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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

Statement

Let $m \in \mathbb{N}$ with $m \ge 1$ , with vector-valued functions, their components $f_i = \pi_i \circ f$ , their limits and their continuity as in Vector-valued functions $f : A \to \mathbb{R}^m$ , their limits and continuity, with the dictionary to the metric notions.

Continuity is componentwise. Let $(X,d_X)$ be a metric space, $A \subseteq X$ , $f : A \to \mathbb{R}^{m}$ and $a \in A$ . Then $f$ is continuous at $a$ if and only if every component $f_i : A \to \mathbb{R}$ $(i<m)$ is continuous at $a$ .
Limits are componentwise. Let $A \subseteq \mathbb{R}$ , let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ), let $f : A \to \mathbb{R}^{m}$ and let $L \in \mathbb{R}^{m}$ . Then $\lim_{x\to c} f(x) = L$ if and only if $\lim_{x\to c} f_i(x) = L_i$ for every $i<m$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ).
Algebra. Let $(X,d_X)$ , $A$ , $a$ be as in clause 1, let $f, g : A \to \mathbb{R}^{m}$ be continuous at $a$ and let $\lambda \in \mathbb{R}$ . Then $f + g$ and $\lambda f$ (defined pointwise) are continuous at $a$ ; the real-valued function $x \mapsto \langle f(x), g(x)\rangle$ is continuous at $a$ (The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ ); and for every norm $N$ on $\mathbb{R}^{m}$ the real-valued function $x \mapsto N(f(x))$ is continuous at $a$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Where $m \ge 1$ is spent. The "if" direction of clauses 1 and 2 divides $\varepsilon$ by $\iota(m)$ , which requires $\iota(m) \ne 0$ ; and clause 3's last part quotes a bound available only for $m \ge 1$ . The "only if" directions hold for every $m$ but say nothing at $m = 0$ , there being no index $i < 0$ .

Facts & Assumptions

Given: A natural $m \ge 1$ ; a metric space $(X,d_X)$ , a subset $A \subseteq X$ , a point $a \in A$ and functions $f, g : A \to \mathbb{R}^{m}$ ; a real $\lambda$ ; and a real $\varepsilon > 0$ .

[L1]

Continuity and limits of vector-valued functions in the $\varepsilon$ - $\delta$ form, the coordinate projections $\pi_i$ , and $|y_i| \le \lVert y\rVert_2$ for $i<m$ (Vector-valued functions $f : A \to \mathbb{R}^m$ , their limits and continuity, with the dictionary to the metric notions, Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded).

[L2]

The comparison $\lVert y\rVert_2 \le \lVert y\rVert_1 = \sum_{i<m}|y_i|$ , and $N(y) \le C\lVert y\rVert_1 \le C\sqrt{\iota(m)}\lVert y\rVert_2$ with $C := \max\{N(e_i) : i<m\} \ge 0$ , together with $|N(y)-N(z)| \le N(y-z)$ , all for $m \ge 1$ (The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$ clauses 1, 2, 3, The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ , 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 $0$ ).

[L3]

The norm axioms for $\lVert\cdot\rVert_2$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Each $\lVert\cdot\rVert_p$ is a norm on $\mathbb{R}^n$ , and the induced metrics are exactly $d_1$ , $d_2$ and $d_\infty$ of the published metric-spaces page), and $\lVert u-v\rVert_2 = d_2(u,v)$ .

[L4]

Cauchy-Schwarz: $|\langle u,v\rangle| \le \lVert u\rVert_2\lVert v\rVert_2$ , together with bilinearity and symmetry of the inner product (Cauchy-Schwarz $\lvert\langle x,y\rangle\rvert \le \lVert x\rVert_2\lVert y\rVert_2$ with its equality case, the triangle inequality for $\lVert\cdot\rVert_2$ , the parallelogram law and polarisation, The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ ).

[L5]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity, $\sum_{i<m}\mu = \iota(m)\mu$ , a sum of nonnegative terms is nonnegative, and each single term is at most such a sum.

