I am looking for a reference/a hint to the following problem:
We are given $f_1(x),f_2(x)$ convex functions (say, on $\mathbb R^d$) such that $f_1(x) \to\infty$ for $\|x\|\to\infty$. Also there is an $\alpha > 0$ such that $f_1(x) - \alpha f_2(x) \geq 0$.
A first consequence is that for any $\epsilon > 0$, $$f_1(x) - (\alpha - \epsilon)f_2(x) \to \infty, \quad \|x\|\to \infty$$ because $f_1(x) - (\alpha - \epsilon)f_2(x) = \frac{\epsilon}{\alpha }f_1(x) + \frac{a-\epsilon}{\alpha }(f_1 - \alpha f_2(x)) \geq \frac{\epsilon}{\alpha}f_1(x) \to \infty$ for $\|x\|\to \infty$.
Now my question is whether (and maybe under which circumstances) it is possible to find a local approximation $g_2(x)$ such that $|g_2(x)-f_2(x)| \to 0$ for $\|x\|\to \infty$ and
$$f_1(x) - \beta g_2(x)$$ is convex for some $\beta \in (0,\alpha]$.
The concrete example I have in mind is the following: $X = \mathbb R^d$ and $f_1(x) = \sum_{k=1}^d \frac{x_k^2}{\sigma_k^2}$ (a weighted squared $l^2$-norm) and $f_2(x) = \|x\|_1^2$ (the squared $l^1$-norm).
It can be shown that $f_1(x) - \alpha f_2(x) \geq 0$ for $\alpha \leq (\sum_{k=1}^d \sigma_k^2)^{-1}$.
Unfortunately, $f_1(x) - \beta\cdot f_2(x)$ is not convex for any $\beta\in (0,\alpha]$. My question in this specific example corresponds to finding a function $g_2(x)$ such that $|g_2(x)-f_2(x)|$ is "asymptotically small", e.g. $g_2(x) \geq f_2(x)$ and $|g_2(x)-f_2(x)|\to 0$ for $\|x\|_1\to\infty$. In addition for some $\beta \in (0,\alpha]$, I want
$$f_1(x) -\beta f_2(x)$$
to be convex. Note again that $f_1(x)$ is convex, and I only need convexity of $f_1 - \beta f_2$ for an arbitrarily small $\beta \in (0,\alpha)$.
PS: I have thought a bit more about this problem and it looks like the following might work (at least in this specific setting):
$g_2(x) := \left(\sum_{k=1}^d (\sqrt{\sigma_k^2 + |x_k|^2} - \sigma_k)\right)^2$