A bounded functional is a measurement: it assigns to each vector a number, linearly and continuously. In \(\mathbb R^n\) the coordinate functionals are enough to identify a vector, to compute its norm, and to tell whether it lies in a given subspace. Does an arbitrary normed space have enough bounded functionals to do the same? The question is not idle. Nothing in the definition of a normed space mentions functionals, and for spaces that fail the triangle inequality, such as the analogue of \(L^p\) with \(0<p<1\) , the only continuous linear functional is zero.

Here is the difficulty in its simplest form. Given a nonzero vector \(x_0\) , we would like a functional \(F\) of norm \(1\) with \(F(x_0)=\lVert x_0\rVert\) . On the line through \(x_0\) there is no choice: \(F(tx_0)=t\lVert x_0\rVert\) . The problem is to extend this functional from the line to the whole space without increasing its norm. In an inner product space one would compose with the orthogonal projection onto the line. A general normed space has no orthogonal projections, and the Hahn–Banach theorem is what replaces them. It is the one major theorem of this book that does not need completeness; it needs convexity.

3.1 Functionals that can be written down

Before the abstract theorem it helps to see what dual spaces look like when they can be computed. For \(1\le p<\infty\) let \(q\) be the conjugate exponent, \(1/p+1/q=1\) , with \(q=\infty\) when \(p=1\) .

So \((\ell^p)^*\) is \(\ell^q\) and \(c_0^*\) is \(\ell^1\) , as normed spaces. The proof used density of \(c_{00}\) , which fails in \(\ell^\infty\) , and indeed \((\ell^\infty)^*\) is strictly larger than \(\ell^1\) (Exercise 12). For function spaces the corresponding fact, that \((L^p)^*\) is \(L^q\) for \(1\le p<\infty\) on a \(\sigma\) -finite measure space, is proved in Measure Theory by means of the Radon–Nikodym theorem; we shall quote it when needed.

3.2 Extending by one dimension

The proof of the Hahn–Banach theorem has an analytic part and a set-theoretic part. The analytic part is the extension of a functional to a subspace one dimension larger, and it is cleanest when the bound on the functional is expressed not by a norm but by something more flexible.

Every norm and every seminorm is sublinear, and so is \(x\mapsto C\lVert x\rVert\) . A sublinear functional need not be symmetric or nonnegative: on the bounded real sequences, \(p(x)=\limsup_nx_n\) is sublinear. This extra generality costs nothing in the proof and is what later yields separation of convex sets.

The lemma is constructive: it tells us precisely which values are allowed, and so it tells us when the extension is unique.

Example 3.2: The interval of extensions, computed

In \(\mathbb R^2\) let \(M\) be the first coordinate axis and \(f(t,0)=t\) . For each of the norms \(\lVert\cdot\rVert_1\) , \(\lVert\cdot\rVert_2\) , \(\lVert\cdot\rVert_\infty\) on \(\mathbb R^2\) the functional \(f\) has norm \(1\) on \(M\) , and an extension \(F(x)=x_1+cx_2\) has norm \(1\) exactly when \(F\le p\) for \(p\) equal to the norm. Take \(x_0=e_2\) and compute the interval of Lemma 3.3, where \(m=(t,0)\) .

For \(\lVert\cdot\rVert_1\) : \(p(m+x_0)-f(m)=\lvert t\rvert+1-t\) , whose infimum over \(t\) is \(b=1\) , attained for all \(t\ge 0\) ; and \(-p(-m-x_0)-f(m)=-\lvert t\rvert-1-t\) , whose supremum is \(a=-1\) , attained for \(t\le 0\) . Every \(c\in[-1,1]\) is allowed. Indeed the norm of \(F\) on \((\mathbb R^2,\lVert\cdot\rVert_1)\) is \(\max\{1,\lvert c\rvert\}\) by Theorem 3.1 in two dimensions.

For \(\lVert\cdot\rVert_2\) : \(b=\inf_t(\sqrt{t^2+1}-t)=0\) , approached as \(t\to\infty\) , and \(a=\sup_t(-\sqrt{t^2+1}-t)=0\) , approached as \(t\to-\infty\) . Only \(c=0\) is allowed. For \(\lVert\cdot\rVert_\infty\) : \(b=\inf_t(\max\{\lvert t\rvert,1\}-t)=0\) , attained for \(t\ge 1\) , and \(a=0\) likewise; again only \(c=0\) .

