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The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

Every linearly independent subset of a finite-dimensional vector space extends to a basis. Starting with its vectors, add a vector outside their current span until the span is the whole space. Each addition preserves linear independence, and the process stops because the dimension is finite. This also lets a linear functional on a vector subspace extend to the ambient space by assigning arbitrary values on the additional basis vectors.

For an algebraic dual space $X^{\ast }={\mathrm{Hom}}_F(X,F)$, the linear map $J:X\to X^{\ast \ast }$ sends $x$ to evaluation at $x$. It is injective because linear functionals separate nonzero vectors: extend any nonzero vector to a basis and use its coordinate functional. If $X$ has a finite basis $e_i$ with dual basis $e_i^{\ast }$, every $B\in X^{\ast \ast }$ equals $J({\sum }_{i}B(e_i^{\ast })e_i)$, so $J$ is an isomorphism. For infinite-dimensional spaces, surjection does not follow from this finite sum argument.

A linear functional on $X^{\ast }$ is continuous for the weak-star topology exactly when it is evaluation at an element of $X$. Continuity makes it vanish on the common kernel of finitely many evaluations, so it is a linear combination of those evaluations. The canonical embedding into the bidual identifies this dual with $X$.

In a Hausdorff locally convex space, a separating family of continuous seminorms has $p(x)>0$ for any chosen nonzero $x$ and some $p$. Define a linear functional on its one-dimensional span attaining $p(x)$ and extend it under that seminorm by the real Hahn-Banach theorem. The extension is continuous because it is seminorm-bounded, and distinguishes $x$ from zero. The Hausdorff condition is essential.

The nth coordinate functional of a Schauder basis is the bounded linear functional $e_n^{\ast }({\sum }_k{a}_k{e}_k)=a_n$.

A covector is an element of the dual space, hence a linear functional on a vector space.

If a linear functional on a normed vector space has $f(v_N)=1$ for a sequence tending to zero in norm, it is discontinuous and unbounded. An explicit example is the algebraic span in ${\ell }^2$ of the coordinate vectors and $w=(2^{-n}{)}_{n\ge 1}$. Define $f(w)=1$ and $f(e_n)=0$. The tails $v_N=w-{\sum }_{n=1}^N{2}^{-n}{e}_n$ have norm $2^{-N}/\sqrt{3}$ but image one. Thus the functional vanishes on the dense finite-support subspace while being nonzero elsewhere.

For a linear map between finite-dimensional vector spaces, the image of its dual map is the annihilator of a vector subspace given by its kernel. Inclusion follows by composition. Conversely a linear functional vanishing on the kernel descends to the image and extends to the target by basis extension. The identity and the rank-nullity theorem show that a linear map and its dual have equal rank.

A norm-closed finite-codimensional vector subspace $Y\subseteq X^{\ast }$ that is weak-star dense is a norming subspace. Its annihilator $F=Y^{\perp }\subseteq X^{\ast \ast }$ is finite-dimensional and disjoint from $JX$. The positive distance $d$ of the unit sphere from $F$ gives a triple-dual linear functional vanishing on $F$ and taking a value at least $d$ at a unit vector. Apply the finite-dimensional interpolation form of Goldstine's theorem to the Banach space $X^{\ast }$, then normalize and take a supremum to obtain norming constant $d$.

After clearing denominators in a matrix, for every disjoint pair of index sets $I,B$ with $B$ nonempty and ${\sum }_{i\in B}{a}_i$ outside the linear span of the $I$-columns, choose an integer linear functional vanishing on those columns but not on the indicated sum. Choose a prime number dividing none of the finitely many nonzero evaluations. The last nonzero digit coloring for this prime, applied to a monochromatic positive solution, groups coordinates into blocks by P-adic valuation. Applying the appropriate functional and reducing modulo the prime shows that each block sum lies in the linear span of all earlier columns. This proves necessity in Rado's theorem without an asymptotic limiting argument.

A finite-dimensional unital associative algebra over a field admitting a linear functional $\epsilon$ whose pairing $(a,b)\mapsto \epsilon (ab)$ is nondegenerate. The dual numbers with functional extracting the nilpotent coefficient are an example; semisimplicity is unnecessary.

Every finite-dimensional complex Lie algebra representation of a nilpotent Lie algebra decomposes into invariant generalized weight spaces. The nilpotence assumption is essential: a general solvable action can have extensions between distinct weights.

Here is a proof. For $h,k\in H$, nilpotence gives $(\mathrm{ad}h{)}^{N}k=0$, hence $(\mathrm{ad}\rho (h){)}^N\rho (k)=0$. By the nilpotent commutator preserves generalized eigenspaces lemma, every generalized eigenspace of $\rho (h)$ is invariant under every $\rho (k)$. Split successively by the acting operators of a basis of $H$. Each resulting block is invariant under $H$, and each basis operator has only one eigenvalue on it. The Lie theorem triangularizes the action on each block; its diagonal entries therefore give one and the same linear functional $\lambda$ on every basis vector of that block. Every $\rho (h)-\lambda (h)I$ is strictly upper triangular there, proving that the block is $V^{\lambda }$. Distinct blocks have distinct weights, so this is a direct sum. On the zero generalized weight space, every acting operator is a nilpotent endomorphism, so the Engel theorem supplies a nonzero common annihilated vector whenever that space is nonzero.

For a finite-dimensional complex Lie algebra representation $\rho$ of a nilpotent Lie algebra $H$, a generalized weight space is the simultaneous generalized eigenspace displayed above, where $\lambda :H\to \mathbb{C}$ is a linear functional. Every occurring $\lambda$ vanishes on the derived algebra of $H$. Ordinary weight vectors satisfy the same equations without the powers.

For finite positive measures $\mu ,\nu$, put $\rho =\mu +\nu$. The linear functional $g\mapsto \int gd\nu$ is bounded on real $L^2(\rho )$ by the Cauchy-Schwarz inequality. The Riesz representation theorem gives a measurable $h$ with $\nu (A)={\int }_{A}hd\rho$. Testing indicators of its negative and greater-than-one level sets shows $0\le h\le 1$ almost everywhere. Subtraction gives $\mu (A)={\int }_A(1-h)d\rho$. This constructs the densities without assuming the Radon-Nikodym theorem as a prior result.

If a linear functional $\ell$ vanishes on the image of a linear map $T$ but $\ell (b)\ne 0$, the equation $Tx=b$ has no solution. In Euclidean coordinates, a vector $n$ with $n^{T}A=0$ provides such an obstruction whenever $n^{T}b\ne 0$. In finite dimensions these annihilating functionals exactly characterize membership in the image: $b\in \mathrm{im}A$ precisely when it is orthogonal to every vector in $\ker A^T$.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

A real linear function on ${\mathbb{R}}^n$ has the form $x\mapsto c^{T}x$.

A matrix coefficient of a finite-dimensional group representation is a scalar function obtained by applying a linear functional to the translate of a fixed vector. In a chosen basis these functions include $x\mapsto \rho (x{)}_{ij}$. For a finite group, the Schur orthogonality relations make the scaled coefficients of all inequivalent unitary irreducible representations an orthonormal basis of its scalar functions.

A nonempty compact face of a convex set in a Hausdorff locally convex space contains an extreme point. Chains of compact faces have nonempty compact intersection. A minimal face supplied by the Zorn lemma must be a singleton, since a separating linear functional would otherwise expose a proper maximizer face. This is the existence step in the Krein-Milman theorem.