Linear functional ($\phi :V\to F$)

A linear functional is a linear map from a vector space $V$ to its scalar field $F$:Its output is one scalar, rather than an arbitrary vector in another space. In ${\mathbb{R}}^2$, $\phi (x,y)=2x-y$ gives $\phi (3,1)=5$, $\phi (1,2)=0$, and $\phi (4,3)=5$: adding the two inputs adds their measured values.

The level sets $\phi (v)=c$ visualize what the measurement retains and discards. For this example they are parallel straight lines $y=2x-c$. Moving parallel to $(1,2)$ changes the input but leaves $\phi$ unchanged. The zero level set is its kernel of a linear map, while the other level sets are translates of that vector subspace.

Inputs need not be geometric arrows. On the vector space of polynomials of polynomial degree at most two, evaluating at zero is a linear functional, as is taking the definite integral from zero to one. For $p(t)=a+bt+c{t}^2$, they give $p(0)=a$ and ${\int }_0^{1}p(t)dt=a+b/2+c/3$. Thus polynomial evaluation and integration are two different measurements of the same polynomial. Coordinate extraction is another measurement, developed under dual basis.

A linear functional is defined by linearity, not by continuity. In finite-dimensional real or complex normed vector spaces it is automatically continuous; in infinite dimension it need not be. The continuous dual space adds this requirement when small changes in the input should give small changes in the output.

For a nonzero Euclidean vector $w$, the Cauchy-Schwarz inequality gives $z\cdot w\ge -\Vert w\Vert$ when $\Vert z\Vert =1$. Equality is attained uniquely at $z=-w/\Vert w\Vert$. On a sphere of radius $r>0$ the minimizer is $-rw/\Vert w\Vert$, and the minimum is $-r\Vert w\Vert$. This turns many constrained vector problems into a normalization of one fixed vector. If $w=0$, every point minimizes the functional.

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.