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Functional Analysis

A rigorous study of infinite-dimensional vector spaces equipped with norms or inner products, their operators, and the powerful theorems that connect algebra, topology and analysis.

Normed & Banach Spaces Hilbert Spaces Bounded Linear Operators Fundamental Theorems Spectral Theory Dual Spaces & Reflexivity

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01 Normed Spaces & Banach Spaces

Normed linear spaces provide the natural framework for studying convergence, completeness, and continuity in infinite-dimensional settings. Banach spaces — complete normed spaces — are the workhorses of modern analysis.

◆ Normed Linear Spaces

Definition — Normed Space A normed linear space is a pair $(X, \|\cdot\|)$ where $X$ is a vector space over $\mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$) and $\|\cdot\|: X \to [0,\infty)$ satisfies:
(i) $\|x\| = 0 \iff x = 0$,
(ii) $\|\alpha x\| = |\alpha|\,\|x\|$ for all $\alpha \in \mathbb{F}$,
(iii) $\|x + y\| \le \|x\| + \|y\|$ (triangle inequality).

Every normed space is a metric space with $d(x,y) = \|x - y\|$. Convergence, open/closed sets, and continuity are all defined via this induced metric.

◆ Completeness and Banach Spaces

Definition — Banach Space A normed space $(X, \|\cdot\|)$ is called a Banach space if every Cauchy sequence in $X$ converges in $X$ (i.e., $X$ is complete with respect to the induced metric).

Key examples:

$\ell^p$ spaces ($1 \le p \le \infty$): sequences $(x_n)$ with $\|(x_n)\|_p = \bigl(\sum |x_n|^p\bigr)^{1/p} < \infty$. Complete for every $p$.
$C[a,b]$: continuous functions on $[a,b]$ with the sup-norm $\|f\|_\infty = \sup_{t \in [a,b]}|f(t)|$.
$L^p(\mu)$ spaces: $p$-integrable functions with $\|f\|_p = \bigl(\int |f|^p\,d\mu\bigr)^{1/p}$.

★ Example

Show that $C[0,1]$ with the $L^1$-norm $\|f\|_1 = \int_0^1 |f(t)|\,dt$ is not a Banach space.

Solution: Consider $f_n(t) = \begin{cases} 0 & t \le \tfrac{1}{2} - \tfrac{1}{n} \\ \text{linear} & \tfrac{1}{2}-\tfrac{1}{n} < t < \tfrac{1}{2} \\ 1 & t \ge \tfrac{1}{2} \end{cases}$. Each $f_n$ is continuous and $(f_n)$ is Cauchy in $\|\cdot\|_1$. However, the $L^1$-limit is the step function $\mathbf{1}_{[1/2,1]}$ which is not continuous. Hence $C[0,1]$ is not complete under $\|\cdot\|_1$.

02 Hilbert Spaces

Hilbert spaces are Banach spaces whose norm arises from an inner product, giving access to orthogonality, projections, and Fourier-like expansions.

◆ Inner Products & Orthonormal Bases

Definition — Inner Product Space An inner product on a vector space $H$ is a map $\langle \cdot, \cdot \rangle : H \times H \to \mathbb{F}$ satisfying: (i) conjugate symmetry $\langle x,y\rangle = \overline{\langle y,x\rangle}$, (ii) linearity in the first argument, (iii) $\langle x,x\rangle \ge 0$ with equality iff $x = 0$. A complete inner product space is a Hilbert space.

Theorem — Cauchy-Schwarz Inequality For all $x, y \in H$: $|\langle x, y \rangle| \le \|x\|\,\|y\|$, with equality iff $x$ and $y$ are linearly dependent.

A set $\{e_\alpha\}$ is orthonormal if $\langle e_\alpha, e_\beta \rangle = \delta_{\alpha\beta}$. An orthonormal set is a basis (or complete orthonormal system) if its closed linear span equals $H$.

◆ Bessel's Inequality & Parseval's Identity

Theorem — Bessel's Inequality If $\{e_n\}$ is an orthonormal sequence in $H$ and $x \in H$, then $$\sum_{n=1}^{\infty} |\langle x, e_n \rangle|^2 \le \|x\|^2.$$

Theorem — Parseval's Identity $\{e_n\}$ is a complete orthonormal system if and only if for every $x \in H$: $$\|x\|^2 = \sum_{n=1}^{\infty} |\langle x, e_n \rangle|^2.$$ Equivalently, $x = \sum_n \langle x, e_n\rangle e_n$ for all $x \in H$.

