Do every linear functional on a Banach space attains its norm? In case of finite-dimension, the argument follows via compactness of the unit ball. What about in infinite-dimensional case? Is there any any general version of it?
Any help will be appreciated!
Since the accepted answer did not give a counter example, it might be good to exhibit one here.
Consider the linear functional on $c_0$ given by $$f(x) =\sum_{n=1}^\infty x_n/2^n.$$
It is easy to show that $\|f\|=1$, but yet there is no unit vector $x$ in $c_0$ such that $|f(x)|=1$.
One would like to take $x=(1,1,1,\ldots)$, but unfortunately this in not a vector in $c_0$.
It is not necessary that an functional on a Banach space $X$ would attain its norm. However, in case of a reflexive Banach space, the unit ball remains compact under the weak topology (the minimum topology needed to maintain the continuity of each member of the $X^*$).
A general version of this is every compact operator on a reflexive Banach space attains its norm. In particular, the continuous linear functionals are compact.
Whenever, $X$ is not reflexive, no such conclusion is true. However, still we can say something about $X^*$. Every member of $X^{**}$, which are image of any member of $X$, under canonical isomorphism, attains their norm. This is because of Banach-Alaoglu Theorem, which says:
Given any Banach space $X$, unit ball of $X^{*}$ is compact with respect to the weak$^*$ topology.
Weak$^*$ topology: The minimum topology on $X^*$, needed to maintain the continuity of each member of $\psi(X)$, where $\psi: X \to X^{**}$ denotes the canonical embedding.
