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Consider a sequence $\{ f_k \}_{k=1}^{\infty}$ of locally-bounded functions $f_k: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$.

Assume the following.

For any sequence $\{X_k\}_{k=1}^{\infty}$ of compact sets $X_k \subset \mathbb{R}^n$ such that $X_k \subseteq X_{k+1}$ and $X_k \rightarrow \mathbb{R}^n$, there exist (a uniform) $M \in \mathbb{R}_{>0}$ such that

$$ \sup_{x \in X_k} f_k(x) \leq M$$

Say if the following claim holds (or find a counterexample).

There exists $K \in \mathbb{Z}_{\geq 1}$ such that

$$ \sup_{x \in \mathbb{R}^n} f_K(x) < \infty$$

Note: can we use this argument?

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  • $\begingroup$ Suppose not: you can find an increasing sequence of compact sets $\{K_j\}$ such that $\sup_{x\in K_j}f_j(x)\geq j$ and $\bigcup_jK_j=\Bbb R^n$. $\endgroup$ – Davide Giraudo Jul 3 '12 at 21:25
  • $\begingroup$ Can you please be more explicit? Why is this a contradiction? I mean even if $K_j$ are "increasing", the functions $f_j$ can be like "decreasing" for each fixed $x$... So I'm not very clear on that... $\endgroup$ – user693 Jul 3 '12 at 21:28
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If it's not the case, then for all $k\geq 1$, we have $\sup_{x\in\Bbb R^n}f_k(x)=+\infty$. There is $x_k\in\Bbb R^n$ such that $f_k(x_k)\geq k$. We define $X_1=\{x_1\}$ and by induction $X_k:=\{x_k\}\cup X_{k-1}\cup \{x,\lVert x\rVert\leq k\}$. Then $\{X_k\}$ is an increasing sequence of compact sets, and for all $k$, $$k\leq \sup_{x\in X_k}f_k(x)\leq M$$ which is not possible.

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If for any sequence $\{X_k\}_{k=1}^{\infty}$ of compact sets $X_k \subset \mathbb{R}^n$ such that $X_k \subseteq X_{k+1}$ and $X_k \rightarrow \mathbb{R}^n$, there exist (a uniform) $M \in \mathbb{R}_{>0}$ such that

$$ \sup_{x \in X_k} f_k(x) \leq M$$

Choosing $\{X^{n}_k\}_k$ such that $X^{n}_1 = \overline{B_n(0)}$ and $X^{n}_k= \overline{B_{n+k}(0)} \ \forall n$. We have that \begin{equation} f_1(x) \le M \quad \mbox{in}\quad \forall n. \end{equation} In fact, \begin{equation} \sup_{X^{n}_{1}} f_1 = \sup_{\overline{B_n(0)}} f_1 \le M \quad \forall n. \end{equation} Then, $\sup_{\mathbb{R}^{n}}f_1\le M$.

Moreover, in same way \begin{equation} \sup_{\mathbb{R}^{n}}f_k \le M \quad \forall k. \end{equation}

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  • $\begingroup$ I'm not clear with this answer. What is $B_n(0)$? what is $f$? $\endgroup$ – user693 Jul 3 '12 at 22:09
  • $\begingroup$ $B_n(0)= \{x\in \mathbb{R}^n : |x| < n\}$ do not exist $f$ is $f_1$. $\endgroup$ – user29999 Jul 3 '12 at 22:10

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