Linear form#Real and imaginary parts of a linear functional

The constant zero function, mapping every vector to zero, is trivially a linear functional. Every other linear functional (such as the ones below) is surjective (that is, its range is all of {{mvar|k}}).

• Indexing into a vector: The second element of a three-vector is given by the one-form $[0,\; 1,\; 0].$ That is, the second element of $[x,\; y,\; z]$ is $$[0,\; 1,\; 0]\; \backslash cdot\; [x,\; y,\; z]\; =\; y.$$
• Mean: The mean element of an $n$-vector is given by the one-form $\backslash left[1/n,\; 1/n,\; \backslash ldots,\; 1/n\backslash right].$ That is, $$\backslash operatorname\{mean\}(v)\; =\; \backslash left[1/n,\; 1/n,\; \backslash ldots,\; 1/n\backslash right]\; \backslash cdot\; v.$$
• Sampling: Sampling with a kernel can be considered a one-form, where the one-form is the kernel shifted to the appropriate location.
• Net present value of a net cash flow, $R(t),$ is given by the one-form $w(t)\; =\; (1\; +\; i)^\{-t\}$ where $i$ is the discount rate. That is, $$\backslash mathrm\{NPV\}(R(t))\; =\; \backslash langle\; w,\; R\backslash rangle\; =\; \backslash int\_\{t=0\}^\backslash infty\; \backslash frac\{R(t)\}\{(1+i)^\{t\}\}\backslash ,dt.$$

Suppose that vectors in the real coordinate space $\backslash R^n$ are represented as column vectors

$$\backslash mathbf\{x\}\; =\; \backslash begin\{bmatrix\}x\_1\backslash \backslash \; \backslash vdots\backslash \backslash \; x\_n\backslash end\{bmatrix\}.$$

For each row vector there is a linear functional $f\_\{\backslash mathbf\{a\}\}$ defined by

$$f\_\{\backslash mathbf\{a\}\}(\backslash mathbf\{x\})\; =\; a\_1\; x\_1\; +\; \backslash cdots\; +\; a\_n\; x\_n,$$

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 $\backslash mathbf\{a\}$ and the column vector $\backslash mathbf\{x\}$:

The trace $\backslash operatorname\{tr\}\; (A)$ of a square matrix $A$ 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 $n\; \backslash times\; n$ matrices. The trace is a linear functional on this space because $\backslash operatorname\{tr\}\; (s\; A)\; =\; s\; \backslash operatorname\{tr\}\; (A)$ and $\backslash operatorname\{tr\}\; (A\; +\; B)\; =\; \backslash operatorname\{tr\}\; (A)\; +\; \backslash operatorname\{tr\}\; (B)$ for all scalars $s$ and all $n\; \backslash times\; n$ matrices $A\; \backslash text\{\; and\; \}\; B.$

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

$$I(f)\; =\; \backslash int\_a^b\; f(x)\backslash ,\; dx$$

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

$$\backslash begin\{align\}$$

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{align}

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

$$\backslash operatorname\{ev\}\_c\; f\; =\; f(c).$$

The mapping $f\; \backslash mapsto\; f(c)$ is linear since

$$\backslash begin\{align\}$$

(f + g)(c) &= f(c) + g(c) \\

(\alpha f)(c) &= \alpha f(c).

\end{align}

If $x\_0,\; \backslash ldots,\; x\_n$ are $n\; +\; 1$ distinct points in $[a,\; b],$ then the evaluation functionals $\backslash operatorname\{ev\}\_\{x\_i\},$ $i\; =\; 0,\; \backslash ldots,\; n$ form a basis of the dual space of $P\_n$ ({{harvtxt|Lax|1996}} proves this last fact using Lagrange interpolation).

A function $f$ having the equation of a line $f(x)\; =\; a\; +\; r\; x$ with $a\; \backslash neq\; 0$ (for example, $f(x)\; =\; 1\; +\; 2\; x$) is {{em|not}} a linear functional on $\backslash R$, since it is not linear.For instance, $f(1\; +\; 1)\; =\; a\; +\; 2\; r\; \backslash neq\; 2\; a\; +\; 2\; r\; =\; f(1)\; +\; f(1).$ It is, however, affine-linear.

File:Gradient 1-form.svgs of constant value, each corresponding to those vectors that α maps to a given scalar value shown next to it along with the "sense" of increase. The {{color box|purple}} zero plane is through the origin.]]

