Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:X\rightarrow\mathbb{R}$ be a continuous convex function over the banach space $X$. (Note that it is everywhere finite.) In particular, it is lower semicontiuous, and by Mazur's theorem [S. Mazur, Studia Math. 4, 70 (1933)] also (sequentally) weakly lower semicontinuous.

$f$ is upper semicontinuous, but is it also (sequentally) weakly upper semicontinuous?

share|cite|improve this question
How do you apply Mazur's Theorem to prove that your function is Sequentially Weakly Lower Semicontinuous? – Tomás Feb 6 '13 at 10:06
For convex sets, closedness implies weak closedness. Hence lower semicontinuity implies weak lower semicontiunity. Theorem by Mazur, S. Mazur, Studia Math. 4, 70 (1933). – Jas Ter Feb 6 '13 at 10:08
Sorry but I can't get it. If $x_n\rightarrow x$ weakly, how do you show that $f(x)\leq\liminf f(x_n)$? – Tomás Feb 6 '13 at 10:12
I cannot prove this theorem by Mazur myself, but it is standard to use in convex analysis. A convex function is lsc in some topology iff its sublevelsets are closed in this topology. This is an easy consequence of the usual def of lsc: "for every $\lambda<f(x)$, there is a neighborhood $U$ of $x$ such that $f(U)>\lambda$". The liminf characterization is equivalent to this. The sublevel sets $\{x:\lambda\leq f(x)\}$ are convex and strongly closed (easy consequence of def). Mazur's theorem then states that closed convex sets are weakly closed. Hence, $f$ is weakly lsc. – Jas Ter Feb 6 '13 at 10:34
Ok you are right, now I understood, thank you. – Tomás Feb 6 '13 at 10:43
up vote 3 down vote accepted

The norm on any normed space is convex and continuous and it is weakly lower semicontinuous, see also Aliprantis-Border, Infinite Dimensional Analysis: A Hitchhiker's Guide, Lemma 6.22, p. 235.

If $X$ is infinite-dimensional, the weak topology and the norm topology are distinct. Therefore the norm is not weakly upper semicontinuous since this would imply that it is weakly continuous, and consequently the weak topology and the norm topology would have to coincide.

For example, the norm is not weakly sequentially upper semicontinuous in a Hilbert space. An orthonormal sequence $\{e_n\}_{n \in \mathbb{N}}$ converges weakly to zero by Bessel's inequality, but $$0 \lt 1 = \liminf\limits_{n \to \infty} \lVert e_n\rVert = \limsup\limits_{n\to\infty} \lVert e_n\rVert.$$

share|cite|improve this answer
Thanks! This was my "gut feeling", but I had no way to express it. I thought maybe there was some property of the convex function that would help, but your counterexample is beautiful. – Jas Ter Feb 6 '13 at 11:08

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.