From Extension Theorems to Dual Spaces
The Hahn-Banach Theorem, established earlier in this course, extends bounded linear functionals from a subspace without increasing their norm. That extension principle has a useful consequence: a normed space has enough continuous linear functionals to detect the size of each of its vectors. To make this precise, we collect those functionals into a new normed space, called the dual.
Dual spaces also have an important completeness property. Even if the original normed space is incomplete, its continuous dual is complete. These two facts—completeness of the dual and the ability of its elements to detect the original norm—are central tools in functional analysis.
The Continuous Dual and Its Norm
Linearity means \(F(ax+by)=aF(x)+bF(y)\) for vectors \(x,y\in X\) and scalars \(a,b\in\mathbb{R}\). The operator norm measures the largest size of \(F(x)\) on the closed unit ball. Equivalently, it is the least constant \(C\geq0\) for which \(|F(x)|\leq C\|x\|\) for every \(x\in X\). Indeed, the definition gives this bound first for nonzero \(x\) by applying it to \(x/\|x\|\), and the bound is automatic at \(x=0\). Conversely, any such bound implies \(|F(x)|\leq C\) whenever \(\|x\|\leq1\).
For a linear functional, continuity and the existence of a finite bound of this form are equivalent. If \(F\) is continuous at \(0\), choose \(\delta>0\) such that \(\|z\|<\delta\) implies \(|F(z)|<1\). For \(x\neq0\), apply this to \(z=\delta x/(2\|x\|)\). Linearity gives \(|F(x)|<2\|x\|/\delta\). In the other direction, a bound \(|F(x)|\leq C\|x\|\) implies continuity, since \(|F(x)-F(y)|\leq C\|x-y\|\).
The operator norm is a norm on \(X^*\). Homogeneity and the triangle inequality follow from the corresponding properties of absolute value and the supremum. If \(\|F\|=0\), then \(|F(x)|\leq\|F\|\|x\|=0\) for every \(x\), so \(F\) is the zero functional. Thus \(X^*\) is itself a normed space.
Worked Example: The Dual Norm for a Weighted Norm on \(\mathbb{R}^3\)
Equip \(\mathbb{R}^3\) with the norm \(\|(x,y,z)\|=2|x|+|y|+3|z|\). Every linear functional on \(\mathbb{R}^3\) has the form \(F(x,y,z)=ax+by+cz\) for some real coefficients \(a,b,c\). We claim that $$ \|F\|=\max\left\{\frac{|a|}{2},\,|b|,\,\frac{|c|}{3}\right\}. $$
Let \(M\) denote the maximum on the right. For every \((x,y,z)\), \[ |F(x,y,z)|\leq |a||x|+|b||y|+|c||z| \leq M(2|x|+|y|+3|z|). \] Consequently, \(\|F\|\leq M\). To see equality, suppose first that \(M=|a|/2\) and \(a\neq0\). The vector \((\operatorname{sgn}(a)/2,0,0)\) has norm \(1\), and \(F\) takes the value \(|a|/2=M\) there. If instead \(M=|b|\) with \(b\neq0\), use \((0,\operatorname{sgn}(b),0)\); if \(M=|c|/3\) with \(c\neq0\), use \((0,0,\operatorname{sgn}(c)/3)\). Each chosen vector has norm \(1\) and gives \(|F|=M\). If \(M=0\), all coefficients vanish and both sides are zero. This verifies the formula in every case.
The Dual Is Complete
A Cauchy sequence in the dual is a sequence of functionals that becomes uniformly close on the unit ball. The limit can be constructed by taking the ordinary real limit at each vector. The key point is then to show that these pointwise limits combine into a bounded linear functional and that convergence holds in operator norm.
Proof. Let \((F_n)\) be a Cauchy sequence in \(X^*\). For each fixed \(x\in X\), \[ |F_n(x)-F_m(x)|\leq\|F_n-F_m\|\|x\|. \] The right-hand side tends to zero as \(m,n\to\infty\), so \((F_n(x))\) is Cauchy in \(\mathbb{R}\). Define \(F(x)=\lim_{n\to\infty}F_n(x)\).
