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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

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Yes for II. Take a unit-norm linear functional $\varphi$ that does not attain its norm on the unit ball and consider the minimization of $f(x)=\max(1,\|x\|)+|\varphi(x)-1|$. The infimum is $1$, but is not attained. –  user53153 Mar 2 '13 at 19:34
    
@5pm, if we assume striclty convex, is it still true? –  Tomás Mar 2 '13 at 19:37
    
The argument I gave applies to every nonreflexive Banach space (it's a theorem of James that every such space has a bounded linear functional that is not norm-attaining.) –  user53153 Mar 2 '13 at 20:12
    
Ok I undertood it. But is it still true for strictly convex functions? I think you function is not strictly convex, am I right? –  Tomás Mar 2 '13 at 20:14
    
Here's an example with a strictly convex function: renorm a nonreflexive space such as $\ell^1$ to make its norm strictly convex. Since it's still nonreflexive, there exists a unit functional $\varphi$ that does not attain its norm. Pick $x_0$ such that $\varphi(x_0)=1$. Then the strictly convex function $f(x)=\|x-x_0\|$ does not attain minimum on the kernel of $f$, which is a Banach space of its own. Indeed, the infimum is $1$, but is not attained. –  user53153 Mar 2 '13 at 23:38

1 Answer 1

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 $.

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