Representations of dual spaces

TL;DR

This paper introduces a nonlinear approach to representing dual spaces of Banach spaces, providing simplified proofs of classical theorems and extending to Orlicz spaces, also deriving a version of the Helly-Hahn-Banach theorem.

Contribution

It presents a new nonlinear representation method for duals of Banach spaces, simplifying proofs and extending to Orlicz spaces, with an additional result on the Helly-Hahn-Banach theorem.

Findings

Simplified proofs of $H'=H$ and $(L^p)'=L^q$ theorems

Extension of representation to Orlicz spaces

Derivation of a version of the Helly-Hahn-Banach theorem

Abstract

We give a nonlinear representation of the duals for a class of Banach spaces. This leads to classroom-friendly proofs of the classical representation theorems $H'=H$ and $(L^p)'=L^q$. Our proofs extend to a family of Orlicz spaces, and yield as an unexpected byproduct a version of the Helly-Hahn-Banach theorem.

Equations

$$\text{if}\Vert x\Vert =\Vert y\Vert =1\text{and}\Vert \frac{x+y}{2}\Vert \ge 1-\delta ,\text{then}\Vert x-y\Vert \le \epsilon .$$

$$F(x+tu)=F(x)+t{F}^{\prime }(x)u+o(t)\text{as}t\to 0.$$

$$\frac{N(x+tz)-N(x)}{t}\le \frac{N(tz)}{t}=N(z)=1,$$

$$\frac{d}{dt}\Vert x+t(z-y(z)x)\Vert {|}_{t=0}=0,\text{i.e.,}{N}^{\prime }(x)(z-y(z)x)=0$$

$$t\mapsto N\left(a+tu\right)-t{N}^{\prime }(a)u$$

$$\Vert N\left(a+u\right)-N(a)-N^{\prime }(a)u\Vert \le \epsilon \Vert u\Vert .$$

$$y(x)=1,\text{and}y(z)<1\text{for all other}z\in S.$$

$$\Vert \frac{x_n+x_m}{2}\Vert \ge y\left(\frac{x_n+x_m}{2}\right)>1-\delta$$

$$\left|y(x_n)-1\right|=\left|y(x_n)-y_n(x_n)\right|\le \Vert y_n-y\Vert \to 0,$$

$$\Phi (x)u:=⟨x,u⟩,x,u\in X$$

$$\Vert x+tu\Vert =\sqrt{{\Vert x\Vert }^2+2t⟨x,u⟩+t^2{\Vert u\Vert }^2}=1+t⟨x,u⟩+o(t)\text{as}t\to 0.$$

$$\Phi (g)f:=\int gfdx,g\in L^q,f\in L^p$$

$$N^{\prime }(h_0)u=\int {\left|h_0\right|}^{\frac{p}{q}}(\mathrm{sign}{h}_0)udx.$$

$$\Phi (g):=⟨g,f⟩\text{and}\Phi (g)f:=\int \overline{g}fdx$$

$$\mathrm{Re}\phi (h)=\int g_{1}hdx\text{and}\mathrm{Re}\phi (ih)=\int g_{2}hdx$$

$$\underset{u\to 0}{\lim }\frac{P(u)}{u}=0\text{and}\underset{u\to \infty }{\lim }\frac{P(u)}{u}=\infty ,$$

$$Q(v):=\underset{u\in \mathbb{R}}{\sup }\left\{u\left|v\right|-P(u)\right\}.$$

$$f\in L^P$$

$$g\in L^Q$$

$${\Vert f\Vert }_{P,\text{Lux}}:=\inf \left\{k>0:\int P(k^{-1}f)dx\le 1\right\}$$

$${\Vert g\Vert }_{Q,\text{Orl}}:=\sup \left\{\int fgdx:\int P(f)dx\le 1\right\},$$

$$\int \left|fg\right|dx\le {\Vert f\Vert }_{P,\text{Lux}}{\Vert g\Vert }_{Q,\text{Orl}}$$

