Dual Spaces

From Functionals to a Dual Space

A single linear functional assigns one scalar to each vector. Often, however, the object of interest is not one measurement but the collection of all linear measurements available on a space. That collection is itself a vector space. When the original space is normed, the bounded linear functionals form a normed space of measurements, called its continuous dual. The dual norm records how large a functional can be on inputs of norm at most one.

Definition: Let $V$ be a real vector space. Its algebraic dual, denoted $V^\#$, is the vector space of all linear functionals $L:V\to\mathbb{R}$, with addition and scalar multiplication defined pointwise. If $V$ is normed, its continuous dual, denoted $V^*$, is the subspace of $V^\#$ consisting of all bounded linear functionals. For $L\in V^*$, define the dual norm by $$ \|L\|_{V^*}=\sup_{\|x\|_V\leq1}|L(x)|. $$

The distinction between $V^\#$ and $V^*$ matters: algebraic linearity alone does not ensure boundedness. For a normed space, the continuity-boundedness equivalence from “Continuity of Linear Maps” says that $V^*$ is also the collection of continuous linear functionals. In this tutorial, “dual” means the continuous dual whenever a norm is specified. The notation $V'$ is also commonly used for this space.

For $L\in V^*$, the operator-norm results from “Linear Functional” give $|L(x)|\leq\|L\|_{V^*}\|x\|_V$ for every $x\in V$. Equivalently, the dual norm is the least nonnegative constant that works in this inequality. Its unit-ball definition also shows what convergence in the dual norm means: two functionals are close when their values are uniformly close on every input of norm at most one.

Computing Dual Norms

Worked Example: The Dual Norm for a Coordinate Functional

Let $V=\mathbb{R}^2$ with $\|(x,y)\|_1=|x|+|y|$. For real numbers $a,b$, define $L_{a,b}(x,y)=ax+by$. The functional is linear, and

|L_{a,b}(x,y)| \leq |a||x|+|b||y| \leq \max\{|a|,|b|\}(|x|+|y|).

Thus $L_{a,b}$ is bounded and $\|L_{a,b}\|_{V^*}\leq\max\{|a|,|b|\}$. To obtain equality, suppose first that $|a|\geq|b|$. If $a\neq0$, choose $x=(\operatorname{sgn}(a),0)$, where $\operatorname{sgn}(a)=a/|a|$. Then $\|x\|_1=1$ and $|L_{a,b}(x)|=|a|$. If $a=0$, the assumption implies $b=0$, and both the functional norm and the claimed maximum are zero. If instead $|b|>|a|$, choose $x=(0,\operatorname{sgn}(b))$, which has norm one and gives $|L_{a,b}(x)|=|b|$. Therefore, in all cases,

\|L_{a,b}\|_{V^*}=\max\{|a|,|b|\}.

Thus the coefficients of a functional on this space are measured by the maximum of their absolute values. This calculation identifies the norm on the collection of these coordinate functionals; it is not the same as the norm used for the inputs.

Worked Example: A Bounded Integral Functional

On $C[0,1]$ with the supremum norm, define

L(f)=\int_0^1 t^2 f(t)\,dt.

Linearity follows from linearity of the Riemann integral. Since $t^2\geq0$ on $[0,1]$, every $f\in C[0,1]$ satisfies

|L(f)| \leq\int_0^1 t^2|f(t)|\,dt \leq\|f\|_\infty\int_0^1t^2\,dt =\frac{1}{3}\|f\|_\infty.

It follows that $\|L\|_{V^*}\leq1/3$. The constant function $\mathbf{1}(t)=1$ has supremum norm one and $L(\mathbf{1})=\int_0^1t^2\,dt=1/3$. Hence $\|L\|_{V^*}\geq1/3$, proving that $\|L\|_{V^*}=1/3$. Here a single input attains the upper bound.

Worked Example: Linearity Without Boundedness

Let $V=C^1[0,1]$ be given the supremum norm $\|f\|_\infty=\sup_{t\in[0,1]}|f(t)|$, and define $D(f)=f'(0)$. This map is linear, but it is not bounded for this norm. For each positive integer $n$, set $f_n(t)=\sin(nt)/n$. Then $f_n\in C^1[0,1]$,

\|f_n\|_\infty\leq\frac{1}{n}, \qquad D(f_n)=f_n'(0)=\cos(0)=1.

If $D$ were bounded, some finite $C\geq0$ would satisfy $|D(f)|\leq C\|f\|_\infty$ for every $f\in V$. Applying this to $f_n$ would give $1\leq C/n$ for every positive integer $n$, which fails whenever $n>C$. Thus $D$ belongs to the algebraic dual but not the continuous dual. The example shows why a norm on the domain is part of the definition of $V^*$.

The Continuous Dual Is a Normed Space

Theorem (Normed-Space Structure of the Continuous Dual): If $V$ is a normed space, then $V^*$, with pointwise addition and scalar multiplication and the dual norm, is a normed vector space.

Proof. If $L,M\in V^*$ and $\alpha,\beta\in\mathbb{R}$, then $\alpha L+\beta M$ is linear. For every $x\in V$,

|(\alpha L+\beta M)(x)| \leq|\alpha||L(x)|+|\beta||M(x)| \leq\bigl(|\alpha|\|L\|_{V^*}+|\beta|\|M\|_{V^*}\bigr)\|x\|_V.

So $\alpha L+\beta M$ is bounded and belongs to $V^*$, making $V^*$ a vector space. The dual norm is nonnegative. If $\|L\|_{V^*}=0$, then $|L(x)|=0$ for every $x$ with $\|x\|_V\leq1$. For any nonzero $x$, the vector $x/\|x\|_V$ is in that unit ball, so $L(x)=\|x\|_V L(x/\|x\|_V)=0$. Also $L(0)=0$, and therefore $L$ is the zero functional. Conversely, the zero functional has norm zero.

