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Summary of Key Theorems

TheoremStatement
Hahn-BanachBounded functionals extend preserving norm
Open MappingSurjective bounded operator between Banach spaces is open
Closed GraphA linear operator between Banach spaces is bounded iff its graph is closed
Uniform BoundednessPointwise-bounded family of operators is uniformly bounded
Riesz RepresentationEvery functional on a Hilbert space is given by inner product
Spectral TheoremCompact self-adjoint operators have orthonormal eigenbasis
Fredholm AlternativeFor compact TT and λ0\lambda \neq 0: either λIT\lambda I - T is invertible or has nontrivial kernel
Banach-AlaogluClosed unit ball of XX^* is weak*-compact

Statement. Let XX be a normed vector space and YXY \subseteq X a subspace. If f:YRf : Y \to \mathbb{R} is a bounded linear functional, then there exists an extension F:XRF : X \to \mathbb{R} such that FY=fF|_Y = f and F=f\|F\| = \|f\|.

Intuition. The theorem guarantees that we can extend linear functionals from a subspace to the whole space without increasing the norm. This is fundamental for duality theory.

Corollaries. For any nonzero xXx \in X there exists fXf \in X^* with f=1\|f\| = 1 and f(x)=xf(x) = \|x\|. The natural embedding XXX \to X^{**} is isometric.

Statement. If T:XYT : X \to Y is a bounded surjective linear operator between Banach spaces, then TT is an open map (the image of every open set is open).

Intuition. Surjectivity plus completeness forces the operator to be well-behaved topologically. A key consequence is that a bijective bounded linear operator has a bounded inverse.

Corollary (Inverse Mapping Theorem). If T:XYT : X \to Y is a bijective bounded linear operator between Banach spaces, then T1T^{-1} is bounded.

Statement. Let T:XYT : X \to Y be a linear operator between Banach spaces. Then TT is bounded if and only if its graph Γ(T)={(x,Tx):xdom(T)}\Gamma(T) = \{(x, Tx) : x \in \mathrm{dom}(T)\} is closed in X×YX \times Y.

Intuition. This theorem provides a powerful way to prove boundedness: instead of checking continuity directly, one checks that whenever xnxx_n \to x and TxnyTx_n \to y, then y=Txy = Tx.

Common Pitfall. The domain and codomain must both be Banach spaces. A linear operator T:C[0,1]C[0,1]T : C[0,1] \to C[0,1] defined by Tf=fTf = f'' has a closed graph in C[0,1]×C[0,1]C[0,1] \times C[0,1], but its domain C2[0,1]C^2[0,1] is not complete under the sup norm. The theorem does not apply.

Statement (Banach-Steinhaus). Let {Tα}αA\{T_\alpha\}_{\alpha \in A} be a family of bounded linear operators from a Banach space XX to a normed space YY. If supαTαx<\sup_\alpha \|T_\alpha x\| < \infty for each xXx \in X, then supαTα<\sup_\alpha \|T_\alpha\| < \infty.

Intuition. Pointwise boundedness implies uniform boundedness. This is a consequence of the Baire category theorem and is often used to prove that certain operators are unbounded.

Application. If a sequence of bounded operators converges pointwise, the limit operator is bounded and the sequence is uniformly bounded in norm.

Statement. Let HH be a Hilbert space. For every bounded linear functional f:HRf : H \to \mathbb{R}, there exists a unique vector yHy \in H such that f(x)=x,yf(x) = \langle x, y \rangle for all xHx \in H, and f=y\|f\| = \|y\|.

Intuition. In a Hilbert space, the dual space is isometrically isomorphic to the space itself. Every functional is just an inner product with some vector.

Consequence. The map y,yy \mapsto \langle \cdot, y \rangle is an isometric isomorphism HHH \cong H^*, so Hilbert spaces are self-dual.

Statement. If T:HHT : H \to H is a compact self-adjoint operator on a Hilbert space HH, then there exists an orthonormal basis of HH consisting of eigenvectors of TT. The corresponding eigenvalues are real and converge to zero.

Intuition. Compact self-adjoint operators are diagonalisable: they behave like infinite-dimensional symmetric matrices. This generalises the finite-dimensional spectral theorem.

Application. Sturm-Liouville problems, integral equations with symmetric kernels, and the quantum mechanical position and momentum operators all involve spectral decompositions.

Statement. Let T:XXT : X \to X be a compact operator on a Banach space XX. For any nonzero λC\lambda \in \mathbb{C}, either λIT\lambda I - T is invertible with bounded inverse, or λ\lambda is an eigenvalue of TT with finite multiplicity.

