# An upper bound for a continuous and positive function

Let $f \colon \mathbb{R}^{n} \to \mathbb{R}_{\geq 0}$ be a continuous function with $f(x) \to + \infty$ as $| x | \to +\infty$. Let $a \in \mathbb{R}^{n}$, a compact set $A \subset \mathbb{R}^{n}$ and $b_{i} := \frac{1}{i}$ for all $i \in \mathbb{N}$ be given. Let $f_{i}(x) := f(x + a b_{i})$ for all $x \in \mathbb{R}^{n}$ and $i \in \mathbb{N}$. My question is: Does there exist some integrable (in the sense of Lebesgue integration) function $g \colon \mathbb{R}^{n} \to \mathbb{R}_{\geq 0}$ such that for all $x \in A$ and for all $i \in \mathbb{N}$, the following holds $$f_{i}(x) \leq g(x) ?$$ If so, how can we prove this claim? Thanks.

• It seems like you should be able to do something with the supremum of $f(x)$ in a closed disk centered at the origin. I haven't the time to think about it right now, but maybe this will give you an idea. Commented Sep 9, 2013 at 18:57
• Can't we define $g(x)=\max\{f(x+ab_i):i\in\mathbb{N}\}$?
– QED
Commented Sep 9, 2013 at 18:58
• I don't understand a thing, surely I am wrong. If $f$ is continuous and infinite at infinity, then surely it isn't Lebesgue integrable. Therefore its translates $f_i$ cannot be Lebesgue integrable either. But then, $g$ cannot be Lebesgue integrable. Commented Sep 9, 2013 at 19:31
• Thanks guys for the helpful answers. I didn't explain the problem correctly. Actually the integration in on a subset of "R^{n}" on which "f" has finite values. Commented Sep 9, 2013 at 19:39

The answer seems to be trivial. For each point $x \in A$ and each index $i \in \mathbb{N}$ there exists a point $y\in A+B$ such that $f_i(x)=f(y)$, where $B$ is a ball centered at the zero of the space $\mathbb R^n$ with radius $\|a\|$. Since the set $A+B$ is compact and the function $f$ is continuous, the function $f$ is bounded on the set $A+B$. So $f_i(x)\le g(x)$, where we can take as $g$ an arbitrary constant function $g(x)=c$ such that $c\ge\max\{f(y):y\in A+B\}$.