\blacktriangleleft

Notions of continuity

Let $(X, d_{X})$ , $(Y, d_{Y})$ and $(Z, d_{Z})$ be metric spaces. For all $c \in X$ and $r > 0$ , let $B_{X}(c, r)$ denote the closed ball of center $c$ and radius $r$ ; that is,

B_{X}(c, r) \doteq \{x \in X, d_{X}(c, x) \leq r\}.

We use the same notation in $Y$ and $Z$ . We always equip $\mathbb{ R}$ with its usual metric $d_{\mathbb{ R}}(x, y) \doteq \left| y - x\right|$ .

Boundedness
Continuity
Uniform continuity
Modulus of continuity
Application

Boundedness

The set of maps from $X$ to $Y$ is denoted $\mathcal{ F}(X, Y)$ .

Definition (Boundedness) We say that $f \in \mathcal{ F}(X, Y)$ is bounded if $f(X)$ is bounded as a subset of $Y$ ; that is,

\operatorname{diam}(f, X, Y) \doteq \sup \{d_{Y}(f(x), f(y)), x \in X, y \in X\} < \infty.

The subset of

\mathcal{ F}(X, Y)

consisting in bounded maps is denoted

\mathcal{ B}(X, Y)

Proposition The map $d_{\mathcal{ B}(X, Y)}: \mathcal{ B}(X, Y) \times \mathcal{ B}(X, Y) \to \mathbb{ R}$ defined by

d_{\mathcal{ B}(X, Y)}(f, g) \doteq \sup \{d_{Y}(f(x), g(x)), x \in X\}

is a metric that turns $\mathcal{ B}(X, Y)$ into a metric space.

Proof By construction, $d_{\mathcal{ B}(X, Y)}$ inherits the non-negativity and symmetry from $d_{Y}$ . Suppose $f, g \in \mathcal{ B}(X, Y)$ are such that $d_{\mathcal{ B}(X, Y)}(f, g) = 0$ . Then $d_{Y}(f(x), g(x)) = 0$ , for all $x \in X$ , which is equivalent to $f(x) = g(x)$ for all $x \in X$ ; that is, $f = g$ . Finally, given $f, g, h \in \mathcal{ B}(X, Y)$ and $x \in X$ , the triangle inequality for $d_{Y}$ implies

d_{Y}(f(x), h(x)) \leq d_{Y}(f(x), g(x)) + d_{Y}(g(x), h(x)).

Taking the supremum on $x \in X$ and using the properties of the supremum, we reach $d_{\mathcal{ B}(X, Y)}(f, h) \leq d_{\mathcal{ B}(X, Y)}(f, g) + d_{\mathcal{ B}(X, Y)}(g, h)$ . This shows $d_{\mathcal{ B}(X, Y)}$ is a metric on $\mathcal{ B}(X, Y)$ . $\blacksquare$

Note that if $f, g \in \mathcal{ B}(X, Y)$ , then

\begin{align*} d_{Y}(f(x), g(x)) & \leq d_{Y}(f(x), f(y)) + d_{Y}(f(y), g(z)) + d_{Y}(g(z), g(x)) \\ & \leq \operatorname{diam}(f, X, Y) + d_{Y}(f(y), g(z)) + \operatorname{diam}(g, X, Y) \end{align*}

for all $x, y, z \in X$ . Taking the supremum on $x$ and the infimum on $y, z$ , we find

d_{\mathcal{ B}(X, Y)}(f, g) \leq \operatorname{diam}(f, X, Y) + \operatorname{diam}(g, X, Y) + \inf \{d_{Y}(f(y), g(z)), y \in X, z \in X\}.

In particular, if $f(X) \cap g(X) \neq \varnothing$ , the infimum above is $0$ .

