# Continuous Coercive Functional not Bounded Below

Let $X$ be a infinite dimensional Banach space. How to construct a example of a continuous function $f:X\rightarrow\mathbb{R}$ such that $f$ is coercive, but is not bounded below.

$f$ is coercive if $\|u\|\rightarrow \infty$ then $f(u)\rightarrow\infty$

$f$ is bounded below if there exist a constant $C$ such that $f(u)\geq C$ for all $u\in X$

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What are the definitions of coercive and bounded below funcion? – Norbert Jan 26 '13 at 18:38
@Norbert, I edited the question with the definitions that you asked. – Tomás Jan 26 '13 at 19:05
do you mean $f(u)\to\infty$ or $f(u)\to+\infty$ ? – Norbert Jan 26 '13 at 19:22
Are you sure that there is such a function? – Dirk Jan 26 '13 at 19:36

$$f(u)=\|u\|-\sum_{n=1}^\infty n(1-\|u-u_n\|)^+$$ where $(u_n)$ is a bounded sequence such that $\|u_m-u_n\|\ge 2$ whenever $m\ne n$.
Please verify if my arguing is correct. In the case of $X$ being reflexive the function that you defined cannot be weakly sequenatially lower semi continuous, because this would implie that $f$ has a global minimum? – Tomás Jan 26 '13 at 20:41
@Tomás Yes, that is correct. In a more concrete way, $f(u_n)\to -\infty$ by construction, there is a weakly convergent subsequence $u_{n_k}\rightharpoonup u$, yet $f(u)$ is not $-\infty$. You can also consider $g=e^f$, which is bounded from below, continuous (indeed, locally Lipschitz), and not WSLSC by the preceding argument. – user53153 Jan 26 '13 at 20:46