Limit of Minima is Finite 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?
 A: 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.