Continuity

Definition (Local continuity) A map

f: X \to Y

is (locally) continuous at

x \in X

if for all

\varepsilon > 0

, there exists

\delta > 0

such that

f

sends

B_{X}(x, \delta)

into

B_{Y}(f(x), \varepsilon)

. More precisely,

d_{Y}(f(x), f(y)) \leq \varepsilon

for all

y \in Y

such that

d_{X}(x, y) \leq \delta

Proposition

If $f \in \mathcal{ F}(X, Y)$ is continuous at $x \in X$ and $g \in \mathcal{ F}(Y, Z)$ is continuous at $f(x)$ , then $g \circ f \in \mathcal{ F}(X, Z)$ is continuous at $x$ .
If $Y$ is a $\mathbb{ K}$ -vector space and $f, g \in \mathcal{ F}(X, Y)$ are continuous at $x \in X$ , then $\alpha f + \beta g \in \mathcal{ F}(X, Y)$ are continuous at $x$ for all $\alpha, \beta \in \mathbb{ K}$ .
If $Y$ is an algebra and $f, g \in \mathcal{ B}(X, Y)$ are continuous at $x \in X$ , then $f g \in \mathcal{ B}(X, Y)$ is continuous at $x$ .

Proof Let $f \in \mathcal{ F}(X, Y)$ be continuous at $x \in X$ and $g \in \mathcal{ F}(Y, Z)$ be continuous at $f(x)$ . Let $\varepsilon > 0$ . Since $g$ is continuous at $f(x)$ , there exists $\delta_{g} > 0$ such that $g(B_{Y}(f(x), \delta_{g})) \subset B_{Z}((g \circ f)(x), \varepsilon)$ . Since $f$ is continuous at $x$ , there exists $\delta_{f} > 0$ such that $f(B_{X}(x, \delta_{f})) \subset B_{Y}(f(x), \delta_{g})$ . Then choosing $\delta = \delta_{f}$ , we have $(g \circ f)(B_{X}(x, \delta)) \subset B_{Z}((g \circ f)(x), \varepsilon)$ so $g \circ f$ is continuous at $x$ .

Suppose now that $Y$ is a vector space. Let $f, g \in \mathcal{ F}(X, Y)$ be continuous at $x \in X$ . Let $\alpha, \beta \in \mathbb{ K}$ . For convenience, denote $h = \alpha f + \beta g$ . Let $\varepsilon > 0$ and define $\varepsilon' = \varepsilon / (\left| \alpha\right| + \left| \beta\right| + 1)$ . Since $f$ and $g$ are continuous at $x$ , there exist $\delta_{f}, \delta_{g} > 0$ such that $f(B_{X}(x, \delta_{f})) \subset B_{Y}(f(x), \varepsilon')$ and similarly for $g$ . Choosing $\delta = \min(\delta_{f}, \delta_{g}) > 0$ , we then have, for all $y \in B_{X}(x, \delta)$ ,

d_{Y}(h(x), h(y)) \leq \left| \alpha\right| d_{Y}(f(x), f(y)) + \left| \beta\right| d_{Y}(g(x), g(y)) \leq (\left| \alpha\right| + \left| \beta\right|) \varepsilon' \leq \varepsilon,

so $h$ is continuous at $x$ .

Finally, suppose $Y$ is an algebra and $f, g \in \mathcal{ B}(X, Y)$ are continuous at $x$ . Let $\epsilon > 0$ and define $\epsilon' = \epsilon / (\left\| f\right\|_{\infty} + \left\| g\right\|_{\infty} + 1)$ . Since $f$ and $g$ are continuous at $x$ , there exist $\delta_{f}, \delta_{g} > 0$ such that $f(B_{X}(x, \delta_{f})) \subset B_{Y}(f(x), \varepsilon')$ and similarly for $g$ . Choosing $\delta = \min(\delta_{f}, \delta_{g}) > 0$ , we have, for all $y \in B_{X}(x, \delta)$ , the triangle inequality yields

\begin{align*} d_{Y}((f g)(x), (f g)(y)) & = d_{Y}(f(x) g(x), f(x) g(y)) + d_{Y}(f(x) g(y), f(y) g(y)) \\ & = \left| f(x)\right| d_{Y}(g(x), g(y)) + \left| g(y)\right| d_{Y}(f(x), f(y)) \\ & \leq (\left\| f\right\|_{\infty} + \left\| g\right\|_{\infty}) \epsilon' \\ & \leq \epsilon, \end{align*}

so $f g$ is continuous at $x$ . $\blacksquare$

There is more to say about maps that are continuous everywhere.

