Linear Functional

Scalar-Valued Linear Maps

A bounded linear operator sends vectors in one normed space to vectors in another. A linear functional is the particular case in which the target is the real numbers. It assigns a single scalar to each input while respecting addition and scalar multiplication. This makes functionals useful for measuring one aspect of a vector or function, and their kernels identify inputs that the measurement cannot distinguish from zero.

Definition: Let $V$ be a real vector space. A linear functional on $V$ is a linear map $L:V\to\mathbb{R}$; that is, for all $x,y\in V$ and $\alpha,\beta\in\mathbb{R}$, $$ L(\alpha x+\beta y)=\alpha L(x)+\beta L(y). $$ If $V$ is normed, $L$ is bounded when there is a finite constant $C\geq0$ such that $|L(x)|\leq C\|x\|_V$ for every $x\in V$.

A functional is an operator with the real line as its target, so the boundedness and operator-norm results for linear maps apply. In particular, a bounded functional has operator norm $\|L\|=\sup_{\|x\|_V\leq1}|L(x)|$, and its norm is the least constant in the inequality $|L(x)|\leq\|L\|\|x\|_V$. We use the continuity-boundedness equivalence established in “Continuity of Linear Maps” without repeating its proof here.

The zero functional, which sends every vector to zero, is linear and bounded, with norm zero. A nonzero functional has at least one input with a nonzero output. The following examples show that the norm can often be found by combining a bound valid for all inputs with a specific input that attains that bound, or approaches it.

Computing Functional Norms

Worked Example: A Coordinate Functional on $\mathbb{R}^3$

Give $\mathbb{R}^3$ the norm $\|(a,b,c)\|_1=|a|+|b|+|c|$, and define $L(a,b,c)=4a-2b+c$. The expression is linear in the coordinates, so $L$ is a linear functional. For every $(a,b,c)$,

|L(a,b,c)| \leq4|a|+2|b|+|c| \leq4(|a|+|b|+|c|) =4\|(a,b,c)\|_1.

Thus $L$ is bounded and $\|L\|\leq4$. To obtain the reverse inequality, take $x=(1,0,0)$. Then $\|x\|_1=1$ and $|L(x)|=4$, so the unit-ball formula gives $\|L\|\geq4$. Therefore $\|L\|=4$. The inputs on which $L$ vanishes form the plane $4a-2b+c=0$, a constraint that will be described algebraically below.

Worked Example: An Integral Functional

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

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

Linearity follows from linearity of the Riemann integral. For every $f\in C[0,1]$, the integral triangle inequality and $|f(t)|\leq\|f\|_\infty$ give

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

Hence $\|L\|\leq 3/2$. For the constant function $\mathbf{1}(t)=1$, we have $\|\mathbf{1}\|_\infty=1$ and $L(\mathbf{1})=\int_0^1(1+t)\,dt=3/2$. Thus $\|L\|\geq3/2$, and $\|L\|=3/2$. This is a scalar-valued map: it returns one real number, rather than a new function as an integral operator does.

Worked Example: An Endpoint-Difference Functional

Define $D:C[0,1]\to\mathbb{R}$ by $D(f)=f(0)-f(1)$, again using the supremum norm on $C[0,1]$. This is linear, and

|D(f)| \leq |f(0)|+|f(1)| \leq2\|f\|_\infty.

Therefore $\|D\|\leq2$. The continuous function $f(t)=1-2t$ has $\|f\|_\infty=1$, since its values range from $1$ to $-1$, and $D(f)=1-(-1)=2$. This proves $\|D\|\geq2$, so $\|D\|=2$. Its kernel consists exactly of the continuous functions with $f(0)=f(1)$.

Point evaluation $f\mapsto f(a)$ on $C(K)$ with the supremum norm was established earlier as a bounded linear functional. The endpoint-difference example is related, but its norm calculation also shows how combining measurements can change the bound: the two endpoint values can contribute with opposite signs and yield a norm of two.

