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I believe I can prove the following result.

Let $f_n:\mathbb{R}^n\to \mathbb{R}$ be a sequence of convex functions that converges almost everywhere to a function $f:\mathbb{R}^n\to\mathbb{R}$. Then $f$ is convex in the sense that there is a convex function $F:\mathbb{R}^n\to\mathbb{R}$ such that $F=f$ a.e.

I am sure, it must be known.

Question. Where can I find a proof?

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3 Answers 3

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It follows from Theorem 10.8 in

R. Tyrrell Rockafellar. Convex analysis. Princeton Mathematical Series, No. 28. Princeton University Press, Princeton, N.J., 1970.

This theorem essentially says the following: if $f_n$ are convex functions on an open domain $\Omega\subset\mathbb{R}^n$ that converge pointwise (to a finite value) on a dense subset of $\Omega$, then the limit exists for every point of $\Omega$, this limit is a convex function, and the convergence is uniform on every compact set inside $\Omega$.

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  • $\begingroup$ Thank you. This is exactly what I was looking for. $\endgroup$ Oct 3 at 18:34
  • $\begingroup$ You're welcome. I remember it took me some time to find a reference when I needed it. $\endgroup$
    – Del
    Oct 4 at 8:29
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I hope it's OK to post with a proof rather than a reference; when I started writing I thought this would be shorter...

Lemma. Let $(f_n \mid n \geq 1)$ be a sequence of convex functions $f_n: \mathbf{R}^d \to \mathbf{R}$, assumed to converge a.e. to a function $f: \mathbf{R}^d \to \mathbf{R}$. Then (i) the convergence is locally uniform, and (ii) $f$ is convex.

  • Throughout let $D_R \subset \mathbf{R}^d$ be the $d$-dimensional disc of radius $R$. We will use the three nested discs $D_1 \subset D_2 \subset D_5$, ultimately showing that the convergence is uniform on the innermost one: we prove first (in Claim 1) that the $f_n$ are bounded above on $D_5$, next (in Claim 2) that they are bounded below on $D_2$, and finally (in Claim 3) that they are equicontinuous on $D_1$. The conclusion follows by Arzela-Ascoli.

  • Recall first that convex functions are continuous (on the interior of their domains); this is true in particular for every $f_n$.

  • For all points $x \in \mathbf{R}^d$ and unit vectors $v \in \mathbf{R}^d$, let $\mathbf{L}_{x,v} \subset \mathbf{R}^d$ be the affine line through $x$ directed by $v$. For example, let $x \neq y \in \mathbf{R}^d$, and $v = y - x/\lvert y - x \rvert$. Any convex function $h: \mathbf{R}^d \to \mathbf{R}$ remains convex when restricted to $\mathbf{L}_{x,v}$, and along this line one has $h(x+tv) \geq h(x) + t(h(y)-h(x))/\lvert y - x \rvert$ for all $t \not\in (0,\lvert y - x \rvert)$. We will use this repeatedly with $h = f_n$.

Claim 1. There is $M > 0$ so that $f_n \leq M$ in $D_5$ for all $n$.

Proof. If not, then there is a sequence of points $x_n \in D_3$ along which $M_n := f_n(x_n) \to + \infty$. Let $\mathbf{L}_{x_n,v}$ be an arbitrary line through $x_n$. Along it, $f_n(x_n + tv) \geq f_n(x_n) + t (f_n(x_n + v) - f_n(x_n))$ for all $t \not \in (0,1)$. Pick $t < -1$ or $t > 1$ depending on the sign of $f_n(x_n + v) - f_n(x_n)$. For one of the half-lines one has $f_n \geq M_n$ along it. By varying $v$ one finds that $f_n \geq M_n$ on at least half of the annulus $D_2(x_n) \setminus D_1(x_n)$: $\mathcal{H}^d( \{ x \in D_2(x_n) \setminus D_1(x_n) \mid f_n(x) \geq M_n \}) \geq \mathcal{H}^d( D_2 \setminus D_1) / 2$. This is absurd because $D_2(x_n) \subset D_5$, and $f_n$ converges a.e. to $f$ in that set. $\blacksquare$

Claim 2. There is $C > 0$ so that $f_n \geq - CM$ in $D_2$ for all $n$.

Proof. We argue by contradiction: as $f_n \leq M$ on $D_5$, if there were a sequence of points $x_n \in D_2$ with $f_n(x_n) \leq -C_n M_n \to \infty$, then $f_n \to -\infty$ on $D_2$. But this is impossible because $f_n$ converges a.e. in $D_2$.

