How can a single function, $F: K ( \subset \mathbb{R}) \to \mathcal{L}(X)$ be uniformly bounded?

Question

I'm working in the book A Short Course on Operator Semigroups by Engel/Nagel. They say:

1.2 Lemma Let $X$ be a Banach space and let $F$ be a function from a compact set $K \subset \mathbb{R}$ into $\mathcal{L}(X)$. Then the following are equivalent:

(a) $F$ is continuous for the strong operator topology; i.e. the mappings $K \ni t \mapsto F(t)x \in X$ are continuous for every $x \in X$.

(b) $F$ is uniformly bounded on $K$, and the mapings $K \ni t \mapsto F(t)x \in X$ are continuous for all $x$ in some dense subset $D$ of $X$.

My question is: I am familiar with the idea that a family of functions can be uniformly bounded, but how is it possible that a single function, $F$, can be uniformly bounded (cf: part (b) of the blockquote)?

My guess is that because $F$ maps into $\mathcal{L}(X)$, the output of $F$ is a family of functions, and it is this family of functions that is uniformly bounded. OR, maybe it means that for each $x \in X$, the family orbit maps $F(t)x$ are uniformly bounded for $t \in K$. But I am not sure. Thanks so much!

Answer 1

Just look at the first lines of the proof.

Answer 2

For a linear operator, being bounded is the same as being continuous or Lipschitz continuous. Or in other words, it has a finite operator norm.

That a family of operators be uniformly bounded means that that there is a common upper bound of the operator norms over the family.

This all is rather trivial in finite dimensions. But in infinite dimensions counter-examples to such trivialities are easy to construct, many on the sequence space. So for instance in finite dimension for a matrix valued function it is equivalent that

• the entries/coefficients are all continuous individually, and
• the function is also continuous in any operator norm (and thus the operator norm is continuous and consequently bounded over all finite intervals).

The theorem in question here tells how that can be generalized to infinite dimensions.