In finite dimensions, a linear functional can be visualized in terms of its level sets, the sets of vectors which map to a given value. In three dimensions, the level sets of a linear functional are a family of mutually parallel planes; in higher dimensions, they are parallel hyperplanes. This method of visualizing linear functionals is sometimes introduced in general relativity texts, such as Gravitation by {{harvtxt|Misner|Thorne|Wheeler|1973}}.

If $x\_0,\; \backslash ldots,\; x\_n$ are $n\; +\; 1$ distinct points in {{closed-closed|a, b}}, then the linear functionals $\backslash operatorname\{ev\}\_\{x\_i\}\; :\; f\; \backslash mapsto\; f\backslash left(x\_i\backslash right)$ defined above form a basis of the dual space of {{math|P_n}}, the space of polynomials of degree $\backslash leq\; n.$ The integration functional {{math|I}} is also a linear functional on {{math|P_n}}, and so can be expressed as a linear combination of these basis elements. In symbols, there are coefficients $a\_0,\; \backslash ldots,\; a\_n$ for which

$$I(f)\; =\; a\_0\; f(x\_0)\; +\; a\_1\; f(x\_1)\; +\; \backslash dots\; +\; a\_n\; f(x\_n)$$

for all $f\; \backslash in\; P\_n.$ This forms the foundation of the theory of numerical quadrature.{{harvnb|Lax|1996}}

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.

File:1-form linear functional.svg Euclidean space. The number of (1-form) hyperplanes intersected by a vector equals the inner product.{{Harvard citation text|Misner|Thorne|Wheeler|1973}} p. 57]]

Every non-degenerate bilinear form on a finite-dimensional vector space {{mvar|V}} induces an isomorphism {{math|V → V{{i sup|∗}} : v ↦ v{{sup|∗}}}} such that

$$v^*(w)\; :=\; \backslash langle\; v,\; w\backslash rangle\; \backslash quad\; \backslash forall\; w\; \backslash in\; V\; ,$$

where the bilinear form on {{mvar|V}} is denoted $\backslash langle\; \backslash ,\backslash cdot\backslash ,\; ,\; \backslash ,\backslash cdot\backslash ,\; \backslash rangle$ (for instance, in Euclidean space, $\backslash langle\; v,\; w\; \backslash rangle\; =\; v\; \backslash cdot\; w$ is the dot product of {{mvar|v}} and {{mvar|w}}).

The inverse isomorphism is {{nowrap|V{{i sup|∗}} → V : v{{sup|∗}} ↦ v}}, where {{mvar|v}} is the unique element of {{mvar|V}} such that

$$\backslash langle\; v,\; w\backslash rangle\; =\; v^*(w)$$

for all $w\; \backslash in\; V.$

The above defined vector {{nowrap|v{{sup|∗}} ∈ V{{i sup|∗}}}} is said to be the dual vector of $v\; \backslash in\; V.$

In an infinite dimensional Hilbert space, analogous results hold by the Riesz representation theorem. There is a mapping {{nowrap|V ↦ V{{i sup|∗}}}} from {{mvar|V}} into its {{em|continuous dual space}} V{{i sup|∗}}.

{{hatnote|Below, we assume that the dimension is finite. For a discussion of analogous results in infinite dimensions, see Schauder basis.}}

Let the vector space {{mvar|V}} have a basis $\backslash mathbf\{e\}\_1,\; \backslash mathbf\{e\}\_2,\backslash dots,\backslash mathbf\{e\}\_n$, not necessarily orthogonal. Then the dual space $V^*$ has a basis $\backslash tilde\{\backslash omega\}^1,\backslash tilde\{\backslash omega\}^2,\backslash dots,\backslash tilde\{\backslash omega\}^n$ called the dual basis defined by the special property that

$$\backslash tilde\{\backslash omega\}^i\; (\backslash mathbf\; e\_j)\; =\; \backslash begin\{cases\}$$

1 &\text{if}\ i = j\\

0 &\text{if}\ i \neq j.

\end{cases}

Or, more succinctly,

$$\backslash tilde\{\backslash omega\}^i\; (\backslash mathbf\; e\_j)\; =\; \backslash delta\_\{ij\}$$

where $\backslash delta\_\{ij\}$ is the Kronecker delta. Here the superscripts of the basis functionals are not exponents but are instead contravariant indices.