So the norm-preserving extension is unique for two of the norms and there is a whole interval of them for the third.

Figure 3.1 translates the lemma into geometry. The set where a norm-one functional \(F\) equals \(1\) is a line that touches the unit ball without cutting it, because \(F\le 1\) on the ball. Counting extensions is counting supporting lines at the point \(e_1\) .

3.3 The Hahn–Banach theorem

Repeating the one-step extension finitely many times handles finite-dimensional spaces. To reach an arbitrary space the standard tool is Zorn’s lemma: if every totally ordered subset of a nonempty partially ordered set has an upper bound, then the set has a maximal element. Zorn’s lemma is equivalent to the axiom of choice, and we take it as part of the set-theoretic background.

Two remarks on the hypotheses. Completeness plays no part: \(X\) may be any normed space, and \(M\) any subspace, closed or not. And the theorem is about scalar-valued maps. For operators into a general Banach space, norm-preserving extension can fail and even bounded extension can fail: it is a theorem of Phillips that the identity map of \(c_0\) has no bounded linear extension from \(\ell^\infty\) to \(c_0\) .

When \(X\) is separable, Zorn’s lemma can be avoided. Extend \(f\) one dimension at a time along a countable dense sequence, which defines a bounded functional on a dense subspace, and then extend to the whole space by Theorem 2.6 (Exercise 7).

3.4 What the theorem buys

The corollaries are used far more often than the theorem. Each one constructs a functional with prescribed behaviour on a small subspace and then extends it.

The formula is dual to the definition of the operator norm: \(\lVert F\rVert\) is a supremum over vectors, which may not be attained (Example 3.1), whereas \(\lVert x\rVert\) is a supremum over functionals, which always is.

The displayed formula turns a minimization over \(M\) into a maximization over functionals. A minimizer need not exist, but a maximizing functional always does, and any admissible \(F\) gives a lower bound on the distance: exhibiting an approximant and a functional with matching values proves optimality with no search.

3.5 The second dual and the completion

Since \(X^*\) is a normed space it has a dual of its own, the second dual \(X^{\ast\ast}=(X^*)^*\) . Each \(x\in X\) defines an element \(Jx\) of \(X^{\ast\ast}\) by evaluation:

Clearly \(J\) is linear and \(\lvert(Jx)(f)\rvert\le\lVert f\rVert\lVert x\rVert\) , so \(\lVert Jx\rVert\le\lVert x\rVert\) . Corollary 3.6 gives a functional of norm \(1\) with \(f(x)=\lVert x\rVert\) , hence \(\lVert Jx\rVert=\lVert x\rVert\) : the canonical embedding \(J:X\to X^{\ast\ast}\) is a linear isometry.

A reflexive space is isometric to the dual space \(X^{\ast\ast}\) and is therefore complete, by Theorem 2.5. For \(1<p<\infty\) the space \(\ell^p\) is reflexive: by Theorem 3.1 a functional on \((\ell^p)^*=\ell^q\) is given by pairing with some \(x\in\ell^p\) , and that functional is exactly \(Jx\) . The space \(c_0\) is not reflexive. Its dual is \(\ell^1\) and its second dual is \((\ell^1)^*=\ell^\infty\) ; under these identifications \(J\) is the inclusion of \(c_0\) into \(\ell^\infty\) , which is not onto. Likewise \(\ell^1\) is not reflexive, because \((\ell^\infty)^*\) contains functionals that do not come from \(\ell^1\) .

The embedding \(J\) also delivers the completion promised in Chapter 1.

3.6 Separating convex sets

The sublinear form of the theorem has a geometric reading. A nonzero \(F\in X^*\) and a real number \(\gamma\) determine the closed hyperplane \(\{x:\operatorname{Re}F(x)=\gamma\}\) , which divides \(X\) into two half-spaces. To separate two sets is to find a hyperplane with one set on each side. In this section \(X\) is a real normed space; for complex scalars one applies the results to the real part of a functional.

