Functional Analysis

Best approximation in Hilbert space

Best approximation asks for the point of a closed subspace M of a Hilbert space H that is nearest to a given vector f. The nearest point is the orthogonal projection P_Mf, characterised by f-P_Mf\perp M, and the minimal distance is d=\|f-P_Mf\|. If e_1,\dots,e_n is an orthonormal basis of a finite-dimensional subspace Y, then P_Yf=\sum_k\langle f,e_k\rangle e_k and d^2=\|f\|^2-\sum_k|\langle f,e_k\rangle|^2, so the distance is determined by the coefficients alone. This is the working tool for polynomial subspaces such as \mathcal P_1 on [-1,1], where the normalised Legendre polynomials form an orthonormal basis. When M is the kernel of a bounded functional \varphi=\langle\cdot,v\rangle, then M=v^\perp and \operatorname{dist}(f,M)=|\langle f,v\rangle|/\|v\|, because the component of f along v is the only part that cannot be removed inside M. A common error is to forget the division by \|v\| or to invert the ratio. If T is self-adjoint with T^2=I, then \ker(I-T) is the set of fixed vectors, and the distance from a vector to this kernel equals the norm of its component in the orthogonal complement, namely \tfrac12(I-T)\varphi. For example, in \mathbb R^2 the nearest point of the line \mathbb R(1,0) to (3,4) is (3,0), at distance 4.

Closed unbounded linear operators

A linear operator T with domain D(T) is closed when its graph \{(x,Tx):x\in D(T)\} is a closed set, which means that x_n\to x and Tx_n\to y together force x\in D(T) and y=Tx. Closedness does not imply boundedness. On its natural domain, a diagonal operator Tx=(d_nx_n) with unbounded entries is closed, but \|Te_k\|/\|e_k\|=|d_k| for the unit vectors, so no constant C satisfies \|Tx\|\le C\|x\|. Conversely, a bounded operator defined on the whole space is automatically closed, since continuity lets limits pass through T. The closed graph theorem gives the key link: if X and Y are Banach spaces and T:X\to Y is linear and closed, then T is bounded. A standard way to apply it is through an auxiliary map: if S is bounded and injective and S\circ T is bounded, then the graph of T is closed, so T is bounded even though boundedness was not assumed. Surjectivity does not follow from this argument, since the zero operator is a counterexample.

Bounded linear functionals on Hilbert spaces

By the Riesz representation theorem, every bounded linear functional f on a Hilbert space H has the form f(x)=\langle x,y\rangle for a unique y\in H, and \|f\|=\|y\|. For an orthonormal basis (e_n), this gives f(e_n)=\langle e_n,y\rangle and hence \sum_n|f(e_n)|^2=\|y\|^2=\|f\|^2<\infty. The values of a bounded functional on an orthonormal basis therefore form a square-summable sequence, and conversely every square-summable sequence (a_n) is realised by a unique bounded functional with f(e_n)=a_n. A sequence such as 1/n is square-summable and is admissible, whereas 1/\sqrt n has a divergent sum of squares and cannot be the value sequence of any bounded functional. The norm of a functional can be read from its representer. For F(f)=\int_{-1}^1 f(x)g(x)\,dx with real g, the Cauchy–Schwarz inequality gives |F(f)|\le\|f\|_2\|g\|_2 with equality at f=g, so \|F\|=\|g\|_2 and \|F\|^2=\int_{-1}^1 g^2\,dx.

Inner product in Hilbert space

An inner product is linear in one variable, conjugate-symmetric, and positive definite, and it induces the norm \|x\|=\langle x,x\rangle^{1/2}. In a complex space, with linearity in the first slot, one has \langle u,\lambda v\rangle=\bar\lambda\langle u,v\rangle and \langle v,u\rangle=\overline{\langle u,v\rangle}. In a real space the form is symmetric and bilinear. The Cauchy–Schwarz inequality |\langle u,v\rangle|\le\|u\|\,\|v\| holds in both cases. The polarisation identity recovers the inner product from norms: in the complex case, \langle u,v\rangle=\tfrac14\sum_{k=0}^{3}i^{k}\|u+i^{k}v\|^2, and the factors i^k must be tracked carefully. When a norm depends on a parameter t, such as \|u+e^{it}v\|^2, expanding it produces terms e^{\pm it}\langle u,v\rangle and their conjugates, and averaging over a period isolates the constant term because \frac{1}{2\pi}\int_0^{2\pi}e^{ikt}\,dt=0 for k\ne0. For an orthonormal basis (e_k), the inner product is computed coefficientwise: \langle\sum a_ke_k,\sum b_ke_k\rangle=\sum a_k\bar b_k. For instance, if g has real coefficients c_m, then \langle g,e_k+e_{k+1}\rangle=c_k+c_{k+1}, and alternating signs make adjacent terms cancel.

