TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable

Statement

Let $m \in N$ with $m \geq 1$ , let $a, b \in R$ with $a \leq b$ and let $f : [a, b] \to R^{m}$ . If $a < b$ , assume that $f$ is integrable (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). Then:

if $a < b$ , the real-valued function $t \mapsto ∥ f (t) ∥_{2}$ is integrable on $[a, b]$ (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ );
$∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2} .$

The hypothesis $a \leq b$ is not decoration. With the orientation convention of The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ and The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, interchanging the limits changes the sign of the right-hand side but not of the left, so for $b < a$ the correct statement is $∥ \int_{a}^{b} f ∥_{2} \leq ∣ \int_{a}^{b} ∥ f ∥_{2} ∣$ ; the displayed inequality as written is false in that case. This is the same trap the scalar inequality of If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ carries.

Clause 1 is a genuine obligation and is discharged before the estimate. That each $f_{i}$ is integrable does not by itself say that $\sqrt{\sum_{i < m} f_{i}^{2}}$ is; the square root has to be brought in through If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable.

Facts & Assumptions

Given: A natural $m \geq 1$ , reals $a \leq b$ , a function $f : [a, b] \to R^{m}$ that is integrable when $a < b$ , with components $f_{0}, \dots, f_{m - 1}$ , and the vector $v : = \int_{a}^{b} f \in R^{m}$ ; write $g (t) : = \sum_{i < m} f_{i} (t)^{2}$ , so that $∥ f (t) ∥_{2} = \sqrt{g (t)}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ).

[L1]

The vector-valued integral is componentwise: when $a < b$ , $f$ is integrable exactly when every $f_{i}$ is bounded and Darboux integrable, and then $(\int_{a}^{b} f)_{i} = \int_{a}^{b} f_{i}$ ; $\int_{a}^{a} f = 0$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , Lower bound, bounded below, bounded set).

[L2]

Linearity of the integral: integrable functions on $[a, b]$ are closed under sums and scalar multiples, and $\int_{a}^{b} (λ u + μ w) = λ \int_{a}^{b} u + μ \int_{a}^{b} w$ (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[L3]

Monotonicity of the integral: for $a < b$ and integrable $u \leq w$ on $[a, b]$ , $\int_{a}^{b} u \leq \int_{a}^{b} w$ ; and an integrable $u \geq 0$ has $\int_{a}^{b} u \geq 0$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L4]

Products and squares: if $u, w$ are integrable on $[a, b]$ then so are $∣ u ∣$ , $u^{2}$ and $u w$ (If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ ).

[L5]

Composition: if $u$ is integrable on $[a, b]$ with values in $[α, β]$ and $φ$ is continuous on $[α, β]$ , then $φ \circ u$ is integrable on $[a, b]$ (If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L6]

Square roots (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives): every $c \geq 0$ has a unique $\sqrt{c} \geq 0$ with $(\sqrt{c})^{2} = c$ , and $s \mapsto s^{2}$ is strictly increasing on the nonnegatives, hence injective there.

[L7]

Continuous inverse theorem: a continuous injective function on an order-convex subset of $R$ is a bijection onto its order-convex image, whose inverse is continuous (Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length); and $s \mapsto s^{2}$ is continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[L8]

Cauchy-Schwarz and the inner product: $⟨ u, w ⟩ = \sum_{i < m} u_{i} w_{i}$ is bilinear and symmetric, $∥ u ∥_{2} = \sqrt{⟨ u, u ⟩}$ , $∥ u ∥_{2} \geq 0$ , and $∣ ⟨ u, w ⟩ ∣ \leq ∥ u ∥_{2} ∥ w ∥_{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L9]

Laws of finite sums and induction (Laws of finite sums and finite products, Finite sums and finite products, by recursion, The principle of mathematical induction).

[L10]

Order arithmetic: $u > 0$ gives $u^{- 1} > 0$ , a product of nonnegatives is nonnegative, and $t \leq ∣ t ∣$ (Inverses of positives are positive, and reciprocation reverses order, Basic properties of the absolute value).

Proof

technique · direct

1.1

If $a = b$ then $\int_{a}^{b} f = 0$ and $\int_{a}^{b} ∥ f ∥_{2} = 0$ by the oriented convention, so clause 2 reads $0 \leq 0$ and holds, while clause 1 says nothing in that case; assume $a < b$ from here on.

1.2

Each component $f_{i}$ is bounded and integrable on $[a, b]$ , so each $f_{i}^{2}$ is integrable.

L1L4

1.3

Pointwise, $⟨ v, f (t) ⟩ \leq ∣ ⟨ v, f (t) ⟩ ∣ \leq ∥ v ∥_{2} ∥ f (t) ∥_{2}$ by Cauchy-Schwarz.

