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GATE Mathematics » Functional Analysis

Functional Analysis

Normed linear spaces, Banach and Hilbert spaces, bounded linear operators, the four fundamental theorems, and spectral theory — with rigorous definitions, worked examples, and interactive practice.

Normed Spaces Inner Product & Hilbert Spaces Bounded Operators Fundamental Theorems Spectral Theory
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01
Normed Linear Spaces & Banach Spaces

A normed linear space equips a vector space with a notion of length. When every Cauchy sequence converges, the space is called a Banach space — the natural setting for analysis in infinite dimensions.

◆ Normed Spaces
Definition — Normed Linear Space A normed linear space is a pair $(V, \|\cdot\|)$ where $V$ is a vector space over $\mathbb{R}$ (or $\mathbb{C}$) and $\|\cdot\|: V \to [0, \infty)$ satisfies:
  • Positive definiteness: $\|x\| = 0 \iff x = 0$.
  • Homogeneity: $\|\alpha x\| = |\alpha|\,\|x\|$ for all scalars $\alpha$.
  • Triangle inequality: $\|x + y\| \le \|x\| + \|y\|$.
The norm induces a metric: $d(x,y) = \|x - y\|$.
Definition — Banach Space A normed linear space is a Banach space if it is complete: every Cauchy sequence in $V$ converges to a limit in $V$.

Important examples:

  • $\mathbb{R}^n$ with any $\ell^p$ norm is a Banach space.
  • $\ell^p = \{(x_n) : \sum |x_n|^p < \infty\}$ for $1 \le p < \infty$, with $\|(x_n)\|_p = \left(\sum |x_n|^p\right)^{1/p}$, is a Banach space.
  • $C[a,b]$ with $\|f\|_\infty = \sup_{x \in [a,b]}|f(x)|$ is a Banach space.
  • $C[a,b]$ with $\|f\|_1 = \int_a^b |f(x)|\,dx$ is not complete, hence not a Banach space.
★ Example
Show that $\ell^\infty$ (the space of bounded sequences with the sup norm) is a Banach space.
Solution: Let $(x^{(k)})$ be a Cauchy sequence in $\ell^\infty$. For each fixed $n$, $|x_n^{(k)} - x_n^{(m)}| \le \|x^{(k)} - x^{(m)}\|_\infty \to 0$, so $(x_n^{(k)})_k$ is Cauchy in $\mathbb{R}$, converging to some $x_n$. Let $x = (x_n)$. Given $\varepsilon > 0$, choose $K$ so that $\|x^{(k)} - x^{(m)}\|_\infty < \varepsilon$ for $k, m \ge K$. Letting $m \to \infty$: $|x_n^{(k)} - x_n| \le \varepsilon$ for all $n$, so $\|x^{(k)} - x\|_\infty \le \varepsilon$. Also $\|x\|_\infty \le \|x^{(K)}\|_\infty + \varepsilon < \infty$, so $x \in \ell^\infty$.
◆ Equivalent Norms and Finite Dimensions
Theorem On a finite-dimensional vector space, all norms are equivalent. In particular, every finite-dimensional normed space is a Banach space (complete).
Riesz's Lemma If $Y$ is a proper closed subspace of a normed space $X$, then for every $\varepsilon > 0$, there exists $x \in X$ with $\|x\| = 1$ and $d(x, Y) \ge 1 - \varepsilon$.

A consequence of Riesz's lemma: the closed unit ball of a normed space is compact if and only if the space is finite-dimensional.

02
Inner Product Spaces & Hilbert Spaces

An inner product space has a richer structure than a normed space — it allows us to define angles, orthogonality, and projections. A complete inner product space is a Hilbert space.

◆ Inner Product Spaces
Definition — Inner Product An inner product on a vector space $V$ over $\mathbb{C}$ is a map $\langle \cdot, \cdot \rangle: V \times V \to \mathbb{C}$ satisfying:
  • Conjugate symmetry: $\langle x, y \rangle = \overline{\langle y, x \rangle}$.
  • Linearity in the first argument: $\langle \alpha x + \beta y, z \rangle = \alpha \langle x, z \rangle + \beta \langle y, z \rangle$.
  • Positive definiteness: $\langle x, x \rangle \ge 0$, with equality iff $x = 0$.
The induced norm is $\|x\| = \sqrt{\langle x, x \rangle}$.
Cauchy-Schwarz Inequality For all $x, y$ in an inner product space: $|\langle x, y \rangle| \le \|x\|\,\|y\|$, with equality iff $x$ and $y$ are linearly dependent.
Parallelogram Law A norm $\|\cdot\|$ comes from an inner product if and only if it satisfies: $$\|x + y\|^2 + \|x - y\|^2 = 2\|x\|^2 + 2\|y\|^2.$$
★ Example
Does $\ell^1$ have an inner product that induces its norm?
Solution: No. Take $x = (1,0,0,\dots)$ and $y = (0,1,0,\dots)$. Then $\|x\|_1 = \|y\|_1 = 1$, $\|x+y\|_1 = 2$, $\|x-y\|_1 = 2$. The parallelogram law requires $4 + 4 = 2(1) + 2(1) = 4$, but $8 \ne 4$. So the $\ell^1$ norm does not come from an inner product.
◆ Hilbert Spaces and Orthogonality
Definition — Hilbert Space A Hilbert space is an inner product space that is complete with respect to the induced norm.

Key examples: $\mathbb{R}^n$, $\mathbb{C}^n$, $\ell^2$, $L^2[a,b]$.

Projection Theorem Let $M$ be a closed subspace of a Hilbert space $H$. For every $x \in H$, there exists a unique $m \in M$ minimising $\|x - m\|$ (the orthogonal projection). Moreover, $H = M \oplus M^\perp$.
Riesz Representation Theorem Every bounded linear functional $f$ on a Hilbert space $H$ can be uniquely represented as $f(x) = \langle x, y \rangle$ for some $y \in H$, and $\|f\| = \|y\|$.
Bessel's Inequality & Parseval's Identity For an orthonormal set $\{e_n\}$ and any $x \in H$: $\sum_n |\langle x, e_n \rangle|^2 \le \|x\|^2$ (Bessel). If $\{e_n\}$ is a complete orthonormal basis: $\sum_n |\langle x, e_n \rangle|^2 = \|x\|^2$ (Parseval).
03
Bounded Linear Operators

Bounded linear operators are the continuous linear maps between normed spaces. Their study reveals the interplay between algebraic and topological structure.

◆ Operator Norm and Continuity
Definition — Bounded Linear Operator A linear map $T: X \to Y$ between normed spaces is bounded if there exists $M \ge 0$ such that $\|Tx\| \le M\|x\|$ for all $x \in X$. The operator norm is: $$\|T\| = \sup_{\|x\| \le 1} \|Tx\| = \sup_{\|x\|=1} \|Tx\| = \sup_{x \ne 0} \frac{\|Tx\|}{\|x\|}.$$
Theorem For a linear operator $T: X \to Y$, the following are equivalent:
  • $T$ is bounded.
  • $T$ is continuous at every point.
  • $T$ is continuous at $0$.
★ Example
Find the operator norm of $T: \ell^2 \to \ell^2$ defined by $T(x_1, x_2, x_3, \dots) = (0, x_1, x_2, x_3, \dots)$ (the right shift).
Solution: We have $\|Tx\|_2^2 = \sum_{n=2}^{\infty}|x_{n-1}|^2 = \sum_{n=1}^{\infty}|x_n|^2 = \|x\|_2^2$, so $\|Tx\|_2 = \|x\|_2$ for all $x$. Hence $\|T\| = 1$. The right shift is an isometry.
◆ Compact Operators
Definition — Compact Operator A linear operator $T: X \to Y$ is compact if $T(B_X)$ has compact closure in $Y$, where $B_X = \{x \in X : \|x\| \le 1\}$ is the closed unit ball. Equivalently, $T$ maps bounded sequences to sequences with convergent subsequences.
Properties of Compact Operators
  • Every compact operator is bounded.
  • The set of compact operators $\mathcal{K}(X,Y)$ is a closed subspace of $\mathcal{B}(X,Y)$ (the bounded operators).
  • Finite-rank operators are compact. The limit of finite-rank operators (in operator norm) is compact.
  • If $X$ is infinite-dimensional, the identity operator $I: X \to X$ is not compact (by Riesz's lemma).
★ Example
Show that $T: \ell^2 \to \ell^2$ defined by $T(x_n) = \left(\frac{x_n}{n}\right)$ is compact.
Solution: Define $T_k(x_n) = \left(\frac{x_1}{1}, \frac{x_2}{2}, \dots, \frac{x_k}{k}, 0, 0, \dots\right)$. Each $T_k$ has finite rank $k$, hence is compact. We have $\|T - T_k\| = \sup_{\|x\|_2 = 1}\left(\sum_{n=k+1}^{\infty}\frac{|x_n|^2}{n^2}\right)^{1/2} \le \frac{1}{k+1} \to 0$. So $T$ is the norm limit of compact operators, hence compact.
04
Fundamental Theorems

The four pillars of functional analysis: Hahn-Banach, Open Mapping, Closed Graph, and Uniform Boundedness. These theorems have profound consequences throughout analysis.

◆ Hahn-Banach Theorem
Theorem — Hahn-Banach (Extension Form) Let $X$ be a normed space, $Y \subseteq X$ a subspace, and $f: Y \to \mathbb{R}$ a bounded linear functional. Then $f$ can be extended to a bounded linear functional $\tilde{f}: X \to \mathbb{R}$ with $\|\tilde{f}\| = \|f\|$.

Consequences:

  • For every $x \ne 0$ in $X$, there exists $f \in X^*$ with $f(x) = \|x\|$ and $\|f\| = 1$.
  • The dual space $X^*$ separates points of $X$.
  • $\|x\| = \sup\{|f(x)| : f \in X^*, \|f\| \le 1\}$.
◆ Open Mapping and Closed Graph Theorems
Theorem — Open Mapping (Banach-Schauder) If $T: X \to Y$ is a surjective bounded linear operator between Banach spaces, then $T$ is an open map (i.e., $T$ maps open sets to open sets).

Corollary (Bounded Inverse): A bijective bounded linear operator between Banach spaces has a bounded inverse. That is, $T^{-1}$ is automatically continuous.

Theorem — Closed Graph Let $T: X \to Y$ be a linear operator between Banach spaces. If the graph $\{(x, Tx) : x \in X\} \subseteq X \times Y$ is closed, then $T$ is bounded.
★ Example
Let $T: C^1[0,1] \to C[0,1]$ be the differentiation operator $Tf = f'$, where $C^1[0,1]$ has the sup norm. Is $T$ bounded?
Solution: No. Consider $f_n(x) = x^n$. Then $\|f_n\|_\infty = 1$ but $\|Tf_n\|_\infty = \|nx^{n-1}\|_\infty = n \to \infty$. So $T$ is unbounded. This does not contradict the Closed Graph Theorem because $C^1[0,1]$ with the sup norm is not a Banach space (it is not closed in $C[0,1]$).
◆ Uniform Boundedness Principle
Theorem — Banach-Steinhaus (Uniform Boundedness) Let $X$ be a Banach space, $Y$ a normed space, and $\{T_\alpha\}_{\alpha \in I} \subseteq \mathcal{B}(X,Y)$. If $\sup_\alpha \|T_\alpha x\| < \infty$ for every $x \in X$ (pointwise boundedness), then $\sup_\alpha \|T_\alpha\| < \infty$ (uniform boundedness).

Application: If a sequence of bounded operators $(T_n)$ satisfies $\sup_n \|T_n x\| < \infty$ for all $x$, then $\sup_n \|T_n\| < \infty$. This is used to show that pointwise convergent sequences of operators are uniformly bounded.

05
Spectral Theory

Spectral theory extends eigenvalue theory to infinite-dimensional spaces. For compact and self-adjoint operators, the spectrum has a particularly elegant structure.

◆ Spectrum and Resolvent
Definition — Spectrum Let $T \in \mathcal{B}(X)$ where $X$ is a Banach space. The resolvent set is $\rho(T) = \{\lambda \in \mathbb{C} : (\lambda I - T) \text{ is bijective with bounded inverse}\}$. The spectrum is $\sigma(T) = \mathbb{C} \setminus \rho(T)$.

The spectrum decomposes into:

  • Point spectrum $\sigma_p(T)$: $\lambda$ is an eigenvalue ($\lambda I - T$ is not injective).
  • Continuous spectrum $\sigma_c(T)$: $\lambda I - T$ is injective with dense (but not all of $X$) range.
  • Residual spectrum $\sigma_r(T)$: $\lambda I - T$ is injective but its range is not dense.
Theorem For $T \in \mathcal{B}(X)$, the spectrum $\sigma(T)$ is a non-empty compact subset of $\{\lambda : |\lambda| \le \|T\|\}$. The spectral radius is $r(T) = \sup\{|\lambda| : \lambda \in \sigma(T)\} = \lim_{n \to \infty} \|T^n\|^{1/n}$.
◆ Spectral Theorem for Compact Operators
Spectral Theorem for Compact Self-Adjoint Operators Let $T$ be a compact self-adjoint operator on a Hilbert space $H$. Then:
  • The spectrum of $T$ is real and consists of at most countably many eigenvalues, with $0$ as the only possible accumulation point.
  • Eigenspaces for distinct eigenvalues are orthogonal.
  • Each non-zero eigenvalue has finite multiplicity.
  • $H$ has an orthonormal basis of eigenvectors of $T$: $Tx = \sum_n \lambda_n \langle x, e_n \rangle e_n$.
★ Example
Find the spectrum of the operator $T: \ell^2 \to \ell^2$ defined by $T(x_1, x_2, x_3, \dots) = \left(\frac{x_1}{1}, \frac{x_2}{2}, \frac{x_3}{3}, \dots\right)$.
Solution: $T$ is a diagonal operator with entries $d_n = 1/n$. The eigenvalues are $\lambda_n = 1/n$ for $n = 1, 2, 3, \dots$ with eigenvectors $e_n$ (standard basis). These accumulate at $0$. Since $T$ is compact (shown earlier), $0 \in \sigma(T) \setminus \sigma_p(T)$ (it is in the continuous spectrum). So $\sigma(T) = \{1/n : n \ge 1\} \cup \{0\}$ and $\sigma_p(T) = \{1/n : n \ge 1\}$.
★ Key Takeaways
✎ Practice Problems
Problem 1
Show that $c_0 = \{(x_n) \in \ell^\infty : x_n \to 0\}$ with the sup norm is a Banach space.
Show Solution ▼
$c_0$ is a closed subspace of $\ell^\infty$ (a Banach space). To see it is closed: if $(x^{(k)}) \subset c_0$ with $x^{(k)} \to x$ in $\ell^\infty$, then for any $\varepsilon > 0$, pick $K$ with $\|x^{(K)} - x\|_\infty < \varepsilon/2$. Since $x^{(K)} \in c_0$, there exists $N$ with $|x_n^{(K)}| < \varepsilon/2$ for $n > N$. Then $|x_n| \le |x_n - x_n^{(K)}| + |x_n^{(K)}| < \varepsilon$ for $n > N$. So $x \in c_0$. A closed subspace of a Banach space is Banach.
Problem 2
Using the parallelogram law, show that $\ell^p$ for $p \ne 2$ cannot be given an inner product inducing the $\ell^p$ norm.
Show Solution ▼
Take $x = e_1 = (1,0,0,\dots)$ and $y = e_2 = (0,1,0,\dots)$. Then $\|x\|_p = \|y\|_p = 1$, $\|x+y\|_p = 2^{1/p}$, $\|x-y\|_p = 2^{1/p}$. The parallelogram law requires $2 \cdot 2^{2/p} = 2 \cdot 1 + 2 \cdot 1 = 4$, giving $2^{2/p} = 2$, hence $p = 2$. For $p \ne 2$, the law fails.
Problem 3
Let $T: \ell^2 \to \ell^2$ be the left shift $T(x_1, x_2, x_3, \dots) = (x_2, x_3, x_4, \dots)$. Find $\|T\|$ and determine whether $T$ is compact.
Show Solution ▼
$\|Tx\|^2 = \sum_{n=2}^\infty |x_n|^2 \le \sum_{n=1}^\infty |x_n|^2 = \|x\|^2$, so $\|T\| \le 1$. Equality is achieved by $x = e_2 = (0,1,0,\dots)$: $\|Te_2\| = \|e_1\| = 1$. So $\|T\| = 1$. $T$ is not compact: $Te_n = e_{n-1}$ for $n \ge 2$, and $(e_n)_{n \ge 2}$ is a bounded sequence whose image $(e_{n-1})$ has no convergent subsequence (since $\|e_i - e_j\| = \sqrt{2}$ for $i \ne j$).
Problem 4
State the Hahn-Banach theorem and use it to show that for any $x_0 \ne 0$ in a normed space $X$, there exists $f \in X^*$ with $f(x_0) = \|x_0\|$ and $\|f\| = 1$.
Show Solution ▼
Define $g: \text{span}\{x_0\} \to \mathbb{R}$ by $g(\alpha x_0) = \alpha\|x_0\|$. Then $|g(\alpha x_0)| = |\alpha|\,\|x_0\| = \|\alpha x_0\|$, so $\|g\| = 1$. By Hahn-Banach, extend $g$ to $f \in X^*$ with $\|f\| = \|g\| = 1$. Then $f(x_0) = g(x_0) = \|x_0\|$.
Problem 5
Find the spectrum of the multiplication operator $M: L^2[0,1] \to L^2[0,1]$ defined by $(Mf)(x) = xf(x)$.
Show Solution ▼
For $\lambda \notin [0,1]$, the operator $(\lambda I - M)$ is invertible: $((\lambda I - M)^{-1}f)(x) = f(x)/(\lambda - x)$, which is bounded since $|\lambda - x|$ is bounded away from $0$ on $[0,1]$. For $\lambda \in [0,1]$, $(\lambda I - M)f(x) = (\lambda - x)f(x)$. This is injective (if $(\lambda - x)f(x) = 0$ a.e., then $f = 0$ a.e.), so $\lambda$ is not an eigenvalue. But the range is not all of $L^2$ (e.g., the constant function $1$ is not in the range). Hence $\sigma(T) = [0,1]$, which is entirely continuous spectrum (no eigenvalues).
🎯 Interactive Quiz
1. Which of the following spaces is a Hilbert space?
A $\ell^1$
B $\ell^2$
C $\ell^\infty$
D $C[0,1]$ with $\|\cdot\|_\infty$
2. The Hahn-Banach theorem guarantees that:
A Every bounded linear functional on a Hilbert space comes from an inner product
B A bounded linear functional on a subspace can be extended to the whole space with the same norm
C A surjective bounded linear map between Banach spaces is an open map
D Pointwise bounded families of operators are uniformly bounded
3. Which operator is compact on $\ell^2$?
A Identity operator $I$
B Right shift $(x_1,x_2,\dots) \mapsto (0,x_1,x_2,\dots)$
C $T(x_n) = (x_n/n)$
D Left shift $(x_1,x_2,\dots) \mapsto (x_2,x_3,\dots)$
4. The spectrum of a bounded linear operator on a Banach space is always:
A Finite
B Open
C Non-empty and compact
D Uncountable
5. A norm on a vector space comes from an inner product if and only if it satisfies:
A Triangle inequality
B Cauchy-Schwarz inequality
C Parallelogram law
D Bessel's inequality