$$\Phi (g)(f):=\int fgdx$$

$$\left|u-v\right|\ge \epsilon (\left|u\right|+\left|v\right|)⟹\frac{P(u)+P(v)}{2}-P\left(\frac{u+v}{2}\right)\ge \rho (\epsilon )\frac{P(u)+P(v)}{2}.$$

$$K:=\left\{(u,v)\in {\mathbb{R}}^2:\left|u\right|+\left|v\right|=1\right\}.$$

$$\int P(h)dx=1,\text{and}\int P(f)dx>1\text{for all other}f\in H,$$

$$t\mapsto \int P\left(h+t[f-\phi (f)h]\right)dx$$

$$\int P^{\prime }(h)\left(f-\phi (f)h\right)dx=0.$$

$$g=\frac{P^{\prime }(h)}{\int P^{\prime }(h)hdx}\in L^Q.\text{qed}$$

$$P\left(\frac{u-v}{2}\right)\le \epsilon \cdot \frac{P(u)+P(v)}{2}+\frac{1}{\rho (\epsilon )}\left(\frac{P(u)+P(v)}{2}-P\left(\frac{u+v}{2}\right)\right).$$

$$P\left(\frac{u-v}{2}\right)\le P\left(\frac{\epsilon }{2}\left|u\right|+\frac{\epsilon }{2}\left|v\right|+(1-\epsilon )\cdot 0\right)\le \epsilon \cdot \frac{P(u)+P(v)}{2}.$$

Full text

1. A bijection between normed spaces and their duals

Let $X$ be a real normed space and $X^{\prime}$ its dual. We recall the following two notions:

Definition**.**

• •

$X$ is uniformly convex if for every $\varepsilon>0$ there exists a $\delta>0$ such that

[TABLE]

• •

A function $F:X\to\mathbb{R}$ is Gâteaux differentiable in $x$ if there exists a continuous linear functional $F^{\prime}(x)\in X^{\prime}$ such that for all $u\in X$,

[TABLE]

Let us denote by $S$ and $S^{\prime}$ the unit spheres of $X$ and $X^{\prime}$, respectively. We recall (see Lemma 3 below) that if $X$ is a uniformly convex Banach space, then the restriction to $S$ of every linear form $y\in S^{\prime}$ achieves its maximum in a unique point $M(y)$; furthermore, the map $M:S^{\prime}\to S$ is continuous.

Theorem 1**.**

Let $X$ be a uniformly convex Banach space. If its norm $N:X\to\mathbb{R}$ is Gâteaux differentiable on the unit sphere $S$, then $M:S^{\prime}\to S$ is a continuous bijection, and it inverse is the derivative $N^{\prime}:S\to S^{\prime}$ of the norm.

Proof.

If $x,z\in S$ and $t>0$, then by the triangular inequality we have

[TABLE]

so that $N^{\prime}(x)z\leq 1$, with equality if $z=x$. It follows that $\left\lVert N^{\prime}(x)\right\rVert=1$.

By uniform convexity, the linear form $N^{\prime}(x)$ achieves its maximum at a unique point (see Lemma 3 below), so that $N^{\prime}(x)z<1$ for all $z\in S$, different from $x$. Hence every $x\in S$ is uniquely determined by $N^{\prime}(x)$, and thus $N^{\prime}$ is injective.

