Let $\Omega$ be a measure space and let $1<p<\infty$ and $p'=p/(p-1)$. Also let us agree to write simply $L^p$ instead of $L^p(\Omega)$.

If we identify the dual space $\left(L^p\right)^\star$ with $L^{p'}$ via the isomorphism $$\left[ v \in L^{p'}\right] \leftrightarrow \left[ L_v\in \left(L^p\right)^\star\ \text{defined to be}\ \langle L_v, u\rangle=\int_{\Omega}vu\, d\mu\right],$$ we can define an interesting map $\mathcal{D}\colon L^p\to \left(L^p\right)^\star$ as follows: $$\mathcal{D}u=\frac{\lvert u \rvert^{p-1}\text{signum}(u)}{\left(\int_{\Omega}\lvert u \rvert^p\, d\mu\right)^{\frac{p-2}{p}}},$$ (with the convention that $\mathcal{D}0=0$). This map $\mathcal{D}$ is non-linear, except for $p=2$ in which case it becomes the Riesz isomorphism of the Hilbert space $L^2$ with its dual. Also, $\mathcal{D}$ is norm-preserving and is characterized by the following property:

$\mathcal{D}u$ is the unique element of $\left(L^p\right)^\star$ such that $$\tag{1} \lVert \mathcal{D}u\rVert_\star=\lVert u \rVert_p\quad \text{and}\quad \langle \mathcal{D}u,u\rangle=\lVert \mathcal{D}u\rVert_{\star}\lVert u\rVert_p.$$

Moreover, $\mathcal{D}$ is invertible: indeed, as it is readily seen, $$\mathcal{D}^{-1}v=\frac{\lvert v \rvert^{p'-1}\text{signum}(v)}{\left(\int_{\Omega}\lvert v \rvert^{p'}\, d\mu\right)^{\frac{p'-2}{p'}}}.$$ We can also characterize $\mathcal{D}^{-1}$ in complete analogy with what we have already done:

$\mathcal{D}^{-1}v$ is the unique element of $L^p$ such that $$\tag{2}\lVert \mathcal{D}^{-1}v\rVert_p=\lVert v\rVert_{\star}\quad \text{and}\quad \langle v, \mathcal{D}^{-1}v\rangle=\lVert v\rVert_\star\lVert\mathcal{D}^{-1}v\rVert_p.$$

The question follows.

Question Suppose that $X$ is an abstract Banach space. Which properties must $X$ have so that a mapping $\mathcal{D}\colon X \to X^\star$ is uniquely defined by property (1)? When is $\mathcal{D}$ bijective?

As already mentioned in this question, the existence and bijectivity of $\mathcal{D}$ in $L^p$ space seem to have something to do with the uniform convexity of $L^p$ and $(L^p)^\star$. I guess there is something more general beneath the surface here.

Thank you for reading.


2 Answers 2


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:

supporting hyperplanes in x

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:

  1. 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.

  1. 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.

  • 1
    $\begingroup$ Very nice answer! Now I believe that I have a much clearer view over the proof of the Riesz Representation Theorem outlined here. Thank you very much! $\endgroup$ Commented Nov 2, 2012 at 21:09

Theorem 1. Let $X$ be a normed space over field $\mathbb{R}$ then the following conditions are equivalent

1) $X^*$ is a strictly convex space.

2) each functional defined on subspace of $X$ have unique norm preserving extension.

Proof. See this article.

Theorem 2. Let $X$ be a Banach space, then the following conditions are equivalent

1) $X$ is reflexive

2) each functional $f\in X^*$ attains its norm

Proof. See this article.

