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A convergent sequence in $\mathbb{R}^{3}$ and the integral $\int_0^1 (1, t, t^{2})$ , computed componentwise

Example

Two parts of the vector-valued toolkit are exercised here, in $\mathbb{R}^{3}$ : componentwise sequence convergence and componentwise integration.

A convergent sequence. For $k \in \mathbb{N}$ put

$x^{(k)} \;:=\; \Bigl(\tfrac{1}{\iota(k+1)},\ 1 - \tfrac{1}{\iota(k+1)},\ 2\Bigr) \;\in\; \mathbb{R}^{3},$

with $\iota$ the canonical natural (The canonical natural $\iota(n) = n \cdot 1_F$ of a field); the shift by one is there because $\mathbb{N}$ contains $0$ and $\iota(0) = 0$ . Then $x^{(k)} \to (0,1,2)$ in $(\mathbb{R}^{3}, d_2)$ .

An integral. Let $f : [0,1] \to \mathbb{R}^{3}$ be $f(t) := (1,\ t,\ t^{2})$ . Then $f$ is integrable and

$\int_0^1 f \;=\; \Bigl(1,\ \tfrac{1}{\iota(2)},\ \tfrac{1}{\iota(3)}\Bigr).$

The norm inequality is strict here. $\bigl\lVert\int_0^1 f\bigr\rVert_2 = \iota(7)/\iota(6)$ , while $\int_0^1\lVert f\rVert_2 \ge \iota(53)/\iota(45) > \iota(7)/\iota(6)$ . The exact value of $\int_0^1\sqrt{1+t^{2}+t^{4}}\,dt$ is not computed: it needs machinery this page does not have, and a crude lower bound is enough to separate the two sides of For $a \le b$ and $f : [a,b] \to \mathbb{R}^m$ integrable when $a<b$ , $\bigl\lVert\int_a^b f\bigr\rVert_2 \le \int_a^b \lVert f\rVert_2$ ; for $a<b$ , $\lVert f\rVert_2$ is integrable.

Facts & Assumptions

Given: The sequence $\bigl(x^{(k)}\bigr)$ and the function $f$ above; the abbreviation $g(t) := \lVert f(t)\rVert_2 = \sqrt{1+t^{2}+t^{4}}$ (The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ , The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ ).

[L1]

Convergence in $\mathbb{R}^{n}$ for $n \ge 1$ is componentwise (For $n \ge 1$ a sequence in $\mathbb{R}^n$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $\mathbb{R}^n$ is complete in every norm clause 1, Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences, The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ ).

[L2]

For every real $\varepsilon>0$ there is $K$ with $1/\iota(K+1) < \varepsilon$ ; $\iota(k+1)>0$ ; and $0<u\le v$ gives $0 < 1/v \le 1/u$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L3]

The vector-valued integral is componentwise, and $f$ is integrable exactly when each $f_i$ is (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ , Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

A continuous function on $[a,b]$ is integrable, and $\int_a^b h = G(b)-G(a)$ for any primitive $G$ of a continuous $h$ (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_a^b f = G(b)-G(a)$ for any primitive $G$ ); polynomial functions are 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) and $x \mapsto x^{p}$ has derivative $\iota(p)x^{\,p-1}$ for $p \ge 1$ (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Integer powers $a^m$ ).

[L5]

Monotonicity and linearity of the integral (If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ , Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$ , Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L6]

$\lVert f(t)\rVert_2$ is integrable on $[0,1]$ and $\bigl\lVert\int_0^1 f\bigr\rVert_2 \le \int_0^1\lVert f\rVert_2$ (For $a \le b$ and $f : [a,b] \to \mathbb{R}^m$ integrable when $a<b$ , $\bigl\lVert\int_a^b f\bigr\rVert_2 \le \int_a^b \lVert f\rVert_2$ ; for $a<b$ , $\lVert f\rVert_2$ is integrable, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L7]

Square roots and squaring: $\sqrt{c}$ is the unique nonnegative $s$ with $s^{2} = c$ , and for $a,b \ge 0$ , $a \le b$ exactly when $a^{2} \le b^{2}$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ , Squaring is monotone on the nonnegatives).

[L8]

Canonical naturals carry sums to sums and products to products and are strictly increasing and positive (Canonical naturals are positive and strictly increasing, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L9]

Continuity of a vector-valued function is componentwise (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions clause 1, Vector-valued functions $f : A \to \mathbb{R}^m$ , their limits and continuity, with the dictionary to the metric notions).

Verification

technique · direct

1.1

The three coordinate sequences of $\bigl(x^{(k)}\bigr)$ are $k \mapsto 1/\iota(k+1)$ , $k \mapsto 1 - 1/\iota(k+1)$ and the constant $2$ . Given a real $\varepsilon>0$ , take $K$ with $1/\iota(K+1)<\varepsilon$ ; then for $k \ge K$ one has $0 < 1/\iota(k+1) \le 1/\iota(K+1) < \varepsilon$ , so the first converges to $0$ , the second to $1$ and the third to $2$ .

