The following is a proposition from Folland:

Suppose $X$ and $Y$ are vector spaces with topologies defined, respectively, by the families $\{p_\alpha\}_{\alpha \in A}$ and $\{q_\beta\}_{\beta \in B}$ of seminorms, and $T: X \rightarrow Y$ is a linear map. Then $T$ is continuous iff for each $\beta \in B$ there exist $\alpha_1,\ldots, \alpha_k \in A$ and $C > 0$ such that $q_\beta(Tx) \leq C \sum_1^k p_{\alpha_j}(x)$.

The proof for one direction is obvious and follows from linearity. It is the converse that is causing me some trouble in following Folland's proof. I've included his proof below in chunks, interrupting in sections where I am unclear.

If $T$ is continuous, for every $\beta \in B$ there is a neighborhood of 0 in $X$ such that $q_\beta(Tx) < 1$ for $x \in u$.

How does continuity ensure such a ball of radius $1$ and centered at $0$? Is it because the open sets of $Y$ are all balls with radius less than $r$ for some $r \geq 0$?

By Theorem 5.14a we may assume that $U = \bigcap_1^k U_{0\alpha_j\epsilon_j}$, where $$U_{x\alpha\epsilon} = \{y \in X : p_\alpha(y-x) < \epsilon\}.$$ Let $\epsilon = \min(\epsilon_1, \ldots, \epsilon_k)$; then $q_\beta(Tx) < 1$ whenever $p_{\alpha_j}(x) < \epsilon$ for all $j$.

Now, given $x \in X$ there are two possibilities. If $p_{\alpha_j}(x) > 0$ for some $j$, let $y = \epsilon x\big/\sum_1^k p_{\alpha_j}(x)$. Then $p_{\alpha_j}(y) < \epsilon$ for all $j$. So, $$q_\beta(Tx) = \sum_1^k \epsilon^{-1} p_{\alpha_j}(x)q_\beta(Ty) \leq \epsilon^{-1}\sum_1^k p_{\alpha_j}(x).$$

I fail to see the motivation behind this definition of $y$, and how Folland concludes that $p_{\alpha_j}(y) < \epsilon$ for all $j$. I also am having trouble following the last equation above.

On the other hand, if $p_{\alpha_j}(x) = 0$ for all $j$, then $p_{\alpha_j}(rx) = 0$ for all $j$ and all $r > 0$, hence $rq_\beta(Tx) = q_\beta(T(rx)) < 1$ for all $r > 0$, hence $q_\beta(Tx) = 0$. Thus $q_\beta(Tx) \leq \epsilon^{-1}\sum_1^k p_{\alpha_j}(x)$ in this case too, and we are done.

This part of the proof is clear, but I have added it for completeness. It is mainly the case where $p_{\alpha_j}(x) > 0$ that is confusing me.


1 Answer 1


If you have topogies given by the seminorms (especially uncountable) don't think of a metric. These are seminorms that define topology, not metric.

Use the following definition of continuity: $T$ is continuous iff

  • for every open set in codomain its inverse is open in domain (for general case)
  • for every open neighbourhood of $0$ in codomain its inverse is an open neighbourhood of 0 in domain (for linear case)
  • for every neighbourhood of $0$ from the basis in codomain its inverse is an open neighbourhood of 0 in domain (for linear case)

Here the set $\{y\in Y: q_\beta(y)<1\}$ is used.

The idea of $y$ is the following. We would like to estimate $q_\beta(Tx)$. We obtained an estimate: $q_\beta(Tx)<1$ but only if $x$ is small, i.e. if $p_{\alpha_j}(x)<\varepsilon$ for all $j$. But our $x$ is any. Therefore we scale the vector $x$, consider $y=tx$ for sufficiently small $t$. If $p_{\alpha_j}(tx)<\varepsilon$ for all $j$ then $q_\beta(T(tx))<1$, that is $q_\beta(Tx)<1/t$. This is our desired estimate. Now we need to find the proper $t$. $$p_{\alpha_j}(tx)<\varepsilon \iff t<\frac{\varepsilon}{p_{\alpha_j}(x)}.\tag{1}$$ And this must be satisfied for all $j$. The easiest way is to define $$t=\frac{\varepsilon}{2\sum_j p_{\alpha_j}(x)}.$$ The factor $2$ is needed to guarantee (1) in case if only one $p_{\alpha_j}(x)>0$ (to avoid equality in (1)). This shows that there's a tiny mistake in your proof.

The above calculations show how to conclude that $p_{\alpha_j}(y)<\varepsilon$, but we can do it directly, without above explanation of the idea behind the choice of $y$. Namely, using the definition without the factor $2$: $$p_{\alpha_j}(y) = p_{\alpha_j}\left(\frac{\varepsilon x}{\sum_i p_{\alpha_i}(x)}\right) = \frac{\varepsilon }{\sum_i p_{\alpha_i}(x)} p_{\alpha_j}(x)\leq \varepsilon.$$

The last problematic estimate is just $$q_\beta(Tx) = \frac 1t q_\beta(T(tx))<\frac 1t.$$

  • $\begingroup$ Thanks that's very helpful! I believe I was thinking of topologies induced by seminorms incorrectly. What do you mean by "for every neighbourhood of 0 from the basis in codomain its inverse is an open neighbourhood of 0 in domain"? $\endgroup$
    – CBBAM
    Commented Jun 10, 2022 at 0:50
  • 1
    $\begingroup$ In many cases of topological vector spaces the topology is defined by the neighbourhood base (see for example: math.stackexchange.com/questions/1438080/…). In case of topology defined by seminorms $p_\alpha$ we have the base sets $\{x:p(x)<\varepsilon\}$. $\endgroup$
    – Mateo
    Commented Jun 10, 2022 at 1:56
  • $\begingroup$ So that statement essentially says that the preimage of each neighborhood base around 0 in the codomain must be an open set containing 0 in the domain? $\endgroup$
    – CBBAM
    Commented Jun 10, 2022 at 2:36
  • 1
    $\begingroup$ @CBBAM Yes. This is a necessary and sufficient condition for continuity of linear operator. $\endgroup$
    – Mateo
    Commented Jun 10, 2022 at 2:40
  • $\begingroup$ Got it, thank you for all your help! $\endgroup$
    – CBBAM
    Commented Jun 10, 2022 at 3:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .