Limit of Minima is Finite

Question

Suppose $f$ and $g$ are non-negative functional from a Hausdorff topological space $X$ to $[0,\infty]$. Let $x_n$ be a sequence of minimizers of the problem $$ x^n\triangleq \operatorname{arginf}_{x \in X} f(x) + ng(x). $$ If $$ \lim_{n \to \infty} f(x^n) + ng(x^n)<\infty, $$ does this necessarily imply that $$ \lim_{n \to \infty}g(x^n)=0? $$

Here's my Argument: Suppose that $\lim_{n \to \infty}g(x^n)\neq 0$. Note that for any $n<m$ \begin{align} f(x)+ng(x) &\leq f(x)+mg(x) \\ \operatorname{inf}_{x \in X}f(x)+ng(x) &\leq \operatorname{inf}_{x \in X}f(x)+mg(x) \\ f(x^n)+ng(x^n) &\leq f(x^m)+mg(x^m) \\ ng(x^n)\leq f(x^n)+ng(x^n) &\leq f(x^m)+mg(x^m), \end{align} the last line follows from the fact that $f$ is non-negative.

Since $g$ is non-negative and it was assumed that $$ c\triangleq \lim_{n \to \infty}g(x^n)\neq 0, $$ then $$ \infty =\lim_{n \to \infty}nc = \lim_{n \to \infty}ng(x^n)\leq \lim_{n \to \infty}f(x^n)+ng(x^n) $$ a contradiction of the finiteness assumption (namely that: $ \lim_{n \to \infty} f(x^n) + ng(x^n)<\infty, $).

Therefore, $g(x^n)=0$.
Did I go haywire somewhere?

Answer 1

Your proof is not correct. Just because $0$ is not the limit of $g(x^n)$ does not mean that $g(x^n)$ converges to something else. It may not converge anywhere.

What you can say is that if $g(x^n)\not\to 0$, then (by non-negativity of $g$) you have some $\varepsilon>0$ and a sequence $(n_k)_k$ such that for all $k$ we have $g(x^{n_k})>\varepsilon$.

Other than that (and $\operatorname{arginf}$ instead of $\inf$ in the second line), your argument seems technically correct.

However, if you look at your proof, it is not really a proof by contradiction, but rather, by contraposition. You don't use the hypothesis about $\lim_n f(x^n)+ng(x^n)$, you simply show (from $g(x^n)\not\to0$) that this sequence is unbounded (which implies that it cannot converge to any finite number). Moreover, it is unnecessary to consider the $m$, or indeed, the value of $f(x)+ng(x)$ at any point other than $g(x^n)$. The only thing that matters is that $ng(x^n)$ is unbounded and that $ng(x^n)\leq f(x^n)+ng(x^n)$.

This is how I would write the proof.

The proof is by contraposition, so we assume that $g(x^n)\not\to 0$. Therefore (since $g(x^n)\geq 0$) for some $\varepsilon>0$ and for some infinite sequence $(n_k)_k$ of natural numbers, we have $g(x^{n_k})\geq \varepsilon$. It follows that $n_k g(x^{n_k})\geq n_k\cdot \varepsilon \to \infty$.

But since $f(x^{n_k})\geq 0$, this implies that $n_kg(x^{n_k})\leq f(x^{n_k})+n_kg(x^{n_k})\to \infty$, and hence $f(x^{n})+ng(x^{n})$ does not converge to any real number, which completes the proof.