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

$$\backslash tilde\{u\}\; =\; \backslash sum\_\{i=1\}^n\; u\_i\; \backslash ,\; \backslash tilde\{\backslash omega\}^i.$$

Then, applying the functional $\backslash tilde\{u\}$ to a basis vector $\backslash mathbf\{e\}\_j$ yields

$$\backslash tilde\{u\}(\backslash mathbf\; e\_j)\; =\; \backslash sum\_\{i=1\}^n\; \backslash left(u\_i\; \backslash ,\; \backslash tilde\{\backslash omega\}^i\backslash right)\; \backslash mathbf\; e\_j\; =\; \backslash sum\_i\; u\_i\; \backslash left[\backslash tilde\{\backslash omega\}^i\; \backslash left(\backslash mathbf\; e\_j\backslash right)\backslash right]$$

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

$$\backslash begin\{align\}$$

\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{align}

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

When the space {{mvar|V}} carries an inner product, then it is possible to write explicitly a formula for the dual basis of a given basis. Let {{mvar|V}} have (not necessarily orthogonal) basis $\backslash mathbf\{e\}\_1,\backslash dots,\; \backslash mathbf\{e\}\_n.$ In three dimensions ({{math|1=n = 3}}), the dual basis can be written explicitly

$$\backslash tilde\{\backslash omega\}^i(\backslash mathbf\{v\})\; =\; \backslash frac\{1\}\{2\}\; \backslash left\backslash langle\; \backslash frac\; \{\; \backslash sum\_\{j=1\}^3\backslash sum\_\{k=1\}^3\backslash varepsilon^\{ijk\}\; \backslash ,\; (\backslash mathbf\; e\_j\; \backslash times\; \backslash mathbf\; e\_k)\}\; \{\backslash mathbf\; e\_1\; \backslash cdot\; \backslash mathbf\; e\_2\; \backslash times\; \backslash mathbf\; e\_3\}\; ,\; \backslash mathbf\{v\}\; \backslash right\backslash rangle\; ,$$

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

In higher dimensions, this generalizes as follows

where $\backslash star$ is the Hodge star operator.

Modules over a ring are generalizations of vector spaces, which removes the restriction that coefficients belong to a field. Given a module {{mvar|M}} over a ring {{mvar|R}}, a linear form on {{mvar|M}} is a linear map from {{mvar|M}} to {{mvar|R}}, where the latter is considered as a module over itself. The space of linear forms is always denoted {{math|Hom_k(V, k)}}, whether {{mvar|k}} is a field or not. It is a right module if {{mvar|V}} is a left module.

The existence of "enough" linear forms on a module is equivalent to projectivity.{{Cite book|last=Clark|first=Pete L.|url=http://alpha.math.uga.edu/~pete/integral2015.pdf|title=Commutative Algebra|publisher=Unpublished|at=Lemma 3.12}}

{{math theorem|math_statement=An {{mvar|R}}-module {{mvar|M}} is projective if and only if there exists a subset $A\backslash subset\; M$ and linear forms $\backslash \{f\_a\backslash mid\; a\; \backslash in\; A\backslash \}$ such that, for every $x\backslash in\; M,$ only finitely many $f\_a(x)$ are nonzero, and

$$x=\backslash sum\_\{a\backslash in\; A\}\{f\_a(x)a\}$$

|name=Dual Basis Lemma

}}

{{anchor|Real and complex linear functionals|Real and imaginary parts of a linear functional}}

{{See also|Linear complex structure|Complexification}}

Suppose that $X$ is a vector space over $\backslash Complex.$ Restricting scalar multiplication to $\backslash R$ gives rise to a real vector space{{sfn|Rudin|1991|pp=57}} $X\_\{\backslash R\}$ called the {{em|realification}} of $X.$

Any vector space $X$ over $\backslash Complex$ is also a vector space over $\backslash R,$ endowed with a complex structure; that is, there exists a real vector subspace $X\_\{\backslash R\}$ such that we can (formally) write $X\; =\; X\_\{\backslash R\}\; \backslash oplus\; X\_\{\backslash R\}i$ as $\backslash R$-vector spaces.

