Linear form

Suppose that vectors in the real coordinate space ${\displaystyle \mathbb {R} ^{n}}$ are represented as column vectors

$${\displaystyle \mathbf {x} ={\begin{bmatrix}x_{1}\\\vdots \\x_{n}\end{bmatrix}}.}$$

For each row vector ${\displaystyle \mathbf {a} ={\begin{bmatrix}a_{1}&\cdots &a_{n}\end{bmatrix}}}$ there is a linear functional ${\displaystyle f_{\mathbf {a} }}$ defined by

$${\displaystyle f_{\mathbf {a} }(\mathbf {x} )=a_{1}x_{1}+\cdots +a_{n}x_{n},}$$

and each linear functional can be expressed in this form.

This can be interpreted as either the matrix product or the dot product of the row vector ${\displaystyle \mathbf {a} }$ and the column vector ${\displaystyle \mathbf {x} }$:

$${\displaystyle f_{\mathbf {a} }(\mathbf {x} )=\mathbf {a} \cdot \mathbf {x} ={\begin{bmatrix}a_{1}&\cdots &a_{n}\end{bmatrix}}{\begin{bmatrix}x_{1}\\\vdots \\x_{n}\end{bmatrix}}.}$$

The trace ${\displaystyle \operatorname {tr} (A)}$ of a square matrix ${\displaystyle A}$ is the sum of all elements on its main diagonal. Matrices can be multiplied by scalars and two matrices of the same dimension can be added together; these operations make a vector space from the set of all ${\displaystyle n\times n}$ matrices. The trace is a linear functional on this space because ${\displaystyle \operatorname {tr} (sA)=s\operatorname {tr} (A)}$ and ${\displaystyle \operatorname {tr} (A+B)=\operatorname {tr} (A)+\operatorname {tr} (B)}$ for all scalars ${\displaystyle s}$ and all ${\displaystyle n\times n}$ matrices ${\displaystyle A{\text{ and }}B.}$

Linear functionals first appeared in functional analysis, the study of vector spaces of functions. A typical example of a linear functional is integration: the linear transformation defined by the Riemann integral

$${\displaystyle I(f)=\int _{a}^{b}f(x)\,dx}$$

is a linear functional from the vector space ${\displaystyle C[a,b]}$ of continuous functions on the interval ${\displaystyle [a,b]}$ to the real numbers. The linearity of ${\displaystyle I}$ follows from the standard facts about the integral:

$${\displaystyle {\begin{aligned}I(f+g)&=\int _{a}^{b}[f(x)+g(x)]\,dx=\int _{a}^{b}f(x)\,dx+\int _{a}^{b}g(x)\,dx=I(f)+I(g)\\I(\alpha f)&=\int _{a}^{b}\alpha f(x)\,dx=\alpha \int _{a}^{b}f(x)\,dx=\alpha I(f).\end{aligned}}}$$

Let ${\displaystyle P_{n}}$ denote the vector space of real-valued polynomial functions of degree ${\displaystyle \leq n}$ defined on an interval ${\displaystyle [a,b].}$ If ${\displaystyle c\in [a,b],}$ then let ${\displaystyle \operatorname {ev} _{c}:P_{n}\to \mathbb {R} }$ be the evaluation functional

$${\displaystyle \operatorname {ev} _{c}f=f(c).}$$

The mapping ${\displaystyle f\mapsto f(c)}$ is linear since

$${\displaystyle {\begin{aligned}(f+g)(c)&=f(c)+g(c)\\(\alpha f)(c)&=\alpha f(c).\end{aligned}}}$$

If ${\displaystyle x_{0},\ldots ,x_{n}}$ are ${\displaystyle n+1}$ distinct points in ${\displaystyle [a,b],}$ then the evaluation functionals ${\displaystyle \operatorname {ev} _{x_{i}},}$ ${\displaystyle i=0,\ldots ,n}$ form a basis of the dual space of ${\displaystyle P_{n}}$ (Lax (1996) proves this last fact using Lagrange interpolation).

