---
tags: metric, memo
---

# Conditions of Convergence

## Claim

Assume $\theta_0$ is the unique parameter that minimizes $Q$. If $Q$ is not continuous and the parameter space $\Theta$ is not compact, then the below statement may not hold:

Given any $\varepsilon>0$, $| \theta_0 - \theta|>\varepsilon \implies Q(\theta_0) > Q(\theta) + \delta$ for some $\delta>0$.

## Counterexamples

1.

$$
Q(x)=
\begin{cases}
x, \text{ if } x \le 1\\
\frac{1}{x}, \text{ if} x > 1\\
\end{cases}
$$

If $x \in [0,\infty)$, then $x=0$ is the minimizer, but the statement is failed.

2.

$$
Q(x)=
\begin{cases}
0 , \text{ if } x=0 \\
\frac{1}{n}, \text{ if } x = \frac{m}{n} \text{ is irreducible fraction}\\
1, \text{ Otherwise}\\
\end{cases}
$$

If $x \in [-\varepsilon, \varepsilon]$, then $x=0$ is the minimizer, but the statement is failed.