GATE Mathematics Functional Analysis Syllabus

Define an orthonormal set and an orthonormal basis of a Hilbert space.

A set $\{e_{\alpha}\}$ is orthonormal if $\langle e_{\alpha},e_{\beta}\rangle=\delta_{\alpha\beta}$. It is an orthonormal basis (complete orthonormal set) if its closed linear span is all of $H$, equivalently if the only vector orthogonal to every $e_{\alpha}$ is $0$.

State Bessel's inequality for an orthonormal set $\{e_{n}\}$.

For any $x\in H$, $\sum_{n}|\langle x,e_{n}\rangle|^{2}\leq\|x\|^{2}$. The numbers $\langle x,e_{n}\rangle$ are the Fourier coefficients of $x$.

State Parseval's identity and what it characterizes.

For an orthonormal set $\{e_{\alpha}\}$, $\|x\|^{2}=\sum_{\alpha}|\langle x,e_{\alpha}\rangle|^{2}$ for all $x\in H$ iff $\{e_{\alpha}\}$ is an orthonormal basis. Equivalently $x=\sum_{\alpha}\langle x,e_{\alpha}\rangle e_{\alpha}$.

Give the Fourier expansion of a vector with respect to an orthonormal basis $\{e_{\alpha}\}$.

$x=\sum_{\alpha}\langle x,e_{\alpha}\rangle\,e_{\alpha}$, where only countably many coefficients are nonzero, and $\langle x,y\rangle=\sum_{\alpha}\langle x,e_{\alpha}\rangle\overline{\langle y,e_{\alpha}\rangle}$ (the generalized Parseval relation).

List the equivalent characterizations of an orthonormal set $\{e_{\alpha}\}$ being an orthonormal basis (maximal).

(1) $\overline{\operatorname{span}}\{e_{\alpha}\}=H$; (2) $\langle x,e_{\alpha}\rangle=0\ \forall\alpha\Rightarrow x=0$; (3) $x=\sum_{\alpha}\langle x,e_{\alpha}\rangle e_{\alpha}$ for all $x$; (4) Parseval: $\|x\|^{2}=\sum_{\alpha}|\langle x,e_{\alpha}\rangle|^{2}$ for all $x$; (5) the set is maximal orthonormal.

Describe the Gram-Schmidt orthonormalization process.

From linearly independent $\{x_{n}\}$ build orthonormal $\{e_{n}\}$: set $v_{1}=x_{1}$, and $v_{n}=x_{n}-\sum_{k=1}^{n-1}\langle x_{n},e_{k}\rangle e_{k}$, then $e_{n}=v_{n}/\|v_{n}\|$. Then $\operatorname{span}\{e_{1},\dots,e_{n}\}=\operatorname{span}\{x_{1},\dots,x_{n}\}$.

When is a Hilbert space separable, in terms of its orthonormal basis?

A Hilbert space is separable iff it has a countable orthonormal basis. All separable infinite-dimensional Hilbert spaces are isometrically isomorphic to $\ell^{2}$.

State the Projection Theorem for a closed subspace of a Hilbert space.

If $M$ is a closed subspace of a Hilbert space $H$, then $H=M\oplus M^{\perp}$: every $x\in H$ has a unique decomposition $x=m+m^{\perp}$ with $m\in M$, $m^{\perp}\in M^{\perp}$. The map $P:x\mapsto m$ is the orthogonal projection onto $M$.

State the closest-point (best approximation) property characterizing the orthogonal projection $m=Px$.

For $x\in H$ and closed convex $M$, there is a unique $m\in M$ minimizing $\|x-m\|$. When $M$ is a closed subspace, $m=Px$ is characterized by $x-m\perp M$, i.e. $\langle x-m,y\rangle=0$ for all $y\in M$.

List the defining algebraic properties of an orthogonal projection $P$ onto a closed subspace.

$P$ is linear, idempotent ($P^{2}=P$), self-adjoint ($P^{*}=P$), with $\|P\|=1$ (if $P\neq 0$), range $M$ closed, kernel $M^{\perp}$, and $I-P$ is the orthogonal projection onto $M^{\perp}$.