Every linear functional on $X$ is complex-valued while every linear functional on $X\_\{\backslash R\}$ is real-valued. If $\backslash dim\; X\; \backslash neq\; 0$ then a linear functional on either one of $X$ or $X\_\{\backslash R\}$ is non-trivial (meaning not identically $0$) if and only if it is surjective (because if $\backslash varphi(x)\; \backslash neq\; 0$ then for any scalar $s,$ $\backslash varphi\backslash left((s/\backslash varphi(x))\; x\backslash right)\; =\; s$), where the image of a linear functional on $X$ is $\backslash C$ while the image of a linear functional on $X\_\{\backslash R\}$ is $\backslash R.$

Consequently, the only function on $X$ that is both a linear functional on $X$ and a linear function on $X\_\{\backslash R\}$ is the trivial functional; in other words, $X^\{\backslash \#\}\; \backslash cap\; X\_\{\backslash R\}^\{\backslash \#\}\; =\; \backslash \{\; 0\; \backslash \},$ where $\backslash ,\{\backslash cdot\}^\{\backslash \#\}$ denotes the space's algebraic dual space.

However, every $\backslash Complex$-linear functional on $X$ is an Linear operator (meaning that it is additive and homogeneous over $\backslash R$), but unless it is identically $0,$ it is not an $\backslash R$-linear {{em|functional}} on $X$ because its range (which is $\backslash Complex$) is 2-dimensional over $\backslash R.$ Conversely, a non-zero $\backslash R$-linear functional has range too small to be a $\backslash Complex$-linear functional as well.

If $\backslash varphi\; \backslash in\; X^\{\backslash \#\}$ then denote its real part by $\backslash varphi\_\{\backslash R\}\; :=\; \backslash operatorname\{Re\}\; \backslash varphi$ and its imaginary part by $\backslash varphi\_i\; :=\; \backslash operatorname\{Im\}\; \backslash varphi.$

Then $\backslash varphi\_\{\backslash R\}\; :\; X\; \backslash to\; \backslash R$ and $\backslash varphi\_i\; :\; X\; \backslash to\; \backslash R$ are linear functionals on $X\_\{\backslash R\}$ and $\backslash varphi\; =\; \backslash varphi\_\{\backslash R\}\; +\; i\; \backslash varphi\_i.$

The fact that $z\; =\; \backslash operatorname\{Re\}\; z\; -\; i\; \backslash operatorname\{Re\}\; (i\; z)\; =\; \backslash operatorname\{Im\}\; (i\; z)\; +\; i\; \backslash operatorname\{Im\}\; z$ for all $z\; \backslash in\; \backslash Complex$ implies that for all $x\; \backslash in\; X,${{sfn|Rudin|1991|pp=57}}

&= \varphi_i(i x) + i \varphi_i(x)\\

\end{alignat}

and consequently, that $\backslash varphi\_i(x)\; =\; -\; \backslash varphi\_\{\backslash R\}(i\; x)$ and $\backslash varphi\_\{\backslash R\}(x)\; =\; \backslash varphi\_i(ix).${{sfn|Narici|Beckenstein|2011|pp=9-11}}

The assignment $\backslash varphi\; \backslash mapsto\; \backslash varphi\_\{\backslash R\}$ defines a bijective{{sfn|Narici|Beckenstein|2011|pp=9-11}} $\backslash R$-linear operator $X^\{\backslash \#\}\; \backslash to\; X\_\{\backslash R\}^\{\backslash \#\}$ whose inverse is the map $L\_\{\backslash bull\}\; :\; X\_\{\backslash R\}^\{\backslash \#\}\; \backslash to\; X^\{\backslash \#\}$ defined by the assignment $g\; \backslash mapsto\; L\_g$ that sends $g\; :\; X\_\{\backslash R\}\; \backslash to\; \backslash R$ to the linear functional $L\_g\; :\; X\; \backslash to\; \backslash Complex$ defined by

$$L\_g(x)\; :=\; g(x)\; -\; i\; g(ix)\; \backslash quad\; \backslash text\{\; for\; all\; \}\; x\; \backslash in\; X.$$

The real part of $L\_g$ is $g$ and the bijection $L\_\{\backslash bull\}\; :\; X\_\{\backslash R\}^\{\backslash \#\}\; \backslash to\; X^\{\backslash \#\}$ is an $\backslash R$-linear operator, meaning that $L\_\{g+h\}\; =\; L\_g\; +\; L\_h$ and $L\_\{rg\}\; =\; r\; L\_g$ for all $r\; \backslash in\; \backslash R$ and $g,\; h\; \backslash in\; X\_\backslash R^\{\backslash \#\}.${{sfn|Narici|Beckenstein|2011|pp=9-11}}

