The Canonical Embedding Into the Bidual

From Vectors to Functionals on the Dual

Every vector in a normed space determines a continuous linear functional on the space of continuous linear functionals: evaluate each functional at that vector. This construction is available for every normed space, and it preserves the norm exactly. It gives a canonical way to regard a space as part of its bidual, but it does not, by itself, say that every element of the bidual comes from a vector. That distinction is the point of the definition of reflexivity.

In this tutorial, \(X\) is a normed vector space over \(\mathbb{K}\), where \(\mathbb{K}\) is either \(\mathbb{R}\) or \(\mathbb{C}\). No measure space is involved in the abstract definitions. The continuous dual \(X^*\) is the vector space of continuous linear maps \(F:X\to\mathbb{K}\). Its norm is the operator norm. The bidual \(X^{**}\) is the continuous dual of \(X^*\), with its own operator norm. A Banach space is a normed space complete in its norm.

Definition: For \(x\in X\), define \(Jx:X^*\to\mathbb{K}\) by \((Jx)(F)=F(x)\). The map \(J:X\to X^{**}\) given by \(x\mapsto Jx\) is called the canonical embedding into the bidual, once it is verified that \(Jx\in X^{**}\) for every \(x\).

The operator norm of \(G\in X^{**}\) is \(\|G\|=\sup_{\|F\|\le1}|G(F)|\). Thus verifying that \(Jx\) belongs to \(X^{**}\) means showing that evaluation at \(x\) is a continuous linear functional on \(X^*\). The central calculation is the exact norm identity \(\|Jx\|=\|x\|\).

The Isometry Theorem

Theorem: For every normed space \(X\) over \(\mathbb{R}\) or \(\mathbb{C}\), the canonical map \(J:X\to X^{**}\) is linear and satisfies \(\|Jx\|=\|x\|\) for every \(x\in X\). In particular, \(J\) is injective and is an isometry onto its image. No completeness assumption on \(X\) is needed for this statement.

Proof. Fix \(x\in X\). For any \(F\in X^*\), linearity of \(F\) gives \((Jx)(F)=F(x)\), which is linear as a function of \(F\). The definition of the norm of \(F\) gives \(|F(x)|\le\|F\|\|x\|\). Therefore \(|(Jx)(F)|\le\|x\|\|F\|\), so \(Jx\) is continuous and \(\|Jx\|\le\|x\|\). For \(x=0\), this already proves \(\|Jx\|=\|x\|=0\). If \(x\ne0\), define \(F_0\) on the one-dimensional subspace spanned by \(x\) by \(F_0(\alpha x)=\alpha\|x\|\), for \(\alpha\in\mathbb{K}\). Then \(|F_0(\alpha x)|=|\alpha|\|x\|=\|\alpha x\|\), so \(\|F_0\|=1\). By the Hahn–Banach theorem, \(F_0\) extends to an element \(F\in X^*\) with \(\|F\|=1\). Consequently \(\|Jx\|\ge |(Jx)(F)|=|F(x)|=\|x\|\). Combining the two inequalities proves the norm identity. Finally, \(J\) is linear because for \(x,y\in X\), \(a,b\in\mathbb{K}\), and \(F\in X^*\), \((J(ax+by))(F)=F(ax+by)=aF(x)+bF(y)=(aJx+bJy)(F)\). If \(Jx=Jy\), then \(\|x-y\|=\|Jx-Jy\|=0\), so \(x=y\). Thus \(J\) is injective and preserves distances. \(\square\)

The only external result used in this proof is Hahn–Banach, in its norm-preserving extension form: a bounded linear functional on a subspace of a normed space extends to the whole space with the same norm. The proof of Hahn–Banach is a standard result of functional analysis and is not repeated here. The other inequalities and the linearity assertion follow directly from the definitions.

Worked Example: The Canonical Map for the Plane with the l1 Norm

Take \(X=\mathbb{R}^2\) with \(\|(x_1,x_2)\|_1=|x_1|+|x_2|\). We compute its dual and bidual norms and then write \(J\) explicitly.