The link between convex sets and sublinear functionals is the gauge. If \(C\) is an open convex set containing \(0\) , define

Because \(C\) contains a ball \(B(0,r)\) , we have \(p_C(x)\le\lVert x\rVert/r\) . Convexity of \(C\) makes \(p_C\) subadditive, by the same convex-combination identity as in Exercise 3 of Chapter 1Chapter 1, and \(p_C\) is positively homogeneous by construction, so \(p_C\) is sublinear. Finally \(C=\{x:p_C(x)<1\}\) : if \(x\in C\) then, \(C\) being open, \((1+\varepsilon)x\in C\) for small \(\varepsilon\) , so \(p_C(x)<1\) ; and if \(p_C(x)<1\) then \(x/t\in C\) for some \(t<1\) , and \(x=t(x/t)+(1-t)0\in C\) .

Exercises

1. Let \(\varphi(x)=\sum_kx_k/k\) . Find \(\lVert\varphi\rVert\) as a functional on \(\ell^1\) and on \(\ell^2\) , say at which unit vectors the norm is attained, and explain why the formula does not define a bounded functional on \(c_0\) .

   Answer

   Here \(y=(1/k)\) . On \(\ell^1\) : \(\lVert\varphi\rVert=\lVert y\rVert_\infty=1\) , attained at \(e_1\) . On \(\ell^2\) : \(\lVert\varphi\rVert=\lVert y\rVert_2=\pi/\sqrt6\approx 1.2825\) , attained at \(y/\lVert y\rVert_2\) . On \(c_0\) a bounded functional must come from some \(y\in\ell^1\) by Theorem 3.1, and \((1/k)\notin\ell^1\) ; concretely the series diverges at \(x=(1/\ln(k+1))\in c_0\) .

2. Let \(\varphi(x)=x_1-2x_2+3x_3\) on \(\mathbb R^3\) . Compute \(\lVert\varphi\rVert\) when \(\mathbb R^3\) carries \(\lVert\cdot\rVert_\infty\) , \(\lVert\cdot\rVert_1\) , and \(\lVert\cdot\rVert_2\) , giving a unit vector that attains the norm in each case.

   Answer

   With \(\lVert\cdot\rVert_\infty\) : \(\lVert\varphi\rVert=1+2+3=6\) , attained at \((1,-1,1)\) . With \(\lVert\cdot\rVert_1\) : \(\lVert\varphi\rVert=\max\{1,2,3\}=3\) , attained at \(e_3\) . With \(\lVert\cdot\rVert_2\) : \(\lVert\varphi\rVert=\sqrt{14}\approx 3.742\) , attained at \((1,-2,3)/\sqrt{14}\) .

3. Write out the proof that every bounded functional \(\varphi\) on \(c_0\) has the form \(\varphi(x)=\sum_kx_ky_k\) with \(y\in\ell^1\) and \(\lVert\varphi\rVert=\lVert y\rVert_1\) , including the step that shows the series for \(\varphi(x)\) converges to \(\varphi(x)\) .

   Solution

   Let \(\varphi\in c_0^*\) and \(y_k=\varphi(e_k)\) . For each \(n\) define \(x^{(n)}\in c_{00}\) by \(x^{(n)}_k=\overline{y_k}/\lvert y_k\rvert\) if \(k\le n\) and \(y_k\ne 0\) , and \(x^{(n)}_k=0\) otherwise. Then \(\lVert x^{(n)}\rVert_\infty\le 1\) and by linearity

   \[ \varphi(x^{(n)})=\sum_{k\le n}x^{(n)}_k\,y_k=\sum_{k\le n}\lvert y_k\rvert . \]

   Hence \(\sum_{k\le n}\lvert y_k\rvert\le\lVert\varphi\rVert\) for every \(n\) , so \(y\in\ell^1\) and \(\lVert y\rVert_1\le\lVert\varphi\rVert\) .

   Now let \(x\in c_0\) and let \(x^{[K]}=(x_1,\dots,x_K,0,\dots)=\sum_{k\le K}x_ke_k\) . Linearity gives \(\varphi(x^{[K]})=\sum_{k\le K}x_ky_k\) . Since \(x\in c_0\) , \(\lVert x-x^{[K]}\rVert_\infty=\sup_{k>K}\lvert x_k\rvert\to 0\) , and \(\varphi\) is continuous, so \(\varphi(x^{[K]})\to\varphi(x)\) . Therefore the series \(\sum_kx_ky_k\) converges and its sum is \(\varphi(x)\) . It converges absolutely, in fact, since \(\lvert x_ky_k\rvert\le\lVert x\rVert_\infty\lvert y_k\rvert\) . The same estimate gives \(\lvert\varphi(x)\rvert\le\lVert x\rVert_\infty\lVert y\rVert_1\) , hence \(\lVert\varphi\rVert\le\lVert y\rVert_1\) . Together with the first paragraph, \(\lVert\varphi\rVert=\lVert y\rVert_1\) . The step that would fail for \(\ell^\infty\) is \(\lVert x-x^{[K]}\rVert_\infty\to 0\) , which is exactly the statement \(x\in c_0\) .

