uniform continuity+$\sup_{t\geq 0,\delta\in[0,1]}\left|\int_t^{t+\delta}u(s)ds\right|<\infty$=compact range

Question

Let $u:\mathbb{R}^+\to X$ be a uniformly continuous function, where $X$ is a Banach space. Assume moreover that $$\sup_{t\geq 0,\delta\in[0,1]}\left|\int_t^{t+\delta}u(s)ds\right|<\infty$$ How to prove that $u$ has a relatively compact range, i.e the set $\{ u(t):t\geq0 \}$ is relatively compact?

Answer 1

Counterexample: $X$ is a Hilbert space with orthonormal basis $\{e_n\}$, $u(n) = e_n$, and on each $[n,n+1]$ the function $u$ is linear. Then the image of $u$ lies in the unit ball, so the integral assumption is satisfied. But on the other hand, the image contains $\{e_n\}$, so it's not precompact.