[L6]

$\iota(m) > 0$ for $m \ge 1$ , and $u>0$ gives $u^{-1}>0$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L7]

A nonempty finite set of reals has a minimum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set); and a family of nonempty sets indexed by a natural number has a choice function, this being a theorem of ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values), which is what licenses picking one $\delta_i$ for each $i<m$ .

[L8]

Absolute value (Basic properties of the absolute value): $|t| \ge 0$ , $|st| = |s||t|$ , and $|s+t| \le |s|+|t|$ .

Proof

technique · direct

1.1

For every $y \in \mathbb{R}^{m}$ : $|y_i| \le \lVert y\rVert_2$ for each $i<m$ , and $\lVert y\rVert_2 \le \sum_{i<m}|y_i|$ .

L1L2

1.2

If $a_i < b_i$ for every $i<m$ and $m \ge 1$ , then $\sum_{i<m}a_i < \sum_{i<m}b_i$ : the list $i \mapsto b_i - a_i$ has positive terms, so its sum is at least its term at index $0$ , hence positive, and additivity gives the strict inequality.

L5L6

1.3

For each $i<m$ the set of positive reals $\delta$ witnessing continuity of $f_i$ at $a$ for a given tolerance is nonempty whenever $f_i$ is continuous at $a$ , so a choice function on the family indexed by $m$ produces $\delta_0,\dots,\delta_{m-1}$ simultaneously, with no choice principle used.

1.4

For $f+g$ : given $\varepsilon > 0$ , pick $\delta_1, \delta_2 > 0$ for the tolerance $\varepsilon/\iota(2)$ at $f$ and at $g$ and put $\delta := \min\{\delta_1,\delta_2\}$ ; then for $d_X(x,a) < \delta$ , $\lVert (f+g)(x)-(f+g)(a)\rVert_2 = \lVert (f(x)-f(a)) + (g(x)-g(a))\rVert_2 \le \lVert f(x)-f(a)\rVert_2 + \lVert g(x)-g(a)\rVert_2 < \varepsilon$ .

L1L3L6L7

1.5

L1L3L6L8

1.6

For $N \circ f$ : by [L2], $|N(f(x)) - N(f(a))| \le N\bigl(f(x)-f(a)\bigr) \le C\sqrt{\iota(m)}\,\lVert f(x)-f(a)\rVert_2$ ; so a $\delta$ for the tolerance $\varepsilon/(C\sqrt{\iota(m)}+1)$ at $f$ serves for $N\circ f$ .

L1L2L6

1.7

For $\langle f,g\rangle$ : first take $\delta_0 > 0$ with $\lVert g(x)-g(a)\rVert_2 < 1$ for $d_X(x,a) < \delta_0$ , so that $\lVert g(x)\rVert_2 \le \lVert g(x)-g(a)\rVert_2 + \lVert g(a)\rVert_2 < B := \lVert g(a)\rVert_2 + 1$ there.

L1L3

2.1

Suppose $f$ is continuous at $a$ and fix $i<m$ . Given $\varepsilon > 0$ , take $\delta$ from the definition; for $x \in A$ with $d_X(x,a) < \delta$ , step 1.1 gives $|f_i(x)-f_i(a)| \le \lVert f(x)-f(a)\rVert_2 < \varepsilon$ . So $f_i$ is continuous at $a$ .

step 1.1L1

2.2

Conversely suppose every $f_i$ is continuous at $a$ . Given $\varepsilon > 0$ , the real $\varepsilon/\iota(m)$ is positive; by step 1.3 choose $\delta_i > 0$ for each $i<m$ with $|f_i(x)-f_i(a)| < \varepsilon/\iota(m)$ whenever $x \in A$ and $d_X(x,a) < \delta_i$ , and put $\delta := \min\{\delta_0,\dots,\delta_{m-1}\} > 0$ .

