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Let $F:(0,\infty) \to [0,\infty)$ be a continuous function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$. Suppose also that $F$ is convex on $[1-\epsilon,1]$, for some $\epsilon>0$.

Let $y_n \in [1-\epsilon,1)$, $y_n \to 1$, and suppose $m_n$ is a slope of a supporting line to $F|_{[1-\epsilon,1]}$ at $y_n$, i.e. $$ F(x) \ge F(y_n)+m_n(x-y_n) \, \, \, \text{ for every } \, \, x \in [1-\epsilon,1]. $$ Is it true that $\lim_{n \to \infty}m_n=0$?

By plugging $x=1$, we get $0=F(1) \ge F(y_n)+m_n(1-y_n)$ so $m_n<0$.

If $F$ is differentiable at $y_n$, then this forces $m_n=F'(y_n)$, so if $F$ is $C^1$ near $x=1$, we have $m_n=F'(y_n) \to F'(1)=0$. However, I don't want to assume that $F$ is differentiable.

Martin R
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Asaf Shachar
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3 Answers3

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Not under the hypotheses given. Consider $F(x) = |x-1|$. In this case, the unique supporting line for $y_n \in [1-\varepsilon, 1]$ is 1, so $m_n$ is a constant sequence.

More generally, I think you're after a sequential lower hemicontinuity condition for the subdifferential of $F\vert_{[1-\varepsilon, 1]}$. As a correspondence the subdifferential will be (cyclically) monotone, hence every selection from it will be nondecreasing, but absent stronger conditions not every value of $\partial F\vert_{[1-\varepsilon, 1]}(0)$ will be obtainable as a limit (from the left) of a selection.

Pete Caradonna
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The slopes $m_n$ increase with increasing $y_n$ so that $\lim_{n \to \infty}m_n$ exists. But $$ F_-'(y_n) \le m_n \le F_+'(y_n) \le F_-'(1) $$ so that limit is strictly negative if the left derivative $F_-'(1)$ is negative.

In fact one has $\lim_{n \to \infty}m_n = F_-'(1)$. If not, then $m_n \le a < F_-'(1)$ for all $n$. Then $$ F(x) \ge F(y_n)+m_n(x-y_n) \ge F(y_n) + a(x-y_n) $$ for $1-\epsilon < x < y_n$. It follows that $F(x) \ge F(1) + a(x-1)$ for $1-\epsilon < x < 1$ and therefore $F_-'(1) \le a$, contrary to the assumption.

Martin R
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  • Thanks, that is a nice argument. I am happy to see you again, you have answered many of my questions in the past. In fact, this question was motivated by an incomplete argument that I had for this problem: https://math.stackexchange.com/questions/3724637/is-there-a-global-convexity-locally-around-a-minimum ...I hope this one will interest you as well. Thanks again. – Asaf Shachar Feb 18 '21 at 17:02
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No. Consider $F(x)=\lvert x-1\rvert+(x-1)^2$, which is strictly convex on $\mathbb{R}$ and differentiable except at $x=1$. We have obviously $\lim_{n\to\infty} m_n=-1$ coming from the $\lvert x-1\rvert$ for $x<1$.

user10354138
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