To prove that $N^{\prime}$ is onto, pick $y\in S^{\prime}$ arbitrarily. Then $x:=M(y)$ is a point of minimal norm of the affine hyperplane $\left\{z\in X\ :\ y(z)=1\right\}$, and therefore

[TABLE]

for every $z\in X$. Since $N^{\prime}(x)x=1$, this is equivalent to $y=N^{\prime}(x)$. ∎

Remarks 2*.*

1. (i)

The above proof is essentially an application of the Lagrange multiplier theorem to maximize a linear functional on the unit sphere. Since the norm is not assumed to be a $C^{1}$ function, we have considered an equivalent extremal problem to minimize the norm on a closed affine hyperplane. 2. (ii)

The homogeneous extension of $N^{\prime}:S\to S^{\prime}$ yields a bijection between $X$ and $X^{\prime}$. 3. (iii)

It follows from the theorem that if $N^{\prime}$ is continuous on $S$, then $M$ is a homeomorphism between $S^{\prime}$ and $S$, and its homogeneous extension is a homeomorphism between $X^{\prime}$ and $X$. 4. (iv)

If $N^{\prime}$ is continuous on $S$, then it is also continuous on $X\setminus\left\{0\right\}$ by homogeneity. Then the norm is in fact Fréchet differentiable in every $a\in X\setminus\left\{0\right\}$, and hence it is a $C^{1}$ function. To see this, for any fixed $\varepsilon>0$ choose a small open ball $B_{r}(a)\subset X\setminus\left\{0\right\}$ such that $\left\lVert N^{\prime}(x)-N^{\prime}(a)\right\rVert<\varepsilon$ for all $x\in B_{r}(a)$. For any fixed $u\in B_{r}(0)$, applying the Lagrange mean value theorem to the function

[TABLE]

in $[0,1]$ we obtain that

[TABLE]

We recall the short proof of the lemma used above:

Lemma 3**.**

Let $X$ be a uniformly convex Banach space, and $y\in S^{\prime}$. There exists a (unique) point $x=M(y)\in S$ such that

[TABLE]

Furthermore, the map $M:S^{\prime}\to S$ is a continuous.

Proof.

Since $X$ is complete and $S$ is closed, it suffices to show that every sequence $(x_{n})\subset S$ satisfying $y(x_{n})\to 1$ is a Cauchy sequence. Given $\varepsilon>0$ arbitrarily, choose $\delta>0$ according to the uniform convexity, and then choose a large integer $n_{0}$ such that $y(x_{n})>1-\delta$ for all $n\geq n_{0}$. If $m,n\geq n_{0}$, then

[TABLE]

and therefore $\left\lVert x_{n}-x_{m}\right\rVert<\varepsilon$.

For the continuity of $M$ we have to show that if $y_{n}\to y$ in $S^{\prime}$, then $M(y_{n})\to M(y)$. Writing $x_{n}:=M(y_{n})$ and $x:=M(y)$ for brevity, since

[TABLE]

$y(x_{n})\to 1$, and therefore $(x_{n})$ converges to a point $\tilde{x}\in S$ satisfying $y(\tilde{x})=1$ as in the first part of the proof. Since $y(x)=1$ we have $\tilde{x}=x$ by uniqueness. ∎

2. Applications of Theorem 1

Theorem 1 has important consequences. We start with a strengthened version of the Helly–Hahn–Banach theorem for special spaces:

Theorem 4**.**

Let $X$ be a Banach space satisfying the hypotheses of Theorem 1, and $y_{1}$ a continuous linear functional defined on some subspace $X_{1}$ of $X$.

Then $y_{1}$ extends to a unique continuous linear functional $y\in X^{\prime}$ with preservation of the norm.

Proof.

We may assume that $\left\lVert y_{1}\right\rVert=1$, and we may assume by a continuous extension to the closure of $X_{1}$ that $X_{1}$ is closed. Then $X_{1}$ also satisfies the hypotheses of Theorem 1, so that $y_{1}=N^{\prime}(x_{1})|_{X_{1}}$ for a unique $x_{1}\in S\cap X_{1}$. It follows that $y:=N^{\prime}(x_{1})$ is an extension of $y_{1}$, and $\left\lVert N^{\prime}(x_{1})\right\rVert=1=\left\lVert y_{1}\right\rVert$.