Definition (Global continuity) If

f: X \to Y

is continuous everywhere in

X

, then

f

is said to be (globally) continuous.

The set of continuous maps from $X$ to $Y$ is denoted $\mathcal{ C}(X, Y)$ .

We now give a few examples to illustrate the difference between locally and globally continuous functions.

Example The identity map

\operatorname{Id}_{X}: X \to X

belongs to

\mathcal{ C}(X, X)

. Let

\varepsilon > 0

and

x \in X

. Let

\delta = \varepsilon

and

y \in B_{X}(x, \delta)

. Then

d_{X}(\operatorname{Id}_{X}(x), \operatorname{Id}_{X}(y)) = d_{X}(x, y) \leq \delta = \varepsilon

Example Let

z \in Y

. The constant map

\tilde{z}: X \to Y

defined by

\tilde{z}(x) = z

belongs to

\mathcal{ C}(X, Y)

. Let

\varepsilon > 0

and

x \in X

. Let

\delta > 0

and

y \in B_{X}(x, \delta)

. Then

d_{X}(\tilde{z}(x), \tilde{z}(y)) = d_{X}(z, z) = 0 \leq \varepsilon

Example Let

z \in X

. The distance map

d_{z}: X \to \mathbb{ R}

defined by

d_{z}(x) = d_{X}(x, z)

belongs to

\mathcal{ C}(X, \mathbb{ R})

. Let

\varepsilon > 0

and

x \in X

. Let

\delta = \varepsilon

and

y \in B_{X}(x, \delta)

. Then the reverse triangle inequality for metric spaces provides

\left| d_{z}(y) - d_{z}(x)\right| = \left| d_{X}(y, z) - d_{X}(x, z)\right| \leq d_{X}(x, y) \leq \delta = \varepsilon

Example Let $\alpha \geq 0$ . The power function $p_{\alpha}: \mathbb{ R}_{+} \to \mathbb{ R}$ defined by $p_{\alpha}(x) = x^{\alpha}$ belongs to $\mathcal{ C}(\mathbb{ R}_{+}, \mathbb{ R})$ . Let $\varepsilon > 0$ and $x \in \mathbb{ R}_{+}$ . We separate two cases depending on the position of $\alpha$ relative to $1$ .

If $\alpha \in [0, 1]$ , define $\delta = \varepsilon^{1/\alpha}$ . The case $y = x$ being trivial, we assume $y \neq x$ and write

\left| y^{\alpha} - x^{\alpha}\right| = \underbrace{\frac{\left| y^{\alpha} - x^{\alpha}\right|}{\left| y - x\right|^{\alpha}}}_{\phi(x, y)} \left| y - x\right|^{\alpha} \leq \delta^{\alpha} = \varepsilon.

Indeed, the function $\phi$ is homogeneous in $x$ and $y$ and since $x \neq y$ , we introduce $t = \min(x, y) / \max(x, y) \in [0, 1)$ and rewrite

\phi(x, y) = \psi(t), \qquad \psi: [0, 1) \ni t \mapsto \frac{1 - t^{\alpha}}{(1 - t)^{\alpha}}.

A quick study of the function $\psi$ shows that it is bounded by $1$ .

If $\alpha > 1$ , define $\delta > 0$ such that $\alpha (x + \delta)^{\alpha - 1} \delta = \varepsilon$ (such a $\delta$ exists because $t \mapsto (x + t)^{\alpha - 1} t$ is increasing). The mean value theorem ensures the existence of $c \in (x, y)$ such that

\left| y^{\alpha} - x^{\alpha}\right| = \alpha \left| \int_{x}^{y} t^{\alpha - 1} dt\right| = \alpha c^{\alpha - 1} \left| y - x\right| \leq \alpha \max(x, y)^{\alpha - 1} \left| y - x\right| \leq \alpha (x + \delta)^{\alpha - 1} \delta = \varepsilon.

