With your nice question you touch upon a classical circle of ideas in the geometry of Banach spaces going back to Banach's book and which was first developed in more depth in the work of Šmulyan, Klee and Day in the forties and fifties. I'll try to give an outline of the basic intuitions:

Let $X$ be a Banach space with unit ball $B_X = \{x : \lVert x \rVert \leq 1\}$, unit sphere $S_X = \{x : \lVert x \rVert = 1\}$ and dual space $X^\ast$. Write $\langle x^\ast, x\rangle = x^{\ast}(x)$ for the duality pairing between $X^\ast$ and $X$.

Your “duality map” $\mathcal{D}$ is uniquely determined by what it does on the unit sphere since $\mathcal{D}(x) = \lVert x \rVert \mathcal{D}\left(\frac{x}{\lVert x \rVert}\right)$ for $x \neq 0$, so let us focus on the unit sphere. I will first look at a variant and relate it to $\mathcal{D}$ afterwards.

For every $x \in S_X$ in the unit sphere define the set of *norming functionals* by
$$
\nu(x) = \{x^\ast \in B_{X^{\ast}}:\langle x^\ast,x\rangle=1\}.
$$
The set $\nu(x)$ is non-empty by Hahn-Banach, it is weak*-closed and convex by definition, so $\nu(x)$ is weak*-compact by Alaoglu's theorem. Moreover, $\nu(x) \subset S_X$ because $\lVert x^\ast\rVert \lt 1$ implies $\lvert \langle x^\ast, x \rangle \rvert \lt 1$.

Geometrically, $\nu(x)$ parameterizes the supporting hyperplanes of $B_X$ in $x \in S_X$: for each $x^\ast \in \nu(x)$ we have that the unit ball is contained in its associated half-space: $B_X \subset \{y \in X : \langle x^\ast, y \rangle \leq 1\}$ and $x$ belongs to its associated bounding hyperplane: $x \in \{y \in X : \langle x^\ast, y\rangle = 1\} \cap S_X$.

The point $x \in S_X$ is said to be a *point of smoothness*, if $\nu(x)$ is a singleton, that is: there is a unique supporting hyperplane of $B_X$ in $x$. The idea is that if the point $x$ lies on a corner or a lower dimensional face of the unit ball then the set of supporting hyperplanes is not unique. Compare these pictures from Wikipedia:

In the picture on the left there is a unique supporting hyperplane in all points. The “corners” on the right have many supporting hyperplanes while the points between two corners have a unique supporting hyperplane.

A Banach space is said to be *smooth* if every $x \in S_X$ is a point of smoothness. This is precisely the situation you ask about: for $x \in S_X$ we must then have $\mathcal{D}(x) = \nu(x)$.

A first easy observation relating smoothness to *strict convexity* is the following duality result:

- If $X^\ast$ is strictly convex then $X$ is smooth.

*Proof.* Let $x \in S_X$ and let $x^\ast,y^\ast \in \nu(x)$. If $x^\ast \neq y^\ast$ then strict convexity of $X^\ast$ implies that $\lVert \frac{1}{2} (x^\ast + y^\ast)\rVert \lt 1$ so that $\frac{1}{2} (x^\ast + y^\ast) \notin \nu(x)$ contradicting the convexity of $\nu(x)$. Therefore $\nu(x)$ contains exactly one point.
2. If $X^\ast$ is smooth then $X$ is strictly convex.

*Proof.* If $X$ is not strictly convex then there are $x,y \in S_X$ with $x \neq y$ such that $\frac{1}{2}(x+y) \in S_X$. Let $x^\ast \in S_{X^\ast}$ be such that $x^\ast\left(\frac{1}{2}(x+y)\right) = 1$. Then we must have $x^\ast(x) = 1 = x^\ast(y)$ and hence we have the two distinct points $\iota_X(x), \iota_X(y) \in \nu(x^\ast)$ where $\iota_X\colon X \to X^{\ast\ast}$ is the canonical inclusion, so $X^\ast$ is not smooth.

The converses of both these assertions are wrong, but if $X$ is reflexive we can conclude: $X$ is strictly convex (resp. smooth) if and only if $X^\ast$ is smooth (resp. strictly convex).

*Added:* The above “duality” between convexity and smoothness is a guiding principle in the entire theory and most of the fundamental result are similar in spirit, if much more subtle to prove.

It is not hard to show that $X$ is strictly convex if and only if $\nu(x) \cap \nu(y) = \emptyset$ for all $x, y \in X$.

Summarizing and expanding all the above:

If $X$ is smooth (in particular, if $X^\ast$ is strictly convex) we have a canonical map $\nu \colon S_X \to S_{X^\ast}$ which is injective if and only if $X$ itself is strictly convex. If $X$ is reflexive then strict convexity of both $X$ and $X^\ast$ implies bijectivity of $\nu$.

Only bijectivity might not be entirely obvious: If $X$ is reflexive and both $X$ and $X^\ast$ are strictly convex then both $X$ and $X^\ast$ are strictly smooth, hence we have well-defined and injective maps $\nu \colon S_X \to S_{X^\ast}$ and $\nu^\ast \colon S_{X^\ast} \to S_X$. The definitions reveal that $\nu^\ast(\nu(x)) = x$ and $\nu(\nu^\ast(x^\ast)) = x^\ast$ yielding bijectivity.

Let me add: A unique supporting hyperplane should be something like a tangent plane, so it is not too surprising that one can prove that the functional $\nu(x) \colon X \to \mathbb{R}$ is determined by the Gâteaux derivative of the norm map $N(x) = \lVert x \rVert$: for all $y \in X$ we have
$$
\nu(x)y = \partial_y N(x) = \lim_{t \to 0} \frac{\lVert x + ty\rVert - \lVert x\rVert}{t}
$$
(since $N$ is convex, $y \mapsto \partial_y N(x)$ is linear). In fact, the norm of a Banach space is Gâteaux differentiable in $x \in S_X$ if and only if $x$ is a point of smoothness.

*Added later:*

Smoothness and continuity of $\nu$ is strongly related to the geometry of the unit sphere: One can easily show that for smooth $X$ the map $\nu\colon S_X \to S_{X^\ast}$ is norm-weak*-continuous. The map $\nu$ is norm-norm continuous if and only if the norm is “uniformly Fréchet differentiable”, so one calls a Banach space uniformly smooth whenever $\nu$ is norm-norm-continuous.

Let me end by stating Šmulyan's theorem clarifying the role of uniform convexity in your question:

A Banach space $X$ is uniformly smooth [resp. uniformly convex] if and only if $X^\ast$ is uniformly convex [resp. uniformly smooth].

Of course, this generalizes and partly explains your observations on $L^p$, $1 \lt p \lt \infty$.

A reference for all this and much more is chapter 5 of Megginson's book (where strict convexity is called rotundity) or alternatively I recommend the book by Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos, Václav Zizler. In both these books you can find thorough expositions, lots of detailed examples and historical references as well as many enlightening exercises.

## Best Answer

Take $x_m(n)=1$ if $n=m$ and $0$ otherwise. This sequence converges weakly to $0$ becasue $\sum x_m(n)y_n =y_m \to 0$ as $ m \to \infty$ for any $y=(y_n)$ in the dual of $\ell^{p}$ (which is $\ell^{q})$. Of course, this sequence does not converge strongly.