How to show that $f : {}^\ast \Bbb R \to {}^\ast \Bbb R$ is bounded, if it obtains a limited value everywhere? Let $f :  {}^\ast \Bbb R \to  {}^\ast \Bbb R$. For all $x \in  {}^\ast \Bbb R $, there exists $y \in \Bbb R $ such that $f(x) \leq y$. How to show that there exists  $r \in \Bbb R$, such that for all $z \in  {}^\ast \Bbb R$  we have $f(z) \leq r$?
I know this can be shown by a direct application of the dual form of idealization principle. Is there any other way to show this, say, using ultrapower construction?
Moreover, if $f$ is continuous, is it true that $f$ necessarily obtain a maximal value?
 A: The set of upper bounds for the function contains all unlimited numbers.  Therefore by the underspill principle, it must contain a finite number, as well.
Of course, this is in the assumption that the function is internal.  Otherwise there are obvious counterexamples.
A: Our goal is to prove the statement
$$ (\forall z \in {}^{*}\Bbb{R} ) (\exists r \in \Bbb{R}) ( f(z) \leq r ) \quad \Longrightarrow \quad (\exists r \in \Bbb{R})(\forall z\in {}^{*} \Bbb{R})( f(z) \leq r). $$
We prove the contrapositive:
$$ (\forall r \in \Bbb{R})(\exists z\in {}^{*} \Bbb{R})( f(z) > r) \quad \Longrightarrow \quad (\exists z \in {}^{*}\Bbb{R} ) (\forall r \in \Bbb{R}) ( f(z) > r ). $$
Assume that for any $r \in \Bbb{R}$, there exists $z \in {}^{*}\Bbb{R} $ such that $f(z) > r$. For each fixed $r \in \Bbb{R}$, applying the transfer principle, we have
$$(\exists z \in {}^{*}\Bbb{R})( f(z) > r ) \quad \Longrightarrow \quad (\exists z \in \Bbb{R})( f(z) > r ). $$
Thus we have
$$ (\forall r \in \Bbb{R})(\exists z \in \Bbb{R})( f(z) > r ). $$
Passing to the transfer principle, we obtain
$$ (\forall r \in {}^{*}\Bbb{R})(\exists z \in {}^{*}\Bbb{R})( f(z) > r ). $$
In particular, for a given infinitely large $R \in {}^{*}\Bbb{R}$, we have $f(z) > R$ for some $z \in {}^{*}\Bbb{R}$. This can be rephrased as
$$ (\exists z \in {}^{*}\Bbb{R} ) (\forall r \in \Bbb{R}) ( f(z) > r ). $$
This is exactly what we wanted, and hence the proof is completed.
To show that $f$ may fail to achieve its maximum, we consider the function
$$ f(x) = \frac{|x| + 1}{|x| + 2}. $$
It is clear that for all $x \in \Bbb{R}$ we have $f(x) < 1$ while for each $r < 1$ there exists $x \in \Bbb{R}$ such that $f(x) > r$. This can be transferred into the corresponding statement in the hyperreal field, and thus $f(x)$ has no maximum on $\Bbb{R}^{*}$.
