I am having difficulty following a proof about balls (subsets) of radially unbounded functionals. Let $U$ be a Banach Space. Let the space of admissible controls $U_{ad}\subset U \ne \emptyset$ be closed and convex. Let the objective functional $f:U \rightarrow \mathbb{R}$ be continuous with $f(u) \ge c > -\infty \: \: \forall u \in U_{ad}$ and radially unbounded, such that: \begin{equation*} \left\Vert u\right\Vert \rightarrow \infty \implies f(u) \rightarrow \infty \end{equation*} Let $j$ be the infimum of $f$ on $U_{ad}$.
Now, radially unboundedness implies that there exists a $r > 0$ such that:
\begin{equation*}
f(u) > j + 1 \: \: \forall u \in U_{ad} \text{ with } \left\Vert u \right\Vert > r
\end{equation*}
Here is where I run into trouble - why there is a $j + 1$ ?
I asked the instructor, and he told me that it could be any arbitrarily small $\varepsilon > 0$, just such that the ball is not an empty set, i.e. $ f(u) > j + \varepsilon$ as the "infimum is not necessarily obtainable."
However, if $\varepsilon$ can be arbitrarily small,why is this not implied already by the strict inequality $f(u) > j$?
Could someone provide an example of why this $\varepsilon$ is required?