The present question regards the proof of the following theorem which is found in Adams' Sobolev spaces, §2.30 - 2.33. Here $(\Omega, \mathcal{M}, \mu)$ denotes some arbitrary measure space and $L^p$ always stands for $L^p(\Omega)$.

Riesz Representation Theorem Let $1<p<\infty$ and let $p'=p/(p-1)$. For any $v \in L^{p'}$ denote $L_v$ to be the linear functional on $L^p$ defined by the following equation: $$\langle L_v, w\rangle =\int_{\Omega} vw\, d\mu.$$ Then the mapping $v\mapsto L_v$ is an isometric isomorphism of $L^{p'}$ onto $[L^p]^\star$.

The most interesting thing to prove is that this mapping $L$ is surjective, which as far as I know is usually done by means of the Radon-Nikodym's theorem of measure theory. On the contrary, Adams' book employs uniform convexity of $L^p$ space and two of the four Clarkson's inequalities, precisely: $$\tag{38} \forall u, v \in L^p, 2\le p <\infty,\quad \left\lVert \frac{u+v}{2}\right\rVert_p^{p'}+\left\lVert \frac{u-v}{2}\right\rVert_p^{p'}\ge\left( \frac{1}{2}\lVert u \rVert_p^p+\frac{1}{2}\lVert v\rVert_p^p\right)^{p'-1}, $$ $$\tag{40} \forall u, v \in L^p, 1< p \le 2,\quad \left\lVert \frac{u+v}{2}\right\rVert_p^{p}+\left\lVert \frac{u-v}{2}\right\rVert_p^{p}\ge \frac{1}{2}\lVert u \rVert_p^p+\frac{1}{2}\lVert v\rVert_p^p. $$ His proof follows those steps:

  1. Because of uniform convexity, there exists a duality mapping $$\left[ F\in \left( L^p\right)^\star \right] \to \left[\text{the unique}\ w\in L^p\ \text{s.t.}\ \lVert w\rVert_p=\lVert F \rVert_\star\ \text{and}\ \langle F, w\rangle=\lVert F\rVert_\star^2\right].$$
  2. The duality mapping has an explicitly known left inverse $$\left[ L_v\in (L^p)^\star\ \text{where}\ v=\frac{\lvert w\rvert^{p-1}\text{signum}(w)}{\left(\int \lvert w\rvert^p\,d\mu\right)^{\frac{p-2}{p}}}\right] \leftarrow \left[ w\in L^p\right].$$
  3. Because of inequalities (38) and (40), the duality mapping is injective.

This means that the duality mapping is bijective and so, in particular, that any linear functional $T$ is of the form $L_v$, which is what Adams wanted to prove.

However, it seems to me that he actually proved much more than that: namely, this proof introduces the duality mapping, which is (as far as I can understand) a generalization of the Riesz isomorphism between a Hilbert space and its dual. Also, it looks like this mapping only depends on some easily generalizable properties of $L^p$ such as uniform convexity. So:

Question How much is it known about the duality mapping in an abstract Banach space? What are the minimal hypotheses that guarantee its existence? In what spaces has it got an explicit analytical expression?

Thank you for reading.

  • 1
    $\begingroup$ A small remark: If you look at von Neumann's proof of the Radon-Nikodym theorem (via the $L^2$-duality), as in Rudin's Real and complex analysis, chapter 6, for example, step 2 your outline and Rudin's argument (compare with 6.10 and 6.17) are actually very close to each other. $\endgroup$
    – commenter
    Oct 31, 2012 at 5:18

1 Answer 1


For every normed space $X$ you have duality with its dual via $$ \langle\cdot,\cdot\rangle: X\times X^*\to\mathbb{C}:(x,f)\mapsto f(x) $$ This duality is bilinear, and what is more $$ \Vert x\Vert=\sup\{|\langle x, f\rangle|: f\in X^*,\;\Vert f\Vert\leq 1\}\qquad \Vert f\Vert=\sup\{|\langle x, f\rangle|: x\in X,\;\Vert x\Vert\leq 1\} $$ Thus good duality always exist, the question is explicit description of $X^*$. The problem of complete description of duals of normed spaces seems to be completely hopeless. But for most common spaces there are some. Here are some of them

Let $(\Omega,\Sigma,\mu)$ be a measure space, then $$ L_p(\Omega,\Sigma,\mu)^*= \begin{cases} L_{p/(p-1)}(\Omega,\Sigma,\mu)\quad&\text{ if }\quad p\in(1,+\infty)\\ L_{\infty}(\Omega,\Sigma,\mu)\quad&\text{ if }\quad p=1\\ \mathrm{ba}(\Omega,\Sigma,\mu)\quad&\text{ if }\quad p=\infty \end{cases} $$ where $\mathrm{ba}(\Omega,\Sigma,\mu)$ is a space of finitely additive signed bounded measures on $(\Omega,\Sigma,\mu)$.

Let $\Omega$ be a normal space, then $$ C(\Omega)^*=\mathrm{rba}(\Omega) $$ where $rba(\Omega)$ is a space of regular bounded finitely-additive complex-valued measures on algebra generated by closed sets.

Let $\Omega$ be locally compact Hausdorff space, then $$ C(\Omega)^*=\mathrm{rca}(\Omega) $$ where $\mathrm{rca}(\Omega)$ is a space of regular bounded $\sigma$-additive complex-valued Borel measures on $\Omega$.

Proofs of these results you can find in Linear Operators, General Theory by N. Dunford, J. T. Schwartz

There is a non-commutative analogue for $L_p$ duality. Let $H$, $K$ be Hilbert spaces, then $$ S_p(H,K)^*= \begin{cases} S_{p/(p-1)}(K,H)\quad&\text{ if }\quad p\in(1,+\infty)\\ \mathcal{B}(K,H)\quad&\text{ if }\quad p=1\\ S_{\infty}(K,H)\quad&\text{ if }\quad p=\infty \end{cases} $$ Where $S_p(H,K)$ is $p$-th Shatten class operators.

Feel free to add some other dualities here.

  • $\begingroup$ Nice and interesting answer, however this is not exactly what I was looking for. Unfortunately my question was too long and unnecessarily complicated to be clear. What I am interested in is the mapping $$[w \in L^p]\to\left[ L_v\in (L^p)^\star,\ v=\frac{\lvert w \rvert^{p-1}\text{signum}(w)}{\left( \int\lvert w\rvert^p\, dx\right)^{\frac{p-2}{p}}}\right]$$ $\endgroup$ Oct 31, 2012 at 14:30
  • $\begingroup$ Ok, I can provide daulization mappings for you in each case. Will it be enough for you? $\endgroup$
    – Norbert
    Oct 31, 2012 at 14:31
  • $\begingroup$ ... which is a kind of nonlinear version of the Riesz isomorphism of a Hilbert space and its dual. I was wondering: what property should we require on a Banach space $X$ so that it possesses such a mapping $X\to X^\star$ and in which cases is this mapping explicitly computable. $\endgroup$ Oct 31, 2012 at 14:35
  • $\begingroup$ Also, sorry for not being clear. $\endgroup$ Oct 31, 2012 at 14:36
  • $\begingroup$ @GiuseppeNegro You are asking too much, and your question is too vague. For example there spaces for which there is no explicit description of their duals (for example $\ell_\infty$). Nonexistence here is strongly correlated with axiom of choice. Another reason is that we don't have rigorous definitions of computable dual space. For example descrition of $(L_\infty)^*$ as $\mathrm{ba}$ is computable? At first sight yes, but in fact no one can say what finitely additive measure is. $\endgroup$
    – Norbert
    Oct 31, 2012 at 14:47

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