}(x)$. Then if $\varepsilon >0$, $\sum_{i=1}^{n}\varepsilon_{i}^{-1}p_{\alpha_{i}}((\alpha + \varepsilon )^{-1}x) =(\alpha + \varepsilon )^{-1}\alpha <1$, so $p((\alpha + \varepsilon )^{-1}x)<1$ and so $p(x) < \alpha + \varepsilon $. As this holds for all $\varepsilon >0$, $p(x) \leq \alpha = \sum_{i=1}^{n}\varepsilon_{i}^{-1}p_{\alpha_{i}}(x)$. As this holds for all $x\in X$, this proves the desired result.

To apply this result to the particular case of interest, let $\tau_{\rho}$ denote the locally convex topology generated by $\{\rho_{\alpha} : \alpha \in A\}$ and let $\tau_{d}$ denote the locally convex topology generated by $\{d_{\beta} : \beta \in B\}$. Suppose that $\tau_{d}$ is contained in $\tau_{\rho}$. By the definition of the topology on $\tau_{d}$, for each $\alpha \in A$, $\rho_{\alpha}$ is a continuous seminorm on $X$. By the result proved above, there exist $\beta_{1}, \ldots , \beta_{n}\in B$ and $C>0$ such that for all $x\in X$,

$$\rho_{\alpha}(x) \leq \sum_{i=1}^{n}C d_{\beta_{i}}(x) \leq C\left(\sum_{i=1}^{n}d_{\beta_{i}}(x) \right).$$

The other corresponding condition is proved in an essentially identical manner.

Answer 2

Fix $\alpha$ and $\epsilon > 0$. Then $\{y\in X\colon \rho_\alpha(y)<\epsilon\}$ is a neighborhood of $0$ in $\tau_\rho$. So it's a neighborhood of $0$ in $\tau_d$. Therefore there exist $\beta_1,\ldots,\beta_n$ and $\delta>0$ such that $$\{y\in X\colon d_{\beta_1}(y)\le\delta\land\ldots\land d_{\beta_n}(y)\le\delta\}\subseteq\{y\in X\colon \rho_\alpha(y)<\epsilon\}\tag{1}$$

For $x\in X$ let $M=\sum_{i=1}^n d_{\beta_i}(x)$. If $M>0$ then $d_{\beta_i}(\frac{\delta}{M}x)=\frac{\delta}{M}d_{\beta_i}(x)\le \frac{\delta}{M}M=\delta$ for all $i\in\{1,\ldots,n\}$. Therefore by $(1)$, $\rho_\alpha(\frac{\delta}{M}x)<\epsilon$ and $\rho_\alpha(x)<\frac{\epsilon}{\delta}M=\frac{\epsilon}{\delta}\sum_{i=1}^n d_{\beta_i}(x)$, so taking $C=\frac{\epsilon}{\delta}$ we get the kind of bound we're looking for.

If $M=0$ I claim that $\rho_\alpha(x)=0$. Because if $M=0$ then $d_{\beta_i}(x)=0$ for all $i\in\{1,\ldots,n\}$. So $d_{\beta_i}(\mu x)=0$ for any $\mu > 0$ and $i$, so by $(1)$, $\rho_\alpha(\mu x)<\epsilon$ for all $\mu>0$ which implies that $\rho_\alpha(x)<\frac{\epsilon}{\mu}$ for all $\mu>0$. By taking $\mu\to +\infty$ we get that $\rho_\alpha(x)=0$. So taking $C=\frac{\epsilon}{\delta}$ works for this case as well.