Evaluation Maps on Function Spaces

Evaluation at a Fixed Point

A function can be studied by examining its value at a single input. Fixing that input turns evaluation into a map from a space of functions to the real numbers. For instance, if the input is \(2\), evaluation sends each polynomial \(p\) to the number \(p(2)\). The map has a simple formula, but its domain is a vector space of functions, and checking its properties means using the operations on that space.

In this tutorial, all functions are real-valued. For an interval \(I\subseteq\mathbb R\), let \(C(I,\mathbb R)\) denote the vector space of continuous functions from \(I\) to \(\mathbb R\), with addition and scalar multiplication defined pointwise. Let \(P_n(\mathbb R)\) denote the vector space of real polynomials of degree at most \(n\), including the zero polynomial. A point \(a\) must belong to a function's domain for evaluation at \(a\) to be defined.

The notation for the map depends on the function space under consideration, even when the fixed point \(a\) is the same. For example, evaluation at \(2\) can be applied to polynomials, or to continuous functions defined on an interval containing \(2\). The formula is always “substitute \(a\) for the input”; the collection of allowable functions can differ.

Why Evaluation Is Linear

Proof. To verify linearity, take any \(f,g\in V\) and \(c\in\mathbb R\). Addition and scalar multiplication in \(V\) are pointwise, so \((f+g)(a)=f(a)+g(a)\) and \((cf)(a)=c f(a)\). Consequently, \(\operatorname{ev}_a(f+g)=\operatorname{ev}_a(f)+\operatorname{ev}_a(g)\) and \(\operatorname{ev}_a(cf)=c\operatorname{ev}_a(f)\). These are the additivity and homogeneity conditions for a linear map. \(\square\)

The theorem requires the functions to be defined at the fixed point, so that all three values in each equality exist. No continuity assumption is needed for linearity: evaluation is linear on any real vector space of real-valued functions with a common domain containing \(a\). Continuity becomes relevant when we specify a particular function space, not when we check the two linearity conditions.

Polynomial Evaluation: Kernel and Image

For \(P_n(\mathbb R)\), every polynomial has a unique expression as a combination of the shifted powers \(1,(x-a),(x-a)^2,\ldots,(x-a)^n\). One way to see this is to substitute \(x=y+a\): a polynomial in \(x\) of degree at most \(n\) becomes a polynomial in \(y\) of degree at most \(n\), with unique coefficients. Substituting \(x=a\) then sets every shifted power except \(1\) to zero.

Proof. Write any \(p\in P_n(\mathbb R)\) in its unique shifted-power expression \(p(x)=b_0+b_1(x-a)+\cdots+b_n(x-a)^n\). Substitution gives \(p(a)=b_0\), because \((a-a)^k=0\) for every \(k\geq1\). Thus \(p(a)=0\) exactly when \(b_0=0\). When \(n\geq1\), the polynomials with \(b_0=0\) are precisely the linear combinations of \((x-a),\ldots,(x-a)^n\). When \(n=0\), the only polynomial is constant, and the only constant with value zero at \(a\) is the zero polynomial. Finally, for any \(r\in\mathbb R\), the constant polynomial \(p(x)=r\) belongs to \(P_n(\mathbb R)\) and satisfies \(p(a)=r\). Therefore every real number is in the image. \(\square\)

Evaluation on Continuous Functions

Now let \(I=[u,v]\) be a closed interval and choose \(a\in I\). The evaluation map \(\operatorname{ev}_a:C(I,\mathbb R)\to\mathbb R\) is defined because every function in this space is defined at every point of \(I\). Its kernel consists of the continuous functions that vanish at \(a\). Its image is all of \(\mathbb R\), since constant functions are continuous and can take any prescribed real value.

Proof. By the definition of the kernel, a function belongs to \(\ker(\operatorname{ev}_a)\) exactly when its evaluation is zero. Since \(\operatorname{ev}_a(f)=f(a)\), this is exactly the condition \(f(a)=0\), giving the stated kernel. For the image, every output is a real number because the functions are real-valued. Conversely, given any \(r\in\mathbb R\), the constant function \(f(t)=r\) is continuous on \(I\) and satisfies \(\operatorname{ev}_a(f)=f(a)=r\). Thus every real number is an output, and the image is \(\mathbb R\). \(\square\)

Common Mistake: Confusing a Value with a Coefficient

Evaluation is not the operation of selecting a coefficient. For example, the polynomial \(p(x)=1+x\) has constant coefficient \(1\), but evaluation at \(2\) gives \(p(2)=1+2=3\), not \(1\). A related error is to claim that \(\operatorname{span}\{1,x\}\) is the kernel of evaluation at \(2\) on \(P_2(\mathbb R)\). That cannot be right: the constant polynomial \(1\) is in this span, but \(\operatorname{ev}_2(1)=1\ne0\). The correct kernel is \(\operatorname{span}\{x-2,(x-2)^2\}\); both generators evaluate to zero, and the polynomial calculation above shows that every polynomial in the kernel is a combination of them.

The kernel describes functions that vanish at the chosen input, not functions whose coefficients happen to vanish. The image describes the possible output values, not the possible functions. For the polynomial and continuous-function spaces considered here, constant functions guarantee that every real output is attainable. These facts make evaluation a useful way to encode a function's value at a point—for example, when a condition specifies a value that a polynomial or continuous function must take.