Similarly for the imaginary part, the assignment $\backslash varphi\; \backslash mapsto\; \backslash varphi\_i$ induces an $\backslash R$-linear bijection $X^\{\backslash \#\}\; \backslash to\; X\_\{\backslash R\}^\{\backslash \#\}$ whose inverse is the map $X\_\{\backslash R\}^\{\backslash \#\}\; \backslash to\; X^\{\backslash \#\}$ defined by sending $I\; \backslash in\; X\_\{\backslash R\}^\{\backslash \#\}$ to the linear functional on $X$ defined by $x\; \backslash mapsto\; I(i\; x)\; +\; i\; I(x).$

This relationship was discovered by Henry Löwig in 1934 (although it is usually credited to F. Murray),{{sfn|Narici|Beckenstein|2011|pp=10-11}} 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 $\backslash varphi\; :\; X\; \backslash to\; \backslash Complex$ is a linear functional on $X$ with real part $\backslash varphi\_\{\backslash R\}\; :=\; \backslash operatorname\{Re\}\; \backslash varphi$ and imaginary part $\backslash varphi\_i\; :=\; \backslash operatorname\{Im\}\; \backslash varphi.$

Then $\backslash varphi\; =\; 0$ if and only if $\backslash varphi\_\{\backslash R\}\; =\; 0$ if and only if $\backslash varphi\_i\; =\; 0.$

Assume that $X$ is a topological vector space. Then $\backslash varphi$ is continuous if and only if its real part $\backslash varphi\_\{\backslash R\}$ is continuous, if and only if $\backslash varphi$'s imaginary part $\backslash varphi\_i$ is continuous. That is, either all three of $\backslash varphi,\; \backslash varphi\_\{\backslash R\},$ and $\backslash varphi\_i$ are continuous or none are continuous. This remains true if the word "continuous" is replaced with the word "bounded". In particular, $\backslash varphi\; \backslash in\; X^\{\backslash prime\}$ if and only if $\backslash varphi\_\{\backslash R\}\; \backslash in\; X\_\{\backslash R\}^\{\backslash prime\}$ where the prime denotes the space's continuous dual space.{{sfn|Rudin|1991|pp=57}}

Let $B\; \backslash subseteq\; X.$ If $u\; B\; \backslash subseteq\; B$ for all scalars $u\; \backslash in\; \backslash Complex$ of unit length (meaning $|u|\; =\; 1$) thenIt is true if $B\; =\; \backslash varnothing$ so assume otherwise. Since $\backslash left|\backslash operatorname\{Re\}\; z\backslash right|\; \backslash leq\; |z|$ for all scalars $z\; \backslash in\; \backslash Complex,$ it follows that $\backslash sup\_\{x\; \backslash in\; B\}\; \backslash left|\backslash varphi\_\{\backslash R\}(x)\backslash right|\; \backslash leq\; \backslash sup\_\{x\; \backslash in\; B\}\; |\backslash varphi(x)|.$ If $b\; \backslash in\; B$ then let $r\_b\; \backslash geq\; 0$ and $u\_b\; \backslash in\; \backslash Complex$ be such that $\backslash left|u\_b\backslash right|\; =\; 1$ and $\backslash varphi(b)\; =\; r\_b\; u\_b,$ where if $r\_b\; =\; 0$ then take $u\_b\; :=\; 1.$Then $|\backslash varphi(b)|\; =\; r\_b$ and because $\backslash varphi\backslash left(\backslash frac\{1\}\{u\_b\}\; b\backslash right)\; =\; r\_b$ is a real number, $\backslash varphi\_\{\backslash R\}\backslash left(\backslash frac\{1\}\{u\_b\}\; b\backslash right)\; =\; \backslash varphi\backslash left(\backslash frac\{1\}\{u\_b\}\; b\backslash right)\; =\; r\_b.$ By assumption $\backslash frac\{1\}\{u\_b\}\; b\; \backslash in\; B$ so $|\backslash varphi(b)|\; =\; r\_b\; \backslash leq\; \backslash sup\_\{x\; \backslash in\; B\}\; \backslash left|\backslash varphi\_\{\backslash R\}(x)\backslash right|.$ Since $b\; \backslash in\; B$ was arbitrary, it follows that $\backslash sup\_\{x\; \backslash in\; B\}\; |\backslash varphi(x)|\; \backslash leq\; \backslash sup\_\{x\; \backslash in\; B\}\; \backslash left|\backslash varphi\_\{\backslash R\}(x)\backslash right|.$ $\backslash blacksquare${{sfn|Narici|Beckenstein|2011|pp=126-128}} $$\backslash sup\_\{b\; \backslash in\; B\}\; |\backslash varphi(b)|\; =\; \backslash sup\_\{b\; \backslash in\; B\}\; \backslash left|\backslash varphi\_\{\backslash R\}(b)\backslash right|.$$