Take thus some fixed $C > 0$, some sequence $C_n \to +\infty$, and two sequences of points $x_n,y_n \in D_2$ so that $f_n(x_n) = -C_n M$ but $f_n(y_n) \geq -CM.$

Let $v_n = y_n - x_n / \lvert y_n - x_n \rvert$. Along the line $\mathbf{L}_{x_n,v_n}$, \begin{align} f(x_n + t v_n) &\geq f(x_n) + t(f(y_n) - f(x_n))/\lvert y_n - x_n \rvert \\ &\geq -C_n M + t(-CM + C_n M)/\lvert y_n - x_n \rvert \\ &\geq MC_n(-1 + t(1 - C/C_n)/\lvert y_n - x_n \rvert) \end{align} for all $t > \lvert y_n - x_n \rvert$. We may assume that $C_n/100 > C \geq 1$ say; therefore if $t \in [2 + \lvert y_n - x_n \rvert, 3 + \lvert y_n - x_n \rvert]$ then \begin{equation} f(x_n + t v_n) \geq MC_n(-1 + 2 (1 - C/C_n)) \geq 99M. \end{equation} But for this range of $t$, $x_n + t v_n \in D_5$, where $f_n \leq M$; this is absurd. $\blacksquare$

Claim 3. The $f_n$ are $2CM$-Lipschitz in $D_1$.

Proof. We argue again by contradiction. Let $x_n,y_n \in D_1$ be two sequences of points along which $f_n(y_n) \geq f_n(x_n)$ and $f_n(y_n) - f_n(x_n) > 2CM \lvert y_n - x_n \rvert$. Let $v_n$ be the unit vector from $x_n$ to $y_n$. Along the line $\mathbf{L}_{x_n,v_n}$, \begin{equation} f_n(x_n + tv_n) > f_n(x_n) + 2tCM \geq CM \end{equation} for all $t \in [1 + \lvert y_n - x_n \rvert,2+\lvert y_n - x_n \rvert]$. This is absurd because $x_n + t v_n \in D_5$ for this range of $t$, where $f_n \leq M$. $\blacksquare$

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  • $\begingroup$ Thank you. This is more or less the proof that I had in mind. It seems quite surprising that nobody could find a reference in the literature. I am actually more interested in a reference rather than a proof, but I voted up your answer as a token of appreciation. $\endgroup$ Oct 2 at 3:25
  • $\begingroup$ Isn't it possible to proceed in the following way: 1/ Reduce to the case in which $(f_n)_n$ is uniformly bounded, considering $(f_n \wedge R)_n \rightarrow f \wedge R$ a.e. 2/ In that case, for any $\rho\in\mathscr{D}$, one has $(f_n\star\rho)_n \rightharpoonup f$ (at least) in $\mathscr{D}'$ by dominated convergence 2/ For a non-negative kernel $\rho$, convexity of $f_n$ implies $\text{Hess}(f_n\star\rho) \geq 0$ (at least) in $\mathscr{D}'$ and this information passes to the limit 3/ If $\text{Hess}(f\star \rho)\geq 0$ for all non-negative kernel, then $\text{Hess}(f)\geq 0$, right ? $\endgroup$ Oct 28 at 14:33
  • $\begingroup$ @AymanMoussa Sorry, I haven't thought about this at all since writing this, and I've kind of lost interest in the problem. In any case, Piotr knows a lot more about these things than I do. Perhaps he'd be willing to answer your question. $\endgroup$
    – Leo Moos
    Oct 28 at 15:03
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This is trickier than it seems. It is easy to see that the limit $f$ satisfies Jensen's inequality a.e., a property sometimes known as almost convexity. In the univariate case, $f$ being almost convex is known to imply that it is equal to some convex function $F$ a.e. (see for example Parnami and Vasudeva On the Stability of Almost Convex Functions). In dimension greater than 1, it is unclear that this is true without extra assumptions. Maybe the following paper could be helpful: La Torre and Rocca, Almost everywhere convex functions on Rn and weak derivatives.

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    $\begingroup$ I think the answer might be a bit more elementary than you assumed, because the convergence can be shown to be uniform locally. $\endgroup$
    – Leo Moos
    Oct 2 at 14:54
  • $\begingroup$ Yes, I realized that after reading your answer. The fact that $f$ is a pointwise limit of convex function gives a lot more information and control than I thought. So the question is much more specific that asking in the abstract whether an almost convex function is almost everywhere equal to a convex function. Thanks for your answer! $\endgroup$
    – Thibaut
    Oct 2 at 18:53
  • $\begingroup$ Forgot to say: welcome to the page! $\endgroup$
    – Leo Moos
    Oct 2 at 19:27

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