Example Define the sign function

s_{a}: \mathbb{ R} \to \mathbb{ R}

defined by

s_{a}(x) = -1

x < 0

s_{a}(x) = +1

x > 0

and

s_{a}(0) = a

for some

a \in \mathbb{ R}

. We will show that

s_{a}

cannot be made continuous at

0

. Let

x = 0

and

y \in \mathbb{ R}

. If

y > 0

, then

s_{a}(y) - s_{a}(x) = 1 - a

, which is independent of

y

. The only way to make this smaller than

\varepsilon

for all

\varepsilon > 0

is to choose

a = 1

. Conversely, if

y < 0

, then

s_{a}(y) - s_{a}(x) = - 1 - a

and the only way to make this smaller than

\varepsilon

for all

\varepsilon > 0

is to choose

a = -1

. There is no choice of

a

that makes

s_{a}

continuous.

The last example would motivate a notion of "directional continuity", which is a generalisation of one-sided continuity to metric spaces.

Proposition (Structure of $\mathcal{ C}(X, Y)$ )

If $f \in \mathcal{ C}(X, Y)$ and $g \in \mathcal{ C}(Y, Z)$ , then $g \circ f \in \mathcal{ C}(X, Z)$ .
If $Y$ is a vector space, then $\mathcal{ C}(X, Y)$ is a subvector space of $\mathcal{ F}(X, Y)$ .
If $f, g \in \mathcal{ C}(X, Y) \cap \mathcal{ B}(X, Y)$ , then $f g \in \mathcal{ C}(X, Y) \cap \mathcal{ B}(X, Y)$ .

Proof The previous proposition on the stability of local continuity under composition, linear combinations and product readily extends to global continuity. We have proved that the constant function

\tilde{0_{Y}}: X \to Y

is continuous in the second example. This shows that

\mathcal{ C}(X, Y)

is a vector space when

Y

is one.

\blacksquare

Uniform continuity

In the definition of continuity, $\delta$ is allowed to depend on $x$ and $\varepsilon$ . The following stronger definition of continuity requires that $\delta$ be independent of $x$ .

Definition (Uniform continuity) A map

f: X \to Y

is uniformly continuous if for all

\varepsilon > 0

, there exists

\delta > 0

such that

f

sends any ball of radius

\delta

X

into a ball of radius

\varepsilon

Y

. More precisely,

d_{Y}(f(x), f(y)) \leq \varepsilon

for all

x, y \in X

such that

d_{X}(x, y) \leq \delta

The set of uniformly continuous functions from $X$ to $Y$ is denoted $\mathcal{ U}(X, Y)$ .

Example From the previous examples, looking at the dependence of

\delta

\varepsilon

, we observe that the identity map, the constant map, the distance map and the power function with

\alpha \in [0, 1]

are uniformly continuous. However, the power function with

\alpha > 1

is not uniformly continuous.

Proposition (Structure of $\mathcal{ U}(X, Y)$ )

If $Y$ is a vector space, then $\mathcal{ U}(X, Y)$ is a subvector space of $\mathcal{ C}(X, Y)$ .
If $f \in \mathcal{ U}(X, Y)$ and $g \in \mathcal{ U}(Y, Z)$ , then $g \circ f \in \mathcal{ U}(X, Z)$ .
If $Y$ is a normed algebra and $f, g \in \mathcal{ U}(X, Y)$ are bounded, then $fg \in \mathcal{ U}(X, Y)$ .

Proof This proof is identical to its counterpart for

\mathcal{ C}(X, Y)

. The only difference is that

\delta

is defined before

x

\blacksquare

Modulus of continuity

It is useful to characterise the dependence of $\delta$ on $\varepsilon$ for a uniformly continuous function. This is what the modulus of continuity measures. We first identify candidate moduli of continuity.

Definition (Modulus of continuity) A function

\omega: \mathbb{ R}_{+} \to \mathbb{ R}_{+}

is a modulus of continuity if

\omega

is non-decreasing,

\omega(0) = 0

and

\omega

is continuous at

0

The set of moduli of continuity is written $M$ . Note that $M$ is stable under non-negative linear combinations and composition. We can now characterise uniform continuity with moduli of continuity.

