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On $C(X)$ for compact Hausdorff $X$, any complete norm making every point evaluation functional continuous is equivalent to the uniform norm. If $f_n$ converges in the new norm to $f$ and uniformly to $g$, evaluation gives $f(x)=g(x)$ for every $x$, so the identity map between the two Banach spaces has closed graph. The closed graph theorem and open mapping theorem make the identity and its inverse bounded. Completeness of the proposed norm is essential to this argument.

For $p(X)=a+bX+c{X}^2$, the three point evaluation functionals at $0,1,-1$ form a basis of the dual space of real polynomials of degree at most two. The displayed recovery formulas express the coordinate functionals, hence the dual basis to $1,X,X^2$, as linear combinations of these evaluations. More generally distinct-node polynomial interpolation identifies values with coefficient data through an invertible Vandermonde matrix.

A finite element degree of freedom is one of the linear functionals chosen to specify a local shape function in a finite-dimensional vector space $P$. Examples are point evaluation functionals and edge or cell integrals. The finite element is an unisolvent finite element when the complete list gives a bijection from $P$ to the corresponding list of real values.

Label the polygon vertices of the nondegenerate triangle $K$ as $a_1,a_2,a_3$, and put $m_{ij}=(a_i+a_j)/2$ for its three edge midpoints. The six monomials $1,x_1,x_2,x_1^2,x_1{x}_2,x_2^2$ form a basis of ${\mathcal{P}}_2(K)$, so its dimension is six. Nondegeneracy matters: $K$ has nonempty interior, and a polynomial vanishing on that interior is the zero polynomial.

Let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the affine functions of the barycentric coordinate system, characterized by ${\lambda }_i(a_j)={\delta }_{ij}$ and ${\sum }_i{\lambda }_i=1$. Define six quadratic polynomialsAt $a_k$, $N_i(a_k)={\delta }_{ik}$ and all $N_{ij}(a_k)$ vanish. At $m_{kl}$, exactly ${\lambda }_k$ and ${\lambda }_l$ equal $1/2$, while the third is zero. Thus every $N_i$ vanishes at all three edge midpoints, and $N_{ij}(m_{kl})$ is one for the matching unordered pair and zero otherwise.

These six polynomials have the cardinal property for the six point evaluation functionals. Evaluating any vanishing linear combination at the nodes forces each of its coefficients to vanish. They are therefore six linearly independent members of a six-dimensional vector space, hence a basis.

In particular, every $p\in {\mathcal{P}}_2(K)$ has the unique expansionThis proves both existence and uniqueness for arbitrary six prescribed nodal values, which is precisely an unisolvent finite element. It also identifies the finite element degrees of freedom and shape functions of the quadratic Lagrange triangle explicitly.

For a continuous function $g$ on $[0,1]$, the L-infinity norm of its restriction to $(0,1)$ equals its supremum norm on $[0,1]$. Indeed, values arbitrarily close to a maximum occur on an interval of positive Lebesgue measure, even if the maximum lies at an endpoint. Also, continuous functions equal almost everywhere on $(0,1)$ are equal everywhere on $[0,1]$. Thus restriction embeds $C([0,1])$ isometrically as a linear subspace of $X$.

The point evaluation functional $E_{x_0}(g)=g(x_0)$ is complex-linear and satisfies $|E_{x_0}(g)|\le \Vert g{\Vert }_{\infty }$. The constant function one attains equality, so $\Vert E_{x_0}\Vert =1$. The complex Hahn-Banach theorem extends this bounded linear functional from that linear subspace to $X$, preserving its operator norm. Therefore

First scale the stated unit-ball approximation. For every nonzero residual $r\in F$, apply the hypothesis to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$. This gives $v\in E$ withFor zero residual use $v=0$. Start with $r_0=y$, choose $v_j$ by this rule for $r_j$, and set $r_{j+1}=r_j-T{v}_j$. Inductively,The Banach space $E$ therefore contains the sum $x={\sum }_{j\ge 0}{v}_j$, withThe partial sums satisfy $T{\sum }_{j=0}^{n-1}{v}_j=y-r_n$. Boundedness of $T$ gives convergence of the left side to $Tx$, while $r_n\to 0$ gives $Tx=y$. Thus $T$ is surjective with the stated lifting bound. This geometric correction for approximate surjectivity used completeness of $E$, not any unproved completeness of $F$.