Similarly, if $\backslash varphi\_i\; :=\; \backslash operatorname\{Im\}\; \backslash varphi\; :\; X\; \backslash to\; \backslash R$ denotes the complex part of $\backslash varphi$ then $i\; B\; \backslash subseteq\; B$ implies $$\backslash sup\_\{b\; \backslash in\; B\}\; \backslash left|\backslash varphi\_\{\backslash R\}(b)\backslash right|\; =\; \backslash sup\_\{b\; \backslash in\; B\}\; \backslash left|\backslash varphi\_i(b)\backslash right|.$$

If $X$ is a normed space with norm $\backslash |\backslash cdot\backslash |$ and if $B\; =\; \backslash \{x\; \backslash in\; X\; :\; \backslash |\; x\; \backslash |\; \backslash leq\; 1\backslash \}$ is the closed unit ball then the supremums above are the operator norms (defined in the usual way) of $\backslash varphi,\; \backslash varphi\_\{\backslash R\},$ and $\backslash varphi\_i$ so that{{sfn|Narici|Beckenstein|2011|pp=126-128}} $$\backslash |\backslash varphi\backslash |\; =\; \backslash left\backslash |\backslash varphi\_\{\backslash R\}\backslash right\backslash |\; =\; \backslash left\backslash |\backslash varphi\_i\; \backslash right\backslash |.$$

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

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

{{see also|Continuous linear operator}}Below, all vector spaces are over either the real numbers $\backslash R$ or the complex numbers $\backslash Complex.$

If $V$ is a topological vector space, the space of continuous linear functionals — the {{em|continuous dual}} — is often simply called the dual space. If $V$ is a Banach space, then so is its (continuous) dual. To distinguish the ordinary dual space from the continuous dual space, the former is sometimes called the {{em|algebraic dual space}}. In finite dimensions, every linear functional is continuous, so the continuous dual is the same as the algebraic dual, but in infinite dimensions the continuous dual is a proper subspace of the algebraic dual.

A linear functional {{mvar|f}} on a (not necessarily locally convex) topological vector space {{mvar|X}} is continuous if and only if there exists a continuous seminorm {{mvar|p}} on {{mvar|X}} such that $|f|\; \backslash leq\; p.${{sfn|Narici|Beckenstein|2011|p=126}}

Continuous linear functionals have nice properties for analysis: a linear functional is continuous if and only if its kernel is closed,{{harvnb|Rudin|1991|loc=Theorem 1.18}} and a non-trivial continuous linear functional is an open map, even if the (topological) vector space is not complete.{{sfn|Narici|Beckenstein|2011 |p=128}}

A vector subspace $M$ of $X$ is called maximal if $M\; \backslash subsetneq\; X$ (meaning $M\; \backslash subseteq\; X$ and $M\; \backslash neq\; X$) and does not exist a vector subspace $N$ of $X$ such that $M\; \backslash subsetneq\; N\; \backslash subsetneq\; X.$ A vector subspace $M$ of $X$ is maximal if and only if it is the kernel of some non-trivial linear functional on $X$ (that is, $M\; =\; \backslash ker\; f$ for some linear functional $f$ on $X$ that is not identically {{math|0}}). An affine hyperplane in $X$ is a translate of a maximal vector subspace. By linearity, a subset $H$ of $X$ is a affine hyperplane if and only if there exists some non-trivial linear functional $f$ on $X$ such that $H\; =\; f^\{-1\}(1)\; =\; \backslash \{\; x\; \backslash in\; X\; :\; f(x)\; =\; 1\; \backslash \}.${{sfn|Narici|Beckenstein|2011|pp=10-11}}