Definition (Uniform continuity with modulus) A function

f: X \to Y

is uniformly continuous with modulus

\omega \in M

d_{Y}(f(x), f(y)) \leq \omega(d_{X}(x, y))

for all

x, y \in X

. We also simply say that

f

\omega

-uniformly continuous,

\omega

-UC for short, and we write

f \in \mathcal{ U}(\omega, X, Y)

We first show that if $f \in \mathcal{ U}(X, Y)$ , then $f \in \mathcal{ U}(\omega, X, Y)$ for some well-chosen $\omega \in M$ .

Proposition (Minimal modulus of continuity) If $f \in \mathcal{ U}(X, Y)$ and define $\omega_{f} \in M$ as

\omega_{f}(\delta) = \sup \{d_{Y}(f(x), f(y)), x \in X, y \in X, d_{X}(x, y) \leq \delta\}.

Then $f \in \mathcal{ U}(\omega_{f}, X, Y)$ .

Proof Let us show that $\omega_{f} \in M$ . When $\delta = 0$ , one must have $x = y$ in the definition of $\omega_{f}(0)$ , so $\omega_{f}(0) = 0$ . Besides, $\omega_{f}$ is non-increasing by construction. Now given $\varepsilon > 0$ , the uniform continuity of $f$ ensures the existence of $\delta > 0$ such that $d_{Y}(f(x), f(y)) \leq \varepsilon$ for all $x, y \in X$ with $d_{X}(x, y) \leq \delta$ . Taking the supremum on that region, we find $\omega(t) \leq \varepsilon$ for all $t \in [0, \delta]$ . This is exactly the definition of the continuity of $\omega_{f}$ at $0$ .

Finally, by construction, for all $\varepsilon > 0$ , choosing $\delta > 0$ such that $\omega(\delta) \leq \varepsilon$ implies that $d_{Y}(f(x), f(y)) \leq \varepsilon$ whenever $d_{X}(x, y) \leq \delta$ . This shows $f \in \mathcal{ U}(\omega_{f}, X, Y)$ . $\blacksquare$

We now show that uniform continuity is equivalent to having a modulus of continuity.

Proposition

f \in \mathcal{ U}(X, Y)

if and only if there exists

\omega \in M

such that

f \in \mathcal{ U}(\omega, X, Y)

Proof Suppose $f \in \mathcal{ U}(\omega, X, Y)$ for some $\omega \in M$ . Let $\varepsilon > 0$ . Since $\omega$ is non-decreasing, there exists $\delta > 0$ such that $\omega(\delta) \leq \varepsilon$ (if $\omega$ is bounded and $\varepsilon$ is larger than the maximum of $\omega$ , then any choice of $\delta$ is acceptable). For all $x, y \in X$ such that $d_{X}(x, y) \leq \delta$ , we have $d_{Y}(f(x), f(y)) \leq \omega(d_{X}(x, y)) \leq \omega(\delta) \leq \varepsilon$ . This shows $f \in \mathcal{ U}(X, Y)$ .

Conversely, we have already seen that $f \in \mathcal{ U}(X, Y)$ implies $f \in \mathcal{ U}(\omega_{f}, X, Y)$ . $\blacksquare$

Note that if a map admits a modulus of continuity, then it admits infinitely many. Indeed, if $\omega, \eta \in M$ are such that $\eta \geq \omega$ and $f \in \mathcal{ U}(\omega, X, Y)$ , then $f \in \mathcal{ U}(\eta, X, Y)$ . Still, a uniformly continuous function $f$ admits a minimal modulus of continuity.

Proposition (Minimal modulus of continuity) Let

\omega \in M

such that

f \in \mathcal{ U}(\omega, X, Y)

. Then

\omega \geq \omega_{f}

Proof Let

\omega \in M

such that

f \in \mathcal{ U}(\omega, X, Y)

. Then given

\delta > 0

, we have

d_{Y}(f(x), f(y)) \leq \omega(\delta)

for all

x, y \in X

with

d_{X}(x, y) \leq \delta

. Taking the supremum on that region, we find

\omega_{f}(\delta) \leq \omega(\delta)

\blacksquare

Proposition (Structure of $\mathcal{ U}(\omega, X, Y)$ )