Every linear functional on \(\mathbb{R}^2\) has the form \(F_a(x)=a_1x_1+a_2x_2\), where \(a=(a_1,a_2)\). For \(\|x\|_1\le1\),

|F_a(x)|\le |a_1||x_1|+|a_2||x_2| \le \max(|a_1|,|a_2|)(|x_1|+|x_2|) \le \max(|a_1|,|a_2|).

If \(a\ne0\), choosing \(x=(1,0)\) when \(|a_1|\ge|a_2|\), or \(x=(0,1)\) when \(|a_2|>|a_1|\), attains the larger coordinate in absolute value. If \(a=0\), the norm is zero. Thus \(\|F_a\|=\max(|a_1|,|a_2|)\), so \(X^*\) is identified with \(\mathbb{R}^2\) carrying the maximum norm.

A linear functional on this dual has the form \(G_b(F_a)=b_1a_1+b_2a_2\). Its norm is \(|b_1|+|b_2|\): the upper bound follows from \(|b_1a_1+b_2a_2|\le(|b_1|+|b_2|)\max(|a_1|,|a_2|)\), and equality is attained by choosing each \(a_i\) to have absolute value one and the sign of \(b_i\) when \(b_i\ne0\). For \(x=(x_1,x_2)\), \(Jx\) acts by \((Jx)(F_a)=a_1x_1+a_2x_2\). Hence \(Jx=G_x\) and \(\|Jx\|=|x_1|+|x_2|=\|x\|_1\).

In fact, this particular \(J\) is onto: every \(G_b\in X^{**}\) equals \(Jb\). Check: for \(x=(1,-2)\), \(\|x\|_1=|1|+|-2|=3\), and the bidual norm of \(Jx\) is \(|1|+|-2|=3\). Taking \(a=(1,-1)\), whose dual norm is \(\max(1,1)=1\), gives \((Jx)(F_a)=1\cdot1+(-1)\cdot(-2)=3\), attaining that norm.

Worked Example: Evaluation on a Dual of L1

Let the measure space be \(([0,1],\mathcal{L},m)\), where \(m\) is Lebesgue measure, and take the real Banach space \(X=L^1(m)\). As established by the duality theorem for \(L^1\), its dual is represented by \(L^\infty(m)\): a function \(g\in L^\infty(m)\) gives \(F_g(f)=\int_0^1 fg\,dm\), with \(\|F_g\|=\|g\|_\infty\). Elements of these spaces are equivalence classes modulo equality \(m\)-a.e.

Let \(E=[0,\tfrac12]\) and \(x=\chi_E\). The endpoint convention does not affect its Lebesgue measure: \(m(E)=\tfrac12\). Therefore \(\|x\|_1=\int_0^1\chi_E\,dm=\tfrac12\). The canonical image acts on a dual functional by

(Jx)(F_g)=F_g(x)=\int_0^1\chi_Eg\,dm=\int_Eg\,dm.

For the specific choice \(g=\chi_E\), \(\|F_g\|=\|\chi_E\|_\infty=1\), because \(E\) has positive measure and \(\chi_E=1\) there. Then \((Jx)(F_g)=\int_E1\,dm=\tfrac12\). For every \(g\) with \(\|F_g\|\le1\), equivalently \(\|g\|_\infty\le1\), we have \(|\int_Eg\,dm|\le\int_E|g|\,dm\le m(E)=\tfrac12\). Thus the supremum defining \(\|Jx\|\) is exactly \(\tfrac12=\|x\|_1\), checked directly in this instance.

Worked Example: The Map on l1 Sequences

Use counting measure on \(\mathbb{N}=\{1,2,\ldots\}\), a sigma-finite measure space. Then \(L^1\) is the sequence space \(\ell^1\), and the duality result for \(L^1\) identifies \((\ell^1)^*\) with \(\ell^\infty\): a bounded sequence \(a=(a_n)\) acts by \(F_a(x)=\sum_{n=1}^\infty a_nx_n\), with \(\|F_a\|=\|a\|_\infty\).