A function ${\displaystyle f}$ having the equation of a line ${\displaystyle f(x)=a+rx}$ with ${\displaystyle a\neq 0}$ (for example, ${\displaystyle f(x)=1+2x}$) is not a linear functional on ${\displaystyle \mathbb {R} }$, since it is not linear.[nb 1] It is, however, affine-linear.

If ${\displaystyle x_{0},\ldots ,x_{n}}$ are ${\displaystyle n+1}$ distinct points in [a, b], then the linear functionals ${\displaystyle \operatorname {ev} _{x_{i}}:f\mapsto f\left(x_{i}\right)}$ defined above form a basis of the dual space of P_n, the space of polynomials of degree ${\displaystyle \leq n.}$ The integration functional I is also a linear functional on P_n, and so can be expressed as a linear combination of these basis elements. In symbols, there are coefficients ${\displaystyle a_{0},\ldots ,a_{n}}$ for which

$${\displaystyle I(f)=a_{0}f(x_{0})+a_{1}f(x_{1})+\dots +a_{n}f(x_{n})}$$

for all ${\displaystyle f\in P_{n}.}$ This forms the foundation of the theory of numerical quadrature.[6]

Linear functionals are particularly important in quantum mechanics. Quantum mechanical systems are represented by Hilbert spaces, which are anti–isomorphic to their own dual spaces. A state of a quantum mechanical system can be identified with a linear functional. For more information see bra–ket notation.

In the theory of generalized functions, certain kinds of generalized functions called distributions can be realized as linear functionals on spaces of test functions.

Let the vector space V have a basis ${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\dots ,\mathbf {e} _{n}}$, not necessarily orthogonal. Then the dual space ${\displaystyle V^{*}}$ has a basis ${\displaystyle {\tilde {\omega }}^{1},{\tilde {\omega }}^{2},\dots ,{\tilde {\omega }}^{n}}$ called the dual basis defined by the special property that

$${\displaystyle {\tilde {\omega }}^{i}(\mathbf {e} _{j})={\begin{cases}1&{\text{if}}\ i=j\\0&{\text{if}}\ i\neq j.\end{cases}}}$$

Or, more succinctly,

$${\displaystyle {\tilde {\omega }}^{i}(\mathbf {e} _{j})=\delta _{ij}}$$

where δ is the Kronecker delta. Here the superscripts of the basis functionals are not exponents but are instead contravariant indices.

A linear functional ${\displaystyle {\tilde {u}}}$ belonging to the dual space ${\displaystyle {\tilde {V}}}$ can be expressed as a linear combination of basis functionals, with coefficients ("components") u_i,

$${\displaystyle {\tilde {u}}=\sum _{i=1}^{n}u_{i}\,{\tilde {\omega }}^{i}.}$$

Then, applying the functional ${\displaystyle {\tilde {u}}}$ to a basis vector ${\displaystyle \mathbf {e} _{j}}$ yields

$${\displaystyle {\tilde {u}}(\mathbf {e} _{j})=\sum _{i=1}^{n}\left(u_{i}\,{\tilde {\omega }}^{i}\right)\mathbf {e} _{j}=\sum _{i}u_{i}\left[{\tilde {\omega }}^{i}\left(\mathbf {e} _{j}\right)\right]}$$

due to linearity of scalar multiples of functionals and pointwise linearity of sums of functionals. Then

$${\displaystyle {\begin{aligned}{\tilde {u}}({\mathbf {e} }_{j})&=\sum _{i}u_{i}\left[{\tilde {\omega }}^{i}\left({\mathbf {e} }_{j}\right)\right]\\&=\sum _{i}u_{i}{\delta }_{ij}\\&=u_{j}.\end{aligned}}}$$

So each component of a linear functional can be extracted by applying the functional to the corresponding basis vector.

When the space V carries an inner product, then it is possible to write explicitly a formula for the dual basis of a given basis. Let V have (not necessarily orthogonal) basis ${\displaystyle \mathbf {e} _{1},\dots ,\mathbf {e} _{n}.}$ In three dimensions (n = 3), the dual basis can be written explicitly

$${\displaystyle {\tilde {\omega }}^{i}(\mathbf {v} )={\frac {1}{2}}\left\langle {\frac {\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsilon ^{ijk}\,(\mathbf {e} _{j}\times \mathbf {e} _{k})}{\mathbf {e} _{1}\cdot \mathbf {e} _{2}\times \mathbf {e} _{3}}},\mathbf {v} \right\rangle ,}$$

for ${\displaystyle i=1,2,3,}$ where ε is the Levi-Civita symbol and ${\displaystyle \langle \cdot ,\cdot \rangle }$ the inner product (or dot product) on V.