If $f \in \mathcal{ U}(\omega_{f}, X, Y)$ and $g \in \mathcal{ U}(\omega_{g}, Y, Z)$ , then $g \circ f \in \mathcal{ U}(\omega_{g} \circ \omega_{f}, X, Y)$ .
If $Y$ is a vector space and $f \in \mathcal{ U}(\omega_{f}, X, Y)$ and $g \in \mathcal{ U}(\omega_{g}, X, Y)$ , then $\alpha f + \beta g \in \mathcal{ U}(\left| \alpha\right| \omega_{f} + \left| \beta\right| \omega_{g}, X, Y)$ .
If $Y$ is a normed algebra and $f \in \mathcal{ U}(\omega_{f}, X, Y)$ is bounded and $g \in \mathcal{ U}(\omega_{g}, X, Y)$ is bounded, then $f g \in \mathcal{ U}(\left\| f\right\|_{\mathcal{ C}(X, Y)} \omega_{g} + \left\| g\right\|_{\mathcal{ C}(X, Y)} \omega_{f})$ .

Application

We conclude with a short application to the error analysis of the approximation of a function $f \in \mathcal{ U}(\omega, I \subset \mathbb{ R}, \mathbb{ R})$ by its Bernstein coefficient, defined by

B_{n}(f)(x) = \sum_{k = 0}^{n} \binom{n}{k} x^{k} (1 - x)^{n - k} f\left(\frac{k}{n}\right),

for all $n \geq 1$ . Let $x \in I$ and $\delta \leq 1/n$ . We compute

\begin{align*} \left| f(x) - B_{n}(f)(x)\right| & \leq \sum_{k = 0}^{n} \binom{n}{k} x^{k} (1 - x)^{n - k} \left| f(x) - f\left(\frac{k}{n}\right)\right| \\ & \leq \sum_{k = 0}^{n} \binom{n}{k} x^{k} (1 - x)^{n - k} \omega\left(f, \left| x - \frac{k}{n}\right|\right) \\ & \leq \sum_{k = 0}^{n} \binom{n}{k} x^{k} (1 - x)^{n - k} \left(\left\lfloor \delta^{-1} \left| x - \frac{k}{n}\right| \right\rfloor + 1\right) \omega\left(f, \delta\right) \\ &= \left(1 + \sum_{k = 0}^{n} \binom{n}{k} x^{k} (1 - x)^{n - k} \left\lfloor \delta^{-1} \left| x - \frac{k}{n}\right| \right\rfloor\right) \omega\left(f, \delta\right) \\ &\leq \left(1 + \delta^{-2} \sum_{k = 0}^{n} \binom{n}{k} x^{k} (1 - x)^{n - k} \left(x - \frac{k}{n}\right)^{2}\right) \omega\left(f, \delta\right) \\ &= \left(1 + \delta^{-2} n^{-1} x(1 - x)\right) \omega\left(f, \delta\right). \end{align*}

In the last inequality, we used the fact that if $k \in \{0, \ldots, n\}$ is such that $\left| x - \frac{k}{n}\right| < \delta$ , then $\left\lfloor \delta^{-1} \left| x - \frac{k}{n}\right| \right\rfloor = 0$ and otherwise, $\left\lfloor \delta^{-1} \left| x - \frac{k}{n}\right| \right\rfloor \leq \delta^{-2} \left| x - \frac{k}{n}\right|^{2}$ . To obtain the last equality, we expand the squares and recognise moments of a binomial random variable.

Choosing $\delta^{2} = x(1 - x) / n$ , we find

\left| f(x) - B_{n}(f)(x)\right| \leq 2 \omega\left(f, \sqrt{\frac{x (1 - x)}{n}}\right) \leq 2 \omega\left(f, \frac{1}{2 \sqrt{n}}\right).

Choosing $\delta = 1 / \sqrt{n}$ , we also find

\left| f(x) - B_{n}(f)(x)\right| \leq (1 + x(1 - x)) \omega\left(f, \frac{1}{\sqrt{n}}\right) \leq \frac{5}{4} \omega\left(f, \frac{1}{\sqrt{n}}\right).

\blacktriangleleft

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