If $f$ is a linear functional and $s\; \backslash neq\; 0$ is a scalar then $f^\{-1\}(s)\; =\; s\; \backslash left(f^\{-1\}(1)\backslash right)\; =\; \backslash left(\backslash frac\{1\}\{s\}\; f\backslash right)^\{-1\}(1).$ This equality can be used to relate different level sets of $f.$ Moreover, if $f\; \backslash neq\; 0$ then the kernel of $f$ can be reconstructed from the affine hyperplane $H\; :=\; f^\{-1\}(1)$ by $\backslash ker\; f\; =\; H\; -\; H.$

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.

{{math theorem|name=Theorem{{sfn|Rudin|1991|pp=63-64}}{{sfn|Narici|Beckenstein|2011|pp=1-18}}|math_statement=

If $f,\; g\_1,\; \backslash ldots,\; g\_n$ are linear functionals on {{mvar|X}}, then the following are equivalent:

1. {{mvar|f}} can be written as a linear combination of $g\_1,\; \backslash ldots,\; g\_n$; that is, there exist scalars $s\_1,\; \backslash ldots,\; s\_n$ such that $sf\; =\; s\_1\; g\_1\; +\; \backslash cdots\; +\; s\_n\; g\_n$;
2. $\backslash bigcap\_\{i=1\}^\{n\}\; \backslash ker\; g\_i\; \backslash subseteq\; \backslash ker\; f$;
3. there exists a real number {{mvar|r}} such that $|f(x)|\; \backslash leq\; r\; g\_i\; (x)$ for all $x\; \backslash in\; X$ and all $i\; =\; 1,\; \backslash ldots,\; n.$

}}

If {{mvar|f}} is a non-trivial linear functional on {{mvar|X}} with kernel {{mvar|N}}, $x\; \backslash in\; X$ satisfies $f(x)\; =\; 1,$ and {{mvar|U}} is a balanced subset of {{mvar|X}}, then $N\; \backslash cap\; (x\; +\; U)\; =\; \backslash varnothing$ if and only if for all $u\; \backslash in\; U.${{sfn |Narici|Beckenstein|2011|p=128}}

{{Main|Hahn–Banach theorem}}

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 $\backslash R.$ 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,

{{math theorem|name=Hahn–Banach dominated extension theorem{{sfn|Narici|Beckenstein|2011|pp=177-220}}{{harv|Rudin|1991|loc=Th. 3.2}}|math_statement=

If $p\; :\; X\; \backslash to\; \backslash R$ is a sublinear function, and $f\; :\; M\; \backslash to\; \backslash R$ is a linear functional on a linear subspace $M\; \backslash subseteq\; X$ which is dominated by {{mvar|p}} on {{mvar|M}}, then there exists a linear extension $F\; :\; X\; \backslash to\; \backslash R$ of {{mvar|f}} to the whole space {{mvar|X}} that is dominated by {{mvar|p}}, i.e., there exists a linear functional {{mvar|F}} such that

$$F(m)\; =\; f(m)$$

for all $m\; \backslash in\; M,$ and

$$|F(x)|\; \backslash leq\; p(x)$$

for all $x\; \backslash in\; X.$

}}

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

For any subset {{math|H}} of $X\text{'},$ the following are equivalent:{{sfn|Narici|Beckenstein|2011|pp=225-273}}

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

If {{math|H}} is an equicontinuous subset of $X\text{'}$ then the following sets are also equicontinuous:

the weak-* closure, the balanced hull, the convex hull, and the convex balanced hull.{{sfn|Narici|Beckenstein|2011|pp=225-273}}

Moreover, Alaoglu's theorem implies that the weak-* closure of an equicontinuous subset of $X\text{'}$ is weak-* compact (and thus that every equicontinuous subset weak-* relatively compact).{{sfn|Schaefer|Wolff|1999|loc=Corollary 4.3}}{{sfn|Narici|Beckenstein|2011|pp=225-273}}

• {{annotated link|Discontinuous linear map}}
• {{annotated link|Locally convex topological vector space}}
• {{annotated link|Positive linear functional}}
• {{annotated link|Multilinear form}}
• {{annotated link|Topological vector space}}

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{{reflist|group=proof}}

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