The Kernel and Its Geometry

For a functional $L$, its kernel is the set of inputs assigned the scalar zero: $\ker(L)=\{x\in V:L(x)=0\}$. The general result that the kernel of a linear map is a subspace was established in “Linear Maps.” For a nonzero functional there is a stronger description: the kernel has exactly one independent direction missing from the whole space. This fact requires no norm or boundedness assumption.

Theorem (Decomposition by a Nonzero Functional): Let $V$ be a real vector space and $L:V\to\mathbb{R}$ a nonzero linear functional. There is a vector $u\in V$ with $L(u)=1$, and every $x\in V$ has a unique representation $$ x=y+L(x)u,\qquad y\in\ker(L). $$ Consequently, $V=\ker(L)+\operatorname{span}\{u\}$, and this sum is direct. In particular, the quotient space $V/\ker(L)$ has dimension one.

Proof. Since $L$ is nonzero, there is a vector $v\in V$ such that $L(v)\neq0$. Set $u=v/L(v)$. By linearity, $L(u)=L(v)/L(v)=1$. For any $x\in V$, define $y=x-L(x)u$. Then $L(y)=L(x)-L(x)L(u)=0$, so $y\in\ker(L)$, and rearranging gives the claimed representation.

To prove uniqueness, suppose $x=y_1+\alpha_1u=y_2+\alpha_2u$, where $y_1,y_2\in\ker(L)$. Applying $L$ to both expressions gives $L(x)=\alpha_1$ and $L(x)=\alpha_2$, because $L(u)=1$ and $L(y_1)=L(y_2)=0$. Thus $\alpha_1=\alpha_2$, and then $y_1=y_2$. This proves that the representation is unique and the sum is direct. Every coset in $V/\ker(L)$ is a scalar multiple of the coset $u+\ker(L)$, since $x+\ker(L)=L(x)u+\ker(L)$. This coset is nonzero because $u\notin\ker(L)$, so the quotient has dimension one. $\square$

The theorem explains the plane in the $\mathbb{R}^3$ example: after fixing one vector $u$ on which the functional equals one, any vector splits uniquely into a part in the zero plane and a scalar multiple of $u$. The scalar coefficient is precisely the value $L(x)$. A functional therefore provides both a measurement and a coordinate transverse to its kernel.

Theorem (Kernel of a Bounded Functional Is Closed): Let $V$ be a normed space and $L:V\to\mathbb{R}$ a bounded linear functional. Then $\ker(L)$ is closed in $V$.

Proof. If $L$ is the zero functional, then $\ker(L)=V$, which is closed. Otherwise $\|L\|>0$. Take any $x\notin\ker(L)$, so $|L(x)|>0$. If $\|y-x\|_V<|L(x)|/(2\|L\|)$, boundedness gives

|L(y)-L(x)| \leq\|L\|\|y-x\|_V <\frac{|L(x)|}{2}.

The reverse triangle inequality on $\mathbb{R}$ now yields $|L(y)|\geq |L(x)|-|L(y)-L(x)|>|L(x)|/2>0$. Thus every point outside the kernel has a neighborhood that does not meet the kernel. The complement of $\ker(L)$ is open, so the kernel is closed. $\square$

Distance from the Kernel

For a subset $A$ of a normed space, the distance from $x$ to $A$ is $\operatorname{dist}(x,A)=\inf\{\|x-y\|_V:y\in A\}$. For a nonzero bounded functional, the value $|L(x)|$ measures exactly how far $x$ lies from its kernel, after scaling by the functional norm. This gives a quantitative version of the decomposition theorem.