step 1.3L1L6L7

2.3

By bilinearity, $\langle f(x),g(x)\rangle - \langle f(a),g(a)\rangle = \langle f(x)-f(a),\, g(x)\rangle + \langle f(a),\, g(x)-g(a)\rangle$ , so Cauchy-Schwarz and step 1.7 give $\bigl|\langle f(x),g(x)\rangle - \langle f(a),g(a)\rangle\bigr| \le B\,\lVert f(x)-f(a)\rVert_2 + \lVert f(a)\rVert_2\,\lVert g(x)-g(a)\rVert_2$ for every $x \in A$ with $d_X(x,a) < \delta_0$ .

step 1.7L4L8

3.1

For $x \in A$ with $d_X(x,a) < \delta$ : each $|f_i(x)-f_i(a)| < \varepsilon/\iota(m)$ , so by steps 1.1 and 1.2, $\lVert f(x)-f(a)\rVert_2 \le \sum_{i<m}|f_i(x)-f_i(a)| < \sum_{i<m}\varepsilon/\iota(m) = \varepsilon$ . Hence $f$ is continuous at $a$ , and clause 1 is proved.

step 1.1step 1.2step 2.2L5L6

3.2

Clause 2 is the same two estimates with $f(a)$ replaced by $L$ , $f_i(a)$ by $L_i$ , and the condition $d_X(x,a) < \delta$ by $0 < |x-c| < \delta$ : step 1.1 gives $|f_i(x)-L_i| \le \lVert f(x)-L\rVert_2$ for the forward direction, and steps 1.1, 1.2 give $\lVert f(x)-L\rVert_2 \le \sum_{i<m}|f_i(x)-L_i| < \varepsilon$ for the converse, with $\delta$ the minimum of $m$ radii obtained as in step 2.2.

step 1.1step 1.2step 1.3L1L5L6L7

3.3

Put $P := B + \lVert f(a)\rVert_2 + 1 > 0$ and take $\delta \le \delta_0$ positive with both $\lVert f(x)-f(a)\rVert_2 < \varepsilon/P$ and $\lVert g(x)-g(a)\rVert_2 < \varepsilon/P$ for $d_X(x,a) < \delta$ ; then step 2.3 bounds the difference by $(B + \lVert f(a)\rVert_2)\varepsilon/P < \varepsilon$ , so $\langle f,g\rangle$ is continuous at $a$ .

step 1.7step 2.3L1L6L7

4.1

Steps 1.4, 1.5, 1.6 and 3.3 are clause 3, and with steps 3.1 and 3.2 all three clauses are proved.

step 3.1step 3.2step 1.4step 1.5step 1.6step 3.3∎

Remarks

Why the algebra is proved here rather than quoted. The published 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 Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero are stated for real-valued functions on a subset of $\mathbb{R}$ , and the domain in clause 3 is a subset of an arbitrary metric space; quoting them for a metric domain would be a citation to an item for a claim it does not make. The estimates in steps 1.4 to 3.3 are the same ones, written out. When the domain is a subset of $\mathbb{R}$ , clause 1 and Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace identify the two readings, and the published theorems may then be used on the components.
Clause 3 is what makes the mean value inequality work. The auxiliary function $t \mapsto \langle u, f(t)\rangle$ of The mean value inequality: if $f : [a,b] \to \mathbb{R}^m$ is continuous and differentiable on $(a,b)$ with $\lVert f'\rVert_2 \le M$ , then $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$ is continuous exactly by the inner-product part of clause 3, applied with the constant function $u$ .
Nothing here is a sequential argument, so no choice principle is used beyond the finitely many simultaneous selections of steps 1.3, 2.2 and 3.3, which are covered by Every natural-number-indexed list of nonempty sets has a choice function on its family of values, a theorem of ZF.

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Sources

Vector-valued function (Wikipedia) (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)
MAT237 Multivariable Calculus, Section 1.2: Limits and continuity (standard reference, not scraped)
APEX Calculus, Section 12.2: Calculus and Vector-Valued Functions (standard reference, not scraped)