★ Example

In $L^2[-\pi,\pi]$ with the orthonormal system $\{e_n = \frac{1}{\sqrt{2\pi}} e^{inx}\}_{n \in \mathbb{Z}}$, verify Parseval's identity for $f(x) = x$.

Solution: We have $\hat{f}(n) = \frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi} x\,e^{-inx}\,dx = \frac{(-1)^{n+1}\sqrt{2\pi}\,i}{n}$ for $n \ne 0$ and $\hat{f}(0) = 0$. Then $\sum_{n \ne 0} |\hat{f}(n)|^2 = \sum_{n \ne 0} \frac{2\pi}{n^2} = 4\pi \sum_{n=1}^{\infty}\frac{1}{n^2} = 4\pi \cdot \frac{\pi^2}{6} = \frac{2\pi^3}{3}$. And $\|f\|^2 = \int_{-\pi}^{\pi} x^2\,dx = \frac{2\pi^3}{3}$. They agree, confirming Parseval.

03 Bounded Linear Operators

Linear maps between normed spaces that respect the topological structure form the building blocks of operator theory. Compactness of an operator is a spectral tool of fundamental importance.

◆ Operator Norm

Definition — Bounded Linear Operator A linear map $T: X \to Y$ between normed spaces is bounded if $$\|T\| := \sup_{\|x\| \le 1} \|Tx\| < \infty.$$ Equivalently, $T$ is bounded iff $T$ is (Lipschitz) continuous.

The space $\mathcal{B}(X,Y)$ of all bounded linear operators from $X$ to $Y$, equipped with the operator norm, is a normed space. If $Y$ is a Banach space, then $\mathcal{B}(X,Y)$ is also a Banach space.

◆ Compact Operators & Fredholm Alternative

Definition — Compact Operator A linear operator $T: X \to Y$ is compact if for every bounded sequence $(x_n)$ in $X$, the sequence $(Tx_n)$ has a convergent subsequence in $Y$. Equivalently, $T$ maps bounded sets to relatively compact sets.

Properties: Every finite-rank operator is compact. The set of compact operators is a closed two-sided ideal in $\mathcal{B}(X)$. The identity operator on an infinite-dimensional space is never compact.

Theorem — Fredholm Alternative Let $T: X \to X$ be a compact operator on a Banach space $X$ and $\lambda \ne 0$. Then exactly one of the following holds:
(a) $(T - \lambda I)$ is bijective (so $(T - \lambda I)^{-1} \in \mathcal{B}(X)$), or
(b) $\lambda$ is an eigenvalue of $T$ and $\ker(T - \lambda I)$ is finite-dimensional.

★ Example

Show that the integral operator $Tf(x) = \int_0^1 K(x,t)f(t)\,dt$ on $C[0,1]$ with continuous kernel $K$ is compact.

04 Fundamental Theorems of Functional Analysis

The “big four” theorems — Hahn-Banach, Open Mapping, Closed Graph, and Uniform Boundedness — are cornerstones that derive deep consequences from completeness alone.

◆ Hahn-Banach Theorem

Theorem — Hahn-Banach (Extension Form) Let $X$ be a normed space, $M \subseteq X$ a subspace, and $f: M \to \mathbb{F}$ a bounded linear functional with $\|f\| = c$. Then there exists a bounded linear functional $\tilde{f}: X \to \mathbb{F}$ extending $f$ with $\|\tilde{f}\| = c$.

Consequences:

For every $x_0 \ne 0$ in $X$, there exists $f \in X^*$ with $\|f\| = 1$ and $f(x_0) = \|x_0\|$.
$X^*$ separates points of $X$: if $f(x) = f(y)$ for all $f \in X^*$, then $x = y$.
The canonical embedding $J: X \to X^{**}$ defined by $J(x)(f) = f(x)$ is an isometric injection.

◆ Open Mapping & Closed Graph Theorems

Theorem — Open Mapping Theorem (Banach-Schauder) If $X$ and $Y$ are Banach spaces and $T: X \to Y$ is a surjective bounded linear operator, then $T$ is an open map (i.e., $T$ maps open sets to open sets).

