Let $f_{i}\colon X\rightarrow \mathbb{R}$ be convex, Lipshitz and positive (X - reflexive Banach space) for $i=1,2$. One may easily prove that $f_{i}$ is weak lower semicontinuous. How to prove the following statement:

$lim\, inf\, f_1(x_{n})[f_{2}(x_{n})-f_{2}(x)]\geq0$, where $\{x_{n}\}$ converges weakly to some $x$?

We can write $lim\, inf\, f_1(x_{n})[f_{2}(x_{n})-f_{2}(x)]\geq\lim\, inf\, f_1(x_{n})\cdot lim\, inf\, [f_{2}(x_{n})-f_{2}(x)]$ (it would solve the problem)


$(f_{2}(x_{n})-f_{2}(x))$ has to be positive and I don't know that. Maybe convexity has something to do with the proof?


1 Answer 1



If the sequence $x_n$ converges weakly, by the Uniform Boundedness Principle it must be bounded. Since $f_1$ is Lipschitz, that implies $f_1(x_n)$ is bounded, let's say by $B$. Since $\liminf_n f_2(x_n) \ge f_2(x)$, for any $\epsilon > 0$ there is $N$ such that $f_2(x_n) \ge f_2(x) - \epsilon/B$ for all $n \ge N$, and then $f_1(x_n) (f_2(x_n) - f_2(x)) \ge -\epsilon$ for such $n$. Thus $\liminf_n f_1(x_n)(f_2(x_n) - f_2(x)) \ge 0$.

  • $\begingroup$ Perfect and no convexity used. Thank you so much! $\endgroup$
    – zorro47
    Mar 22, 2017 at 19:20

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