$f:[0,1) \to X$ is s uniformly continuous if $\lim _{t \rightarrow 1} f(t)$ exists

Question

I'm doing Problem III.3.11 from textbook Analysis I by Amann.

My attempt:

We have $[0,a]$ is compact for all $a \in (0,1)$. Moreover, the function that's continuous on a compact domain is uniformly continuous. As such, $$\sup \{a \in (0,1) \mid f:[0,a] \to X \, \text{is uniformly continuous}\} =1$$

1. Since $1$ is not in the domain of $f$, I think $\lim _{t \rightarrow 1} f(t)$ doesn't make sense. Instead, $\lim _{t \rightarrow 1^-} f(t)$ is more appropriate. Please confirm if my understanding is correct!

2. After that, I'm stuck. Could you please shed me some lights? Thank you so much!

Answer 1

If $c:=\lim_{t\to 1^{-}} f(t)$ exists, then $\bar{f}:[0,1]\to X$ defined by

$$ \bar{f}(t)=\begin{cases} f(t) & t\in [0,1) \\ c & t=1 \end{cases} $$ is continuous on $[0,1]$ and hence, uniformly continuous.

Since $f$ is the restriction of a uniformly continuous function on a larger space, $f$ is uniformly continuous.