Corollary (Bounded Inverse Theorem): A bijective bounded linear operator between Banach spaces has a bounded inverse.

Theorem — Closed Graph Theorem Let $X, Y$ be Banach spaces and $T: X \to Y$ a linear operator. Then $T$ is bounded if and only if its graph $\{(x, Tx) : x \in X\}$ is closed in $X \times Y$.

◆ Uniform Boundedness Principle

Theorem — Banach-Steinhaus (Uniform Boundedness) Let $X$ be a Banach space, $Y$ a normed space, and $\{T_\alpha\}_{\alpha \in A} \subset \mathcal{B}(X,Y)$. If $\sup_{\alpha}\|T_\alpha x\| < \infty$ for each $x \in X$ (pointwise bounded), then $\sup_{\alpha}\|T_\alpha\| < \infty$ (uniformly bounded).

★ Example

Use the Uniform Boundedness Principle to show that there exists a continuous $2\pi$-periodic function whose Fourier series diverges at a point.

Solution: The $n$-th partial sum of the Fourier series at $x = 0$ defines $S_n(f) = \int_{-\pi}^{\pi} f(t) D_n(t)\,dt$ where $D_n$ is the Dirichlet kernel. We have $\|S_n\| = \frac{1}{2\pi}\int_{-\pi}^{\pi}|D_n(t)|\,dt \sim \frac{4}{\pi^2}\ln n \to \infty$. If the Fourier series converged at $0$ for every $f \in C[-\pi,\pi]$, then $\{S_n\}$ would be pointwise bounded, hence uniformly bounded by Banach-Steinhaus — contradicting $\|S_n\| \to \infty$. Therefore some continuous function has a divergent Fourier series at $0$.

05 Spectral Theory

The spectrum of an operator generalises the notion of eigenvalues to infinite dimensions, providing deep structural information about the operator.

◆ Spectrum & Resolvent

Definition — Spectrum Let $T \in \mathcal{B}(X)$. The resolvent set is $\rho(T) = \{\lambda \in \mathbb{C} : (T - \lambda I)^{-1} \in \mathcal{B}(X)\}$. The spectrum is $\sigma(T) = \mathbb{C} \setminus \rho(T)$. It decomposes into:
• Point spectrum $\sigma_p(T)$: eigenvalues ($\ker(T - \lambda I) \ne \{0\}$).
• Continuous spectrum $\sigma_c(T)$: $(T - \lambda I)$ injective, dense range, unbounded inverse.
• Residual spectrum $\sigma_r(T)$: $(T - \lambda I)$ injective but range not dense.

Theorem For $T \in \mathcal{B}(X)$ where $X$ is a Banach space: $\sigma(T)$ is a non-empty compact subset of $\{|\lambda| \le \|T\|\}$ and the spectral radius satisfies $r(T) = \lim_{n \to \infty} \|T^n\|^{1/n}$.

◆ Spectral Theorem for Compact Self-Adjoint Operators

Theorem — Spectral Theorem (Compact Self-Adjoint) Let $T$ be a compact self-adjoint operator on a Hilbert space $H$. Then:
(i) All eigenvalues of $T$ are real.
(ii) Eigenvectors corresponding to distinct eigenvalues are orthogonal.
(iii) The eigenvalues form a sequence $\lambda_n \to 0$ (if infinitely many).
(iv) $H$ has an orthonormal basis of eigenvectors of $T$, and $$Tx = \sum_{n=1}^{\infty} \lambda_n \langle x, e_n \rangle e_n.$$

★ Example

Find the eigenvalues of $T: \ell^2 \to \ell^2$ defined by $T(x_1, x_2, x_3, \ldots) = (x_1, \tfrac{x_2}{2}, \tfrac{x_3}{3}, \ldots)$.

Solution: $T$ is diagonal with $Te_n = \frac{1}{n}e_n$ where $\{e_n\}$ is the standard basis. So the eigenvalues are $\lambda_n = \frac{1}{n}$ for $n = 1, 2, 3, \ldots$. Note $\lambda_n \to 0$, $T$ is self-adjoint (since all $\lambda_n$ are real), and $T$ is compact (since it is the limit in operator norm of finite-rank operators $T_N$ that agree with $T$ on the first $N$ coordinates and are zero otherwise).

