# Image of a bounded sequence by a convex continuous function in a Banach space

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.

• Does continuity imply weak continuity? – TZakrevskiy May 14 '15 at 10:13
• @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. – Guillaume Garrigos May 14 '15 at 10:48
• Ok, thank you; the term is a bit counterintuitive, though. – TZakrevskiy May 14 '15 at 10:51
• 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). – PhoemueX May 14 '15 at 15:49

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.