If $\tilde{y}\in S^{\prime}$ is an arbitrary extension of $y_{1}$, then $\tilde{y}=N^{\prime}(\tilde{x}_{1})$ for a unique $\tilde{x}_{1}\in S$, characterized by the equality $\tilde{y}(\tilde{x}_{1})=1$. Since $x_{1}\in S$ and $\tilde{y}(x_{1})=y_{1}(x_{1})=1$, $\tilde{x}_{1}=x_{1}$ by uniqueness. ∎

Remark 5*.*

Geometrically the proof is based on the observation that the unique hyperplane separating $x_{1}$ from the unit ball is the tangent hyperlane at $x_{1}$.

We know that every uniformly convex Banach space is reflexive. This can be seen easily under the further assumption that $X^{\prime}$ satisfies the hypotheses of Theorem 1:

Proposition 6**.**

If $X,X^{\prime}$ are uniformly convex Banach spaces and the norm of $X^{\prime}$ is Gâteaux differentiable on $S^{\prime}$, then $X$ is reflexive.

Proof.

Given $\Phi\in X^{\prime\prime}$ arbitrarily, we have to find $x\in X$ satisfying $\Phi(y)=y(x)$ for all $y\in X^{\prime}$.

We may assue that $\left\lVert\Phi\right\rVert=1$. By Lemma 3 there exists a unique $y\in S^{\prime}$ satisfying $\Phi(y)=1$, and then a unique $x\in S$ satisfying $y(x)=1$. Defining $\Phi_{x}\in X^{\prime\prime}$ by $\Phi_{x}(\tilde{y}):=\tilde{y}(x)$ for all $\tilde{y}\in X^{\prime}$, we have $\Phi,\Phi_{x}\in S^{\prime\prime}$ and $\Phi(y)=1=\Phi_{x}(y)$. Applying Theorem 1 to $X^{\prime}$ we conclude that $\Phi=\Phi_{x}$. ∎

Now we turn to the description of the duals of Hilbert and $L^{p}$ spaces.

Theorem 7** (Riesz–Fréchet [1, 10]).**

If $X$ is a Hilbert space with scalar product $\left\langle\cdot,\cdot\right\rangle$, then the formula

[TABLE]

defines an isometric isomorphism $\Phi$ of $X$ onto $X^{\prime}$.

Proof.

Since $\Phi$ is an isometric isomorphism of $X$ into $X^{\prime}$ by the Cauchy–Schwarz inequality, it suffices to show that $\Phi$ maps $S$ onto $S^{\prime}$.

Every Hilbert space is uniformly convex by the parallelogramma law, and its norm is Gâteaux differentiable in every $x\in S$ with $N^{\prime}(x)=\Phi(x)$ because the following relation holds for each $u\in X$:

[TABLE]

Applying Theorem 1 we conclude that $\Phi$ is a bijection between $S$ and $S^{\prime}$. ∎

As we shall see in the more general context of Orlicz spaces (Lemma 12), the $L^{p}$ norms satisfy the hypotheses of Theorem 1.

Theorem 8** (Riesz [11, 13]).**

If $X=L^{p}$ with $1<p<\infty$ on some measure space and $q=p/(p-1)$ is the conjugate exponent, then the formula

[TABLE]

defines a linear isomorphism $\Phi$ of $L^{q}$ onto $(L^{p})^{\prime}$.

Proof.

Since $\Phi$ is an isometric isomorphism of $L^{q}$ into $(L^{p})^{\prime}$ by the Hölder inequality, it suffices to show that each $\varphi\in(L^{p})^{\prime}$ of norm one has the form $\varphi=\Phi(g)$ with some $g\in L^{q}$.

