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1.1 Continuous functions

To start with, let us recall some basic notation. We write $\mathbb{N}$ for the natural numbers (without $0$ ), $\mathbb{R}$ for the real numbers. We use the curly parentheses for sets: $\{a,b,c\}$ is the set consisting of three elements $a, b, c$ . Sequences are denoted by round parentheses, e.g. $(x_{n})_{n\in\mathbb{N}}$ .

Recall also the following distinction of intervals:

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open: $(a,b)$ as well as $(a,\infty)$ , $(-\infty,b)$ and $(-\infty,\infty)$ ;
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half-open: $(a,b]$ and $[a,b)$ as well as $(-\infty,b]$ and $[a,\infty)$ ;
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closed: $[a,b]$ as well as $[a,\infty)$ , $(-\infty,b]$ and $(-\infty,\infty)$ ;
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bounded: $(a,b),(a,b],[a,b),[a,b]$ ;
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compact = closed and bounded: $[a,b]$ ;

where $a,b\in\mathbb{R}$ and $a<b$ .

A function is written as

f:X\rightarrow Y,

where $X$ is a set, the domain of the function, $Y$ another set, the codomain, and $f$ the prescription that assigns a unique element in $Y$ to each element in $X$ . Typical examples for $X$ and $Y$ are $\mathbb{R}$ or intervals $I\subset\mathbb{R}$ .

If $f:X\rightarrow Y$ is a function and $X^{\prime}\subset X$ a subset then we can define the restriction of $f$ to that subset,

f\restriction_{X^{\prime}}:X^{\prime}\rightarrow Y,

where the domain changes, but codomain and prescription remain the same.

We recall the definition of continuity in its most convenient form:

Definition 1.1.1 (Continuity).

Let $I\subset\mathbb{R}$ be an interval and $x_{0}\in I$ . A function $f:I\rightarrow\mathbb{R}$ is called continuous at $x_{0}$ if

for every $\varepsilon>0$ there is $\delta>0$ such that for all $x\in I$ with $|x-x_{0}|<\delta$

we have $|f(x)-f(x_{0})|<\varepsilon$ .

The function $f$ is continuous if it is continuous at every $x_{0}\in I$ , i.e.,

for every $\varepsilon>0$ and $x_{0}\in I$ there is $\delta>0$ such that for all $x\in I$ with $|x-x_{0}|<\delta$

we have $|f(x)-f(x_{0})|<\varepsilon$ .

Here $\varepsilon>0$ is a “test parameter”, which is varied in order to “test out” whether a suitable $\delta>0$ , possibly depending on $x_{0}$ and $\varepsilon$ , can be constructed and the condition of continuity at $x_{0}$ be fulfiled; $\varepsilon$ cannot be “calculated”, this does not make sense conceptually. One asks, “if I want a variation of at most $\varepsilon$ in y-direction around $f(x_{0})$ , how much can I vary in x-direction around $x_{0}$ ?” The answer, if it exists, is $\delta$ . So $\delta$ can be calculated or estimated in some way if the function is continuous at $x_{0}$ . However, $\delta$ is not unique: once you have found one $\delta>0$ that works for the function and given $x_{0}$ and $\varepsilon$ , any positive number smaller than that $\delta$ would also work instead of $\delta$ . Similarly, if for some $\varepsilon>0$ and $x_{0}$ you can find $\delta>0$ then for any number greater than that $\varepsilon$ you can use the same $\delta$ to satisfy the condition of continuity at $x_{0}$ . The crucial point is that for every possible $\varepsilon>0$ we must be able to find $\delta>0$ in order to prove continuity at $x_{0}$ .

Example 1.1.2.

Let $I=(0,\infty)$ and $f:I\rightarrow\mathbb{R}$ defined by $f(x)=\frac{1}{x}$ and let us verify continuity.

Let us consider the point $x_{0}=2\in I$ . Then for $\varepsilon=\frac{1}{2}$ , we find by graphical inspection $\delta=1$ fulfiling the above criterion of continuity. For $\varepsilon=\frac{1}{4}$ , we find that $\delta=\frac{2}{3}$ works – and thus any $\delta<\frac{2}{3}$ would actually be fine as well: remember that if one $\delta$ fulfils the continuity criterion, then $\delta$ may be replaced by any positive number less than $\delta$ and again fulfil the criterion.

