Let $X$ be a infinite dimensional real Banach space. If $X$ is reflexive, then any continuous, convex coervive function $f:X\rightarrow\mathbb{R}$ has a minimum value, that is assumed for some point $x\in X$. This happens because of two motives:
1 – The ball is compact in the weak topology.
2 – $f$ is weakly sequentially lower semi-continuous.
Note that we can only assume $f$ lower semi continuous to get this result.
On the other hand, if $X$ is not reflexive, we don't have 1 and in the same conditions for $f$, it is possible for $f$ not satisfy 2. So my question is: How to construct $f: X\rightarrow 0$ continuous, convex and coercive, with $X$ non-reflexive, and in such a way that $f$ don't attain a minimum value in some point of $X$?
More specifically:
I- It is possible for $f$ to be unbounded below?
II- It is possible for $f$ to be bounded below, but the minimum is not assumed for any point in $x$?
Thanks
Best Answer
Answer to (II) only: such functions exist. First, recall that every non-reflexive space $X$ has a unit-norm functional $\varphi$ such that $\varphi\ne 1$ on the closed unit ball.
Example 1: $f(x)=\max(\|x\|,1)+|\varphi(x)-1|$. Here $\inf_X f=1$, but $f(x)>1$ for all $x$.
Example 2, strictly convex. Take a nonreflexive Banach space with a strictly convex norm $\|\cdot\|$, for example $\ell_1$ with the norm $\|x\|=\|x\|_1+\|x\|_2$. Let $\varphi$ be as above and pick $x_0$ such that $\varphi(x_0)=1$. The kernel of $\varphi$ is a Banach space on its own right, and the function $f(x)=\|x−x_0\|$ does not attain minimum on it. Indeed, $\inf_{\ker\varphi} f=1 $ but $f(x)>1$ for all $x\in \ker\varphi $.