In higher dimensions, this generalizes as follows

$${\displaystyle {\tilde {\omega }}^{i}(\mathbf {v} )=\left\langle {\frac {\sum _{1\leq i_{2}<i_{3}<\dots <i_{n}\leq n}\varepsilon ^{ii_{2}\dots i_{n}}(\star \mathbf {e} _{i_{2}}\wedge \cdots \wedge \mathbf {e} _{i_{n}})}{\star (\mathbf {e} _{1}\wedge \cdots \wedge \mathbf {e} _{n})}},\mathbf {v} \right\rangle ,}$$

where ${\displaystyle \star }$ is the Hodge star operator.

Every linear functional on ${\displaystyle X}$ is complex-valued while every linear functional on ${\displaystyle X_{\mathbb {R} }}$ is real-valued. If ${\displaystyle \dim X\neq 0}$ then a linear functional on either one of ${\displaystyle X}$ or ${\displaystyle X_{\mathbb {R} }}$ is non-trivial (meaning not identically ${\displaystyle 0}$) if and only if it is surjective (because if ${\displaystyle \varphi (x)\neq 0}$ then for any scalar ${\displaystyle s,}$ ${\displaystyle \varphi \left((s/\varphi (x))x\right)=s}$), where the image of a linear functional on ${\displaystyle X}$ is ${\displaystyle \mathbb {C} }$ while the image of a linear functional on ${\displaystyle X_{\mathbb {R} }}$ is ${\displaystyle \mathbb {R} .}$ Consequently, the only function on ${\displaystyle X}$ that is both a linear functional on ${\displaystyle X}$ and a linear function on ${\displaystyle X_{\mathbb {R} }}$ is the trivial functional; in other words, ${\displaystyle X^{\#}\cap X_{\mathbb {R} }^{\#}=\{0\},}$ where ${\displaystyle \,{\cdot }^{\#}}$ denotes the space's algebraic dual space. However, every ${\displaystyle \mathbb {C} }$-linear functional on ${\displaystyle X}$ is an ${\displaystyle \mathbb {R} }$-linear operator (meaning that it is additive and homogeneous over ${\displaystyle \mathbb {R} }$), but unless it is identically ${\displaystyle 0,}$ it is not an ${\displaystyle \mathbb {R} }$-linear functional on ${\displaystyle X}$ because its range (which is ${\displaystyle \mathbb {C} }$) is 2-dimensional over ${\displaystyle \mathbb {R} .}$ Conversely, a non-zero ${\displaystyle \mathbb {R} }$-linear functional has range too small to be a ${\displaystyle \mathbb {C} }$-linear functional as well.

If ${\displaystyle \varphi \in X^{\#}}$ then denote its real part by ${\displaystyle \varphi _{\mathbb {R} }:=\operatorname {Re} \varphi }$ and its imaginary part by ${\displaystyle \varphi _{i}:=\operatorname {Im} \varphi .}$ Then ${\displaystyle \varphi _{\mathbb {R} }:X\to \mathbb {R} }$ and ${\displaystyle \varphi _{i}:X\to \mathbb {R} }$ are linear functionals on ${\displaystyle X_{\mathbb {R} }}$ and ${\displaystyle \varphi =\varphi _{\mathbb {R} }+i\varphi _{i}.}$ The fact that ${\displaystyle z=\operatorname {Re} z-i\operatorname {Re} (iz)=\operatorname {Im} (iz)+i\operatorname {Im} z}$ for all ${\displaystyle z\in \mathbb {C} }$ implies that for all ${\displaystyle x\in X,}$[9]

$${\displaystyle {\begin{alignedat}{4}\varphi (x)&=\varphi _{\mathbb {R} }(x)-i\varphi _{\mathbb {R} }(ix)\\&=\varphi _{i}(ix)+i\varphi _{i}(x)\\\end{alignedat}}}$$

and consequently, that ${\displaystyle \varphi _{i}(x)=-\varphi _{\mathbb {R} }(ix)}$ and ${\displaystyle \varphi _{\mathbb {R} }(x)=\varphi _{i}(ix).}$[10]

