Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can one give a reference to a result like this:

  • If a sequence of convex functions $f_{n}$ on $\mathbb{R}$ converges pointwise to a non-monotonic function $f$, then $\displaystyle\inf_{\mathbb{R}} f_n$ converges to $\displaystyle\inf_{\mathbb{R}} f$?

Thank you.

share|improve this question

2 Answers 2

I don't know a reference to a result. But the statement is simple enough to just give a proof here.

First observe that $f$ is necessarily convex (taking the limit of $f_i(tx + (1-t)y) \leq tf_i(x) + (1-t)f_i(y)$), and non-monotonic. This implies that $f$ attains its infimum [Roc, Section 27]: there exists $x_0$ such that $f(x_0) \leq f(x)$ for all $x\in \mathbb{R}$.

We now prove the claim by contradiction. Assume that $\inf f_i \not\to \inf f$. Then at least one of $\limsup_i \inf_{\mathbb{R}} f_i > \inf f$ or $\liminf_i \inf_{\mathbb{R}} f_i < \inf f$ is true, we call them cases 1 and 2.

Case 1: after taking a subsequence $f_{i'}$ we have that there exists a fixed constant $\epsilon$ such that $f_{i'}(x) > \inf_{\mathbb{R}} f_{i'} > \inf f + \epsilon$. This contradicts $f_{i'}(x_0) \to f(x_0) = \inf f$.

Case 2: after taking a subsequence $f_{i'}$ we have that there exists a fixed constant $\epsilon$ such that $\inf_{\mathbb{R}} f_{i'} < \inf f - \epsilon$. By definition there exists $x_{i'}$ such that $f_{i'}(x_{i'}) \leq \inf f_{i'} + \epsilon / 2$. Now, if $x_{i'}$ accumulates at $x_{\infty}$, then $f(x_{\infty}) \leq \inf f_{i'} + \epsilon / 2 < \inf f$ is a contradiction. So $x_{i'}$ cannot accumulate, so that only finitely many can belong to every closed interval. Without loss of generality we can assume that a subsequence $x_{i''} \to +\infty$. By convexity $f_{i''}|_{[x_0,x_{i''}]} \leq \inf f$ hence pointwise convergence implies that $f |_{[x_0,\infty)} = f(x_0)$. This contradicts non-monotonicity of $f$.

[Roc]: Rockafeller, Convex Analysis

Note that I've assumed that non-monotonicity means non-weak-monotonicity. If non-monotonicity allows functions that are weakly, but not strongly, monotonic, then the statement is false: if you require $\inf f_n > -\infty$, you can let $f(x) = 0$ if $x < 0$ and $f(x) = x$ for $x\geq 0$. And define $f_n$ by

$$ f_n(x) = \begin{cases} -1 & x < -n \\ x / n & -n \leq x < 0 \\ x & 0 \leq x \end{cases} $$

share|improve this answer

I have since found a few results in the literature, which are somewhat similar to what I had in mind, but not quite the same.

Of course, if the functions $f$ and $f_n$ are assumed to be real-valued, then various simple proofs of the result that I stated can be given; in particular, then the convex functions $f$ and $f_n$ are necessarily continuous, and so, provided that the functions are not monotonic, their infima are attained and hence can be replaced by the corresponding minima. However, in the applications that I had in mind, it is necessary to allow the functions $f$ and $f_n$ to take the value $\infty$ as well, and then the proof becomes significantly more complicated.

Details on the above statements can be found in the note ``A necessary and sufficient condition on the stability of the infimum of convex functions'' that I have just posted at http://arxiv.org/pdf/1307.3806v3.pdf .

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.