06 Dual Spaces & Reflexivity

The dual space $X^*$ of bounded linear functionals encodes crucial information about the structure of $X$. Reflexivity is the condition that the canonical embedding into the bidual is surjective.

◆ Dual Spaces & Riesz Representation

Definition — Dual Space The dual space $X^*$ of a normed space $X$ is $\mathcal{B}(X, \mathbb{F})$, the space of all bounded linear functionals on $X$, equipped with the operator norm. $X^*$ is always a Banach space.

Theorem — Riesz Representation (Hilbert spaces) Let $H$ be 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 $\|f\| = \|y\|$. Thus $H^* \cong H$ (conjugate-linear isometric isomorphism).

Classical dual identifications:

$(\ell^p)^* \cong \ell^q$ where $\frac{1}{p} + \frac{1}{q} = 1$, for $1 \le p < \infty$.
$(L^p)^* \cong L^q$ for $1 \le p < \infty$.
$(\ell^1)^* \cong \ell^\infty$ and $(c_0)^* \cong \ell^1$.

◆ Reflexive Spaces

Definition — Reflexivity A Banach space $X$ is reflexive if the canonical embedding $J: X \to X^{**}$ defined by $J(x)(f) = f(x)$ is surjective (hence an isometric isomorphism).

Facts:

Every Hilbert space is reflexive (by the Riesz theorem).
$\ell^p$ and $L^p$ are reflexive for $1 < p < \infty$.
$\ell^1$, $\ell^\infty$, $L^1$, $L^\infty$, $C[0,1]$ are not reflexive.
A Banach space is reflexive iff its dual is reflexive.

★ Example

Show that $\ell^1$ is not reflexive.

Solution: We have $(\ell^1)^* = \ell^\infty$. If $\ell^1$ were reflexive, then $(\ell^1)^{**} = (\ell^\infty)^* \cong \ell^1$. But $(\ell^\infty)^*$ contains functionals that are not representable by $\ell^1$ sequences — for instance, Banach limits, which are bounded linear functionals on $\ell^\infty$ that extend the ordinary limit but are not given by any element of $\ell^1$. Hence $J: \ell^1 \to (\ell^1)^{**}$ is not surjective.

★ Key Takeaways

Banach spaces are complete normed spaces; Hilbert spaces are Banach spaces with an inner product.
Parseval's identity characterises complete orthonormal systems and underlies all Fourier analysis in Hilbert spaces.
The four pillars — Hahn-Banach, Open Mapping, Closed Graph, Uniform Boundedness — all rely on completeness (Baire category).
Compact operators on Hilbert spaces have a clean spectral theory: eigenvalues accumulate only at $0$.
Reflexivity ($X \cong X^{**}$ canonically) holds for $L^p$ / $\ell^p$ with $1 < p < \infty$ but fails for $p = 1, \infty$.
The Riesz representation theorem identifies $H^*$ with $H$ for every Hilbert space.

✎ Practice Problems

Problem 1

Prove that the closed unit ball in an infinite-dimensional normed space is never compact.

Show Solution ▼

By Riesz's lemma, for any finite-dimensional subspace $M$ and any $\varepsilon > 0$ there exists $x$ with $\|x\| = 1$ and $d(x, M) \ge 1 - \varepsilon$. Inductively construct a sequence $(x_n)$ in the unit ball with $\|x_m - x_n\| \ge \frac{1}{2}$ for $m \ne n$. This sequence has no convergent subsequence, so the closed unit ball is not sequentially compact, hence not compact.

Problem 2

Let $T: \ell^2 \to \ell^2$ be the right shift operator $T(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$. Find $\sigma(T)$, $\sigma_p(T)$, and $\sigma_p(T^*)$.

Show Solution ▼

$\|T\| = 1$ so $\sigma(T) \subseteq \overline{D}(0,1)$. $T$ is an isometry so $Tx = \lambda x$ implies $\|x\| = |\lambda|\,\|x\|$, giving $|\lambda| = 1$ or $x = 0$. But $T(x_1, x_2, \ldots) = \lambda(x_1, x_2, \ldots)$ forces $0 = \lambda x_1$, $x_1 = \lambda x_2$, etc., giving $x = 0$. So $\sigma_p(T) = \emptyset$. The adjoint $T^*$ is the left shift: $T^*(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$. Then $T^* x = \lambda x$ gives $x = (1, \lambda, \lambda^2, \ldots)$ which is in $\ell^2$ iff $|\lambda| < 1$. So $\sigma_p(T^*) = D(0,1)$ (open disk). Since $\sigma(T^*) = \overline{\sigma(T)}$ and must be closed, $\sigma(T) = \overline{D}(0,1)$.