Put $C=R/(1-k)$. Given a Cauchy sequence $(y_n)$ in $F$, choose a subsequence $(y_{n_j})$ such that $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$ for $j\ge 1$. Lift each difference to $u_j\in E$ with $T{u}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert u_j\Vert \le C{2}^{-j}$. Lift $y_{n_1}$ to $u_0$. The sum $u=u_0+{\sum }_{j\ge 1}{u}_j$ exists in $E$, and its image is the limit of the subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit: use the triangle inequality between an arbitrary late term, a later subsequence term, and the limit. This proves $F$ is complete, the completeness forced by uniformly bounded lifting principle.

The open mapping theorem states that a bounded surjective linear operator between Banach spaces maps open sets to open sets. In particular a bounded linear bijection between Banach spaces has a bounded inverse. To deduce the closed graph theorem, let $S:E_1\to F_1$ be an everywhere-defined linear map between Banach spaces with closed graph. Its graph is a closed subspace of the Banach product $E_1\times F_1$, with norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$, so it is itself Banach. The projectionis a bounded linear bijection. The open mapping theorem makes its inverse bounded. Composing that inverse with the bounded second projection proves that $S$ is bounded. This is the required closed graph conclusion, rather than a continuity assumption on $S$.

Finally consider the identity map $I:(C(X),\Vert \cdot \Vert )\to (C(X),\Vert \cdot {\Vert }_{\infty })$. If $f_n\to f$ in the new norm and $f_n\to g$ uniformly, each assumed continuous point evaluation functional gives $f_n(x)\to f(x)$, while uniform convergence gives $f_n(x)\to g(x)$. Hence $f=g$. Since both spaces are metric, this sequential argument proves the graph is closed. Both norms are complete, so the closed graph theorem makes $I$ bounded. Its inverse is bounded by the open mapping theorem. Consequently there are constants $c,C^{\prime }>0$ such thatThis proves Banach norm rigidity from continuous point evaluations. Pointwise continuity alone would not give a common bound on all evaluations; completeness supplies that through the closed graph argument.

Fix $z_0\in S^1$, and let ${\delta }_{z_0}\in C(S^1{)}^{\ast }$ be the point evaluation functional ${\delta }_{z_0}(f)=f(z_0)$. Define orbit averages in the dual space byEach ${\mu }_n$ has norm at most one and satisfies ${\mu }_n(1_{S^1})=1$. The space $C(S^1)$ is separable, so part (a) gives a subsequence converging pointwise to some $\mu \in C(S^1{)}^{\ast }$. In particular,For every $f\in C(S^1)$, the Cesàro average telescopes at the endpoints:whose absolute value is at most $2\Vert f{\Vert }_{\infty }/n$. Passing to the convergent subsequence provesThus $\mu$ is an invariant functional of a composition operator.

It need not be unique when $T$ is not the identity operator. For the nonidentity homeomorphism $\tau (z)=\overline{z}$, both $1$ and $-1$ are fixed. Hence the two distinct point evaluationshave value one on $1_{S^1}$ and are invariant under $T$. Therefore nonidentity does not imply uniqueness.

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.

A polynomial moment functional is a linear functional on polynomials defined by their integrals against a specified function. For example, suppose the required integrals exist and setFor a polynomial $p(t)={\sum }_{j=0}^n{c}_j{t}^j$, linearity givesThus the finitely many moments ${\mu }_0,\dots ,{\mu }_n$ determine $L$ on the vector space of polynomials of degree at most $n$. Existence of these moments is part of the hypothesis; an arbitrary choice of $w$ need not give finite integrals.

For example, take $w=1$ on $[-1,1]$. Then ${\mu }_j=0$ for odd $j$ and ${\mu }_j=2/(j+1)$ for even $j$. On quadratic polynomials,The zero first moment records cancellation of the odd function $t$ on a symmetric interval. This finite-coordinate viewpoint is useful in Gaussian quadrature: a weighted sum of point evaluation functionals integrates every polynomial up to degree $n$ exactly if it agrees with $L$ on the monomials $1,t,\dots ,t^n$.