4. Show that \(\varphi(f)=\int_0^{1/2}f-\int_{1/2}^1f\) is a bounded functional on \(C[0,1]\) with the supremum norm, that \(\lVert\varphi\rVert=1\) , and that there is no \(f\) with \(\lVert f\rVert_\infty=1\) and \(\lvert\varphi(f)\rvert=1\) .

   Solution

   For any \(f\in C[0,1]\) ,

   \[ \lvert\varphi(f)\rvert\le\int_0^{1/2}\lvert f\rvert+\int_{1/2}^1\lvert f\rvert=\int_0^1\lvert f\rvert\le\lVert f\rVert_\infty , \]

   so \(\varphi\) is bounded and \(\lVert\varphi\rVert\le 1\) . For the lower bound let \(f_n\) equal \(1\) on \([0,\tfrac12-\tfrac1n]\) , equal \(-1\) on \([\tfrac12+\tfrac1n,1]\) , and be linear in between, passing through \(0\) at \(x=\tfrac12\) . Then \(\lVert f_n\rVert_\infty=1\) , and on each half the integral falls short of \(\tfrac12\) by the area of a triangle of base \(\tfrac1n\) and height \(1\) :

   \[ \varphi(f_n)=\Bigl(\tfrac12-\tfrac1{2n}\Bigr)+\Bigl(\tfrac12-\tfrac1{2n}\Bigr)=1-\tfrac1n . \]

   Hence \(\lVert\varphi\rVert\ge 1-\tfrac1n\) for all \(n\) , and \(\lVert\varphi\rVert=1\) .

   Suppose \(\lVert f\rVert_\infty=1\) and \(\lvert\varphi(f)\rvert=1\) . Multiplying \(f\) by a scalar of modulus \(1\) we may assume \(\varphi(f)=1\) . Let \(u=\operatorname{Re}f\) , so \(-1\le u\le 1\) . Taking real parts, \(\int_0^{1/2}u-\int_{1/2}^1u=1\) , which can be rewritten as

   \[ \int_0^{1/2}(1-u)+\int_{1/2}^1(1+u)=0 . \]

   Both integrands are continuous and nonnegative, so both vanish identically: \(u=1\) on \([0,\tfrac12]\) and \(u=-1\) on \([\tfrac12,1]\) . At \(x=\tfrac12\) this demands \(u(\tfrac12)=1\) and \(u(\tfrac12)=-1\) , a contradiction. The extremal function would be a step, and steps are not in \(C[0,1]\) .

5. In \(\mathbb R^2\) let \(M\) be the span of \((1,1)\) and \(f(t,t)=t\) . Find \(\lVert f\rVert\) and all norm-preserving extensions \(F(x)=\alpha x_1+\beta x_2\) of \(f\) when \(\mathbb R^2\) carries \(\lVert\cdot\rVert_\infty\) , and again when it carries \(\lVert\cdot\rVert_1\) .

   Solution

   Supremum norm. On \(M\) , \(\lVert(t,t)\rVert_\infty=\lvert t\rvert=\lvert f(t,t)\rvert\) , so \(\lVert f\rVert=1\) . An extension \(F(x)=\alpha x_1+\beta x_2\) must satisfy \(F(1,1)=1\) , that is \(\alpha+\beta=1\) . The norm of \(F\) on \((\mathbb R^2,\lVert\cdot\rVert_\infty)\) is the row sum \(\lvert\alpha\rvert+\lvert\beta\rvert\) (Proposition 2.3). We need \(\lvert\alpha\rvert+\lvert\beta\rvert=1=\alpha+\beta\) , which holds exactly when \(\alpha\ge 0\) and \(\beta\ge 0\) . The norm-preserving extensions are

   \[ F(x)=\alpha x_1+(1-\alpha)x_2,\qquad 0\le\alpha\le 1 , \]

   a whole segment of them.

