Math Camp

Lecture 2: Real Analysis in $\mathbb{R}^{n}$

In economics, we often want to maximize a function subject to constraints: $\max_{x\in D} f(x).$ Here $f$ is the objective function and $D$ is the feasible set: the set of choices satisfying the constraints. There is little point in trying to solve this problem if no solution exists. This lecture develops the machinery to guarantee that a solution exists, in $\mathbb{R}^{n}$ .

1 Distance and sequences

We begin by defining Euclidean distance in $\mathbb{R}^{n}$ . Note that there are other ways to measure and define distance, even in $\mathbb{R}^{n}$ . We opt to only use Euclidean distance for most of this class.

Definition 1.1 (Inner product, norm, and distance). For $x,y\in\mathbb{R}^{n}$ ,

the inner product (also known as the dot product) of $x$ and $y$ is $x \cdot y = \sum_{i=1}^n x_i y_i,$
the (Euclidean) norm (or length) of $x$ is $\lVert x \rVert=\sqrt{x \cdot x}=\sqrt{\sum_{i=1}^n x_i^2},$
the distance between $x$ and $y$ is $d(x,y)=\lVert x-y \rVert = \sqrt{\sum_{i=1}^n (x_i-y_i)^2}.$

For example, $x=(3,4)\in\mathbb{R}^2$ has $x\cdot x=25$ and $\lVert x \rVert=5$ , and the distance between $(1,2)$ and $(4,6)$ is $5$ as well. In $\mathbb{R}^1$ the norm is the absolute value, so $d(x,y)=\lvert x-y \rvert$ .

The inner product and the norm are related by the Cauchy–Schwarz inequality:

Theorem 1.2 (Cauchy–Schwarz). For every $x,y\in\mathbb{R}^{n}$ , $(x\cdot y)^2\leq(x\cdot x)(y\cdot y),$ and hence $\lvert x\cdot y \rvert\leq\lVert x \rVert\lVert y \rVert$ .

Proof (optional). If $y=0$ , both sides of the first inequality vanish.

Next suppose $y\ne0$ , so that $y\cdot y=\sum_{i=1}^n y_i^2>0$ . For every $\lambda\in\mathbb{R}$ , consider a vector $z = x-\lambda y$ . By Definition 1.1, $z \cdot z = \sum_{i=1}^n z_i^2 \geq 0$ . Hence, we have $0 \leq z\cdot z=(x-\lambda y)\cdot(x-\lambda y) =x\cdot x-2\lambda\,(x\cdot y)+\lambda^2\,(y\cdot y).$ Next, choose $\lambda=(x\cdot y)/(y\cdot y)$ so that $0 \leq x\cdot x-2\frac{(x\cdot y)}{(y\cdot y)}\,(x\cdot y)+\left(\frac{(x\cdot y)}{(y\cdot y)}\right)^2\,(y\cdot y) = x\cdot x-\frac{(x\cdot y)^2}{y\cdot y}.$ Multiplying it by $y\cdot y>0$ gives the first inequality in Theorem 1.2. Taking square roots gives the second inequality in Theorem 1.2, since $(x\cdot x)^{1/2}=\lVert x \rVert$ by Definition 1.1. ◻

Here are some useful properties of the norm.

Theorem 1.3 (Norm properties). For $x,y\in\mathbb{R}^{n}$ and $a\in\mathbb{R}$ ,

$\lVert x \rVert\geq0$ , and $\lVert x \rVert=0$ if and only if $x=0$ (positivity),
$\lVert ax \rVert=\lvert a \rvert\lVert x \rVert$ (homogeneity),
$\lVert x+y \rVert\leq\lVert x \rVert+\lVert y \rVert$ (triangle inequality).

Proof. Parts (1) and (2) follow from Definition 1.1. The triangle inequality is shown as follows: $\lVert x+y \rVert^2 =\lVert x \rVert^2+2\,x\cdot y+\lVert y \rVert^2 \leq\lVert x \rVert^2+2\lVert x \rVert\lVert y \rVert+\lVert y \rVert^2 =(\lVert x \rVert+\lVert y \rVert)^2,$ where the first inequality uses the Cauchy–Schwarz inequality. Taking square roots gives the triangle inequality. ◻

Read in terms of the distance function $d$ , the three norm properties say that the distance behaves as a distance should: for all $x,y,z\in\mathbb{R}^{n}$ , $\begin{gathered} d(x,y)\geq0,\quad\text{with equality exactly when }x=y,\\ d(x,y)=d(y,x), \qquad d(x,z)\leq d(x,y)+d(y,z). \end{gathered}$ Here, nonnegativity is part (1) applied to $x-y$ ; symmetry is part (2) with $a=-1$ ; and the triangle inequality is part (3) applied to $x-z=(x-y)+(y-z)$ . A function satisfying these three conditions is called a metric.

Equipped with a concept of distance, we can name the set of points lying within a given distance of a point.

Definition 1.4 (Open ball). For $x\in\mathbb{R}^{n}$ and $r>0$ , the open ball with center $x$ and radius $r$ is $B(x,r)=\{y\in\mathbb{R}^{n}:\lVert y-x \rVert<r\}.$

Definition 1.5 (Interior points, open sets, and limit points).

Let $D\subseteq\mathbb{R}^{n}$ .