For any scalar $\alpha$, the definition gives $\|\alpha L\|_{V^*}=|\alpha|\|L\|_{V^*}$. Finally, for every $x$ with $\|x\|_V\leq1$, \[ |(L+M)(x)|\leq|L(x)|+|M(x)| \leq\|L\|_{V^*}+\|M\|_{V^*}. \] Taking the supremum over the unit ball proves the triangle inequality. The dual norm therefore satisfies all norm axioms. $\square$

Completeness of the Dual

A normed space is a Banach space when every Cauchy sequence in its norm converges to an element of the space. The dual has a useful completeness property: its completeness does not require completeness of the original space. The proof constructs a candidate limit by taking the limit of the functionals at each fixed input, and then verifies convergence uniformly over the unit ball.

Theorem (The Continuous Dual Is Banach): If $V$ is any normed space, then $V^*$, equipped with the dual norm, is complete.

Proof. Let $(L_n)$ be a Cauchy sequence in $V^*$. For each fixed $x\in V$, the operator-norm inequality gives

|L_n(x)-L_m(x)| \leq\|L_n-L_m\|_{V^*}\|x\|_V.

Because $(L_n)$ is Cauchy in the dual norm, the right-hand side tends to zero as $n,m$ tend to infinity. Thus $(L_n(x))$ is a Cauchy sequence of real numbers for each $x$, and so it has a limit. Define $L:V\to\mathbb{R}$ by $L(x)=\lim_{n\to\infty}L_n(x)$.

For $x,y\in V$ and $\alpha,\beta\in\mathbb{R}$, each $L_n$ is linear, so $L_n(\alpha x+\beta y)=\alpha L_n(x)+\beta L_n(y)$. Taking real limits on both sides gives $L(\alpha x+\beta y)=\alpha L(x)+\beta L(y)$. Hence $L$ is linear.

It remains to prove boundedness and norm convergence. Since $(L_n)$ is Cauchy, it is bounded in norm. Indeed, choose $N$ so that $\|L_n-L_N\|_{V^*}<1$ for $n\geq N$. Then $\|L_n\|_{V^*}\leq\|L_N\|_{V^*}+1$ for $n\geq N$, and the finitely many earlier norms are also bounded. Let $M$ be a finite bound for all $\|L_n\|_{V^*}$. For every $x\in V$, \[ |L_n(x)|\leq M\|x\|_V. \] Taking the limit in $n$ shows $|L(x)|\leq M\|x\|_V$. Thus $L\in V^*$.

Now fix $\varepsilon>0$. Since $(L_n)$ is Cauchy in $V^*$, there is an index $N$ such that $\|L_n-L_m\|_{V^*}<\varepsilon$ whenever $n,m\geq N$. Fix $n\geq N$ and $x\in V$. For every $m\geq N$, \[ |L_n(x)-L_m(x)|\leq\|L_n-L_m\|_{V^*}\|x\|_V <\varepsilon\|x\|_V. \] Letting $m$ tend to infinity gives $|L_n(x)-L(x)|\leq\varepsilon\|x\|_V$. Taking the supremum over all $x$ with $\|x\|_V\leq1$ yields $\|L_n-L\|_{V^*}\leq\varepsilon$ for every $n\geq N$. Since this holds for every positive $\varepsilon$, $L_n\to L$ in the dual norm. Every Cauchy sequence in $V^*$ therefore converges in $V^*$, proving completeness. $\square$

The construction uses completeness of the real numbers to define $L(x)$, but it never requires a Cauchy sequence of vectors in $V$ to converge in $V$. This explains why an incomplete normed space can still have a complete continuous dual. The dual norm measures uniform control of scalar outputs, and the Cauchy condition in that norm is strong enough to preserve that control in the limit.

Interpreting the Dual Norm

The continuous dual gathers all bounded scalar measurements on $V$. Its norm compares measurements by their largest possible difference on the unit ball. In particular, $\|L-M\|_{V^*}\leq r$ means \[ |L(x)-M(x)|\leq r\|x\|_V \] for every $x\in V$. Thus convergence in $V^*$ is not merely convergence of $L_n(x)$ for each fixed $x$; it is convergence with one error bound that works uniformly for all inputs of norm at most one.

Keep the domain norm in view whenever identifying a dual. The same algebraic functional can be bounded for one norm and unbounded for another, and its dual norm can change when the domain norm changes. Also, completeness of $V^*$ should not be confused with completeness of $V$: the theorem proves the former without assuming the latter.

Takeaway: The continuous dual $V^*$ consists of the bounded linear functionals on a normed space $V$. The dual norm is the operator norm, makes $V^*$ a normed space, and is complete for every normed $V$, whether or not $V$ itself is complete.

Check Your Understanding

Use the definitions and proofs in this tutorial to answer the following questions.

How does the algebraic dual differ from the continuous dual of a normed space?
For the functional $L_{a,b}(x,y)=ax+by$ on $\mathbb{R}^2$ with the $1$-norm, why is its dual norm $\max\{|a|,|b|\}$?
What feature of the sequence $f_n(t)=\sin(nt)/n$ shows that $f\mapsto f'(0)$ is not bounded in the supremum norm?
In the completeness proof, why is the pointwise limit $L(x)=\lim_n L_n(x)$ linear?
Where does the Cauchy condition in the dual norm ensure that the limit is a continuous linear functional rather than merely a pointwise limit?

What You'll Learn