For general $x_{0}$ and $\varepsilon>0$ , we can determine $\delta$ depending on $x_{0}$ and $\varepsilon$ as follows: if $x=x_{0}-\delta$ then we want

\varepsilon\geq\frac{1}{x}-\frac{1}{x_{0}}=\frac{1}{x_{0}-\delta}-\frac{1}{x_{% 0}},

so $x_{0}\cdot(x_{0}-\delta)\varepsilon\geq x_{0}-(x_{0}-\delta)=\delta$ , so

\delta\leq\frac{x_{0}^{2}\varepsilon}{1+x_{0}\varepsilon}.

If $x=x_{0}+\delta$ then we want

\varepsilon\geq\frac{1}{x_{0}}-\frac{1}{x}=\frac{1}{x_{0}}-\frac{1}{x_{0}+% \delta},

so $x_{0}\cdot(x_{0}+\delta)\varepsilon\geq x_{0}+\delta-x_{0}=\delta$ , so

\delta\leq\frac{x_{0}^{2}\varepsilon}{1-x_{0}\varepsilon}.

Comparing the two upper bounds on $\delta$ , we realise that

\delta:=\frac{x_{0}^{2}\varepsilon}{1+x_{0}\varepsilon}

satisfies bothy of them, so works to prove continuity in $x_{0}$ for given $\varepsilon$ , as requested in Definition 1.1.1.

We see that as $\varepsilon$ gets smaller, so does that $\delta$ – which is not very surprising. For fixed $\varepsilon$ , is there a $\delta$ that works well for all $x_{0}\in I$ ?

The answer is no because $\displaystyle\frac{x_{0}^{2}\varepsilon}{1+x_{0}\varepsilon}\rightarrow 0$ as $x_{0}\rightarrow 0$ .

Instead let’s study $g:(1,\infty)\rightarrow\mathbb{R}$ , with $g(x)=\frac{1}{x}$ . For given $\varepsilon>0$ , can we find a common $\delta>0$ here?

The answer is yes, $\displaystyle\delta=\frac{\varepsilon}{1+\varepsilon}$ .

Notice: since the relevance of $\varepsilon$ lies in “small” values with $\varepsilon$ and possibly $\delta$ close to $0$ and positive, we always assume $\varepsilon$ to be “small enough” for all expressions to make sense, e.g. here such that the denominator does not vanish.

Remark 1.1.3.

The opposite of continuity would be as follows: $f:I\rightarrow\mathbb{R}$ is not continuous or is discontinuous at $x_{0}$ if

there is $\varepsilon>0$ such that for all $\delta>0$ there is $x\in I$ with $|x-x_{0}|<\delta$

and $|f(x)-f(x_{0})|>\varepsilon$ .

Example 1.1.4.

Consider

g:[0,\infty)\rightarrow\mathbb{R},\quad g(x)=\left\{\begin{array}[]{l@{\; :\:}% l}\sin(\frac{\pi}{x})\hfil\;:\>&x\in(0,\infty)\\ 0\hfil\;:\>&x=0.\end{array}\right.

Then $x_{0}=0$ lies in the domain of $g$ but $g$ is not continuous at $x_{0}$ . To see this, take $\varepsilon=\frac{1}{2}$ . Then for any $\delta>0$ , we can find $n\in\mathbb{N}$ such that $\displaystyle x=\frac{2}{1+2n}<\delta$ , which means $|x-x_{0}|<\delta$ but

|g(x)-g(x_{0})|=\Big{|}\sin\Big{(}\frac{1+2n}{2}\pi\Big{)}-0\Big{|}=1>\varepsilon,

proving the criterion in Remark 1.1.3. One can also show that $g$ is continuous at $x_{0}\in(0,\infty)$ .

Let us investigate whether we can find a common $\delta$ for given $\varepsilon>0$ , independent of $x_{0}$ . One can show that restricting $g$ to the interval $(a,\infty)$ or $[a,\infty)$ , with $a>0$ , one can choose

\delta=\frac{a^{2}}{\pi}\varepsilon,

independent of $x_{0}\in[a,\infty)$ . As $a\rightarrow 0$ , we see that $\delta\rightarrow 0$ , and it is impossible to find a common $\delta>0$ for $x_{0}\in(0,\infty)$ .

