Range of gradient map for coercive function I am given a differentiable (and therefore continuous) function $f: \mathbf{E} \to \mathbb{R}$ which satisfies the following growth condition:
$$
\lim_{||x|| \to \infty} \frac{f(x)}{||x||} \to +\infty
$$
Just for the sake of notation, $\mathbf{E}$ is some arbitrary Euclidean space.
The point is to prove that the gradient map of $f$ has range $\mathbf{E}$. The exercise has a hint: "try minimizing the function $f(x) - \langle a, x \rangle $"
Since $f$ is coercive, we know that it has bounded level sets
$$
\mathcal{L}_c = \{ x | f(x) \leq c \}
$$
 for otherwise we would be able to find a sequence $\{x^n \}_n$ with $||x_n|| \to \infty$ so that $f(x_n) \leq c$ which contradicts the given growth condition. Also $f$ is continuous, so its level sets are also closed. Therefore $f$ has compact level sets which implies that it has a (global) minimizer. 
Question: how do I prove that $g(x) = f(x) - a^T x$ also has a global minimizer for every choice $a \in \mathbf{E}$? This would suffice to prove that the gradient map of $f$ is $\mathbf{E}$, since the first-order optimality condition for $g$ would imply $\nabla f = a$.
Edit: An attempt to show that $g$'s level sets are compact: closedness follows from continuity. As for boundedness:
$$
\begin{align*}
    \mathcal{L}^{g}_c &= \{x: f(x) - a^Tx \leq c \} \\
    &= \bigcup_{\substack{c_1, c_2 :\\ c_1 + c_2 = c}} {\underbrace{\{ x: f(x) \leq c_1 \}}_{\mbox{bounded}} \cap \{x: -a^T x \leq c_2 \}}
\end{align*}
$$
The problem is that the above is an infinite union of sets which are bounded (every term is an intersection of a bounded with an unbounded set), for which I am not aware of any way to show boundedness.

The assumption 
$$\lim_{\|x\| \to \infty} \frac{f(x)}{\|x\|} \to +\infty$$
implies that the function $g(x) = f(x) - a^T x$ satisfies 
$$\lim_{\|x\| \to \infty} \frac{g(x)}{\|x\|} \to +\infty$$
as the contribution of $a^T x/\|x\|$ is bounded. 
Hence $g(x)\to +\infty$ as $\|x\|\to \infty$. A continuous function with this property attains global minimum. Indeed, if $\{x_n\}$ is a sequence such that $g(x_n)\to \inf g$, then $\{x_n\}$ must be bounded  and therefore has a convergent subsequence.
