For convex $f$, how to construct smooth $f_\varepsilon$ with $f- \varepsilon \leq f_\varepsilon \leq f\;$?

Question

Question

Let $f : \mathbb R_+ \to \mathbb R$ be a convex function such that $f(x) \to -\infty$ as $x \to \infty$.

For $\varepsilon > 0$, is there a (standard) way to construct a convex function $f_\varepsilon \in C^\infty$ (or at least in $C^1$) such that $f - \varepsilon \leq f_\varepsilon \leq f\,$?

Put differently: Is there an $\varepsilon$-exact smoothing that underestimates $f$ and preserves convexity?

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Thoughts

I'm not even sure that the conjecture I'd like to prove here is true. But, intuitively, it seems like it should be.

My first instinct was to use mollifiers. But (to my knowledge) these only yield pointwise convergence of $f_\varepsilon \to f$convergence as $\varepsilon \downarrow 0$. Perhaps the extra structure demanded of $f$ makes uniform convergence hold?

My guess is that the conjecture is true and somehow results from $f$ being convex and bounded from above. (If it's true, the limit $f(x) \to -\infty$ as $x \to \infty$ is likely a stronger condition than necessary.)

Hand waving (what follows is not rigorous): Since $f$ is defined on a closed set and is bounded from above, it ought to get progressively flatter. I think it follows that $\sup\Big\{\Big|\frac{f(y)-f(x)}{y-x} \Big| : 0 \leq x < y \Big\} < \infty$ and is, therefore, Lipschitz. It seems like this should ought to give us enough regularity that $f_\varepsilon$ and "hug" close enough to $f$ on all $\mathbb R_+$.

Answer 1

Fix $\varepsilon>0$.

Define $g_{\varepsilon}(x):=f- \varepsilon /2$. Then it is clear that $f-\frac{2\varepsilon}{3} < g_{\varepsilon} < f-\frac{\varepsilon}{3}$ and $g_{\varepsilon}$ is convex.

Note that every convex function defined on $\mathbb R$ is continuous (cf. here), which is clearly locally integrable. Thus it makes sense to define mollifications.

Define $f_{\varepsilon, n}:=g_{\varepsilon}*m_n$, where $m_n$ is a sequence of standard mollifiers (i.e. a sequence of $C^\infty$ bump function, cf. here).

For each $n$, the convexity of $f_{\varepsilon, n}$ follows from the convexity of $g_{\varepsilon}$ by verifying the definition directly.

$\color{darkred}{\text{There is a problem of uniformity in $n$ in the following paragraph, but I am too lazy to edit the post.} \\ \text{Please refer to the comments below my post for more details.}}$

A well known result (cf. Theorem $7$ in Appendix C$.5.$ of Evans' classic PDE textbook) tells us that $f_{\varepsilon, n} \to g_{\varepsilon}$ as $n\to \infty$ almost everywhere. However, since $f_{\varepsilon, n}$ and $g_{\varepsilon}$ are continuous, we have $f_{\varepsilon, n} \to g_{\varepsilon}$ everywhere (cf. my quick proof in the remark $3$ below). This implies that, for all large $n$, $f_{\varepsilon, n}$ satisfies our demands (here I give you strict inequalities which are strong than your needs), i.e. $f-\varepsilon <f_{\varepsilon, n}<f$ and $f_{\varepsilon, n}$ is convex and $C^\infty$.

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Remark on Notation: People (and Evans) use $\eta_\varepsilon$ to denote mollifiers. However, $\varepsilon$ has been used here and therefore I use $m_n$ instead.

Remark $2$: It seems like we don't need any asymptotic property of $f$, i.e. we don't really care its behavior when $x$ approaches $\infty$.

Remark $3$: Suppose $f_n,f\in C(\mathbb R)$, $\forall n$, and $f_n\to f$ almost everywhere. We will show that $f_n\to f$ everywhere.

Suppose not. Then $\exists x\in\mathbb R, \varepsilon_0 >0$, $\forall N$, $\exists n_0\ge N$ such that $|f_{n_0}(x)-f(x)|>\varepsilon_0$. Note that $f_{n_0}-f$ is continuous. We may therefore find an open neighborhood of $x$ in which $f_{n_0}-f>\varepsilon_0$. Since Lebesgue measure of an open neighborhood must be positive, it contradicts with our assumption that $f_n\to f$ almost everywhere.

$\tag*{$\square$}$

My reasoning in Remark $3$ works for any measure space whose measure is Lebesgue measure.