Theorem 3. Let $X$ be a normed space, then there exist a map $\mathcal{D}:X\to X^*$ such that $$ \Vert\mathcal{D}(x)\Vert=\Vert x\Vert\qquad\langle \mathcal{D}(x),x\rangle=\Vert\mathcal{D}(x)\Vert\Vert x\Vert\qquad\tag{1} $$

Proof. For arbitrary $x\in X$, by Hahn-Banach theorem there exist $f\in X^*$ such that $f(x)=\Vert x\Vert$ and $\Vert f\Vert=1$. Now we define $$ \mathcal{D}:X\to X^*:x\mapsto \Vert x\Vert f $$ Then we can easily check that $\mathcal{D}$ satisfies $(1)$. Let's prove $\mathcal{D}$ is injective. Assume $\mathcal{D}$ is not injective, then there exist $x\in X\setminus\{0\}$, such that $\mathcal{D}(x)=0$, then for all $x'\in X$ holds $\mathcal{D}(x)(x')=\Vert x\Vert f(x')=0$. Since $x\neq 0$, then for all $x'\in X$ holds $f(x')=0$, and in particular $f(x)=0$. By construction $\Vert x\Vert=f(x)$, so $\Vert x\Vert=0$ and $x=0$. Contradiction, so $\mathcal{D}$ is injective.

Theorem 4. Let $X$ be a normed space with strictly convex dual, then the map $\mathcal{D}$ with properties $(1)$ is uniquely defined and and as the consequence the following additional property holds $$ \mathcal{D}(\lambda x)=|\lambda| \mathcal{D}(x)\tag{2} $$

Proof. Since $X^*$ is strictly convex, then from theorem 1 it follows that map $\mathcal{D}$ satisfying $(1)$ is uniquely defined. Moreover $\mathcal{D}$ have property $(2)$. Indeed, note that if we have $f\in X^*$ such that $f(x)=\Vert f\Vert\Vert x\Vert$ and $\Vert f\Vert=\Vert x\Vert$, then $(|\lambda| f)(x)=\Vert f\Vert\Vert \lambda x\Vert$ and $\Vert |\lambda| f\Vert=\Vert \lambda x\Vert$. From unique definitness of $\mathcal{D}$ we conclude that $(2)$ holds.

Theorem 5. Let $X$ be a reflexive, strictly convex space with strictly convex dual. Then the map $\mathcal{D}$ satisfying $(1)$ is unique, and moreover it is bijective and it satisfies $(2)$.

Proof. Take $f\in X^*$. If $f=0$, then $\mathcal{D}(0)=f$, otherwise consider $\hat{f}=\Vert f\Vert^{-1}f$. Since $X$ is reflexive, there exist $\hat{x}\in X$ such that $\hat{f}(\hat{x})=\Vert \hat{f}\Vert=1$ and $\Vert\hat{x}\Vert=1$. In other words $\Vert \hat{f}\Vert=\Vert \hat{x}\Vert$, $\hat{f}(\hat{x})=\Vert\hat{f}\Vert\Vert\hat{x}\Vert$. Since $X^*$ is strictly convex from theorem 1 we see that such a functional is unique and $\mathcal{D}(\hat{x})=\hat{f}$. Hence for $x=\Vert f\Vert \hat{x}$ we have $\mathcal{D}(x)=\mathcal{D}(\Vert f\Vert \hat{x})=\Vert f\Vert \mathcal{D}(\hat{x})=\Vert f\Vert\hat{f}=f$. This $\mathcal{D}$ is surjective.

