# 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 '20 at 14:58
• I think I got it. I'll edit my post with what I understood from your comment. Apr 7 '20 at 15:04
• I did it. Is this what you meant? Apr 7 '20 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 '20 at 15:30
• Perfect!! Thank you so much!! Apr 7 '20 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 '20 at 19:47