   The \(1\) -norm. Now \(\lVert(t,t)\rVert_1=2\lvert t\rvert\) , so \(\lVert f\rVert=\tfrac12\) . The norm of \(F\) on \((\mathbb R^2,\lVert\cdot\rVert_1)\) is \(\max\{\lvert\alpha\rvert,\lvert\beta\rvert\}\) . We need \(\alpha+\beta=1\) and \(\max\{\lvert\alpha\rvert,\lvert\beta\rvert\}=\tfrac12\) ; then \(\alpha\le\tfrac12\) and \(\beta\le\tfrac12\) with sum \(1\) force \(\alpha=\beta=\tfrac12\) . The extension is unique.

   The geometry explains the difference. The point \((1,1)\) is a corner of the unit ball of \(\lVert\cdot\rVert_\infty\) , where many supporting lines exist, whereas \((\tfrac12,\tfrac12)\) lies in the interior of an edge of the unit ball of \(\lVert\cdot\rVert_1\) , where the edge itself is the only supporting line. This is Example 3.2 with the roles of the two norms exchanged.

6. On the real space \(\ell^\infty\) let \(p(x)=\limsup_n\frac1n(x_1+\dots+x_n)\) . Show that \(p\) is sublinear, and use Theorem 3.4 to prove that there is a linear functional \(\Lambda\) on \(\ell^\infty\) with \(\liminf_nx_n\le\Lambda(x)\le\limsup_nx_n\) and \(\Lambda(Lx)=\Lambda(x)\) for all \(x\) , where \(L\) is the left shift; then compute \(\Lambda(0,1,0,1,\dots)\) .

   Solution

   Write \(\sigma_n(x)=\frac1n(x_1+\dots+x_n)\) . Each \(\sigma_n\) is linear, so \(\sigma_n(x+y)=\sigma_n(x)+\sigma_n(y)\) and the limit superior of a sum is at most the sum of the limits superior; also \(\sigma_n(tx)=t\sigma_n(x)\) and \(\limsup\) commutes with multiplication by \(t\ge 0\) . Hence \(p\) is sublinear, and it is finite because \(\lvert\sigma_n(x)\rvert\le\lVert x\rVert_\infty\) .

   Apply Theorem 3.4 to the subspace \(\{0\}\) and the zero functional, which is dominated by \(p\) there. We get a linear \(\Lambda\) on \(\ell^\infty\) with \(\Lambda\le p\) . Averages of a sequence cannot exceed its limit superior in the limit: if \(x_n\le s+\varepsilon\) for \(n\ge N\) , then \(\sigma_n(x)\le\frac1n\sum_{k<N}x_k+(s+\varepsilon)\) up to a term tending to \(0\) , so \(p(x)\le\limsup_nx_n\) . Thus \(\Lambda(x)\le\limsup_nx_n\) . Replacing \(x\) by \(-x\) gives \(-\Lambda(x)\le\limsup_n(-x_n)=-\liminf_nx_n\) , which is the lower bound.

   For shift invariance, the averages of \(x-Lx\) telescope:

   \[ \sigma_n(x-Lx)=\frac{x_1-x_{n+1}}{n}\to 0 , \]

   so \(p(x-Lx)=0\) and likewise \(p(Lx-x)=0\) . Therefore \(\Lambda(x-Lx)\le 0\) and \(-\Lambda(x-Lx)\le 0\) , that is, \(\Lambda(Lx)=\Lambda(x)\) .

   For \(x=(0,1,0,1,\dots)\) the averages \(\sigma_n(x)\) equal \(\tfrac12\) for even \(n\) and \(\tfrac12-\tfrac1{2n}\) for odd \(n\) , so \(p(x)=\tfrac12\) and \(p(-x)=-\tfrac12\) . Hence \(\Lambda(x)\le\tfrac12\) and \(-\Lambda(x)\le-\tfrac12\) : \(\Lambda(0,1,0,1,\dots)=\tfrac12\) . Such a functional is called a Banach limit. It assigns a “limit” to every bounded sequence, agrees with the ordinary limit when that exists, and is not given by any element of \(\ell^1\) .