Assume that $\mathcal{D}$ is not injective, then there exist $x_1,x_2\in X$ such that $x_1\neq x_2$ and $\mathcal{D}(x_1)=\mathcal{D}(x_2)$. Denote $f=\mathcal{D}(x_1)=\mathcal{D}(x_2)$. Since $\mathcal{D}$ satisfies $(1)$ we have $\Vert f\Vert=\Vert x_1\Vert$, $f(x_1)=\Vert f\Vert\Vert x_1\Vert$ and $\Vert f\Vert=\Vert x_2\Vert$, $f(x_2)=\Vert f\Vert\Vert x_2\Vert$. Fix $t\in(0,1)$ and consider vector $x=t x_1+(1-t)x_2$, then $f(x)=\Vert f\Vert(t\Vert x_1\Vert+(1-t)\Vert x_2\Vert)$. Which implies $f\Vert(t\Vert x_1\Vert+(1-t)\Vert x_2\Vert)=|f(x)|\leq\Vert f\Vert\Vert x\Vert$ and as the consequence $t\Vert x_1\Vert+(1-t)\Vert x_2\Vert\leq\Vert x\Vert$. On the other hand from triangle inequality $\Vert x\Vert\leq t\Vert x_1\Vert +(1-t)\Vert x_2\Vert$. This means that $\Vert tx_1+(1-t)x_2\Vert=t\Vert x_1\Vert+(1-t)\Vert x_2\Vert$ for all $t\in(0,1)$. Consider vectors $\hat{x}_!=\Vert x_1\Vert^{-1}x_1$, $\hat{x}_2=\Vert x_2\Vert^{-1}x_2$. Since $\Vert x_1\Vert=\Vert x_2\Vert=\Vert f\Vert$ we have $\Vert\hat{x}_1\Vert=\Vert\hat{x}_2\Vert=1$ and $\Vert t\hat{x}_1+(1-t)\hat{x}_2\Vert=\Vert t\Vert\hat{x}_1\Vert+(1-t)\Vert\hat{x}_2\Vert$ and $x_1\neq x_2$. This means that unit sphere of $X$ contains segment, and as the consequence $X$ is not strictly convex. Contradiction, so $\mathcal{D}$ is injective.

Since $\mathcal{D}$ is surjective and injective it is bijective. The rest follows from theorem 4.

Theorem 6. Let $X$ be a normed space with uniquely defined bijective map $\mathcal{D}:X\to X^*$ satisfying $(1)$, then $\mathcal{D}$ satisfy $(2)$, and what is more $X$ is reflexive, strictly convex with strictly convex dual.

Proof. Since $\mathcal{D}$ satisfy $(2)$ and uniquely defined then from theorem $4$ we have that $\mathcal{D}$ satisfy $(2)$. Now take arbitrary $f\in X^*$. If $f=0$, then it is obviously attains its norm, otherwise consider $\hat{f}=\Vert f\Vert^{-1} f$. Obviously, $\Vert\hat{f}\Vert=1$ Since $\mathcal{D}$ is bijective, we get some $\hat{x}\in X$ such that $\mathcal{D}(\hat{x})=\hat{f}$. Hence $\Vert\hat{x}\Vert=\Vert\hat{f}\Vert=1$ and $\hat{f}(\hat{x})=\Vert \hat{f}\Vert\Vert\hat{x}\Vert=\Vert \hat{f}\Vert$. Then, $f(\hat{x})=\Vert f\Vert$ while $\Vert\hat{x}\Vert=1$. This means that norm of $f$ is attained. Since for all $f\in X^*$ its norm is attained, then by theorem 2 the space $X$ is reflexive.

Assume that $X^*$ is not strictly convex, then unit sphere of $X^*$ contains some segmet, hence there exist $f_1,f_2\in X^*$ such that $f_1\neq f_2$ and $\Vert f_1\Vert=\Vert f_2\Vert=\Vert(f_1+f_2)/2\Vert=1$. Since $\mathcal{D}$ is bijective then there exist $x\in X$ such that $\mathcal{D}(x)=(f_2+f_2)/2$, and what is more $\Vert x\Vert=\Vert\mathcal{D}\Vert=\Vert(f_1+f_2)/2\Vert=1$. Since $\Vert(f_1(x)+f_2(x))/2\Vert=\mathcal{D}(x)(x)=\Vert\mathcal{D}(x)\Vert x\Vert=1$ and $\Vert f_1\Vert=\Vert f_2\Vert=1$, then $f_1(x)=f_2(x)=1$. Thus $f_1(x)=\Vert f_1\Vert\Vert x\Vert$, $\Vert f_1\Vert=\Vert x\Vert$ and $f_2(x)=\Vert f_2\Vert\Vert x\Vert$, $\Vert f_2\Vert=\Vert x\Vert$. This means that $\mathcal{D(x)}=f_1$ and $\mathcal{D}(x)=f_2$. This contradicts assumption $f_1\neq f_2$. Hence $X^*$ is strictly convex.