Orthogonal complement of a subspace

For a subset M of a Hilbert space, M^\perp=\{y:\langle x,y\rangle=0\text{ for all }x\in M\} is always a closed subspace, and for a closed subspace M the whole space decomposes as H=M\oplus M^\perp. Moreover (M^\perp)^\perp equals the closure of M. In finite dimensions \dim M+\dim M^\perp=n, so the dimension of the complement equals the codimension of M. A single linear constraint \langle x,a\rangle=0 defines a hyperplane whose orthogonal complement is \operatorname{span}\{a\}, so the complement of a hyperplane is one-dimensional in a finite-dimensional space. The same principle holds in infinite dimensions. If M=\{x\in\ell^2:\langle x,v\rangle=0\} for a fixed v\in\ell^2, then M is closed, M^\perp=\operatorname{span}\{v\}, and the projection of u onto M^\perp is \frac{\langle u,v\rangle}{\|v\|^2}\,v.

Continuity of linear operators on c00

The space c_{00} of finitely supported sequences is a normed space under the sup norm or an \ell^p norm, and a linear map on it is continuous exactly when it is bounded, that is, when \sup_{\|x\|\le1}\|Tx\|<\infty. For a diagonal map x\mapsto(d_nx_n) under the sup norm, the unit vectors e_n have images of norm |d_n|, so the map is bounded if and only if \sup_n|d_n|<\infty, and then its norm equals this supremum. A diagonal map with unbounded entries is therefore discontinuous even though it is defined everywhere on c_{00}. If all entries lie in (0,\tfrac12], the norm is at most \tfrac12 and the map is a contraction. For functionals, f(x)=\sum_n a_nx_n is continuous on c_{00} with the \ell^1 norm if and only if \sup_n|a_n|<\infty, because the dual of \ell^1 is \ell^\infty. With the \ell^2 norm it is continuous if and only if \sum_n|a_n|^2<\infty, because \ell^2 is self-dual. The sequence a_n=n^{-1/2} therefore gives a continuous functional for \ell^1 but not for \ell^2, since \sum 1/n diverges.

Operators on the Banach space C[0,1]

The space C[0,1] with the sup norm is a Banach space, and an operator on it can be bounded, injective, or surjective independently of the others. The integration operator (Vf)(x)=\int_0^x f(t)\,dt is the standard example. It satisfies \|Vf\|_\infty\le\|f\|_\infty, so it is bounded with norm at most 1. It is injective, because Vf=0 forces f=(Vf)'=0. Its range consists of C^1 functions vanishing at 0, which is not all of C[0,1] and is not closed in the sup norm. A bounded bijection between Banach spaces has a bounded inverse by the open mapping theorem, so a non-closed range forces the inverse on the range to be unbounded. Thus boundedness and injectivity of an operator do not by themselves yield a bounded inverse; closedness of the range is the decisive condition.

Geometry of unit spheres in normed spaces

The unit sphere of a normed space is S=\{x:\|x\|=1\}, and its shape shows which directions of the norm are flat. Suppose two distinct points x,y of S are joined by a segment lying entirely in S. Then the midpoint \tfrac12(x+y) has norm 1, so \|x+y\|=2=\|x\|+\|y\|, which is the equality case of the triangle inequality. If x and y were linearly dependent, then y=cx with |c|=1; the choice c=1 gives x=y, which is excluded, and c=-1 gives x+y=0, which has norm 0 rather than 2. Hence x and y are linearly independent. A segment inside the unit sphere exists in \ell^\infty and \ell^1, whereas in a Hilbert space the parallelogram law gives \|\tfrac12(x+y)\|^2=1-\tfrac14\|x-y\|^2<1 for distinct unit vectors, so no such segment exists.

Dilation operators on L2(R)

For a>0, the dilation (D_af)(x)=a^{-1/2}f(x/a) is a linear map on L^2(\mathbb R). The change of variables u=x/a gives \|D_af\|_2^2=\int|f(x/a)|^2\,a^{-1}\,dx=\|f\|_2^2, so D_a is a linear isometry, bounded with norm 1. An isometry is also bounded below, meaning \|D_af\|\ge c\|f\| with c=1. The inverse map is D_{1/a}, so D_a is a bijection of L^2(\mathbb R) onto itself. The composition rule is multiplicative: (D_aD_bf)(x)=(ab)^{-1/2}f\big(x/(ab)\big), so D_aD_b=D_{ab}, and not D_{a+b}. The normalising factor a^{-1/2} is exactly what preserves the norm, so it should be checked first whenever a scaling operator is studied.

Norm of linear functionals on C[0,1]

For a bounded linear functional F on a normed space, \|F\|=\sup\{|F(f)|:\|f\|\le1\}. For integration on C[0,1] with the sup norm, |\int_0^1 f\,dt|\le\|f\|_\infty, so \|F\|\le1, and the constant function 1 gives equality, so \|F\|=1. The supremum is attained there, but attainment depends on the subspace over which one maximises. If X is a subspace with boundary conditions such as f(0)=0, then a norming element f\in X with \|f\|_\infty=1 and \int_0^1 f=1 would satisfy \int_0^1(1-f)\,dt=0 with 1-f\ge0 continuous, forcing f\equiv1, which contradicts the boundary condition. So the supremum is not attained on X. The same integral on the unit ball of L^2 attains its supremum at the constant function, because \|1\|_2=1 and the Cauchy–Schwarz inequality is sharp at f=1. Thus the norm of a functional must always be checked in the norm and subspace actually given.