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Volterra operator is Hilbert Schmidt and quasinilpotent

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $H : = L^{2} ([0, 1], C)$ with the integral pairing linear in the first argument ( $L^{2}$ with the integral pairing is a Hilbert space, The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz) and let $V f (x) : = \int_{0}^{x} f (t) d t (0 \leq x \leq 1)$ be the Volterra operator, that is, the integral operator with kernel $k (x, t) : = 1_{{0 \leq t \leq x}}$ on $[0, 1]^{2}$ . Then:

$V$ is Hilbert–Schmidt with $∥ V ∥_{H S}^{2} = \frac{1}{2}$ , hence compact (Hilbert–Schmidt operator and Hilbert–Schmidt norm, Hilbert–Schmidt operators are compact);
Choosing the integral representatives of the iterates, $V^{n} f (x) = \frac{1}{(n - 1)!} \int_{0}^{x} (x - t)^{n - 1} f (t) d t$ for every $n \geq 1$ , and $∥ V^{n + 1} ∥ \leq \frac{1}{n! \sqrt{2 (n + 1) (2 n + 1)}} \to 0$ ;
$V$ has no nonzero eigenvalue: $\ker (V - λ I) = {0}$ for every $λ \neq 0$ ;
the spectrum of $V$ is $σ (V) = {0}$ (Spectrum and resolvent of a bounded operator); thus $V$ is quasinilpotent, and in particular $V$ is not self-adjoint though it is compact, showing that the compact self-adjoint spectral theorem does not apply.

Facts & Assumptions

Given: AC, the complex Hilbert space $L^{2} ([0, 1])$ , the kernel $k (x, t) = 1_{{0 \leq t \leq x}}$ , and the operator $V$ .

[A1]

The kernel is square-integrable. The function $k$ is measurable on the completed product measure and $\int_{[0, 1]^{2}} ∣ k ∣^{2} d (λ \times λ) = \int_{0}^{1} λ ([0, x]) d x = \int_{0}^{1} x d x = \frac{1}{2}$ , by Tonelli, the measure of intervals and the polynomial integral; the class of $k$ lies in $L^{2}$ of the completed product measure with squared norm $\frac{1}{2}$ (The completed product measure, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, Newton–Leibniz remains valid across finitely many exceptional interior points when the primitive is continuous, For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[A2]

Kernel operators. For a kernel class $k$ of finite square norm, the operator $T_{k} f (x) = \int_{0}^{1} k (x, t) f (t) d t$ is bounded with $∥ T_{k} ∥ \leq ∥ k ∥_{2}$ , is Hilbert–Schmidt with $∥ T_{k} ∥_{H S} = ∥ k ∥_{2}$ , and is therefore compact (L two kernels give Hilbert–Schmidt operators, Hilbert–Schmidt operators are compact, Hilbert–Schmidt operator and Hilbert–Schmidt norm, Compact linear operator, A bounded linear operator between normed spaces).

[A3]

Norm and integral bounds. For $f \in L^{2} [0, 1]$ : Cauchy–Schwarz gives $∣ \int_{0}^{x} g ∣ \leq \sqrt{x} ∥ g ∥_{2}$ and $(\int_{0}^{1} ∣ \int_{0}^{x} g ∣^{2} d x)^{1 / 2} \leq ∥ g ∥_{2}$ ; the operator norm is the unit-ball supremum; and $\int_{0}^{1} x^{m} d x = 1 / (m + 1)$ for integers $m \geq 0$ , computed by the Newton–Leibniz formula applied to the primitive $x^{m + 1} / (m + 1)$ of $x^{m}$ , whose derivative is given by the derivative-of-a-power lemma, with the Riemann integral agreeing with the Lebesgue integral (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Newton–Leibniz remains valid across finitely many exceptional interior points when the primitive is continuous, For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[A4]

Spectrum of a compact operator. Under AC, a nonzero spectral value of a compact operator on a complex Banach space is an eigenvalue of finite algebraic multiplicity, and if the space is infinite dimensional then $0$ belongs to the spectrum (Riesz schauder spectrum of a compact operator, Spectrum and resolvent of a bounded operator, Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism, Banach space).

[A5]

Small reciprocal bounds. For every positive real $ε$ , some natural $N \geq 1$ satisfies $1 / N < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ). In particular $1 / N \to 0$ , and $2^{j} \geq j + 1$ by induction, so a tail bounded by $C 2^{- j}$ tends to zero. The latter implication follows from $C 2^{- j} \leq C / (j + 1)$ and the reciprocal bound.

[A6]

Self-adjointness test. The adjoint is characterized by $⟨ T f, g ⟩ = ⟨ f, T^{*} g ⟩$ , and $T$ is self-adjoint when $T^{*} = T$ (The Hilbert-space adjoint of a bounded operator, Self-adjoint, positive, unitary and normal operators).

[A7]

Complex Fubini. A complex product-measurable function with integrable absolute value on a sigma-finite product has equal double and iterated integrals (Fubini's theorem for L^1 functions on a sigma-finite product). Here both factors are finite Lebesgue measure on $[0, 1]$ .

Verification

technique · direct

Given: AC, the space $L^{2} ([0, 1], C)$ , the Volterra kernel and operator, and the bounds above.

