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At $(-1,0)$ the gradient is $(1/2,0{)}^T$. The total derivative is therefore the linear functional

The Riesz-Markov-Kakutani representation theorem identifies the continuous dual of $C(\mathbb{T})$ with the finite complex Radon measures on $\mathbb{T}$: every bounded complex-linear functional $L$ has a unique representationThe right side is the total variation norm of the measure. On this compact metric space Radon measures are regular Borel measures. A functional is positive precisely when its representing measure is positive; then $\Vert L\Vert =L(1)=\mu (\mathbb{T})$.

Factor $A+H=A(I+A^{-1}H)$. The determinant expansion near the identity isthe linear terms in the permutation formula come from choosing one diagonal increment and all remaining identity entries. ThereforeThe required Fréchet derivative is the linear functionalThe quadratic remainder proves full differentiability, rather than only the existence of each directional derivative. This is the derivative of the determinant.

At a constrained critical point, the directional derivative of $f$ vanishes along every admissible tangent velocity:For a regular zero locus, $T_{a}M=\ker D{g}_a$. A linear functional vanishing on this kernel lies in the row space of $D{g}_a$. Equivalently there are Lagrange multipliers ${\lambda }_1,\dots ,{\lambda }_k$ such thatGeometrically, $\nabla f$ is orthogonal to the tangent space, so it lies in its orthogonal complement, spanned by the independent constraint gradients. Thus the level set of $f$ has no first-order change in a tangent direction. The equations provide candidates for constrained extrema; they must still be classified and compared, and boundary or singular cases must be treated separately. The condition is necessary for a regular constrained extremum, not a guarantee of a maximum.

The real projective plane is the set of one-dimensional linear subspaces of ${\mathbb{R}}^3$. Equivalently,A projective point is written in homogeneous coordinates $[x:y:z]$. A projective line is the set of projective points whose representatives belong to one fixed two-dimensional linear subspace of ${\mathbb{R}}^3$. Such a linear subspace is the kernel of a linear map of a nonzero linear functional, so its equation has the form $\alpha x+\beta y+\gamma z=0$ with $(\alpha ,\beta ,\gamma )\ne 0$.

Two distinct projective points have linearly independent representatives $p,q$. Their linear span has dimension two and hence defines a projective line containing both. Conversely, any two-dimensional linear subspace containing both representatives contains their entire linear span and therefore equals it. Thus exactly one projective line passes through two distinct projective points.

The Petrov-Galerkin method chooses an approximate solution $u_h$ in a finite-dimensional trial vector space $V_h$ and tests a weak formulation against a possibly different test space $W_h$. For a bilinear form $a$ and a linear functional $\ell$, it imposes $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in W_h$. Taking $W_h=V_h$ recovers the usual Galerkin method. Different test functions can alter stability and accuracy; the Galerkin orthogonality is then against the test space and need not be an orthogonal projection in an energy norm.

In three dimensions let $f\in L^{6/5}({\mathbb{R}}^3)$ have compact support. Let $v_R\in H_0^1(B_R)$ solve $-\Delta v_R=f$ and extend it by zero. There is a unique $v\in {\dot{H}}^1({\mathbb{R}}^3)$ with $\int \nabla v\cdot \nabla \varphi =\int f\varphi$ for all $\varphi \in {\dot{H}}^1$. Indeed Holder inequality and the homogeneous Sobolev inequality make the right side a continuous linear functional in the gradient norm, and the Riesz representation theorem applies to this Hilbert space.

The weak formulation says that $v_R$ is the orthogonal projection of $v$ onto the closed subspace consisting of zero extensions of $H_0^1(B_R)$. These subspaces increase with $R$ and their union contains every test function, which is dense by the definition of the homogeneous Sobolev space. The best-approximation property of orthogonal projection therefore gives $\Vert \nabla (v_R-v){\Vert }_2\to 0$. The homogeneous Sobolev inequality also gives $\Vert v_R-v{\Vert }_6\to 0$. This convergence does not generally extend to the inhomogeneous Sobolev space $H^1$, as the spherical mean of a Poisson solution outside its source shows for a source of nonzero total mass.

Polynomial evaluation replaces the indeterminate of a polynomial by an input value. For a fixed scalar $a$, the rule $E_a:p\mapsto p(a)$ is a linear functional on a vector space of polynomials, because evaluating a linear combination gives the same linear combination of the values. For $p(t)=1+2t+3t^2$, evaluation at zero gives one and evaluation at two gives seventeen. Polynomial interpolation uses these measurements to reconstruct a polynomial subject to a polynomial degree bound.

For a line $\ell =\{\mathbf{a}+t\mathbf{v}:t\in \mathbb{R}\}$ with $\mathbf{v}\ne 0$, restricting a degree-at-most-$d$ multivariate polynomial gives a univariate polynomial in $t$ of degree at most $d$. Its $d+1$ coefficients are linear functionals of the original coefficient vector. Vanishing on the whole line is therefore equivalent to at most $d+1$ homogeneous linear conditions. Alternatively, $d+1$ distinct roots force the restriction to be identically zero.