1.2

Each component of $f$ is a polynomial function, hence continuous on $[0,1]$ and integrable there; so $f$ is integrable, and $f$ is continuous.

L3L4L9

1.3

The function $t \mapsto t^{\,p+1}/\iota(p+1)$ is a primitive of $t \mapsto t^{p}$ for $p \in \mathbb{N}$ , so $\int_0^1 t^{p}\,dt = 1/\iota(p+1)$ ; at $p = 0,1,2$ this gives $1$ , $1/\iota(2)$ and $1/\iota(3)$ .

L4L8

1.4

For $t \in [0,1]$ put $u := t^{2}+t^{4}$ , so $0 \le u \le \iota(2) \le \iota(3)$ and $g(t) = \sqrt{1+u}$ . Since $(1+u/\iota(3))^{2} = 1 + \iota(2)u/\iota(3) + u^{2}/\iota(9) \le 1+u$ exactly when $u^{2}/\iota(9) \le u/\iota(3)$ , that is exactly when $u \le \iota(3)$ , and both $1+u/\iota(3)$ and $\sqrt{1+u}$ are nonnegative, monotonicity of squaring gives $g(t) \ge 1 + \bigl(t^{2}+t^{4}\bigr)/\iota(3)$ .

L7L8

1.5

Finally $\iota(53)/\iota(45) > \iota(7)/\iota(6)$ , since cross-multiplying by the positive $\iota(45)\iota(6)$ turns the claim into $\iota(318) > \iota(315)$ , which holds because $\iota$ is strictly increasing.

2.1

Hence $x^{(k)} \to (0,1,2)$ in $(\mathbb{R}^{3},d_2)$ , the $\varepsilon$ for the vector being obtained from the three coordinate tolerances exactly as in the proof of [L1].

step 1.1L1

2.2

Therefore $\int_0^1 f = \bigl(1,\ 1/\iota(2),\ 1/\iota(3)\bigr)$ , the coordinates of the vector integral being the integrals of the coordinates.

step 1.2step 1.3L3

2.3

The right-hand side of step 1.4 is continuous, hence integrable, and by step 1.3 and linearity $\int_0^1\bigl(1+(t^{2}+t^{4})/\iota(3)\bigr)dt = 1 + \bigl(1/\iota(3)+1/\iota(5)\bigr)/\iota(3) = 1 + \iota(8)/\iota(45) = \iota(53)/\iota(45)$ .

step 1.3L4L5L8

3.1

Its Euclidean norm satisfies $\bigl\lVert\int_0^1 f\bigr\rVert_2^{2} = 1 + 1/\iota(4) + 1/\iota(9) = \iota(49)/\iota(36)$ , so $\bigl\lVert\int_0^1 f\bigr\rVert_2 = \iota(7)/\iota(6)$ , since $\bigl(\iota(7)/\iota(6)\bigr)^{2} = \iota(49)/\iota(36)$ and both numbers are nonnegative.

step 2.2L7L8

3.2

By monotonicity of the integral, using that $g$ is integrable, $\int_0^1 g \ge \iota(53)/\iota(45)$ .

step 1.4step 2.3L5L6

4.1

So $\bigl\lVert\int_0^1 f\bigr\rVert_2 = \iota(7)/\iota(6) < \iota(53)/\iota(45) \le \int_0^1\lVert f\rVert_2$ : the inequality of [L6] holds on this example and is strict.

step 3.1step 3.2step 1.5L6∎

Remarks

Why the inequality is strict here. Equality in For $a \le b$ and $f : [a,b] \to \mathbb{R}^m$ integrable when $a<b$ , $\bigl\lVert\int_a^b f\bigr\rVert_2 \le \int_a^b \lVert f\rVert_2$ ; for $a<b$ , $\lVert f\rVert_2$ is integrable would require the integrand to point in a fixed direction, and $f(t) = (1,t,t^{2})$ changes direction as $t$ runs over $[0,1]$ . That heuristic is not what is proved above: the proof separates the two sides by an explicit numerical bound, which is the only argument available at this point in the reading order.
The lower bound is deliberately crude. The inequality $\sqrt{1+u} \ge 1 + u/\iota(3)$ used in step 1.4 holds for $0 \le u \le \iota(3)$ and is far from sharp; it is chosen because it is polynomial, so that step 2.3 is an application of Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_a^b f = G(b)-G(a)$ for any primitive $G$ and nothing more. The exact value of $\int_0^1\sqrt{1+t^{2}+t^{4}}\,dt$ is not a value this page can name.
The sequence and the integral are independent computations, put in one item because they exercise componentwise arguments on the same space. Neither uses the other.

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Sources

Vector-valued function (Wikipedia) (standard reference, not scraped)
Riemann integral (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, Section 7.3 (standard reference, not scraped)
APEX Calculus, Section 12.2 (standard reference, not scraped)

A convergent sequence in R3\mathbb{R}^{3}R3 and the integral ∫01(1,t,t2)\int_0^1 (1, t, t^{2})∫01​(1,t,t2), computed componentwise