7. Let \(X\) be a separable normed space, \(M\) a subspace, and \(f\in M^*\) . Prove Theorem 3.5 for real scalars without Zorn’s lemma, using Lemma 3.3 countably many times and then Theorem 2.6.

   Hint

   Let \(\{z_1,z_2,\dots\}\) be dense in \(X\) and put \(M_0=M\) , \(M_n=M_{n-1}+\mathbb Rz_n\) . Extend \(f\) from \(M_{n-1}\) to \(M_n\) by Lemma 3.3 with \(p(x)=\lVert f\rVert\lVert x\rVert\) (if \(z_n\in M_{n-1}\) do nothing). The extensions are consistent, so they define a functional of norm \(\lVert f\rVert\) on \(\bigcup_nM_n\) , which is dense. Finish with Theorem 2.6, whose target \(\mathbb R\) is complete.

8. Prove that if the dual \(X^*\) of a normed space is separable then \(X\) is separable, by choosing for each member \(f_n\) of a dense sequence in \(X^*\) a unit vector \(x_n\) with \(\lvert f_n(x_n)\rvert\ge\tfrac12\lVert f_n\rVert\) and applying Corollary 3.8 to the span of the \(x_n\) .

   Hint

   Let \(N\) be the closed span of the \(x_n\) ; it is separable, since rational combinations of the \(x_n\) are dense in it. If \(N\ne X\) , Corollary 3.7 gives \(F\) with \(\lVert F\rVert=1\) vanishing on \(N\) . Choose \(f_n\) with \(\lVert F-f_n\rVert<\tfrac14\) ; then \(\lVert f_n\rVert>\tfrac34\) but \(\lvert f_n(x_n)\rvert=\lvert(f_n-F)(x_n)\rvert<\tfrac14\) , contradicting \(\lvert f_n(x_n)\rvert\ge\tfrac12\lVert f_n\rVert\) .

9. Compute the distance from \(e_1\) to the hyperplane \(\{x:\sum_kx_k=0\}\) in \(\ell^1\) , and the distance from \(e_1\) to the hyperplane \(\{x:\sum_kx_k/2^k=0\}\) in \(\ell^2\) , using Corollary 3.8.

   Answer

   In \(\ell^1\) the functional is \(\varphi_y\) with \(y=(1,1,1,\dots)\) , of norm \(\lVert y\rVert_\infty=1\) , and its value at \(e_1\) is \(1\) ; the distance is \(1\) . In \(\ell^2\) the functional has norm \(1/\sqrt3\) (Example 3.1) and value \(\tfrac12\) at \(e_1\) ; the distance is \(\tfrac12\sqrt3\approx 0.866\) .

10. For \(f\in C[0,1]\) real-valued, prove that the distance in the supremum norm from \(f\) to the subspace of constant functions is \(\tfrac12(\max f-\min f)\) , by exhibiting a best constant and a functional of norm \(1\) that vanishes on constants.

    Solution

    Let \(m=\min f\) and \(M=\max f\) , attained at points \(s\) and \(t\) of \([0,1]\) , and let \(c_0=\tfrac12(m+M)\) . For every \(x\) , \(m-c_0\le f(x)-c_0\le M-c_0\) , and both extremes have absolute value \(\tfrac12(M-m)\) . So \(\lVert f-c_0\rVert_\infty=\tfrac12(M-m)\) , and the distance from \(f\) to the constants is at most \(\tfrac12(M-m)\) .

    For the lower bound define

    \[ F(g)=\tfrac12\bigl(g(t)-g(s)\bigr),\qquad g\in C[0,1]. \]

    Then \(\lvert F(g)\rvert\le\tfrac12(\lvert g(t)\rvert+\lvert g(s)\rvert)\le\lVert g\rVert_\infty\) , so \(\lVert F\rVert\le 1\) ; \(F\) vanishes on constant functions; and \(F(f)=\tfrac12(M-m)\) . By Corollary 3.8 the distance from \(f\) to the subspace of constants is at least \(\lvert F(f)\rvert=\tfrac12(M-m)\) . The two bounds coincide. If \(f\) is constant the formula gives \(0\) , correctly. For \(f(x)=x\) this gives distance \(\tfrac12\) , best constant \(\tfrac12\) , and certificate \(F(g)=\tfrac12(g(1)-g(0))\) . Notice that the proof never minimized over constants; one candidate and one functional were enough.

