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In my research on convergence properties of certain Bayesian methods I have encountered $\mathop{\mathrm{arg\,min}}$ limits of the forms \begin{equation} \lim_{n\to\infty} \mathop{\mathrm{arg\,min}}_{\mathbf{x}\in\mathbb{R}^d}\frac{f_n(\mathbf{x})}{g_n(\mathbf{x})} \end{equation} and \begin{equation} \lim_{n\to\infty} \mathop{\mathrm{arg\,min}}_{\mathbf{x}\in\mathbb{R}^d}\big(f_n(\mathbf{x}) + g_n(\mathbf{x})\big) \end{equation} where it is assumed that $f_n,g_n\to 0$ plus $f_n/g_n \to 0$ in the former case and $g_n/f_n \to 0$ in the latter, all in pointwise manner. The functions are real-valued and the minimizers can be assumed unique for each $n$.

I find it fairly intuitive that, at least with some additional assumptions involving maybe monotonicity of the function sequences or convexity, one should be able to discard the effect of $g_n$ at the limit. However, I have been unable to find any references dealing with limits of this type but am quite certain that someone must have addressed the topic. Is anyone able provide references? The only vaguely related question that has caught my eye is this.

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  • $\begingroup$ The answer is similar to the answer in comment to the linked question: Gamma convergence. $\endgroup$
    – Dirk
    May 9, 2017 at 15:30

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If $f_n$ is an equicontinuous sequence of functions has pointwise limit $f$ and the latter has the unique minimizer $x_0$, then $\lim_{n\to\infty} \arg\min f_n(x)=\{x_0\}$.

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I am quite confident that the answer to the first question is no (for this exact version):

I think you need to be careful the relativity of your assumptions at least in the first case. Consider two sequences of functions $\hat{f}_n$ with $|\hat{f}_n(x)| < C$ and $\hat{g}_n$ with $c < |\hat{g}_n(x)| < C$ for all $n \in \mathbb{N}$ and $x \in \mathbb{R}^d$.

Now define $f_n = \frac{1}{n^2} \hat{f}_n$ and $g_n = \frac{1}{n} \hat{g}_n$. For sure, $f_n, g_n \rightarrow 0$ and $$|\frac{f_n}{g_n}| < \frac{C}{nc} \rightarrow 0.$$ At the same time, this rescaling does not change minimizers, so $f_n/g_n$ has the same as $\hat{f}_n/\hat{g}_n$. The only condition however that remains with the hat functions is boundedness.

Now you can surely construct two sequences of hat functions that are bounded, but for which the influence of $g$ does not vanish.

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