The assignment ${\displaystyle \varphi \mapsto \varphi _{\mathbb {R} }}$ defines a bijective[10] ${\displaystyle \mathbb {R} }$-linear operator ${\displaystyle X^{\#}\to X_{\mathbb {R} }^{\#}}$ whose inverse is the map ${\displaystyle L_{\bullet }:X_{\mathbb {R} }^{\#}\to X^{\#}}$ defined by the assignment ${\displaystyle g\mapsto L_{g}}$ that sends ${\displaystyle g:X_{\mathbb {R} }\to \mathbb {R} }$ to the linear functional ${\displaystyle L_{g}:X\to \mathbb {C} }$ defined by

$${\displaystyle L_{g}(x):=g(x)-ig(ix)\quad {\text{ for all }}x\in X.}$$

The real part of ${\displaystyle L_{g}}$ is ${\displaystyle g}$ and the bijection ${\displaystyle L_{\bullet }:X_{\mathbb {R} }^{\#}\to X^{\#}}$ is an ${\displaystyle \mathbb {R} }$-linear operator, meaning that ${\displaystyle L_{g+h}=L_{g}+L_{h}}$ and ${\displaystyle L_{rg}=rL_{g}}$ for all ${\displaystyle r\in \mathbb {R} }$ and ${\displaystyle g,h\in X_{\mathbb {R} }^{\#}.}$[10] Similarly for the imaginary part, the assignment ${\displaystyle \varphi \mapsto \varphi _{i}}$ induces an ${\displaystyle \mathbb {R} }$-linear bijection ${\displaystyle X^{\#}\to X_{\mathbb {R} }^{\#}}$ whose inverse is the map ${\displaystyle X_{\mathbb {R} }^{\#}\to X^{\#}}$ defined by sending ${\displaystyle I\in X_{\mathbb {R} }^{\#}}$ to the linear functional on ${\displaystyle X}$ defined by ${\displaystyle x\mapsto I(ix)+iI(x).}$

This relationship was discovered by Henry Löwig in 1934 (although it is usually credited to F. Murray),[11] and can be generalized to arbitrary finite extensions of a field in the natural way. It has many important consequences, some of which will now be described.

Suppose ${\displaystyle \varphi :X\to \mathbb {C} }$ is a linear functional on ${\displaystyle X}$ with real part ${\displaystyle \varphi _{\mathbb {R} }:=\operatorname {Re} \varphi }$ and imaginary part ${\displaystyle \varphi _{i}:=\operatorname {Im} \varphi .}$

Then ${\displaystyle \varphi =0}$ if and only if ${\displaystyle \varphi _{\mathbb {R} }=0}$ if and only if ${\displaystyle \varphi _{i}=0.}$

Assume that ${\displaystyle X}$ is a topological vector space. Then ${\displaystyle \varphi }$ is continuous if and only if its real part ${\displaystyle \varphi _{\mathbb {R} }}$ is continuous, if and only if ${\displaystyle \varphi }$'s imaginary part ${\displaystyle \varphi _{i}}$ is continuous. That is, either all three of ${\displaystyle \varphi ,\varphi _{\mathbb {R} },}$ and ${\displaystyle \varphi _{i}}$ are continuous or none are continuous. This remains true if the word "continuous" is replaced with the word "bounded". In particular, ${\displaystyle \varphi \in X^{\prime }}$ if and only if ${\displaystyle \varphi _{\mathbb {R} }\in X_{\mathbb {R} }^{\prime }}$ where the prime denotes the space's continuous dual space.[9]