11. Prove that a reflexive normed space is a Banach space, and deduce that \(c_{00}\) with any norm that makes it a normed space is not reflexive if it is incomplete in that norm.

    Hint

    If \(J\) is onto, \(X\) is isometrically isomorphic to \(X^{\ast\ast}\) , which is complete by Theorem 2.5; use Exercise 17 of Chapter 2Chapter 2. An incomplete space therefore cannot be reflexive, whatever its norm.

12. Use Corollary 3.7 with \(M=c_0\) and \(x_0=(1,1,1,\dots)\) to produce a bounded functional on \(\ell^\infty\) that is not of the form \(\varphi_y\) for any \(y\in\ell^1\) .

    Solution

    The subspace \(c_0\) is closed in \(\ell^\infty\) , and by Exercise 5 of Chapter 1Chapter 1 the vector \(\mathbf 1=(1,1,1,\dots)\) satisfies \(\operatorname{dist}(\mathbf 1,c_0)=1\) : every \(y\in c_0\) has coordinates tending to \(0\) , so \(\lVert\mathbf 1-y\rVert_\infty\ge 1\) , and \(y=0\) gives equality. Corollary 3.7 yields \(F\in(\ell^\infty)^*\) with \(\lVert F\rVert=1\) , \(F=0\) on \(c_0\) , and \(F(\mathbf 1)=1\) .

    Suppose \(F=\varphi_y\) for some \(y\in\ell^1\) , meaning \(F(x)=\sum_kx_ky_k\) for all \(x\in\ell^\infty\) . Each \(e_k\) belongs to \(c_0\) , so \(y_k=F(e_k)=0\) for every \(k\) . Then \(y=0\) and \(F=0\) , contradicting \(F(\mathbf 1)=1\) . Hence \(F\) is a bounded functional on \(\ell^\infty\) that is not represented by any element of \(\ell^1\) . It follows that the canonical embedding of \(\ell^1\) into \((\ell^1)^{\ast\ast}=(\ell^\infty)^*\) is not onto, so \(\ell^1\) is not reflexive.

13. A student claims: “By the Hahn–Banach theorem the identity operator on \(c_0\) extends to a bounded operator from \(\ell^\infty\) to \(c_0\) of norm \(1\) .” Identify the hypothesis of Theorem 3.5 that does not hold, and say what is actually true.

    Answer

    Theorem 3.5 extends linear maps with values in the scalar field. The identity of \(c_0\) takes values in \(c_0\) . Extending each coordinate functional \(x\mapsto x_k\) separately is possible, but the extensions assemble into a map from \(\ell^\infty\) to \(\ell^\infty\) , not into \(c_0\) . In fact no bounded linear map from \(\ell^\infty\) to \(c_0\) restricts to the identity on \(c_0\) (Phillips).

14. In \(\mathbb R^2\) with the Euclidean norm, let \(K=\{x:x_2\ge x_1^2\}\) and \(x_0=(0,-1)\) . Find a functional \(F\) and numbers \(\gamma_1<\gamma_2\) that separate \(x_0\) from \(K\) strictly, and determine the largest possible gap \(\gamma_2-\gamma_1\) for \(F\) of norm \(1\) .

    Answer

    \(F(x)=x_2\) has norm \(1\) , \(F(x_0)=-1\) , and \(F\ge 0\) on \(K\) ; take \(\gamma_1=-1\) and \(\gamma_2=0\) . The gap of a norm-one separating functional cannot exceed \(\operatorname{dist}(x_0,K)\) , and the squared distance from \(x_0\) to a point \((s,s^2)\) of the parabola is \(s^2+(s^2+1)^2\ge 1\) , so the distance is \(1\) and the gap \(1\) is the largest possible.

15. Let \(A_1=\{x\in\mathbb R^2:x_2\le 0\}\) and \(B=\{x:x_1>0,\ x_1x_2\ge 1\}\) . Show that these closed convex sets are disjoint, that \(\inf\{\lVert a-b\rVert_2:a\in A_1,b\in B\}=0\) , and that no functional separates them strictly; which hypothesis of Theorem 3.10(b) is absent?

    Hint

    For \(b=(t,1/t)\in B\) the point \(a=(t,0)\in A_1\) has \(\lVert a-b\rVert_2=1/t\) . A strictly separating \(F\) of norm \(1\) would give \(\gamma_2-\gamma_1\le F(b)-F(a)\le\lVert b-a\rVert_2\) for all such pairs. Neither set is compact.

