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In the infinite dimensional setting, we know the unit ball is not compact so in general the $\sup_{\|x\|\leq 1}$ can not be replaced by $\max_{\|x\|\leq 1}$.

However, when the Banach space $X$ is reflexive, we could replace $\sup$ by $\max$. Due to Hahn-Banach theorem, for each $f\in X^*$, we know there exists an $u^{**} \in X^{**}$ with norm one such that $$\sup_{\|x\|\leq 1} f(x) = \|f\|_* =u^{**}(f)$$ and since $X$ is reflexive, we know there exists an $u \in X$ with norm one such that $$\sup_{\|x\|\leq 1}f(x) = f(u).$$

Now I want to give a counter example when the space is non-reflexive. Let us take $l^1$ and its dual $l^\infty$, take $$f=(0.9, 0.99, 0.999, \cdots ) \in l^\infty$$ then $\sup_{\|x\|\leq 1}f(x) = 1$ but there is no element in the unit ball would give $f(x) = \sum_i f_i x_i = 1$.

Is this correct? And I think if we could replace $\sup$ by $\max$ in the dual norm, can we conclude that the space has to be reflexive.

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Looks good to me. The converse that you are suggesting is also true, according to Brezis' Functional Analysis (Chapter 1 Section 1):

In general, the “sup” in (5) is not achieved; see, e.g., Exercise 1.3. However, the “sup” in (5) is achieved if $E$ is a reflexive Banach space (see Chapter 3); a deep result due to R. C. James asserts the converse: if $E$ is a Banach space such that for every $f \in E^\ast$ the sup in (5) is achieved, then $E$ is reflexive; see, e.g., J. Diestel [1, Chapter 1] or R. Holmes [1].

The references from his appendix are:

Geometry of Banach spaces: Selected Topics, Springer, 1975.

Geometric Functional Analysis and Its Applications, Springer, 1975.

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