1.1

$V$ is Hilbert–Schmidt with norm $1 / \sqrt{2}$ . By [A1] the class of $k$ has square norm $\frac{1}{2}$ in the completed product measure; the kernel operator of [A2] is $T_{k} f (x) = \int_{0}^{1} k (x, t) f (t) d t = \int_{0}^{x} f (t) d t = V f (x)$ for $f \in L^{2}$ and almost every $x$ , by the definition of $k$ ; Under AC choose a Hilbert basis as supplied by the kernel theorem; hence $V = T_{k}$ is Hilbert–Schmidt with $∥ V ∥_{H S} = ∥ k ∥_{2} = 1 / \sqrt{2}$ and is compact.

A1A2

1.2

Iterated integration and norm decay. By induction on $n$ : for $n = 1$ the formula is the definition of $V$ ; assuming it for $n$ , $V^{n + 1} f (x) = \int_{0}^{x} \frac{1}{(n - 1)!} \int_{0}^{t} (t - s)^{n - 1} f (s) d s d t = \frac{1}{n!} \int_{0}^{x} (x - s)^{n} f (s) d s$ as follows. Cauchy–Schwarz applied to $∣ f ∣$ and $1$ gives $\int_{0}^{1} ∣ f ∣ \leq ∥ f ∥_{2}$ . For each fixed $x$ , the integrand $1_{0 \leq s \leq t \leq x} (t - s)^{n - 1} f (s)$ is product-measurable (put its value zero outside the triangle) and its absolute value is bounded by $∣ f (s) ∣$ . Tonelli thus bounds its double absolute integral by $∥ f ∥_{2} < \infty$ . Complex Fubini [A7] therefore permits reversing the integrals, and the inner integration $\int_{s}^{x} (t - s)^{n - 1} d t = (x - s)^{n} / n$ [A1, A3] gives the displayed identity. Cauchy--Schwarz and Tonelli give $∥ V^{n + 1} f ∥_{2}^{2} \leq \frac{∥ f ∥_{2}^{2}}{(n!)^{2}} \int_{0}^{1} \int_{0}^{x} (x - s)^{2 n} d s d x = \frac{∥ f ∥_{2}^{2}}{(n!)^{2} (2 n + 1) (2 n + 2)},$ so $∥ V^{n + 1} ∥ \leq 1 / (n! \sqrt{(2 n + 1) (2 n + 2)})$ , and the right side tends to $0$ because it is at most $1 / \sqrt{2 n + 2}$ , which tends to zero by [A5]. Changes to $f$ on a null set do not change any integral, so this also identifies the operator classes.

A1A3A5A7algebra

1.3

$V$ is not self-adjoint. Let $f (x) = 1$ and $g (x) = x$ . Then $V f (x) = x$ and $V g (x) = x^{2} / 2$ , so [A3] gives $⟨ V f, g ⟩ = \int_{0}^{1} x^{2} d x = 1 / 3$ but $⟨ f, V g ⟩ = \int_{0}^{1} x^{2} / 2 d x = 1 / 6$ . These values are unequal, whereas [A6] would make them equal if $V = V^{*}$ .

A3A6

2.1

There is no nonzero eigenvalue. Let $V f = λ f$ with $λ \neq 0$ . Iterating, $V^{n} f = λ^{n} f$ for every $n \geq 1$ , so if $f \neq 0$ then step 1.2 gives $1 \leq \frac{∥ V^{n} ∥}{∣ λ ∣^{n}} \leq b_{n} : = \frac{1}{∣ λ ∣^{n} (n - 1)! \sqrt{(2 n - 1) (2 n)}} .$ But $\frac{b_{n + 1}}{b_{n}} = \frac{1}{∣ λ ∣ n} \sqrt{\frac{(2 n - 1) (2 n)}{(2 n + 1) (2 n + 2)}} ⟶ 0,$ Choose $N \geq 1$ with $1 / (∣ λ ∣ N) \leq 1 / 2$ by [A5]. For $n \geq N$ the displayed ratio is at most $1 / 2$ , hence induction gives $b_{N + j} \leq b_{N} 2^{- j} \to 0$ by [A5], contradicting $1 \leq b_{n}$ for every $n$ . Hence $f = 0$ : the kernel of $V - λ I$ is trivial for every $λ \neq 0$ .

step 1.2A5algebra

3.1

The spectrum is ${0}$ . By step 1.1, $V$ is compact on the complex Hilbert, hence Banach, space $L^{2} [0, 1]$ . Thus every nonzero spectral value would be an eigenvalue by [A4], ruled out by step 2.1. To prove $0 \in σ (V)$ directly, for $0 < δ \leq 1$ put $f_{δ} = 1_{[0, δ]} / \sqrt{δ}$ . Then $∥ f_{δ} ∥_{2} = 1$ and $∣ V f_{δ} (x) ∣ = \min (x, δ) / \sqrt{δ} \leq \sqrt{δ}$ , so $∥ V f_{δ} ∥_{2} \leq \sqrt{δ}$ . A bounded inverse with norm $C$ would give $1 \leq C \sqrt{δ}$ for all such $δ$ ; taking $δ = 1 / (j + 1)^{2}$ and using [A5] contradicts this. By the resolvent definition in [A4], $0$ lies in the spectrum. Therefore $σ (V) = {0}$ .

step 1.1step 2.1A1A3A4A5

4.1

Conclusion. Claims 1–4 are [step 1.1], [step 1.2], [step 2.1] and [step 3.1], and [step 1.3] proves the final non-self-adjointness assertion directly.

step 1.1step 1.2step 2.1step 3.1step 1.3∎

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Sources

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.1 and §3.6, the Volterra operator (standard reference, not scraped)