Intuition. For compact perturbations of the identity, the only possible obstructions to invertibility are eigenvalues. The null space and range have complementary dimensions.

Application. This theorem is essential for solving integral equations of the second kind: fλKf=gf - \lambda K f = g, where KK is a compact integral operator.

Statement. The closed unit ball of the dual space XX^* of a normed space XX is compact in the weak* topology.

Intuition. Even though the unit ball in an infinite-dimensional space is never compact in the norm topology, it becomes compact in the weak* topology. This provides a powerful existence principle for optimisation problems.

Application. Existence of minimisers in the calculus of variations: every bounded sequence in the dual has a weak* convergent subsequence.

These six core theorems are deeply interrelated. The Hahn-Banach, Open Mapping, Closed Graph, and Uniform Boundedness principles are often called the four pillars of functional analysis. All rely on the Baire category theorem and completeness.

The Riesz Representation and Spectral theorems are specific to Hilbert spaces, where the geometry of inner products provides stronger structure. The Fredholm Alternative bridges compact operator theory with spectral theory. Banach-Alaoglu uses weak* topology to recover compactness lost in infinite dimensions.

  1. Let X=C[0,1]X = C[0,1] and f(x)=x(0)f(x) = x(0). Use the Hahn-Banach theorem to extend ff to a functional on L[0,1]L^\infty[0,1] with the same norm.
  2. Suppose T:XYT : X \to Y is a linear operator between Banach spaces with closed graph. Show that TT is bounded. Why is the converse trivial?
  3. Give an example of a family of operators {Tn}\{T_n\} such that Tnx\|T_n x\| is bounded for each xx but Tn\|T_n\| \to \infty.
  4. Verify that the operator (Tf)(x)=01sin(xy)f(y)dy(Tf)(x) = \int_0^1 \sin(xy) f(y)\,dy on L2[0,1]L^2[0,1] is compact and self-adjoint. What does the spectral theorem tell you about its eigenfunctions?

The six core theorems of functional analysis give the subject its power: extension of functionals (Hahn-Banach), stability of surjectivity (Open Mapping), automatic continuity from graph closure (Closed Graph), uniform bounds from pointwise bounds (Uniform Boundedness), concrete dual representation (Riesz), and diagonalisation of compact operators (Spectral).

TheoremStatementApplication
Lax-MilgramBounded coercive bilinear form on Hilbert space has unique solutionElliptic PDEs
Hille-YosidaMaximal accretive operators generate contraction semigroupsEvolution equations
Krein-MilmanCompact convex set in locally convex space is closed hull of extreme pointsOptimisation
Stone-WeierstrassSubalgebra separating points and containing constants is denseApproximation theory
Arzelà-AscoliBounded equicontinuous sequence in C(K)C(K) has convergent subsequenceCompactness in function spaces
Space XXDual XX^*Norm on XX^*
Lp(μ)L^p(\mu), 1<p<1 < p < \inftyLq(μ)L^q(\mu), 1/p+1/q=11/p + 1/q = 1gq\|g\|_q
L1(μ)L^1(\mu)L(μ)L^\infty(\mu) (if σ\sigma-finite)g\|g\|_\infty
C(K)C(K), KK compact HausdorffSpace of Radon measures M(K)\mathcal{M}(K)Total variation
c0c_0 (sequences converging to 0)1\ell^1x1\|x\|_1
p\ell^p, 1p<1 \leq p < \inftyq\ell^qxq\|x\|_q
Hilbert space HHHH (Riesz isomorphism)Same norm
TopologyConvergence conditionUse case
NormTnT0\|T_n - T\| \to 0Strongest; uniform approximation
Strong (SOT)TnxTx0\|T_n x - Tx\| \to 0 for each xxPointwise convergence
Weak (WOT)$\langle (T_n - T)x, y \rangle
Weak*$x^*(T_n - T)
Operator typeSpectrumProperties
Compact self-adjointPoint spectrum only, λn0\lambda_n \to 0Orthonormal eigenbasis
UnitaryσS1\sigma \subseteq S^1 (unit circle)U=1\|U\| = 1, U=U1U^* = U^{-1}
NormalT=r(T)\|T\| = r(T) (spectral radius)Spectral theorem applies
Bounded self-adjointσR\sigma \subseteq \mathbb{R}supσ=T\sup\sigma = \|T\| if T0T \geq 0