Take \(x=(\tfrac12,\tfrac14,0,0,\ldots)\). Its norm is \(\|x\|_1=\tfrac12+\tfrac14=\tfrac34\). The canonical image is the functional on \(\ell^\infty\) given by

(Jx)(F_a)=F_a(x)=\frac12a_1+\frac14a_2.

For \(\|a\|_\infty\le1\), the absolute value is at most \(\tfrac12|a_1|+\tfrac14|a_2|\le\tfrac34\). The constant sequence \(a=(1,1,\ldots)\) has \(\|a\|_\infty=1\) and gives \((Jx)(F_a)=\tfrac12+\tfrac14=\tfrac34\). Thus \(\|Jx\|=\tfrac34\), as required. This calculation specifies the canonical image of \(x\); it does not claim that every functional on \(\ell^\infty\) has this form.

Reflexivity Means Surjectivity

Definition: A normed space \(X\) is reflexive if its canonical map \(J:X\to X^{**}\) is surjective. Equivalently, for every \(G\in X^{**}\), there exists \(x\in X\) such that \(G(F)=F(x)\) for every \(F\in X^*\).

This is a definition, not a consequence of the isometry theorem. The theorem proves that \(J\) is one-to-one and norm-preserving, so each vector has a unique image in \(X^{**}\). Reflexivity asks the additional question whether those images exhaust the entire bidual. In the plane example, we checked that every bidual functional is evaluation at a vector, so that space is reflexive. In general, the isometry proof supplies no argument for surjectivity; it only describes the norm and the image of \(J\).

For a Banach space \(X\), the bidual is itself a Banach space. Indeed, the dual of any normed space \(Y\) is complete in its operator norm. To see the relevant argument, let \((G_n)\) be Cauchy in \(Y^*\). For each \(y\in Y\), \(|G_n(y)-G_m(y)|\le\|G_n-G_m\|\|y\|\), so \((G_n(y))\) converges to a scalar. Define \(G(y)=\lim_nG_n(y)\). Taking limits in the linearity identities shows that \(G\) is linear. A Cauchy sequence of norms is bounded, say \(\|G_n\|\le C\); therefore \(|G(y)|\le C\|y\|\), so \(G\in Y^*\). Given \(\varepsilon>0\), choose \(N\) so that \(\|G_n-G_m\|\le\varepsilon\) for \(n,m\ge N\). Fixing \(n\ge N\), letting \(m\to\infty\), and then taking the supremum over \(\|y\|\le1\) gives \(\|G_n-G\|\le\varepsilon\). Thus \(G_n\to G\) in \(Y^*\). Applying this fact to \(Y=X^*\) shows that \(X^{**}\) is complete. Since \(J\) is isometric and \(X\) is complete, its image is also complete and hence closed in \(X^{**}\); closedness still does not imply that the image is all of \(X^{**}\).

A Common Norm Mistake

For \(X=\mathbb{R}^2\) with the \(l^1\) norm, it is incorrect to use the same coordinate formula for the dual norm and the bidual norm. For example, with \(x=(1,-2)\), one might wrongly calculate \(\|Jx\|\) as \(\max(|1|,|-2|)=2\), copying the dual norm formula. But \(Jx\) is a functional on the dual, whose norm is the operator norm on the dual's maximum-norm unit ball. At \(a=(1,-1)\), that ball contains \(a\), and \(|a_1x_1+a_2x_2|=|1+2|=3\). The correct bidual norm is \(|1|+|-2|=3\), exactly the original \(l^1\) norm. Always identify the space on which the functional acts before selecting its unit ball.

Check Your Understanding

Use the definitions and norm identities above to answer the following.

For \(x=0\), what is \(Jx\), and why does the isometry identity hold without applying Hahn–Banach?
Explain why the estimate \(|F(x)|\le\|F\|\|x\|\) proves that \(Jx\) belongs to \(X^{**}\).
In \(\mathbb{R}^2\) with the \(l^1\) norm, compute \(\|J(2,-1)\|\) and give a dual unit-ball element that attains the norm.
State the extra condition, beyond being isometric and injective, that makes a normed space reflexive.
Why does showing that \(J(X)\) is closed in \(X^{**}\) not, by itself, prove reflexivity?