Let $(X, \Vert \cdot \Vert)$ be a Banach space, and $F : X \longrightarrow \mathbb{R}$ a convex function, continuous for the norm topology.

Suppose that $x_n$ is a sequence which weakly converges to some $x \in X$.

Can we say that $F(x_n)$ is bounded in $\mathbb{R}$?

It is true in the following spacial cases:

1) If $X$ is finite dimensional. Indeed $(x_n)$ becomes relatively compact so its image under $F$ is compact, whence bounded.

2) If $F$ is Lipschitz continuous on bounded sets.

3) If $F$ is weakly continuous (equivalently, weakly upper-semicontinuous). In that case we benefit from the inequality $\limsup\limits_{n \in \mathbb{N}} F(x_n) \leq F(x)$.

I feel that this statement is not true, but I am not aware of a counter-example.

  • $\begingroup$ Does continuity imply weak continuity? $\endgroup$ Commented May 14, 2015 at 10:13
  • $\begingroup$ @TZakrevskiy Nope, it is weak continuity which implies norm-continuity. In this particular case, the convexity of $F$ makes it weakly lower-semicontinuous, but no more. $\endgroup$
    – Guillaume
    Commented May 14, 2015 at 10:48
  • $\begingroup$ Ok, thank you; the term is a bit counterintuitive, though. $\endgroup$ Commented May 14, 2015 at 10:51
  • $\begingroup$ Do you want boundedness (i.e. from above and below) or only boundedness from above (as case (3) seems to indicate)? I can show boundedness from below, but I believe that boundedness from above is false in general (though I don't have a counterexample). $\endgroup$
    – PhoemueX
    Commented May 14, 2015 at 15:49

1 Answer 1


Inspired by this question (Nonlinear function continuous but not bounded), one can construct a counterexample as follows:

Let $X =c_0 (\Bbb{N})$, i.e. the space of null-sequences, equipped with the sup-norm. Define

$$ F((x_n)_n )= \sum_n x_n^{2n}. $$

Since the function $x \mapsto x^{2n}$ is convex, it is easy to see that every summand of the series defines a convex, continuous function. Hence, $F$ will be continuous and convex, once we show that the series converges locally uniformly.

But let $x=(x_n)_n\in c_0$ be arbitrary. Then $|x_n|<1/4$ for all $n \geq N$ for a suitable $N$. For $y=(y_n)_n \in c_0$ with $\Vert x-y\Vert <1/4$, this yields $|y_n|<1/2$ for $n \geq N$, so that the Weierstrass M-test shows that the series defining $F$ converges uniformly on the ball $B_{1/4}(x)$.

Now define $x_n 2\delta_n$, with $(\delta_n)_m =0$ for $n\neq m$ and $(\delta_n)_n =1$. Using $(c_0)^\ast \cong \ell^1$, we see $x_n \to 0$ weakly, but we have $F(x_n)=2^{2n}\to\infty$.

EDIT: Exactly the same construction (even with a slightly easier proof) also works if one replaces $c_0$ by $\ell^2$. This shows that failure of the desired property even happens for Hilbert spaces, which are the nicest class of infinite dimensional Banach spaces one could hope for.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .