Suppose that $f:(-\omega,\omega) \rightarrow \mathbb{R}$ is a continuous function, $\omega>0$ and $$\sup_{|\lambda|<\omega} |f(\lambda)|=\infty.$$ Can we conclude that $$\lim_{|\lambda| \rightarrow \omega} |f(\lambda)|=\infty?$$
My attempt: Since $f$ is continuous, then for all compact $K \subset (-\omega,\omega)$ we have $\sup_{\lambda \in K} |f(\lambda)| \leq C_K$. In a way, this tells us that the function cannot explode "in the middle" of the interval. However, I could not use the other definitions to obtain the desired.
The limit at any interior point cannot be infinity, since the function is continuous, but the limit at the endpoints may not exist. For example, consider the function
$$f(x) = \frac{\sin\left(\frac{1}{1-x}\right)}{1-x}$$
on the interval $x\in (-1,1)$. The function blows up as $x\to 1$ in an oscillatory way, so the limit as $x\to 1$ does not exist.
If you know what $\limsup$ is, then note that it is true that
$$\limsup_{|\lambda|\to \omega} |f(\lambda)| = \infty.$$
This directly follows from assumptions, since the definition of the limsup is
$$\limsup_{|\lambda|\to \omega} |f(\lambda)| := \lim_{t\to \omega} \sup_{|\lambda|\geq t}|f(\lambda)|.$$
If the limit on the right hand side is not $\infty$, then it contradicts you assumption that the supremum of $|f|$ is infinity.