I have a question that I encountered during my internship:

Consider a convergent sequence of continuous, convex functions $\{f_n(x)\}_n$ defined in $\mathbb{R}^M$. These functions are uniformly Lipschitz continuous, that is, $\exists C\in\mathbb{R}$ such that:

$$\forall x,y \in \mathbb{R^M},\forall n\ge1\quad |f_n(x)-f_n(y)|\le C|x-y|.$$

Furthermore, each function $f_n(x)$ has a minimizer. The properties of simple convergence and uniform Lipschitz continuity allow us to prove that the convergence is uniform in any compact of $\Bbb R^M$.

My question is:

Can we demonstrate that $\inf_{\Bbb R^M}f_n(x)$ converges to $\inf_{\Bbb R^M}f_{\infty}(x)$ as $n\rightarrow\infty$, where $f_{\infty}(x)$ is the limit of $f_n(x)$ and it is supposed that $\inf_{\mathbb{R^M}}f(x)$ is finite?

Thanks a lot!

  • $\begingroup$ It would make it easier to read if you formatted your formulae in Latex (mathjax). $\endgroup$ – copper.hat Jul 10 '12 at 7:08

No. Take the sequence $$ f_n : x\in \mathbb R \to \frac 1 n \lvert x - n \rvert -1 \in \mathbb R $$ We have $$ \lim_n f_n(x) = 0 \quad\forall x\in \mathbb R\\ \inf_{\mathbb R} f_n(x) = f_n(n) = -1\\ \lvert f_n(x) - f_n(y) \rvert = \frac 1 n \big\lvert \lvert x - n \rvert - \lvert y - n\rvert \big\rvert \leq \frac 1 n \lvert x - y\rvert \leq \lvert x -y \rvert $$ the functions $f_n$ converge punctually to $f_\infty = 0$, they are convex and uniformly Lipschitz continuous but $$ \lim_n \left(\inf_{\mathbb R} f_n(x) \right) = -1 \neq \inf_{\mathbb R} f_\infty(x) = 0 $$

  • $\begingroup$ Very nice counterexample. Entirely contrary to my intuition. $\endgroup$ – copper.hat Jul 10 '12 at 17:12
  • $\begingroup$ Thanks AlbertH for his reply $\endgroup$ – Higgs88 Jul 11 '12 at 7:54

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