Assume that $X$ is not strictly convex, then unit sphere of $X$ contains some segment, i.e. there exist $x_1,x_2\in X$ such that $x_1\neq x_2$, $\Vert x_1\Vert=\Vert x_1\Vert=\Vert (x_1+x_2)/2\Vert=1$. By Hahn-Banach theorem there exist $f\in X^*$ such that with $f((x_1+x_2)/2)=\Vert(x_1+x_2)/2\Vert=1$. The last equality with $\Vert x_1\Vert=\Vert x_2\Vert=1$ implies $f(x_1)=f(x_2)=1$. Thus $f(x_1)=\Vert f\Vert\Vert x_1\Vert$, $\Vert f\Vert=\Vert x_1\Vert$ and $f(x_2)=\Vert f\Vert\Vert x_2\Vert$, $\Vert f\Vert=\Vert x_2\Vert$ and from uniqueness of $\mathcal{D}$ we conclude $\mathcal{D}(x_1)=f$ and $\mathcal{D}(x_2)=f$. Since $\mathcal{D}$ is bijective $x_1=x_2$, which contradicts assumption $x_1\neq x_2$. Hence $X$ is strictly convex.

Theorem 7. Let $X$ be a normed space, then the following conditions are equivalent

1) $X$ is reflexive, strictly convex with strictly convex dual

2) there exist unique bijective map $\mathcal{D}:X\to X^*$ satisfying $(1)$

Proof. Follows from theorem 5 and 6.

  • $\begingroup$ I would like to point out that Theorem 2 is the only one here that is really deep. Theorem 1 is easy: If a functional $f$ has two HB-extensions $g$ and $h$ then $(g+h)/2$ is also a HB-extension, so $X^\ast$ isn't strictly convex. The other direction (a few lines) is due to Foguel see here. // You may be interested in this paper where Troyanski shows that $X = C[0,\omega_1] \mathbin{\oplus_2} \ell^1$ is smooth but $X^\ast$ is not strictly convex. $\endgroup$
    – commenter
    Commented Nov 3, 2012 at 9:16
  • $\begingroup$ I think Theorem 3 is not correct: you don't know that $\mathcal{D}$ is linear (which you use in the proof). You need to assume $X$ to be strictly convex in order for $\mathcal{D}$ to be injective (the problem in the proof of theorem 3 is that $\mathcal{D}$ is not linear, so injectivity is not equivalent to "$\mathcal{D}(x) = 0$ iff $x = 0$"): Suppose $X$ is not strictly convex. There are $x \neq y$ such that $\lVert x \rVert = \lVert y \rVert = \lVert (x+y)/2\rVert = 1$. Choose $f$ such that $\lVert f \rVert = 1$ and $f((x+y)/2) = 1$. Then $f(x) = 1 = f(y)$ so $\mathcal{D}$ is not injective. $\endgroup$
    – commenter
    Commented Nov 3, 2012 at 10:52
  • $\begingroup$ @commenter So for theorem 7 I need to assume that $X$ is strictly convex? $\endgroup$
    – Norbert
    Commented Nov 3, 2012 at 10:59
  • $\begingroup$ This is not a big problem. We know $\mathcal{D}$ with (1) is uniquely defined iff $X^\ast$ is strictly convex and $\mathcal{D}$ is injective iff $X$ is strictly convex. Your theorem 7 then becomes: TFAE 1) $X$ is reflexive, strictly convex and $X^\ast$ is strictly convex. 2) there is a uniquely defined bijective $\mathcal{D}$ satisfying (1). (I wrote this before seeing your last comment. So: yes.) $\endgroup$
    – commenter
    Commented Nov 3, 2012 at 10:59
  • 1
    $\begingroup$ @TomaszKania, indeed this article doesn't mention the space $C[0,\omega_1]\oplus_2 \ell_1$ at all. Maybe this result somehow follows from Troyanski's paper. This paper shows that $(\ell_1,\Vert\cdot\Vert_1)$ is smooth while its dual $(\ell_\infty, \Vert\cdot\Vert')$ is not strictly convex. If you need more details on this paper, feel free to ask. $\endgroup$
    – Norbert
    Commented Feb 28, 2021 at 8:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .