Which of the following normed spaces are Banach spaces?

C[0, 1] with ‖f‖ = ∫₀¹ |f|

the polynomials on [0, 1] with the sup norm

Which statements about separability, duals and reflexivity are true?

the dual of c₀ is isometrically isomorphic to ℓ¹

Which of the following is a consequence of the Hahn–Banach theorem in a normed space X?

for every x ≠ 0 there is f ∈ X∗ with ‖f‖ = 1 and f(x) = ‖x‖

every bounded linear bijection between normed spaces has a bounded inverse

every linear functional on X is bounded

Which statements about the role of completeness are true?

the differentiation map D: (C¹[0, 1], ‖·‖_{∞}) → (C[0, 1], ‖·‖_{∞}) has a closed graph but is unbounded

the identity map from (C[0, 1], ‖·‖_{∞}) to (C[0, 1], ‖·‖₁) is a bounded bijection whose inverse is unbounded

a bounded linear bijection between Banach spaces has a bounded inverse

the differentiation map D: (C¹[0, 1], ‖·‖_{∞}) → (C[0, 1], ‖·‖_{∞}) has a closed graph but is unbounded

the identity map from (C[0, 1], ‖·‖_{∞}) to (C[0, 1], ‖·‖₁) is a bounded bijection whose inverse is unbounded

a bounded linear bijection between Banach spaces has a bounded inverse

every linear functional on an infinite-dimensional normed space is bounded

On c₀₀, the space of finitely supported real sequences with the sup norm, let fₙ(x) = n xₙ. These functionals

are bounded at each x but have ‖fₙ‖ = n, showing that uniform boundedness can fail on a normed space that is not complete

are uniformly bounded, as the uniform boundedness principle predicts

are unbounded linear functionals

Which of the following operators on ℓ² are compact?

the diagonal operator (x₁, x₂, x₃, …) ↦ (x₁, x₂/2, x₃/3, …)

any bounded operator of finite rank

the diagonal operator (x₁, x₂, x₃, …) ↦ (x₁, x₂/2, x₃/3, …)

the right shift (x₁, x₂, …) ↦ (0, x₁, x₂, …)

any bounded operator of finite rank

By the Riesz representation theorem, every bounded linear functional f on a Hilbert space H

is f(x) = ⟨x, y⟩ for a unique y ∈ H, with ‖f‖ = ‖y‖

is f(x) = ⟨x, y⟩ for some y, not necessarily unique

has ‖f‖ = ‖y‖² for the representing vector y

is representable only when H is finite-dimensional

Which statements about Hilbert spaces are true?

ℓᵖ with p ≠ 2 is not a Hilbert space under any inner product inducing its norm

every orthonormal set in a separable Hilbert space is countable

ℓᵖ with p ≠ 2 is not a Hilbert space under any inner product inducing its norm

every orthonormal set in a separable Hilbert space is countable

H = M ⊕ M^{⊥} holds only for finite-dimensional closed subspaces M

Let T be a compact self-adjoint operator on an infinite-dimensional Hilbert space. Which statements are true?

every non-zero eigenvalue of T has a finite-dimensional eigenspace

T has uncountably many distinct eigenvalues

Functional Analysis: Banach Spaces, the Four Big Theorems, Hilbert Spaces and the Spectral Theorem

Functional Analysis is Section 7 of the GATE Mathematics (MA) paper: linear algebra in infinite dimensions, where the new ingredient is that linear maps need not be continuous and closed bounded sets need not be compact. The section names normed spaces, bounded and compact operators, Banach spaces, separability, dual spaces, the Hahn–Banach theorem, the open mapping and closed graph theorems, the principle of uniform boundedness, and then inner-product and Hilbert spaces with orthonormal bases, the projection theorem, the Riesz representation theorem and the spectral theorem for compact self-adjoint operators. Three of the four big theorems need completeness, and the paper tests exactly that: each is paired here with the incomplete space on which it fails. The numerical questions are operator and functional norms, distances computed by projection, and eigenvalues of compact self-adjoint operators.

1. Normed spaces, Banach spaces and bounded operators

A Banach space is a complete normed space. A linear T: X → Y is bounded if ‖Tx‖ ≤ C‖x‖; bounded ⇔ continuous ⇔ continuous at 0, and the least C is the operator norm ‖T‖ = sup_‖x‖≤1 ‖Tx‖. B(X, Y) with this norm is a Banach space whenever Y is. In finite dimensions all norms are equivalent, every linear map is bounded and the closed unit ball is compact; each of these fails in every infinite-dimensional normed space.

Operator and functional norms to compute
Operator	Norm	Why
matrix A on (ℝⁿ, ‖·‖_∞)	largest absolute row sum	attained at a vector of ±1 matching the signs of that row
matrix A on (ℝⁿ, ‖·‖₁)	largest absolute column sum	attained at a standard basis vector
matrix A on (ℝⁿ, ‖·‖₂)	largest singular value √λ_max(AᵀA)	for symmetric A, the largest \|eigenvalue\|
f ↦ ∫₀¹ f − f(1/2) on (C[0, 1], sup)	2, not attained	bounded by 1 + 1; approached by f = 1 with a narrow dip to −1 at 1/2
multiplication by g on L²[0, 1]	‖g‖_∞	\|gf\| ≤ ‖g‖_∞\|f\|; test on f supported where \|g\| is near its sup

⚠️ An unbounded linear map

Differentiation D on the polynomials, or on C¹[0, 1], with the sup norm is linear and unbounded: ‖xⁿ‖_∞ = 1 while ‖Dxⁿ‖_∞ = n. On an infinite-dimensional normed space unbounded linear functionals always exist (extend a Hamel basis), so “linear” never implies “continuous” there.

2. Separability, dual spaces and the Hahn–Banach theorem

The dual X∗ = B(X, 𝕂) is always a Banach space. X is separable if it has a countable dense subset, and reflexive if the canonical isometry X → X∗∗ is onto. If X∗ is separable then X is, but not conversely: ℓ¹ is separable while its dual ℓ^∞ is not.

Duals and separability of the sequence and function spaces
Space	Dual	Separable?	Reflexive?
c₀	ℓ¹	yes	no
ℓ¹	ℓ^∞	yes	no
ℓᵖ, 1 < p < ∞	ℓ^q, 1/p + 1/q = 1	yes	yes
ℓ^∞	larger than ℓ¹	no: the 0–1 sequences are uncountably many at mutual distance 1	no
C[a, b], sup norm	regular Borel measures (functions of bounded variation)	yes (polynomials with rational coefficients)	no

Hahn–Banach (normed form). If M is a subspace of a normed space X and f ∈ M∗, then f extends to F ∈ X∗ with ‖F‖ = ‖f‖. Consequences: for each x ≠ 0 there is f ∈ X∗ with ‖f‖ = 1 and f(x) = ‖x‖; X∗ separates points; ‖x‖ = max_‖f‖≤1 |f(x)|; and a closed subspace M ≠ X can be separated from a point x ∉ M by a functional vanishing on M. The theorem needs no completeness.

3. Uniform boundedness, open mapping and closed graph

The three Baire theorems and the incomplete spaces that break them
Theorem	Statement (X, Y Banach)	Fails without completeness
Uniform boundedness (Banach–Steinhaus)	if {Tₐ} ⊆ B(X, Y) and sup_a ‖Tₐx‖ < ∞ for every x, then sup_a ‖Tₐ‖ < ∞	on c₀₀ (finitely supported sequences, sup norm), fₙ(x) = n xₙ is pointwise bounded but ‖fₙ‖ = n
Open mapping; bounded inverse	a bounded linear surjection is open; a bounded linear bijection has a bounded inverse	the identity (C[0, 1], ‖·‖_∞) → (C[0, 1], ‖·‖₁) is a bounded bijection with unbounded inverse; the target is not complete
Closed graph	a linear T: X → Y with closed graph is bounded	D: (C¹[0, 1], ‖·‖_∞) → (C[0, 1], ‖·‖_∞) has a closed graph (uniform limits of fₙ and fₙ′) and is unbounded; the domain is not complete

🧠 Two complete norms, one inequality

If a space is complete under two norms and ‖x‖₁ ≤ C‖x‖₂ for all x, the identity from the second to the first is a bounded bijection between Banach spaces, so its inverse is bounded and the norms are equivalent. That is why ‖·‖_∞ and ‖·‖₁ on C[0, 1], which satisfy ‖f‖₁ ≤ ‖f‖_∞ and are not equivalent, cannot both be complete.

4. Compact operators

T ∈ B(X, Y) is compact if it maps the unit ball to a relatively compact set (equivalently, bounded sequences to sequences with convergent image subsequences). Finite-rank bounded operators are compact; norm limits of compact operators are compact (for Y Banach); compact operators form a two-sided ideal; and the identity of an infinite-dimensional space is not compact. On ℓ² a diagonal operator (xₙ) ↦ (λₙxₙ) is compact iff λₙ → 0, and an integral operator with a continuous (or square-integrable) kernel on [0, 1] is compact.

Spectrum of a compact T on an infinite-dimensional space: 0 ∈ σ(T); every non-zero point of σ(T) is an eigenvalue with a finite-dimensional eigenspace; the eigenvalues are countable and can accumulate only at 0.
The shifts on ℓ² are not compact: the right shift is an isometry, so it maps the orthonormal eₙ to the orthonormal eₙ₊₁, which has no convergent subsequence.

5. Hilbert spaces: orthonormal bases, projection and Riesz representation

A Hilbert space is a complete inner-product space. A norm comes from an inner product iff it satisfies the parallelogram law ‖x + y‖² + ‖x − y‖² = 2‖x‖² + 2‖y‖²; in ℓᵖ with e₁, e₂ the left side is 2 · 2^2/p and the right side 4, so ℓᵖ is a Hilbert space only for p = 2. An orthonormal basis (eₐ) gives x = Σ⟨x, eₐ⟩eₐ and Parseval ‖x‖² = Σ|⟨x, eₐ⟩|² (Bessel’s inequality ≤ for any orthonormal set). A Hilbert space is separable iff it has a countable orthonormal basis, and every infinite-dimensional separable Hilbert space is isometrically isomorphic to ℓ².

Projection theorem. A non-empty closed convex set C in a Hilbert space has a unique point nearest to any x; for a closed subspace M, H = M ⊕ M^⊥ and x − Px ⊥ M. In L²[−1, 1] the nearest element of span{1, x} to x² is 1/3, and the squared distance is ‖x² − 1/3‖² = 8/45.
Riesz representation. Every f ∈ H∗ is f(x) = ⟨x, y⟩ for a unique y ∈ H, and ‖f‖ = ‖y‖. So f(x) = Σ xₙ/2ⁿ⁻¹ on ℓ² has norm (Σ 4¹⁻ⁿ)^1/2 = √(4/3). Hilbert spaces are reflexive.
The adjoint T∗ is defined by ⟨Tx, y⟩ = ⟨x, T∗y⟩, with ‖T∗‖ = ‖T‖; T is self-adjoint if T = T∗, and then ‖T‖ = sup_‖x‖=1 |⟨Tx, x⟩|.

6. The spectral theorem for compact self-adjoint operators

If T is compact and self-adjoint on a Hilbert space H, then its non-zero eigenvalues are real, form a finite set or a sequence λₙ → 0, have finite-dimensional eigenspaces, and there is an orthonormal set of eigenvectors (eₙ) with Tx = Σ λₙ⟨x, eₙ⟩eₙ for all x; (eₙ) together with an orthonormal basis of ker T is an orthonormal basis of H. At least one of ±‖T‖ is an eigenvalue, so ‖T‖ = max |λₙ|.

ℹ️ A worked integral operator

(Tf)(x) = ∫₀¹ min(x, t) f(t) dt on L²[0, 1] has a continuous symmetric kernel, so it is compact and self-adjoint. u = Tf satisfies −u″ = f with u(0) = 0, u′(1) = 0, so Tf = μf leads to f″ + f/μ = 0 with the same boundary conditions: f = sin((n − 1/2)πx) and μₙ = 1/((n − 1/2)²π²). The eigenfunctions form an orthonormal basis after scaling by √2, the eigenvalues decrease to 0, and ‖T‖ = μ₁ = 4/π² ≈ 0.405.

Key takeaways

Bounded ⇔ continuous for linear maps; ‖T‖ = sup_‖x‖≤1 ‖Tx‖ — row sums for ‖·‖_∞, column sums for ‖·‖₁, singular values for ‖·‖₂.
c₀∗ = ℓ¹, (ℓ¹)∗ = ℓ^∞, (ℓᵖ)∗ = ℓ^q; ℓ^∞ is not separable; ℓᵖ is reflexive only for 1 < p < ∞.
Hahn–Banach needs no completeness; uniform boundedness, open mapping and closed graph do, and c₀₀, (C[0, 1], ‖·‖₁) and (C¹[0, 1], sup) break them.
Compact operators: finite rank and their limits; diagonal operators with λₙ → 0; not the identity or the shifts in infinite dimensions.
Riesz: f = ⟨·, y⟩ with ‖f‖ = ‖y‖; projection gives nearest points; a compact self-adjoint operator is Σλₙ⟨·, eₙ⟩eₙ with ‖T‖ = max |λₙ|.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which of the following normed spaces are Banach spaces?
1. C[0, 1] with ‖f‖ = sup |f|
2. C[0, 1] with ‖f‖ = ∫₀¹ |f|
3. ℓ¹ with ‖x‖ = Σ|xₙ|
4. the polynomials on [0, 1] with the sup norm
Show answer
Answer: A — C[0, 1] with ‖f‖ = sup |f|; C — ℓ¹ with ‖x‖ = Σ|xₙ|
(1) A uniformly Cauchy sequence of continuous functions has a continuous uniform limit. (3) ℓ¹ is complete (Riesz–Fischer for counting measure). (2) Continuous ramps converging in L¹ to a step function are Cauchy with no continuous limit. (4) The Taylor polynomials of eˣ are Cauchy in sup norm and converge to a non-polynomial; a Banach space of countable dimension cannot exist (Baire).
Let A = [[1, 2], [3, 4]] act on ℝ² with the norm ‖(x, y)‖_∞ = max(|x|, |y|). The operator norm of A is ____.
Numerical answer — type the value.
Show answer
Answer: 7
For the sup norm, ‖A‖ is the largest absolute row sum: max(1 + 2, 3 + 4) = 7, attained at (1, 1), whose image (3, 7) has sup norm 7. The column-sum answer 6 belongs to the ‖·‖₁ norm.
Let φ(f) = ∫₀¹ f(t) dt − f(1/2) on C[0, 1] with the sup norm. The norm ‖φ‖ is ____.
Numerical answer — type the value.
Show answer
Answer: 2
|φ(f)| ≤ ‖f‖ + ‖f‖ = 2‖f‖. For the lower bound take fₙ = 1 except for a continuous dip to −1 on an interval of length 1/n around 1/2: ‖fₙ‖ = 1, ∫fₙ ≥ 1 − 2/n and fₙ(1/2) = −1, so φ(fₙ) ≥ 2 − 2/n → 2. The norm is 2, not attained, since attaining it would need f = 1 a.e. and f(1/2) = −1.
Which statements about separability, duals and reflexivity are true?
1. ℓ^∞ is separable
2. the dual of c₀ is isometrically isomorphic to ℓ¹
3. ℓ¹ is reflexive
4. ℓ² is reflexive
Show answer
Answer: B — the dual of c₀ is isometrically isomorphic to ℓ¹; D — ℓ² is reflexive
(1) False: the 0–1 sequences are uncountably many and pairwise at distance 1. (2) Each a ∈ ℓ¹ gives x ↦ Σaₙxₙ with norm ‖a‖₁, and all functionals arise so. (3) False: ℓ¹∗∗ = (ℓ^∞)∗ is strictly bigger than ℓ¹ (a separable space whose bidual is non-separable). (4) Hilbert spaces are reflexive by Riesz.
Which of the following is a consequence of the Hahn–Banach theorem in a normed space X?
1. for every x ≠ 0 there is f ∈ X∗ with ‖f‖ = 1 and f(x) = ‖x‖
2. every bounded linear bijection between normed spaces has a bounded inverse
3. every linear functional on X is bounded
4. X is complete
Show answer
Answer: A — for every x ≠ 0 there is f ∈ X∗ with ‖f‖ = 1 and f(x) = ‖x‖
Define f₀(tx) = t‖x‖ on span{x}, of norm 1, and extend by Hahn–Banach keeping the norm. The bounded inverse theorem needs both spaces complete and fails otherwise; unbounded functionals exist on every infinite-dimensional space; and Hahn–Banach says nothing about completeness of X (X∗ is complete regardless).
Which statements about the role of completeness are true?
1. the differentiation map D: (C¹[0, 1], ‖·‖_∞) → (C[0, 1], ‖·‖_∞) has a closed graph but is unbounded
2. the identity map from (C[0, 1], ‖·‖_∞) to (C[0, 1], ‖·‖₁) is a bounded bijection whose inverse is unbounded
3. a bounded linear bijection between Banach spaces has a bounded inverse
4. every linear functional on an infinite-dimensional normed space is bounded
Show answer
Answer: A — the differentiation map D: (C¹[0, 1], ‖·‖_{∞}) → (C[0, 1], ‖·‖_{∞}) has a closed graph but is unbounded; B — the identity map from (C[0, 1], ‖·‖_{∞}) to (C[0, 1], ‖·‖₁) is a bounded bijection whose inverse is unbounded; C — a bounded linear bijection between Banach spaces has a bounded inverse
(1) If fₙ → f and fₙ′ → g uniformly then f ∈ C¹ with f′ = g, so the graph is closed; ‖xⁿ‖ = 1 but ‖Dxⁿ‖ = n. No contradiction: the domain is not complete. (2) ‖f‖₁ ≤ ‖f‖_∞, but tall thin spikes have ‖·‖₁ → 0 with sup 1. (3) is the bounded inverse theorem. (4) is false: extend a Hamel basis and send its n-th element eₙ (normalised) to n.
On c₀₀, the space of finitely supported real sequences with the sup norm, let fₙ(x) = n xₙ. These functionals
1. are bounded at each x but have ‖fₙ‖ = n, showing that uniform boundedness can fail on a normed space that is not complete
2. are uniformly bounded, as the uniform boundedness principle predicts
3. are unbounded linear functionals
4. are not pointwise bounded
Show answer
Answer: A — are bounded at each x but have ‖fₙ‖ = n, showing that uniform boundedness can fail on a normed space that is not complete
For fixed x ∈ c₀₀, xₙ = 0 for n beyond the support, so sup_n |fₙ(x)| is a maximum over finitely many n: pointwise bounded. Each fₙ is bounded with ‖fₙ‖ = n (test on eₙ). So the conclusion of Banach–Steinhaus fails, which is possible only because c₀₀ is not complete.
Which of the following operators on ℓ² are compact?
1. the identity operator
2. the diagonal operator (x₁, x₂, x₃, …) ↦ (x₁, x₂/2, x₃/3, …)
3. the right shift (x₁, x₂, …) ↦ (0, x₁, x₂, …)
4. any bounded operator of finite rank
Show answer
Answer: B — the diagonal operator (x₁, x₂, x₃, …) ↦ (x₁, x₂/2, x₃/3, …); D — any bounded operator of finite rank
(2) It is the norm limit of its finite-rank truncations, since the tail has norm 1/(N + 1) → 0 (diagonal with λₙ → 0). (4) Its range is finite-dimensional, where bounded sets are relatively compact. (1) and (3) map the orthonormal eₙ to an orthonormal sequence, with no convergent subsequence.
By the Riesz representation theorem, every bounded linear functional f on a Hilbert space H
1. is f(x) = ⟨x, y⟩ for a unique y ∈ H, with ‖f‖ = ‖y‖
2. is f(x) = ⟨x, y⟩ for some y, not necessarily unique
3. has ‖f‖ = ‖y‖² for the representing vector y
4. is representable only when H is finite-dimensional
Show answer
Answer: A — is f(x) = ⟨x, y⟩ for a unique y ∈ H, with ‖f‖ = ‖y‖
If f ≠ 0, ker f is a closed subspace of codimension 1; choosing a unit z ⊥ ker f gives y = conj(f(z)) z. Uniqueness: ⟨x, y − y′⟩ = 0 for all x forces y = y′. Cauchy–Schwarz gives ‖f‖ ≤ ‖y‖ and x = y gives equality. Completeness is what the proof uses, not finite dimension.
The norm of the functional f(x) = x₁ + x₂/2 + x₃/4 + x₄/8 + … on ℓ², correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.15
f(x) = ⟨x, y⟩ with y = (1, 1/2, 1/4, …) ∈ ℓ², so by Riesz ‖f‖ = ‖y‖₂ = (1 + 1/4 + 1/16 + …)^1/2 = (4/3)^1/2 = 1.1547, i.e. 1.15. On ℓ¹ the same formula would give sup |yₙ| = 1, and on ℓ^∞ it would give Σ|yₙ| = 2.
The minimum of ∫₋₁¹ (x² − a − bx)² dx over real a, b, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.18
This is the squared distance in L²[−1, 1] from x² to span{1, x}. By the projection theorem the minimiser is the orthogonal projection: ⟨x², x⟩ = 0 gives b = 0, and a = ⟨x², 1⟩/⟨1, 1⟩ = (2/3)/2 = 1/3. The error x² − 1/3 = (2/3)P₂ has squared norm (4/9)(2/5) = 8/45 = 0.178, i.e. 0.18.
Which statements about Hilbert spaces are true?
1. ℓᵖ with p ≠ 2 is not a Hilbert space under any inner product inducing its norm
2. every orthonormal set in a separable Hilbert space is countable
3. every infinite-dimensional separable Hilbert space is isometrically isomorphic to ℓ²
4. H = M ⊕ M^⊥ holds only for finite-dimensional closed subspaces M
Show answer
Answer: A — ℓᵖ with p ≠ 2 is not a Hilbert space under any inner product inducing its norm; B — every orthonormal set in a separable Hilbert space is countable; C — every infinite-dimensional separable Hilbert space is isometrically isomorphic to ℓ²
(1) With e₁, e₂: ‖e₁ + e₂‖ₚ² + ‖e₁ − e₂‖ₚ² = 2 · 2^2/p, which equals 2‖e₁‖² + 2‖e₂‖² = 4 only for p = 2. (2) Distinct orthonormal vectors are √2 apart, so an uncountable set would defeat a countable dense set. (3) Map a countable orthonormal basis to (eₙ), using Parseval. (4) is false: the projection theorem holds for every closed subspace.
Let T be a compact self-adjoint operator on an infinite-dimensional Hilbert space. Which statements are true?
1. every non-zero eigenvalue of T has a finite-dimensional eigenspace
2. every eigenvalue of T is real
3. 0 belongs to the spectrum of T
4. T has uncountably many distinct eigenvalues
Show answer
Answer: A — every non-zero eigenvalue of T has a finite-dimensional eigenspace; B — every eigenvalue of T is real; C — 0 belongs to the spectrum of T
(1) An infinite orthonormal set of eigenvectors for λ ≠ 0 would have images λeₙ with no convergent subsequence. (2) ⟨Tx, x⟩ is real for self-adjoint T. (3) If 0 ∉ σ(T), T⁻¹ is bounded and I = T⁻¹T is compact, impossible in infinite dimensions. (4) is false: the non-zero eigenvalues form at most a sequence tending to 0.
The operator (Tf)(x) = ∫₀¹ min(x, t) f(t) dt on L²[0, 1] is compact and self-adjoint. Its norm ‖T‖, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.41
u = Tf satisfies u′(x) = ∫ₓ¹ f, u″ = −f, u(0) = 0, u′(1) = 0. So Tf = μf gives μf″ + f = 0, f(0) = 0, f′(1) = 0: f = sin(kx) with cos k = 0, k = (n − 1/2)π, μ = 1/k². The largest is μ₁ = 4/π² = 0.405, and for a compact self-adjoint operator ‖T‖ = max |μ| = 0.41.