Point evaluation functional

For a function space on a set $X$ and $x\in X$, the point evaluation functional is ${\delta }_x(f)=f(x)$. On $C(K)$ for a nonempty compact space $K$, it is a bounded linear functional for the supremum norm:The constant function one attains the bound. The same argument works for bounded continuous functions on any $X$.

Boundedness depends on the chosen norm. On $C([0,1])$ with the L2 norm, fix $x_0\in (0,1)$ and take $f_n(t)=\max (1-n|t-x_0|,0)$ for sufficiently large $n$. Then $f_n(x_0)=1$ but $\Vert f_n{\Vert }_2^2=2/(3n)$, so point evaluation is unbounded. On L2 space itself, it is not even well-defined on equivalence classes: changing a representative at the single point $x_0$ does not change its Lebesgue almost everywhere class.

View $C([0,1])$ inside the complex Lp space $L^{\infty }((0,1))$. Fix $x_0\in [0,1]$. If a unit function $\psi$ satisfies $\Vert g+\psi {\Vert }_{\infty }\ge |g(x_0)|+1$ for every continuous function $g$, then the possible values at $\psi$ of norm-one extensions of point evaluation at $x_0$ form exactly the closed complex unit disc.

The estimate first excludes $\psi \in C([0,1])$. On the direct sum $C([0,1])\oplus ⟨\psi ⟩$, prescribe $L_a(g+c\psi )=g(x_0)+ca$ for $|a|\le 1$. Scaling the estimate gives $\Vert g+c\psi {\Vert }_{\infty }\ge |g(x_0)|+|c|\ge |L_a(g+c\psi )|$, and $L_a(1)=1$ proves operator norm one. The complex Hahn-Banach theorem extends $L_a$ to the whole Lp space. No value outside the disc is possible because $\Vert \psi {\Vert }_{\infty }=1$.