# Limits on locally convex spaces

A curve on a locally convex space is a function $$\gamma : I \to F$$ where $$F$$ is a locally convex space and $$I \subseteq \mathbb{R}$$ is an interval. The curve is differentiable if the following limit exists: $$\gamma'(x) := \lim_{t \to 0}\frac{\gamma(x+t)-\gamma(x)}{t}$$ but what does this limit mean? I mean...elements $$\gamma(x+t)$$ and $$\gamma(x)$$ are in a lcs and this is not (necessarily) a normed space. I'm really stuck at this definition.

If $$F$$ is locally convex, then it is a topological vector space (with, say, a topology given by a family of seminorms). The notion of a limit is replaced by the following.

Definition: Let $$f: I \subseteq \mathbb{R} \to F$$. We write $$\lim_{x \to a}f(x) = L$$ if for every neighborhood $$V$$ of the origin there exists $$\delta > 0$$ such that $$0 \lt |x-a| \lt \delta$$ implies $$f(x) - L \in V$$.

Is this the right definition?

• Here you say $F$ is a locally convex space, meaning that a topology is given based on neighborhoods of the origin which are absolutely convex absorbent sets (the notion can also be defined in terms of seminorms). So the $\epsilon-\delta$ definition of limit is replaced by one in which (instead of small $\epsilon \gt 0$) a small neighborhood of the origin is prescribed. Apr 7, 2020 at 14:58
• I think I got it. I'll edit my post with what I understood from your comment. Apr 7, 2020 at 15:04
• I did it. Is this what you meant? Apr 7, 2020 at 15:14
• Right. In the case of differentiability you apply that general notion of limit to get the meaning of the limit you give at the outset of your Question. Apr 7, 2020 at 15:30
• Perfect!! Thank you so much!! Apr 7, 2020 at 15:31

1. For all continuous seminorm $$||\cdot||$$ on $$F$$, $$\lim_{x\rightarrow a}||f(x)-L||=0$$.
2. For all seminorm $$||\cdot||$$ in $$\mathscr{A}$$, $$\lim_{x\rightarrow a}||f(x)-L||=0$$.
Here $$\mathscr{A}$$ is your favorite set of seminorms on $$F$$ which define the given locally convex topology on $$F$$.
• And, as a variant on this approach to the notion of "differentiability", there are also results (from Schwartz and Grothendieck) that (on locally convex, quasi-complete TVSs) infinite-differentiability of the scalar-valued functions $x\to \lambda(f(x))$, for all $\lambda$ in the dual, implies ("strong") differentiability of the vector-valued function. Jun 26, 2020 at 19:47