Let ${\displaystyle B\subseteq X.}$ If ${\displaystyle uB\subseteq B}$ for all scalars ${\displaystyle u\in \mathbb {C} }$ of unit length (meaning ${\displaystyle |u|=1}$) then[proof 1][12]

$${\displaystyle \sup _{b\in B}|\varphi (b)|=\sup _{b\in B}\left|\varphi _{\mathbb {R} }(b)\right|.}$$

Similarly, if ${\displaystyle \varphi _{i}:=\operatorname {Im} \varphi :X\to \mathbb {R} }$ denotes the complex part of ${\displaystyle \varphi }$ then ${\displaystyle iB\subseteq B}$ implies

$${\displaystyle \sup _{b\in B}\left|\varphi _{\mathbb {R} }(b)\right|=\sup _{b\in B}\left|\varphi _{i}(b)\right|.}$$

If ${\displaystyle X}$ is a normed space with norm ${\displaystyle \|\cdot \|}$ and if ${\displaystyle B=\{x\in X:\|x\|\leq 1\}}$ is the closed unit ball then the supremums above are the operator norms (defined in the usual way) of ${\displaystyle \varphi ,\varphi _{\mathbb {R} },}$ and ${\displaystyle \varphi _{i}}$ so that [12]

$${\displaystyle \|\varphi \|=\left\|\varphi _{\mathbb {R} }\right\|=\left\|\varphi _{i}\right\|.}$$

This conclusion extends to the analogous statement for polars of balanced sets in general topological vector spaces.

• If ${\displaystyle X}$ is a complex Hilbert space with a (complex) inner product ${\displaystyle \langle \,\cdot \,|\,\cdot \,\rangle }$ that is antilinear in its first coordinate (and linear in the second) then ${\displaystyle X_{\mathbb {R} }}$ becomes a real Hilbert space when endowed with the real part of ${\displaystyle \langle \,\cdot \,|\,\cdot \,\rangle .}$ Explicitly, this real inner product on ${\displaystyle X_{\mathbb {R} }}$ is defined by ${\displaystyle \langle x|y\rangle _{\mathbb {R} }:=\operatorname {Re} \langle x|y\rangle }$ for all ${\displaystyle x,y\in X}$ and it induces the same norm on ${\displaystyle X}$ as ${\displaystyle \langle \,\cdot \,|\,\cdot \,\rangle }$ because ${\displaystyle {\sqrt {\langle x|x\rangle _{\mathbb {R} }}}={\sqrt {\langle x|x\rangle }}}$ for all vectors ${\displaystyle x.}$ Applying the Riesz representation theorem to ${\displaystyle \varphi \in X^{\prime }}$ (resp. to ${\displaystyle \varphi _{\mathbb {R} }\in X_{\mathbb {R} }^{\prime }}$) guarantees the existence of a unique vector ${\displaystyle f_{\varphi }\in X}$ (resp. ${\displaystyle f_{\varphi _{\mathbb {R} }}\in X_{\mathbb {R} }}$) such that ${\displaystyle \varphi (x)=\left\langle f_{\varphi }|\,x\right\rangle }$ (resp. ${\displaystyle \varphi _{\mathbb {R} }(x)=\left\langle f_{\varphi _{\mathbb {R} }}|\,x\right\rangle _{\mathbb {R} }}$) for all vectors ${\displaystyle x.}$ The theorem also guarantees that ${\displaystyle \left\|f_{\varphi }\right\|=\|\varphi \|_{X^{\prime }}}$ and ${\displaystyle \left\|f_{\varphi _{\mathbb {R} }}\right\|=\left\|\varphi _{\mathbb {R} }\right\|_{X_{\mathbb {R} }^{\prime }}.}$ It is readily verified that ${\displaystyle f_{\varphi }=f_{\varphi _{\mathbb {R} }}.}$ Now ${\displaystyle \left\|f_{\varphi }\right\|=\left\|f_{\varphi _{\mathbb {R} }}\right\|}$ and the previous equalities imply that ${\displaystyle \|\varphi \|_{X^{\prime }}=\left\|\varphi _{\mathbb {R} }\right\|_{X_{\mathbb {R} }^{\prime }},}$ which is the same conclusion that was reached above.