Due to Lemma 12, the hypotheses of Theorem 1 are fulfilled, and the Gâteaux derivative of the norm of $L^{p}$ in any $h_{0}\in S$ is given by the formula

[TABLE]

This implies $\varphi=\Phi(g_{0})$ with $g_{0}=\left\lvert h_{0}\right\rvert^{\frac{p}{q}}\operatorname{sign}(h_{0})$. ∎

Remarks 9*.*

1. (i)

The formula $F(h):=\left\lvert h\right\rvert^{\frac{p}{q}}\operatorname{sign}h$ defines a bijection between $L^{p}$ and $L^{q}$ whose inverse is $F^{-1}(g):=\left\lvert g\right\rvert^{\frac{q}{p}}\operatorname{sign}g$. Since $\Phi=N^{\prime}\circ F^{-1}$ and $N^{\prime},\Phi$ are homeomorphisms by Theorems 1, 8 and Remark 2 (iii), we conclude that $F$ is a homeomorphism between $L^{p}$ and $L^{q}$. This is a special case of Mazur’s theorem [4] stating that the spaces $L^{p}$ are homeomorphic for all finite $p$. 2. (ii)

We recall the extensions of Theorems 7 and 8 to complex Hilbert and $L^{p}$ spaces: the formulas

[TABLE]

define conjugate linear isometries of of $X$ onto $X^{\prime}$ and of $L^{q}$ onto $(L^{p})^{\prime}$, respectively.

Only the surjectivity needs an additional argument, and this follows by the classical method of Murray [7]. For example, if $\varphi\in(L^{p})^{\prime}$, then $\Phi(g)=\varphi$ with $g:=g_{1}+ig_{2}$ by a direct computation, where the real-valued functions $g_{1},g_{2}\in L^{q}$ are defined by the equalities

[TABLE]

for all real-valued functions $h\in L^{p}$. 3. (iii)

Another proof of Theorem 8 was given recently in [14].

3. Duals of Orlicz spaces

In this section we generalize the proof of Theorem 8 to a class of Orlicz spaces [2, 3, 8, 9, 15]. First we briefly recall some basic facts. Let $P:\mathbb{R}\to\mathbb{R}$ be an even convex function satisfying the conditions

[TABLE]

and $Q:\mathbb{R}\to\mathbb{R}$ its convex conjugate, defined by the formula

[TABLE]

For example, if $P(u)=p^{-1}\left\lvert u\right\rvert^{p}$ for some $1<p<\infty$, then $Q(v)=q^{-1}\left\lvert v\right\rvert^{q}$ with the conjugate exponent $q=p/(p-1)$.

Given a measure space $(\Omega,\mathcal{M},\mu)$, henceforth we consider only measurable functions $f,g,h:\Omega\to\mathbb{R}$. The Orlicz spaces $L^{P}$ and $L^{Q}$ are defined by

[TABLE]

If we endow $L^{P}$ with the Luxemburg norm

[TABLE]

and $L^{Q}$ with the Orlicz norm

[TABLE]

then they become Banach spaces,

[TABLE]

for all $f\in L^{P}$ and $g\in L^{Q}$ because $\left\lVert f\right\rVert_{P,\text{Lux}}\leq 1\Longleftrightarrow\int P(f)\ dx\leq 1$, and the formula

[TABLE]

defines a linear isometry $\Phi:L^{Q}\to(L^{P})^{\prime}$.

Theorem 10**.**

Assume that $P$ is differentiable, and satisfies the following conditions:

1. (i)

there exists a constant $\alpha$ such that $P(2u)\leq\alpha P(u)$ for all $u\in\mathbb{R}$; 2. (ii)

for each $\varepsilon>0$ there exists a $\rho(\varepsilon)>0$ such that

[TABLE]

Then $\Phi:L^{Q}\to(L^{P})^{\prime}$ is an isometrical isomorphism of $L^{Q}$ onto $(L^{P})^{\prime}$.