Theorem (Distance Formula for a Bounded Functional): Let $V$ be a normed space and $L:V\to\mathbb{R}$ a nonzero bounded linear functional. Then, for every $x\in V$, $$ \operatorname{dist}(x,\ker(L))=\frac{|L(x)|}{\|L\|}. $$

Proof. First, if $y\in\ker(L)$, then $L(x)=L(x-y)$. Boundedness therefore implies \[ |L(x)|\leq\|L\|\|x-y\|_V. \] Taking the infimum over all $y\in\ker(L)$ proves $\operatorname{dist}(x,\ker(L))\geq |L(x)|/\|L\|$.

If $L(x)=0$, then $x\in\ker(L)$, so the distance and the stated right-hand side are both zero. Now suppose $L(x)\neq0$. By the definition of $\|L\|$ as a supremum over the unit ball, for every sufficiently large positive integer $n$ there is a vector $v_n$ with $\|v_n\|_V\leq1$ and $|L(v_n)|>\|L\|-1/n$. Choose $n$ large enough that $1/n<\|L\|$; then $L(v_n)\neq0$. Set \[ y_n=x-\frac{L(x)}{L(v_n)}v_n. \] Linearity gives $L(y_n)=L(x)-L(x)=0$, so $y_n\in\ker(L)$. Hence

\operatorname{dist}(x,\ker(L)) \leq\|x-y_n\|_V =\frac{|L(x)|}{|L(v_n)|}\|v_n\|_V \leq\frac{|L(x)|}{|L(v_n)|}.

As $n$ tends to infinity through these sufficiently large integers, $|L(v_n)|$ tends to $\|L\|$: it is at most $\|L\|$ and exceeds $\|L\|-1/n$. The displayed upper bounds consequently tend to $|L(x)|/\|L\|$. This proves the reverse inequality and hence the formula. $\square$

The construction does not assume that the norm of the functional is attained by a unit vector. It uses vectors whose functional values approach the supremum, which is enough to determine the infimum distance. In finite-dimensional spaces a supremum may be attained under suitable compactness conditions, but the distance formula is valid in any normed space.

Why Boundedness Matters

A linear functional can have an algebraically well-defined kernel without being bounded. Boundedness adds compatibility with the norm: small changes to the input force small changes to the scalar output, and the kernel becomes closed. The distance formula makes this compatibility especially clear. If $|L(x)|$ is small relative to $\|L\|$, then $x$ is close to an input that the functional sends to zero.

When estimating a functional, keep the domain norm in view. A coordinate formula may be bounded under one norm and require a different estimate under another. Also distinguish the kernel from the set where a functional takes a prescribed nonzero value: the kernel is a subspace, while a nonzero level set is generally a translate of that subspace. The decomposition theorem and distance formula concern the kernel, not an arbitrary level set.

Takeaway: A linear functional maps a vector space to the scalars. Its norm is the least uniform bound on its output. A nonzero functional splits the space into its kernel and one additional direction; when it is bounded, the kernel is closed and distance to it is exactly $|L(x)|/\|L\|$.

Check Your Understanding

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

What estimate proves that the integral functional in the example is bounded, and which input proves its norm is exact?
Why does a nonzero linear functional have a vector $u$ with $L(u)=1$?
In the decomposition theorem, why must the coefficient of $u$ equal $L(x)$?
Where does boundedness enter the proof that the kernel is closed?
Why can the distance formula be proved without assuming the functional norm is attained?

What You'll Learn

Scalar-Valued Linear Maps

Computing Functional Norms

Worked Example: A Coordinate Functional on \(\mathbb{R}^3\)

Worked Example: An Integral Functional

Worked Example: An Endpoint-Difference Functional

The Kernel and Its Geometry

Distance from the Kernel

Why Boundedness Matters

Check Your Understanding

Linear Functional

What You'll Learn

Scalar-Valued Linear Maps

Computing Functional Norms

Worked Example: A Coordinate Functional on \(\mathbb{R}^3\)

Worked Example: An Integral Functional

Worked Example: An Endpoint-Difference Functional

The Kernel and Its Geometry

Distance from the Kernel

Why Boundedness Matters

Check Your Understanding

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