Show a sequence generating a nonincreasing sequence of function values is bounded if sublevel sets are bounded

Question

Suppose a real-valued function $f(x)$ is such that for some constant $\alpha$, the sub-level set $S_{\alpha}=\{x \mid f(x) \leq \alpha\}$ is nonempty and bounded, then for any sequence $(x^k)$ with $( f(x^k) )$ being monotonically decreasing and converging to $f_* < \alpha$, the sequence $(x^k)$ is bounded.

Answer 1

Yes, this is true, and we can even dispense with the monotone decreasing part.

Because $f(x^k) \to f_*$, we know that, for all $\varepsilon > 0$, there exists some $N \in \Bbb{N}$ such that $$k \ge N \implies |f(x^k) - f_*| < \varepsilon.$$ If we take $\varepsilon = \alpha - f_*$, then we get an $N \in \Bbb{N}$ such that \begin{align*} k \ge N &\implies |f(x^k) - f_*| < \alpha - f_* \\ &\implies f(x^k) - f_* < \alpha - f_* \\ &\implies f(x^k) < \alpha \\ &\implies f(x^k) \le \alpha. \end{align*} That is, the tail end of the sequence lies in the bounded sublevel set.

Let $M \ge 0$ be such that $f(x) \le \alpha \implies |x| \le M$, which exists by assumption. Further, let $$M' = \max\{M, |x^1|, |x^2|, \ldots, |x^{N-1}|\} \ge 0.$$ I claim that $|x^k| \le M'$ for all $k$. If $k < N$, then $|x^k|$ appears in the maximum formula above, hence $|x^k| \le M'$. If $k \ge N$, then as we showed above, $f(x^k) \le \alpha$, which means $|x^k| \le M \le M'$. Either way, the sequence is bounded by $M'$.

Answer 2

As the hypothesis states, let $f$ be a lower bounded function, i.e., for some $\beta\in \Bbb R$, $$f(\mathbf{x}) \ge \beta$$ for all $\mathbf{x}\in \Bbb R^n$. Now, consider the sub-level set $S_{\beta - \varepsilon}$ for any $\varepsilon > 0$. Clearly, $$S_{\beta-\varepsilon} = \{\mathbf{x}\in \Bbb R^n: f(\mathbf{x}) \le \beta - \varepsilon\}= \varnothing$$ which contradicts the hypothesis that all sub-level sets are non-empty.

Therefore, the problem in its current form does not make sense, as the hypotheses are not "consistent."