Let $Kx=a\phi (x)$ on a vector space, where $\phi$ is a linear functional. Its image of a linear map lies in the span of $a$. Applying $K$ twice gives $K^{2}x=a\phi (a)\phi (x)=\phi (a)Kx$. Mathematical induction then proves the displayed power formula. If $\phi (a)=0$, $K^2=0$, even when $K\ne 0$; this is a nilpotent operator. If $\phi (a)\ne 0$, then $K/\phi (a)$ is an idempotent linear map. In a real inner product space, the special case $\phi (x)=b\cdot x$ gives $\phi (a)=a\cdot b$.

A quadratic functional on a real vector space combines a quadratic form $B(v,v)$, a linear functional $\ell (v)$ and a constant. If $B$ is a symmetric bounded bilinear form, its first variation is $DI(v)[w]=2B(v,w)-2\ell (w)$. For a strictly positive $B$, an existing stationary point is the unique minimizer by the quadratic variational principle for a symmetric positive operator.

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

Take independent uniform samples from a $d$-dimensional vector space over the two-element field. At current matrix rank $r$, the existing span contains $2^r$ of the $2^d$ possible vectors, so a new sample increases the matrix rank with probability $1-2^{r-d}$. The mean waiting time for that increase is $1/(1-2^{r-d})$ by the geometric distribution. Adding the waiting times gives $\mathbb{E}M=d+{\sum }_{j=1}^d(2^j-1{)}^{-1}<d+2$. For $d\ge 1$, the bound follows by separating $j=1$ and using $(2^j-1{)}^{-1}<2^{1-j}$ for $j\ge 2$; for $d=0$, no samples are needed.

After $m=d+t$ samples, failure to span implies that some nonzero linear functional vanishes on all samples. Each fixed nonzero functional does so with probability $2^{-m}$; the union bound over the $2^d-1$ possibilities gives failure less than $2^{-t}$. This provides both an expected sample count and a fixed-budget high-success version of Simon's algorithm.

The closed unital algebra generated by a bounded operator and all its resolvents is commutative, because the generators commute. It has exactly the original operator spectrum at that operator: all required inverses outside that spectrum were included, and an inverse in the smaller algebra is also an operator inverse. This permits commutative-algebra linear functional calculus without replacing the spectrum by the possibly larger spectrum of a polynomial-generated algebra.

Every continuous self-map of a nonempty compact convex subset of a locally convex space has a fixed point. Compactness is in the locally convex topology, which may be weaker than a norm topology. This version permits fixed-point arguments on a weak-star topology compact set of linear functionals. See Hartman's fixed-point chapter, Theorem 0.2.

If $Y$ is a closed linear subspace of a Hausdorff locally convex space and $x_0\notin Y$, choose a continuous seminorm $q$ with $q(x_0-y)\ge 1$ for all $y\in Y$, using a basic neighborhood disjoint from $Y$. The linear functional $g(y+t{x}_0)=t$ is bounded by $q$. The Hahn-Banach theorem extends it to a continuous linear functional with the displayed values. This is the linear-subspace form of separation, sharper than merely distinguishing the two sets.

Let $T$ be a singular endomorphism of a finite-dimensional vector space. Choose $0\ne u\in \ker T$ and a nonzero linear functional $\lambda$ vanishing on $\mathrm{im}T$, possible because that image of a linear map is proper. Then $S(x)=\lambda (x)u$ is a nonzero rank-one endomorphism with $TS=ST=0$.

Suppose $(u_n)$ is a normalized block basic sequence in the l-p sequence space, $1\le p<\infty$, and ${\sum }_n\Vert x_n-u_n{\Vert }_p<1$. Choose norm-one linear functionals ${\varphi }_n$ supported on the respective blocks with ${\varphi }_n(u_n)=1$. The bounded linear operator $A=I+{\sum }_n{\varphi }_n(\cdot )(x_n-u_n)$ is invertible by the Neumann series. The block projection $Qz={\sum }_n{\varphi }_n(z)u_n$ has norm at most one, and $AQ{A}^{-1}$ projects onto the closed span of $(x_n)$. Thus that span is a complemented subspace isomorphic to the l-p sequence space.

A spline quasi-interpolant synthesizes a B-spline expansion from local bounded linear functionals, instead of solving a global interpolation system. Suppose each functional is bounded by $c_k\Vert f{\Vert }_{C[t_i,t_{i+k}]}$ and the operator reproduces every spline. On a knot cell $[t_j,t_{j+1}]$, only indices $j+1-k\le i\le j$ contribute. Their supports lie in $[t_{j+1-k},t_{j+k}]$, and subpartition of unity for B-splines proves local operator norm at most $c_k$. Reproduction of degree-$k-1$ polynomials and a local Taylor theorem remainder then give error at most $(1+c_k)(2k-1{)}^k{h}^k\Vert f^{(k)}{\Vert }_{\infty }/k!$, on the basic knot domain with the usual completed boundary basis.

A continuous symmetric coercive bilinear form $a$ on a real Hilbert space and a continuous linear functional $\ell$ give a unique minimum of $I$. The parallelogram identity makes any minimizing sequence Cauchy in the energy norm; coercivity and completeness give its limit. Differentiation gives $a(u,w)=\ell (w)$, and $I(v)-I(u)=a(v-u,v-u)$ proves uniqueness. A nonsymmetric operator instead differentiates to its symmetric part. Mere pointwise positivity without a lower bound in the chosen norm need not yield a minimizing vector.