Exercise W1.2.

Consider $h:(0,10)\rightarrow\mathbb{R}$ , with $h(x)=x^{3}$ . For given $x_{0}\in(0,10)$ and $\varepsilon>0$ , find $\delta>0$ in order to prove that $h$ is continuous. Can you choose $\delta$ depending only on $\varepsilon$ but not on $x_{0}$ ?

Lemma 1.1.5 (Triangle inequality).

Let $a,b\in\mathbb{R}$ , then

|a\pm b|\leq|a|+|b|.

Proof.

Exercise X. ∎

Theorem 1.1.6.

Suppose $I$ is a compact interval and $f:I\rightarrow\mathbb{R}$ a continuous function. Then

for every $\varepsilon>0$ there is $\delta>0$ such that for all $w,x\in I$ with $|w-x|<\delta$

we have $|f(w)-f(x)|<\varepsilon$ .

Proof.

We would like to prove this by contraposition, namely we have to show the following. Suppose that

there is $\varepsilon>0$ such that for every $\delta>0$ there are $w,x\in I$ with $|w-x|<\delta$ $(*)$

and $|f(w)-f(x)|>\varepsilon$ .

Then $f$ is not continuous!

Notice that $(*)$ is true if and only if the following condition is true (taking half of the parameter $\varepsilon$ in $(*)$ ):

there is $\varepsilon>0$ such that for every $\delta>0$ there are $w,x\in I$ with $|w-x|<\delta$ $(**)$

and $|f(w)-f(x)|>2\varepsilon$ .

In order to prove that $f$ is not continuous, take an $\varepsilon>0$ as in $(**)$ . Then for all $n\in\mathbb{N}$ , there are $w_{n},x_{n}\in I$ with $|w_{n}-x_{n}|<\frac{1}{n}$ and $|f(w_{n})-f(x_{n})|>2\varepsilon$ . The fractions $\frac{1}{n}$ take the role of $\delta$ in $(**)$ ; we can use all $n\in\mathbb{N}$ because we said “for every $\delta$ ” in $(**)$ . Then $(x_{n})_{n\in\mathbb{N}}$ forms a sequence in $I$ . The interval $I$ is compact, i.e., bounded and closed, so that the theorem of Bolzano-Weierstrass from MATH113 implies that there is a convergent subsequence $(x_{n_{k}})_{k\in\mathbb{N}}$ . Its limit point lies in $I$ because $I$ is closed (that was the definition of closedness in MATH113), and we denote it by $x_{0}\in I$ (not part of the sequence though).

Similarly, we have the sequence $(w_{n})_{n\in\mathbb{N}}$ , and by the triangle inequality in Lemma 1.1.5, the corresponding subsequence $(w_{n_{k}})_{k\in\mathbb{N}}$ satisfies

	$\displaystyle\|w_{n_{k}}-x_{0}\|\leq$	$\displaystyle\;\|w_{n_{k}}-x_{n_{k}}\|+\|x_{n_{k}}-x_{0}\|$
	$\displaystyle\leq$	$\displaystyle\;\frac{1}{n_{k}}+\|x_{n_{k}}-x_{0}\|\rightarrow 0+0=0,\quad k% \rightarrow\infty,$

so $(w_{n_{k}})_{k\in\mathbb{N}}$ is also convergent with the same limit point $x_{0}$ . Again by the triangle inequality, we have

|f(x_{n_{k}})-f(x_{0})|+|f(x_{0})-f(w_{n_{k}})|\geq|f(x_{n_{k}})-f(w_{n_{k}})|% >2\varepsilon,

for all $k\in\mathbb{N}$ . This means that, for each $k\in\mathbb{N}$ ,

|f(x_{n_{k}})-f(x_{0})|>\varepsilon\mbox{ or }|f(x_{0})-f(w_{n_{k}})|>\varepsilon.

But for every $\delta>0$ there is $k\in\mathbb{N}$ such that $|x_{n_{k}}-x_{0}|<\delta$ and $|w_{n_{k}}-x_{0}|<\delta$ because the two sequence $(x_{n_{k}})_{k\in\mathbb{N}},(w_{n_{k}})_{k\in\mathbb{N}}$ both converge to $x_{0}$ . Hence, we have derived the following statement:

there is $\varepsilon>0$ such that for all $\delta>0$ there is $x\in I$ with $|x-x_{0}|<\delta$

and $|f(x)-f(x_{0})|>\varepsilon$ .