Example 11*.*

If $1<p<\infty$ and $P(u):=p^{-1}\left\lvert u\right\rvert^{p}$, then Theorem 10 reduces to Riesz’s theorem $(L^{p})^{\prime}=L^{q}$ with $q=p/(p-1)$. Indeed, $P$ is differentiable with $P^{\prime}(u)=\left\lvert u\right\rvert^{p-1}\operatorname{sign}u$, and the condition (i) is satisfied with $\alpha=2^{p}$. To check (ii) we may assume by homogeneity that $(u,v)$ belongs to the compact set

[TABLE]

The inequality follows by observing that the left hand side has a positive minimum on $K$ by continuity and strict convexity, while $P(u)+P(v)$ has a finite maximum here. Finally, on the classical spaces $L^{p}$ the Orlicz and Luxemburg norms coincide.

For the proof of Theorem 10 we admit temporarily the following

Lemma 12**.**

Assume that the conditions of Theorem 10 are satisfied. Then

1. (i)

$L^{P}$* is uniformly convex;* 2. (ii)

$P^{\prime}(h)\in L^{Q}$* for every $h\in L^{P}$;* 3. (iii)

for any fixed $f,h\in L^{P}$ the function $t\mapsto\int P(h+tf)\ dx$ is differentiable in zero, and its derivative is equal to $\int P^{\prime}(h)f\ dx$.

Proof of Theorem 10.

As before, it suffices to show that every $\varphi\in(L^{P})^{\prime}$ of unit norm is of the form $\Phi(h)$ for a suitable $h\in L^{P}$.

As in the proof of Theorem 1, by the uniform convexity of $L^{P}$ there exists a function $h\in L^{P}$ such that $\varphi(h)=1$, and $\left\lVert f\right\rVert>1$ for all other functions $f$ in the closed affine hyperplane $H:=\left\{f\in L^{P}\ :\ \varphi(f)=1\right\}$.

By the definition of the Luxemburg norm we have

[TABLE]

so that $h$ also minimizes the functional $f\mapsto\int P(f)\ dx$ on $H$. Hence for any fixed $f\in L^{P}$ the function

[TABLE]

has a minimum in [math], and therefore its derivative vanishes here:

[TABLE]

This is equivalent to $\varphi=\varphi_{g}$ with

[TABLE]

Proof of Lemma 12.

(i) Following McShane [5] first we prove for all $\varepsilon\in(0,1)$ the inequality

[TABLE]

In case $\left\lvert u-v\right\rvert\leq\varepsilon(\left\lvert u\right\rvert+\left\lvert v\right\rvert)$ this readily follows from the convexity and evenness of $P$ and from the equality $P(0)=0$ because

[TABLE]

If $\left\lvert u-v\right\rvert\geq\varepsilon(\left\lvert u\right\rvert+\left\lvert v\right\rvert)$, then we infer from the convexity of $P$ and from the condition (ii) of Theorem 10 that

[TABLE]

We complete the proof of the lemma by showing that if

[TABLE]

then

[TABLE]

with $\alpha$ given by the condition (i) of Theorem 10. This follows by using the inequality (2):

[TABLE]

(ii) By definition we have to show that $P^{\prime}(h)f$ is integrable for every $f\in L^{P}$. Setting $g:=\left\lvert h\right\rvert+\left\lvert h\right\rvert\in L^{P}$, this follows from the inequalities

[TABLE]

The last inequality is a consequence of the condition (i) of Theorem 10: we have

[TABLE]

for all $u\geq 0$, and $uP^{\prime}(u)$ is an even function. (The equality holds because, as a convex function, $P$ is absolutely continuous.)

(iii) We have

[TABLE]

almost everywhere (where both $f$ and $h$ are defined and are finite). Furthermore, if $0<\left\lvert t\right\rvert\leq 1$, then

[TABLE]

with $g:=\left\lvert f\right\rvert+\left\lvert h\right\rvert\in L^{P}$ by the Lagrange mean value theorem with some $t^{\prime}$ between [math] and $t$ (depending on $x$). The last function is integrable because

[TABLE]

Applying Lebesgue’s dominated convergence theorem we conclude that

[TABLE]

Remark 13*.*

A complete, but necessarily rather technical characterization of uniformly convex Orlicz space was given by different tools in [6].