16. Let \(M\) be the span of the vectors \(e_k+e_{k+1}\) , \(k\ge 1\) . Use Corollary 3.8 and Theorem 3.1 to show that \(M\) is dense in \(\ell^2\) but not dense in \(\ell^1\) , and find the distance from \(e_1\) to \(M\) in \(\ell^1\) .

    Solution

    By Corollary 3.8, \(M\) is dense in a normed space \(X\) exactly when no nonzero \(F\in X^*\) vanishes on \(M\) , and \(F\) vanishes on \(M\) exactly when \(F(e_k+e_{k+1})=0\) for every \(k\) .

    In \(\ell^2\) . By Theorem 3.1 every \(F\in(\ell^2)^*\) is \(\varphi_y\) with \(y\in\ell^2\) , and \(F(e_k+e_{k+1})=y_k+y_{k+1}\) . If this vanishes for all \(k\) then \(y_k=(-1)^{k-1}y_1\) , so \(\lvert y_k\rvert=\lvert y_1\rvert\) for all \(k\) . A sequence of constant modulus is square-summable only if it is zero. Hence \(F=0\) and \(M\) is dense in \(\ell^2\) .

    In \(\ell^1\) . Now \(F=\varphi_y\) with \(y\in\ell^\infty\) , and the same computation allows the bounded sequence \(y=(1,-1,1,-1,\dots)\) . So \(\varphi_y\) is a nonzero bounded functional on \(\ell^1\) vanishing on \(M\) , and \(M\) is not dense in \(\ell^1\) .

    The distance. The functionals on \(\ell^1\) that vanish on \(M\) are the multiples \(\lambda\varphi_y\) of this one, with norm \(\lvert\lambda\rvert\lVert y\rVert_\infty=\lvert\lambda\rvert\) . By Corollary 3.8,

    \[ \operatorname{dist}(e_1,M)=\max_{\lvert\lambda\rvert\le 1}\lvert\lambda\varphi_y(e_1)\rvert=\lvert y_1\rvert=1 . \]

    So \(\operatorname{dist}(e_1,M)=1\) in \(\ell^1\) : the zero vector is already a best approximation to \(e_1\) from \(M\) . In \(\ell^2\) , by contrast, the distance from \(e_1\) to the span of the first \(n\) generators is \(1/\sqrt{n+1}\) , which tends to \(0\) .

17. Let \(x_1,\dots,x_n\) be linearly independent vectors in a normed space \(X\) and \(c_1,\dots,c_n\) scalars. Prove that there is \(F\in X^*\) with \(F(x_j)=c_j\) for each \(j\) , using Exercise 17 of Chapter 1Chapter 1 and Theorem 3.5.

    Hint

    On the finite-dimensional subspace \(N\) spanned by the \(x_j\) define \(f(\sum_ja_jx_j)=\sum_ja_jc_j\) . It is linear, and bounded by Exercise 17 of Chapter 1Chapter 1. Extend it by Theorem 3.5.

18. Let \(x=(1-\tfrac1k)_{k\ge 1}\in\ell^\infty\) . Show that \(\sup\{\lvert\varphi_y(x)\rvert:y\in\ell^1,\ \lVert y\rVert_1\le 1\}=1=\lVert x\rVert_\infty\) but that no \(y\in\ell^1\) with \(\lVert y\rVert_1\le 1\) attains the supremum, and reconcile this with Corollary 3.6, which asserts that a maximizing functional exists.

    Answer

    For \(\lVert y\rVert_1\le 1\) , \(\lvert\varphi_y(x)\rvert\le\sum_k(1-\tfrac1k)\lvert y_k\rvert<\sum_k\lvert y_k\rvert\le 1\) unless \(y=0\) , so the value \(1\) is never attained, while \(y=e_k\) gives \(1-\tfrac1k\to 1\) . Corollary 3.6 asserts a maximum over all of \((\ell^\infty)^*\) , which is larger than \(\ell^1\) : a norming functional for \(x\) exists but is not of the form \(\varphi_y\) . A Banach limit (Exercise 6) is one, since \(\Lambda(x)=\lim_k(1-\tfrac1k)=1\) .