For a subset $S$ of a Hilbert space, what is $S^{\perp}$ and what is $(S^{\perp})^{\perp}$?

$S^{\perp}=\{x\in H:\langle x,s\rangle=0\ \forall s\in S\}$ is always a closed subspace. $(S^{\perp})^{\perp}=\overline{\operatorname{span}}\,S$. In particular if $M$ is a closed subspace, $M^{\perp\perp}=M$.

State the Riesz Representation Theorem for bounded linear functionals on a Hilbert space.

For every $f\in H^{*}$ there exists a unique $y\in H$ such that $f(x)=\langle x,y\rangle$ for all $x\in H$, and $\|f\|=\|y\|$. The map $f\mapsto y$ is a conjugate-linear isometric bijection of $H^{*}$ onto $H$.

What does the Riesz Representation Theorem imply about a Hilbert space and its dual (reflexivity, self-duality)?

$H$ is (conjugate-linearly) isometrically isomorphic to its dual $H^{*}$, so $H$ is self-dual and every Hilbert space is reflexive.

State the Lax-Milgram theorem (Riesz-type representation for bounded coercive sesquilinear forms).

If $a:H\times H\to\mathbb{C}$ is a bounded ($|a(x,y)|\leq C\|x\|\|y\|$) and coercive ($\operatorname{Re}a(x,x)\geq c\|x\|^{2}$, $c>0$) sesquilinear form, then for each $f\in H^{*}$ there is a unique $u\in H$ with $a(u,v)=f(v)$ for all $v\in H$.

Define the adjoint $T^{*}$ of a bounded operator $T$ on a Hilbert space and give its key norm identity.

$T^{*}$ is the unique bounded operator satisfying $\langle Tx,y\rangle=\langle x,T^{*}y\rangle$ for all $x,y$. Its existence follows from Riesz representation. Key identities: $\|T^{*}\|=\|T\|$ and $\|T^{*}T\|=\|T\|^{2}$ (the C*-identity).

Define a self-adjoint operator and state the key fact about its inner-product values.

A bounded operator $T$ is self-adjoint if $T=T^{*}$, i.e. $\langle Tx,y\rangle=\langle x,Ty\rangle$ for all $x,y$. Then $\langle Tx,x\rangle\in\mathbb{R}$ for all $x$, and $\|T\|=\sup_{\|x\|=1}|\langle Tx,x\rangle|$.

Define a compact operator between Banach spaces.

A linear operator $T:X\to Y$ is compact if the image $T(B_{X})$ of the closed unit ball has compact closure in $Y$; equivalently, every bounded sequence $\{x_{n}\}$ has a subsequence with $\{Tx_{n}\}$ convergent.

State the Spectral Theorem for a compact self-adjoint operator on a Hilbert space.

If $T$ is a compact self-adjoint operator on a Hilbert space $H$, then $H$ has an orthonormal basis $\{e_{n}\}$ consisting of eigenvectors of $T$, with real eigenvalues $\lambda_{n}$, and $Tx=\sum_{n}\lambda_{n}\langle x,e_{n}\rangle e_{n}$. The nonzero eigenvalues form a finite set or a sequence $\lambda_{n}\to 0$, each of finite multiplicity.

For a compact self-adjoint operator, what are the properties of its eigenvalues and eigenvectors?

Eigenvalues are real; eigenvectors for distinct eigenvalues are orthogonal; each nonzero eigenvalue has finite-dimensional eigenspace; the nonzero eigenvalues form at most a sequence converging to $0$; $\|T\|=\max_{n}|\lambda_{n}|$ equals the largest $|\lambda|$.

State the result that guarantees a nonzero compact self-adjoint operator has an eigenvalue.

For a nonzero compact self-adjoint $T$, at least one of $\pm\|T\|$ is an eigenvalue; equivalently there is an eigenvalue $\lambda$ with $|\lambda|=\|T\|$. This is the starting point of the spectral theorem's proof.