L8L10

2.1

By induction on $p \leq m$ , every finite sum $\sum_{i < p} f_{i}^{2}$ is integrable, the empty sum being the constant $0$ and each successor step adding one integrable function. Hence $g = \sum_{i < m} f_{i}^{2}$ is integrable.

step 1.2L2L9

2.2

The real-valued function $t \mapsto ⟨ v, f (t) ⟩ = \sum_{i < m} v_{i} f_{i} (t)$ is integrable, being a finite sum of scalar multiples of the integrable $f_{i}$ , and by linearity applied $m$ times $\int_{a}^{b} ⟨ v, f ⟩ = \sum_{i < m} v_{i} \int_{a}^{b} f_{i} = \sum_{i < m} v_{i} v_{i} = ⟨ v, v ⟩ = ∥ v ∥_{2}^{2}$ .

step 1.2L1L2L8L9

3.1

$g (t) \geq 0$ for every $t$ , being a finite sum of squares, and $g$ is bounded above: each $∣ f_{i} ∣$ is bounded by some $B_{i}$ , so $g (t) \leq \sum_{i < m} B_{i}^{2} = : K$ . Thus $g$ takes its values in $[0, K]$ .

step 2.1L1L9L10

4.1

The map $s \mapsto s^{2}$ is continuous and injective on the order-convex set $[0, \sqrt{K}]$ , with image $[0, K]$ ; by the continuous inverse theorem its inverse $φ : [0, K] \to [0, \sqrt{K}]$ , $φ (u) = \sqrt{u}$ , is continuous on $[0, K]$ .

step 3.1L6L7

5.1

$∥ f (t) ∥_{2} = \sqrt{g (t)} = φ (g (t))$ for every $t \in [a, b]$ , so $∥ f ∥_{2} = φ \circ g$ is integrable on $[a, b]$ ; this is clause 1.

step 2.1step 3.1step 4.1L5L8

6.1

Both sides of step 1.3 are integrable on $[a, b]$ , so monotonicity and linearity give $∥ v ∥_{2}^{2} = \int_{a}^{b} ⟨ v, f ⟩ \leq \int_{a}^{b} ∥ v ∥_{2} ∥ f ∥_{2} = ∥ v ∥_{2} \int_{a}^{b} ∥ f ∥_{2}$ .

step 5.1step 2.2step 1.3L2L3

6.2

If $v = 0$ then $∥ v ∥_{2} = 0$ , while $\int_{a}^{b} ∥ f ∥_{2} \geq 0$ because $∥ f ∥_{2} \geq 0$ pointwise and $a < b$ ; so clause 2 holds in this case.

step 5.1L3L8

7.1

If $v \neq 0$ then $∥ v ∥_{2} > 0$ , so multiplying the inequality of step 6.1 by the positive $1 / ∥ v ∥_{2}$ gives $∥ v ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ , which is clause 2 in this case.

step 6.1L8L10

8.1

The two cases of steps 6.2 and 7.1 exhaust the possibilities for $v$ , so clause 2 holds; with step 5.1 both clauses are proved.

step 5.1step 6.2step 7.1∎

Remarks

The case split at $v = 0$ is mandatory. Step 6.1 delivers only $∥ v ∥_{2}^{2} \leq ∥ v ∥_{2} \int_{a}^{b} ∥ f ∥_{2}$ , and dividing by $∥ v ∥_{2}$ is illegitimate when that number is $0$ . Many textbook presentations divide without comment; the missing case is genuinely separate, and it is the one where the right-hand side has to be shown nonnegative on its own.
Why the inner-product route rather than a componentwise estimate. Bounding each coordinate of $\int_{a}^{b} f$ separately and reassembling gives a constant depending on $m$ ; the argument above gives the sharp inequality with no constant, and it uses only bilinearity, Cauchy-Schwarz and monotonicity of the integral. The companion page checks the inequality numerically on an explicit curve and shows it is strict there.
Clause 1 is where the hypotheses of If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable are checked, one by one: $g$ is integrable, its values lie in a closed bounded interval, and the outer function is continuous on that interval. The order of that theorem's hypotheses matters — continuous after integrable — and it is respected here.

Depends on

Used by

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Sources

Riemann integral (Wikipedia) (standard reference, not scraped)
Cauchy-Schwarz inequality (Wikipedia) (standard reference, not scraped)
Stephen Semmes, Some Basic Topics in Analysis, Sections 8.1.2–8.1.3 (standard reference, not scraped)
Robert Gressman, Advanced Analysis, Integrating Vector-Valued Functions; Jensen's Inequality (standard reference, not scraped)

For a≤ba≤b and f:[a,b]→Rmf:[a,b]→Rm integrable when a<ba<b, ∥∫abf∥2≤∫ab∥f∥2​∫ab​f​2​≤∫ab​∥f∥2​; for a<ba<b, ∥f∥2∥f∥2​ is integrable