Continuous linear functionals have nice properties for analysis: a linear functional is continuous if and only if its kernel is closed,[14] and a non-trivial continuous linear functional is an open map, even if the (topological) vector space is not complete.[15]

A vector subspace ${\displaystyle M}$ of ${\displaystyle X}$ is called maximal if ${\displaystyle M\subsetneq X}$ (meaning ${\displaystyle M\subseteq X}$ and ${\displaystyle M\neq X}$) and does not exist a vector subspace ${\displaystyle N}$ of ${\displaystyle X}$ such that ${\displaystyle M\subsetneq N\subsetneq X.}$ A vector subspace ${\displaystyle M}$ of ${\displaystyle X}$ is maximal if and only if it is the kernel of some non-trivial linear functional on ${\displaystyle X}$ (that is, ${\displaystyle M=\ker f}$ for some linear functional ${\displaystyle f}$ on ${\displaystyle X}$ that is not identically 0). An affine hyperplane in ${\displaystyle X}$ is a translate of a maximal vector subspace. By linearity, a subset ${\displaystyle H}$ of ${\displaystyle X}$ is a affine hyperplane if and only if there exists some non-trivial linear functional ${\displaystyle f}$ on ${\displaystyle X}$ such that ${\displaystyle H=f^{-1}(1)=\{x\in X:f(x)=1\}.}$[11] If ${\displaystyle f}$ is a linear functional and ${\displaystyle s\neq 0}$ is a scalar then ${\displaystyle f^{-1}(s)=s\left(f^{-1}(1)\right)=\left({\frac {1}{s}}f\right)^{-1}(1).}$ This equality can be used to relate different level sets of ${\displaystyle f.}$ Moreover, if ${\displaystyle f\neq 0}$ then the kernel of ${\displaystyle f}$ can be reconstructed from the affine hyperplane ${\displaystyle H:=f^{-1}(1)}$ by ${\displaystyle \ker f=H-H.}$

Any two linear functionals with the same kernel are proportional (i.e. scalar multiples of each other). This fact can be generalized to the following theorem.

If f is a non-trivial linear functional on X with kernel N, ${\displaystyle x\in X}$ satisfies ${\displaystyle f(x)=1,}$ and U is a balanced subset of X, then ${\displaystyle N\cap (x+U)=\varnothing }$ if and only if ${\displaystyle |f(u)|<1}$ for all ${\displaystyle u\in U.}$[15]

Any (algebraic) linear functional on a vector subspace can be extended to the whole space; for example, the evaluation functionals described above can be extended to the vector space of polynomials on all of ${\displaystyle \mathbb {R} .}$ However, this extension cannot always be done while keeping the linear functional continuous. The Hahn–Banach family of theorems gives conditions under which this extension can be done. For example,

Hahn–Banach dominated extension theorem[18](Rudin 1991, Th. 3.2) — If ${\displaystyle p:X\to \mathbb {R} }$ is a sublinear function, and ${\displaystyle f:M\to \mathbb {R} }$ is a linear functional on a linear subspace ${\displaystyle M\subseteq X}$ which is dominated by p on M, then there exists a linear extension ${\displaystyle F:X\to \mathbb {R} }$ of f to the whole space X that is dominated by p, i.e., there exists a linear functional F such that

$${\displaystyle F(m)=f(m)}$$

for all ${\displaystyle m\in M,}$ and

$${\displaystyle |F(x)|\leq p(x)}$$

for all ${\displaystyle x\in X.}$

Let X be a topological vector space (TVS) with continuous dual space ${\displaystyle X'.}$

For any subset H of ${\displaystyle X',}$ the following are equivalent:[19]

1. H is equicontinuous;
2. H is contained in the polar of some neighborhood of ${\displaystyle 0}$ in X;
3. the (pre)polar of H is a neighborhood of ${\displaystyle 0}$ in X;

If H is an equicontinuous subset of ${\displaystyle X'}$ then the following sets are also equicontinuous: the weak-* closure, the balanced hull, the convex hull, and the convex balanced hull.[19] Moreover, Alaoglu's theorem implies that the weak-* closure of an equicontinuous subset of ${\displaystyle X'}$ is weak-* compact (and thus that every equicontinuous subset weak-* relatively compact).[20][19]