Problem 3

Let $M$ be a closed subspace of a Hilbert space $H$. Prove that $H = M \oplus M^\perp$.

Show Solution ▼

For $x \in H$, let $m_0 = \text{proj}_M(x)$ be the unique closest point in $M$ (exists by completeness and convexity). Set $z = x - m_0$. For any $m \in M$ and $t \in \mathbb{R}$, $\|z\|^2 \le \|z - tm\|^2 = \|z\|^2 - 2t\,\text{Re}\langle z,m\rangle + t^2\|m\|^2$, so $2t\,\text{Re}\langle z,m\rangle \le t^2\|m\|^2$ for all $t$, forcing $\text{Re}\langle z,m\rangle = 0$. Similarly for the imaginary part, giving $z \in M^\perp$. Thus $x = m_0 + z \in M + M^\perp$. Uniqueness: if $x = m_1 + z_1 = m_2 + z_2$, then $m_1 - m_2 = z_2 - z_1 \in M \cap M^\perp = \{0\}$.

Problem 4

Using the Open Mapping Theorem, show that if $T: X \to Y$ is a bounded bijective linear operator between Banach spaces, then $T^{-1}$ is bounded.

Show Solution ▼

Since $T$ is surjective and bounded between Banach spaces, the Open Mapping Theorem implies $T$ is an open map. Thus for every open set $U \subseteq X$, $T(U)$ is open in $Y$. Now $T^{-1}: Y \to X$ is linear (easy check). For the preimage of an open set $U \subseteq X$ under $T^{-1}$, we get $(T^{-1})^{-1}(U) = T(U)$, which is open. Hence $T^{-1}$ is continuous, i.e., bounded.

Problem 5

Show that the Volterra operator $Vf(x) = \int_0^x f(t)\,dt$ on $L^2[0,1]$ has $\sigma(V) = \{0\}$ and $\sigma_p(V) = \emptyset$.

Show Solution ▼

$V$ is compact (it maps bounded sets to equicontinuous sets by Arzelà-Ascoli). If $Vf = \lambda f$ with $\lambda \ne 0$, differentiating gives $f(x) = \lambda f'(x)$ with $f(0) = 0$. The solution is $f(x) = Ce^{x/\lambda}$; the condition $f(0) = 0$ forces $C = 0$, so $f = 0$. Thus $\sigma_p(V) = \emptyset$. Since $V$ is compact and has no nonzero eigenvalues, by the spectral theorem for compact operators, $\sigma(V) = \{0\}$. (Alternatively, one can show $\|V^n\| \le \frac{1}{n!}$ so the spectral radius $r(V) = \lim \|V^n\|^{1/n} = 0$.)

🎯 Interactive Quiz

1. Which of the following spaces is not a Banach space?

A $C[0,1]$ with $\|f\|_\infty$

B The space of polynomials on $[0,1]$ with $\|f\|_\infty$

C $\ell^2$

D $\ell^\infty$

2. Parseval's identity $\|x\|^2 = \sum |\langle x, e_n\rangle|^2$ holds for every $x \in H$ if and only if:

A $\{e_n\}$ is orthonormal

B $\{e_n\}$ is a complete orthonormal system

C $\{e_n\}$ is linearly independent

D $\{e_n\}$ is a finite orthonormal set

3. The Hahn-Banach theorem guarantees that for any $x_0 \ne 0$ in a normed space $X$, there exists $f \in X^*$ with $f(x_0) = \|x_0\|$ and:

A $\|f\| = 1$

B $\|f\| = \|x_0\|$

C $f$ is defined only on $\text{span}\{x_0\}$

D $f$ is compact

4. Let $T$ be a compact operator on an infinite-dimensional Banach space. Which statement is always true?

A All eigenvalues of $T$ are real

B Every nonzero $\lambda \in \sigma(T)$ is an eigenvalue of finite multiplicity

C $\sigma(T)$ is finite

D $T$ is invertible

5. Which of the following Banach spaces is reflexive?

A $\ell^1$

B $\ell^\infty$

C $L^2[0,1]$

D $C[0,1]$