Linearity follows by passing to the limit in the linearity identities. More explicitly, for \(x,y\in X\) and \(a,b\in\mathbb{R}\), \[ F(ax+by)=\lim_{n\to\infty}F_n(ax+by) =\lim_{n\to\infty}\bigl(aF_n(x)+bF_n(y)\bigr) =aF(x)+bF(y). \] We next check that \(F\) is bounded. Since \((F_n)\) is Cauchy, there is an index \(N\) such that \(\|F_n-F_N\|\leq1\) whenever \(n\geq N\). For every such \(n\) and every \(x\in X\), \[ |F_n(x)|\leq |F_N(x)|+\|F_n-F_N\|\|x\| \leq(\|F_N\|+1)\|x\|. \] Taking the limit in \(n\) gives \(|F(x)|\leq(\|F_N\|+1)\|x\|\). Thus \(F\) is continuous and belongs to \(X^*\).
Finally, let \(\varepsilon>0\). Choose \(N_\varepsilon\) such that \(\|F_n-F_m\|<\varepsilon/2\) whenever \(n,m\geq N_\varepsilon\). Fix \(n\geq N_\varepsilon\) and \(x\in X\). For every \(m\geq N_\varepsilon\), \[ |(F_n-F_m)(x)|\leq\frac{\varepsilon}{2}\|x\|. \] Letting \(m\to\infty\) yields \(|(F_n-F)(x)|\leq(\varepsilon/2)\|x\|\), and therefore \(\|F_n-F\|\leq\varepsilon/2<\varepsilon\). Hence \(F_n\to F\) in \(X^*\), proving completeness. \(\square\)
Notice that the proof did not assume that \(X\) is complete. The real numbers provide the pointwise limits, and the Cauchy condition in operator norm supplies the uniform bound needed to show that the limit is continuous.
Worked Examples of Continuous Functionals
Worked Example: Evaluation and Integration on Continuous Functions
Let \(X=C([0,1])\) with the supremum norm \(\|f\|_\infty=\sup_{0\leq t\leq1}|f(t)|\). Fix \(t_0\in[0,1]\) and define the evaluation functional \(\delta_{t_0}(f)=f(t_0)\). It is linear, and \[ |\delta_{t_0}(f)|=|f(t_0)|\leq\|f\|_\infty, \] so \(\|\delta_{t_0}\|\leq1\). The constant function \(f(t)=1\) has supremum norm \(1\) and satisfies \(\delta_{t_0}(f)=1\). Therefore \(\|\delta_{t_0}\|=1\).
Another functional is \(I(f)=\int_0^1 t f(t)\,dt\). It is linear, and \[ |I(f)|\leq\int_0^1 t|f(t)|\,dt \leq\|f\|_\infty\int_0^1t\,dt =\tfrac12\|f\|_\infty. \] Thus \(\|I\|\leq1/2\). For the constant function \(f(t)=1\), we have \(\|f\|_\infty=1\) and \(I(f)=\int_0^1t\,dt=1/2\). Hence \(\|I\|=1/2\). Both examples show how the operator norm records the size of a functional relative to the chosen norm on its domain.
Worked Example: Coordinate Functionals on a Finite-Dimensional Space
On \(\mathbb{R}^2\) with the Euclidean norm, define \(F(x,y)=3x-4y\). The Cauchy–Schwarz inequality gives \[ |F(x,y)|\leq\sqrt{3^2+(-4)^2}\sqrt{x^2+y^2} =5\sqrt{x^2+y^2}, \] so \(\|F\|\leq5\). At the unit vector \((3/5,-4/5)\), \[ F(3/5,-4/5)=3(3/5)-4(-4/5)=9/5+16/5=5. \] Therefore \(\|F\|=5\).
More generally, a linear functional on \(\mathbb{R}^2\) is determined by its values on the coordinate vectors: \(F(x,y)=xF(1,0)+yF(0,1)\). The calculation above is one instance of representing a functional by its coefficients and measuring those coefficients in the appropriate dual norm.
Hahn–Banach Functionals Detect the Norm
In the weighted example, the dual norm was calculated directly from coefficients. For an arbitrary normed space there may be no coordinate system or explicit formula. The Hahn-Banach Theorem nevertheless guarantees that continuous linear functionals can attain the norm of any chosen vector.