Define the spectrum $\sigma(T)$ and the resolvent set of a bounded operator on a complex Banach space.

$\sigma(T)=\{\lambda\in\mathbb{C}:T-\lambda I\text{ is not invertible in }B(X)\}$. The resolvent set is its complement $\rho(T)$, on which $(T-\lambda I)^{-1}$ exists and is bounded. For $X\neq\{0\}$ complex, $\sigma(T)$ is nonempty, compact, and contained in $\{|\lambda|\leq\|T\|\}$.

Classify the spectrum into point, continuous, and residual parts.

Point spectrum: $\lambda$ with $T-\lambda I$ not injective (eigenvalues). Continuous spectrum: $T-\lambda I$ injective with dense, non-closed range. Residual spectrum: $T-\lambda I$ injective with range not dense.

Describe the spectrum of a compact operator on an infinite-dimensional Banach space (Riesz-Schauder theory).

The spectrum is a countable set with $0$ as its only possible limit point; every nonzero spectral value is an eigenvalue of finite multiplicity, and $0\in\sigma(T)$ when $\dim X=\infty$.

Compare the spectrum of a self-adjoint operator with that of a unitary operator on a Hilbert space.

For a bounded self-adjoint operator $\sigma(T)\subseteq\mathbb{R}$ (and residual spectrum is empty). For a unitary operator $U$ ($U^{*}U=UU^{*}=I$), $\sigma(U)\subseteq\{z\in\mathbb{C}:|z|=1\}$, the unit circle.

State the spectral radius and Gelfand's spectral radius formula.

The spectral radius is $r(T)=\max\{|\lambda|:\lambda\in\sigma(T)\}$, and $r(T)=\lim_{n\to\infty}\|T^{n}\|^{1/n}\leq\|T\|$. For a self-adjoint (or normal) operator, $r(T)=\|T\|$.

Compare strong and weak convergence with norm convergence in a Hilbert space.

Norm (strong) convergence $\|x_{n}-x\|\to 0$ implies weak convergence $\langle x_{n},y\rangle\to\langle x,y\rangle$ for all $y$, but not conversely. With weak convergence plus $\|x_{n}\|\to\|x\|$, one recovers norm convergence (Radon-Riesz property). An orthonormal sequence $e_{n}\rightharpoonup 0$ weakly but not in norm.

State the relationship $\ker(T^{*})=(\operatorname{ran} T)^{\perp}$ and the resulting orthogonal decomposition.

For bounded $T$ on a Hilbert space, $\ker(T^{*})=(\operatorname{ran} T)^{\perp}$ and $\overline{\operatorname{ran} T}=(\ker T^{*})^{\perp}$, giving $H=\overline{\operatorname{ran} T}\oplus\ker T^{*}$.

State the Fredholm Alternative for $T=I-K$ with $K$ compact.

For compact $K$ and $\lambda\neq 0$, either $(\lambda I-K)x=y$ has a unique solution for every $y$ (so $\lambda I-K$ is bijective), or the homogeneous equation $(\lambda I-K)x=0$ has a nontrivial solution; in the latter case $\dim\ker(\lambda I-K)=\dim\ker(\lambda I-K^{*})<\infty$.

Compare the Hahn-Banach Theorem with the Open Mapping, Closed Graph, and Uniform Boundedness theorems in terms of completeness requirements.

Hahn-Banach requires no completeness — it holds in any normed (even vector) space. The Open Mapping, Closed Graph, and Uniform Boundedness theorems are 'Baire category' theorems requiring completeness (Banach spaces) of the relevant space(s).

Define a normal operator and relate the classes normal, self-adjoint, and unitary.

$T$ is normal if $TT^{*}=T^{*}T$. Self-adjoint ($T=T^{*}$) and unitary ($T^{*}=T^{-1}$) operators are both special cases of normal operators. A bounded operator is normal iff $\|Tx\|=\|T^{*}x\|$ for all $x$.