This is the criterion in Remark 1.1.3, showing that $f$ is discontinuous at $x_{0}$ . Hence $f$ is not continuous, which concludes the proof by contraposition. ∎

Key idea of proof.

by contraposition, using Bolzano-Weierstrass and the triangle inequality

The proof is a bit challenging, so try to go through it again carefully and make sure you understand all steps and everything is correct and complete. You do not have to memorise all the details though.

Definition 1.1.7 (Uniform continuity).

Let $I\subset\mathbb{R}$ be an interval. A function $f:I\rightarrow\mathbb{R}$ is called uniformly continuous if

for every $\varepsilon>0$ there is $\delta>0$ such that for all $w,x\in I$ with $|w-x|<\delta$

we have $|f(x)-f(w)|<\varepsilon$ .

Corollary 1.1.8.

If $I$ is a compact interval and $f:I\rightarrow\mathbb{R}$ continuous then $f$ is (automatically) uniformly continuous.

Remark 1.1.9.

If $f$ is uniformly continuous then it is continuous. This follows directly by comparing the definitions of continuity and of uniform continuity: for continuity, $\delta$ may depend on $\varepsilon$ and $x_{0}$ , while for uniform continuity, $\delta$ may depend only on $\varepsilon$ .

Example 1.1.10.

Does the converse of Corollary 1.1.8 hold as well? Prove your answer or find a counterexample. What about contraposition or the inverse statement?

The converse would be as follows: If $f:I\rightarrow\mathbb{R}$ is uniformly continuous then $f$ is continuous and $I$ compact.

We know from Remark 1.1.9 that if $f$ is uniformly continuous then it is continuous. However, $I$ need not be compact. Consider the constant function

f:\mathbb{R}\rightarrow\mathbb{R},\quad f(x)=5,\quad x\in\mathbb{R}.

This is uniformly continuous, e.g. for $\varepsilon>0$ we can choose $\delta=1$ to satisfy the condition of uniform continuity; but $\mathbb{R}$ is not bounded and hence not compact. Hence the converse statement is not true in general.

Recall from MATH111: the contrapositive statement is automatically always true if the original statement is true, while the inverse statement is always false if the original statement is true.

Example 1.1.11.

Let us return to Example 1.1.2 and consider $f:[1,2]\rightarrow\mathbb{R}$ with $f(x)=\frac{1}{x}.$ We found that, for $\varepsilon>0$ we can take $\delta=\frac{x_{0}^{2}\varepsilon}{1+x_{0}\varepsilon}$ , and the smallest such value is

\min_{x_{0}\in[1,2]}\frac{x_{0}^{2}\varepsilon}{1+x_{0}\varepsilon}=\frac{% \varepsilon}{1+\varepsilon},

so $\delta=\frac{\varepsilon}{1+\varepsilon}$ would make $f$ uniformly continuous – as to be expected by Corollary 1.1.8. We could even consider $f:[1,\infty)\rightarrow\mathbb{R}$ with $f(x)=\frac{1}{x}$ . Again $\delta=\frac{\varepsilon}{1+\varepsilon}$ works for this $f$ and establishes uniform continuity although $[1,\infty)$ is not compact. Instead, choosing as domain $(0,1)$ or $[0,1]$ does not yield uniform continuity as was shown in Example 1.1.2.

Exercise W1.1.

State 3 substantially different examples of uniformly continuous functions and 3 substantially different examples of continuous but not uniformly continuous functions. Provide a precise definition of the functions, with domain and codomain etc., but no detailed proof required. They should not be taken from the lecture notes.

Exercise X.

Decide and prove whether a continuous function on a compact interval is bounded.

	$\displaystyle\|w_{n_{k}}-x_{0}\|\leq$	$\displaystyle\;\|w_{n_{k}}-x_{n_{k}}\|+\|x_{n_{k}}-x_{0}\|$
	$\displaystyle\leq$	$\displaystyle\;\frac{1}{n_{k}}+\|x_{n_{k}}-x_{0}\|\rightarrow 0+0=0,\quad k% \rightarrow\infty,$