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Semigroup law for fractional integrals

Statement

Let $T > 0$ , $a, b > 0$ , and $g \in L^{1} (0, T)$ . Then, almost everywhere on $(0, T)$ ,

$I^{a} (I^{b} g) = I^{a + b} g .$

Moreover,

$∥ I^{a} g ∥_{L^{1} (0, T)} \leq \frac{T^{a}}{Γ (a + 1)} ∥ g ∥_{L^{1} (0, T)} .$

If $g \in C ([0, T])$ , the semigroup identity holds at every $t \in [0, T]$ . For every $c > 0$ and $t \in [0, T]$ , $I^{c} 1 (t) = \frac{t^{c}}{Γ (c + 1)},$ where $1$ is the constant function with value $1$ .

Facts & Assumptions

Given: $T > 0$ , positive orders $a, b$ , and $g \in L^{1} (0, T)$ .

[F1]

Fractional integral and Caputo derivative: $I^{c} g (t) = Γ (c)^{- 1} \int_{0}^{t} (t - s)^{c - 1} g (s) d s$ for every $c > 0$ at almost every $t$ where the integral is finite.

[F2]

Euler's real Beta integral: $B (p, q) = \int_{0}^{1} y^{p - 1} (1 - y)^{q - 1} d y$ for $p, q > 0$ .

[F3]

The real Beta--Gamma identity: for $p, q > 0$ , $B (p, q) = Γ (p) Γ (q) / Γ (p + q)$ .

[F4]

The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ : for $x > 0$ , $Γ (x + 1) = x Γ (x)$ .

[F5]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: a nonnegative product-measurable function on a product of sigma-finite measure spaces has equal product and iterated integrals, with either iterated order permitted.

[F6]

Integral of the fractional power kernel: $\int_{0}^{t} x^{c - 1} d x = t^{c} / c$ for $c, t > 0$ .

[F7]

The real Gamma function by Euler's integral: $Γ (c) > 0$ for every $c > 0$ .

[F8]

Change of variable in an improper integral: a monotone differentiable change of variable preserves an improper integral when the transformed integral converges.

Proof

1.1givenF1F4F5F6F7algebra

The integrands are measurable. Applying Tonelli [F5] to their absolute values on $0 \leq r \leq s \leq t \leq T$ , for each fixed $r$ set $x = t - s$ and $y = s - r$ ; the region lies in $[0, T]^{2}$ , so the kernel integral is at most $\int_{0}^{T} x^{a - 1} d x \int_{0}^{T} y^{b - 1} d y = T^{a + b} / (a b)$ by [F6]. Hence $\int_{0}^{T} \int_{0}^{t} \int_{0}^{s} (t - s)^{a - 1} (s - r)^{b - 1} ∣ g (r) ∣ d r d s d t \leq T^{a + b} ∥ g ∥_{1} / (a b) < \infty$ , so the double integral is absolutely integrable for almost every fixed $t$ and Tonelli permits swapping $r, s$ . The same theorem gives $∥ I^{a} g ∥_{1} \leq Γ (a)^{- 1} \int_{0}^{T} ∣ g (r) ∣ \int_{r}^{T} (t - r)^{a - 1} d t d r \leq T^{a} ∥ g ∥_{1} / Γ (a + 1)$ using [F4]–[F6] and positivity [F7].

2.1step 1.1F2F5F6F8algebra

After swapping the order, the inner kernel integral is $\int_{r}^{t} (t - s)^{a - 1} (s - r)^{b - 1} d s = (t - r)^{a + b - 1} \int_{0}^{1} (1 - y)^{a - 1} y^{b - 1} d y = (t - r)^{a + b - 1} B (b, a)$ by the substitution $y = (s - r) / (t - r)$ using [F8] and the definition [F2]. This applies for almost every $t$ by step 1.1; if $g$ is continuous, Tonelli and this formula bound the nonnegative double integral for each fixed $t$ by $∥ g ∥_{\infty} B (b, a) \int_{0}^{t} (t - r)^{a + b - 1} d r < \infty$ using [F5]–[F6], so Fubini applies for every $t$ .

3.1step 1.1step 2.1F1F3algebra

Substituting [F3] into step 2.1 cancels $Γ (a) Γ (b)$ in [F1], leaving $I^{a} (I^{b} g) (t) = Γ (a + b)^{- 1} \int_{0}^{t} (t - r)^{a + b - 1} g (r) d r = I^{a + b} g (t)$ almost everywhere, and everywhere for continuous $g$ . This proves the semigroup identity and the $L^{1}$ bound in step 1.1.

4.1givenF1F4F6F7algebra∎

For the constant function, [F1] and [F6] give $I^{c} 1 (t) = t^{c} / (c Γ (c))$ for $t > 0$ . Positivity from [F7] makes the quotient well-defined, and [F4] rewrites it as $t^{c} / Γ (c + 1)$ ; both sides vanish at $t = 0$ .

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Sources

Jeffrey R. L. Webb, Initial Value Problems for Caputo Fractional Equations with Singular Nonlinearities, Electronic Journal of Differential Equations 2019(117), §3, Lemma 3.4 (standard reference, not scraped)