Proof. On the one-dimensional subspace \(M=\mathbb{R}x\), define \(f_0(tx)=t\|x\|\) for \(t\in\mathbb{R}\). This is well-defined because \(x\neq0\). It is linear, and \[ |f_0(tx)|=|t|\|x\|=\|tx\|, \] so its norm on \(M\) is \(1\). The Norm-Preserving Extension Corollary of the Hahn-Banach Theorem gives an extension \(F\in X^*\) with \(\|F\|=1\). Since \(F\) extends \(f_0\), \(F(x)=f_0(x)=\|x\|\).
For any \(G\in X^*\) with \(\|G\|\leq1\), the definition of operator norm gives \(|G(x)|\leq\|x\|\). Thus the supremum in the displayed formula is at most \(\|x\|\). If \(x\neq0\), the functional just constructed has norm \(1\) and satisfies \(|F(x)|=\|x\|\), so the supremum is at least \(\|x\|\). For \(x=0\), every linear functional has value zero, and both sides are zero. The formula follows for all \(x\). \(\square\)
This result says more than that the dual separates distinct vectors: if \(x\neq y\), apply the theorem to \(x-y\) to obtain an \(F\) with \(F(x-y)=\|x-y\|>0\), hence \(F(x)\neq F(y)\). In fact, the family of unit-norm functionals recovers the entire norm, not just whether a vector is zero.
The Bidual and the Canonical Embedding
Since \(X^*\) is a normed space, it has a continuous dual of its own, denoted \(X^{**}=(X^*)^*\). This is called the bidual of \(X\). Each vector \(x\in X\) defines a functional on \(X^*\) by evaluating functionals at \(x\).
For fixed \(x\), the map \(Jx\) is linear in \(F\). It is also continuous, because \(|(Jx)(F)|=|F(x)|\leq\|F\|\|x\|\). Thus \(Jx\in X^{**}\). The next result says that this construction preserves all distances from the origin.
Proof. By the definition of the norm in \(X^{**}\), \[ \|Jx\|=\sup_{\substack{F\in X^*\\\|F\|\leq1}}|(Jx)(F)| =\sup_{\substack{F\in X^*\\\|F\|\leq1}}|F(x)| =\|x\|, \] where the final equality is the norming-functional theorem. The definition also gives \(J(ax+by)(F)=F(ax+by)=aF(x)+bF(y)\), so \(J(ax+by)=aJx+bJy\). If \(Jx=0\), then \(\|x\|=\|Jx\|=0\), which implies \(x=0\). Therefore \(J\) is injective. \(\square\)
The image \(J(X)\) is a subspace of \(X^{**}\) with exactly the same norm geometry as \(X\). The isometry need not, by itself, assert that every element of \(X^{**}\) comes from a vector in \(X\). When it does, the space is called reflexive; that additional property will be considered later.
What Duality Does—and Does Not—Say
The dual records how continuous linear measurements act on vectors. The norming-functional theorem ensures that these measurements are rich enough to recover the norm, while completeness makes \(X^*\) a Banach space regardless of whether \(X\) is complete. These conclusions are useful even when no explicit description of every functional is available.
A common pitfall is to confuse the continuous dual with the algebraic dual, which consists of all linear maps \(X\to\mathbb{R}\) without a continuity requirement. On an infinite-dimensional normed space, algebraic linear functionals need not be bounded. They do not belong to \(X^*\), and the operator norm is not defined for them as a finite norm. In finite-dimensional normed spaces, every linear functional is continuous, but that fact should not be assumed in general.
The norming result also depends on the Hahn-Banach extension theorem: the one-dimensional functional first records the size of a selected vector, and Hahn-Banach extends that measurement to the whole space without increasing its norm. This is why the theorem is a structural foundation for the study of dual spaces, rather than merely an extension technique.
Check Your Understanding
Use the definitions and results above to answer the following questions.
- What bound on \(|F(x)|\) is equivalent to continuity of a linear functional \(F\)?
- Why does a Cauchy sequence in \(X^*\) have a pointwise limit at every \(x\in X\)?
- How does the Hahn-Banach Theorem produce a unit-norm functional \(F\) satisfying \(F(x)=\|x\|\) when \(x\neq0\)?
- Why is the norm of the evaluation functional on \(C([0,1])\) equal to \(1\), rather than merely at most \(1\)?
- What does it mean for the canonical map \(J:X\to X^{**}\) to be an isometry?
- How does the continuous dual differ from the algebraic dual?