A point $x\in D$ is an interior point of $D$ if $B(x,r)\subseteq D$ for some $r>0$ . The set of all interior points of $D$ is its interior, written $\operatorname{int}D$ .
The set $D$ is open if every point of $D$ is an interior point.
A point $a\in\mathbb{R}^{n}$ is a limit point of $D$ if every ball $B(a,r)$ contains a point of $D$ different from $a$ ; equivalently, $B(a,r)\cap\left(D\setminus\{a\}\right)\neq\emptyset \qquad\text{for every }r>0.$

For example, let $a<b$ and take $x\in(a,b)$ . The number $r=\min\{x-a,b-x\}$ is positive and $B(x,r)\subseteq(a,b)$ , so $(a,b)$ is open. The endpoints of $[a,b]$ are not interior points, and therefore $\operatorname{int}[a,b]=(a,b).$ The limit-point condition says that points of $D$ different from $a$ can be found arbitrarily close to $a$ ; the point $a$ need not itself belong to $D$ .

Geometrically, in $\mathbb{R}$ the ball $B(x,r)$ is the interval $(x-r,x+r)$ , and in $\mathbb{R}^2$ it is the disk of radius $r$ around $x$ without its bounding circle.

Definition 1.6 (Sequence and convergence). A sequence in $\mathbb{R}^{n}$ is a list $(x_k)_{k=1}^{\infty}$ with $x_k\in\mathbb{R}^{n}$ . It converges to $x\in\mathbb{R}^{n}$ , and $x$ is its limit, written $x_k\to x$ , if for every $\varepsilon>0$ there is $K\in\mathbb N$ such that $k\geq K\quad\Longrightarrow\quad\lVert x_k-x \rVert<\varepsilon.$ A sequence that has no limit in $\mathbb{R}^{n}$ diverges.

In the notation of Definition 1.4, the condition says that every ball $B(x,\varepsilon)$ , however small its radius, contains all terms of the sequence from some index on.

We check the definition on two sequences in $\mathbb{R}$ .

Example 1.7 (Convergence and divergence).

The sequence $x_k=1/k$ converges to $0$ . Let $\varepsilon>0$ and take an integer $K>1/\varepsilon$ . Then every $k\geq K$ satisfies $\lvert x_k-0 \rvert=\frac1k\leq\frac1K<\varepsilon,$ which is the implication Definition 1.6 requires of $K$ . Since $\varepsilon>0$ was arbitrary, $x_k\to0$ .
The sequence $x_k=(-1)^k$ diverges. Let $x\in\mathbb{R}$ be a candidate limit and take $\varepsilon=1$ . By part (3) of Theorem 1.3, $\lvert 1-x \rvert+\lvert -1-x \rvert\geq\lvert (1-x)-(-1-x) \rvert=2,$ so at least one of the two numbers is at least $1$ . The values $1$ and $-1$ both occur at arbitrarily large indices, so every $K\in\mathbb N$ admits some $k\geq K$ with $\lvert x_k-x \rvert\geq1$ , and the implication in Definition 1.6 fails for $\varepsilon=1$ . Since $x$ was arbitrary, no limit exists.

A subsequence is exactly what is sounds like: a sequence drawn from another sequence by skipping some terms.

Definition 1.8 (Subsequence). If $k_1<k_2<\cdots$ is a strictly increasing sequence of indices, then $(x_{k_j})_{j=1}^{\infty}$ is a subsequence of $(x_k)$ .

Here are some useful properties of sequences and their limits.

Theorem 1.9 (Basic limit facts). Let $(x_k)$ be a sequence in $\mathbb{R}^{n}$ .

If $x_k\to x$ and $x_k\to y$ , then $x=y$ .
If $x_k\to x$ , then $(x_k)$ is bounded: there is $M>0$ with $\lVert x_k \rVert\leq M$ for every $k$ .
The convergence $x_k\to x$ holds if and only if $x_{k,i}\to x_i$ in $\mathbb{R}$ for every coordinate $i=1,\ldots,n$ .
If $x_k\to x$ , then every subsequence of $(x_k)$ also converges to $x$ .

Proof (optional). Uniqueness. Let $\varepsilon>0$ . By Definition 1.6 there is $K_1$ with $\lVert x_k-x \rVert<\varepsilon/2$ for $k\geq K_1$ , and $K_2$ with $\lVert x_k-y \rVert<\varepsilon/2$ for $k\geq K_2$ . Fix one $k\geq\max\{K_1,K_2\}$ and write $x-y=(x-x_k)+(x_k-y)$ ; parts (2) and (3) of Theorem 1.3 give $\lVert x-y \rVert\leq\lVert x_k-x \rVert+\lVert x_k-y \rVert<\varepsilon.$ So the nonnegative number $\lVert x-y \rVert$ lies below every positive $\varepsilon$ , hence equals $0$ , and part (1) of Theorem 1.3 gives $x=y$ .

Boundedness. Left as an exercise: Definition 1.6 with $\varepsilon=1$ leaves all but finitely many terms in $B(x,1)$ , and a bound is the largest of $\lVert x \rVert+1$ and the norms of those remaining terms.

Coordinates. Write $a_j=x_{k,j}-x_j$ . Since $a_i^2$ is one of the squares summed in Definition 1.1, and since that sum is at most $\left(\sum_{j=1}^n\lvert a_j \rvert\right)^2$ , taking square roots gives $\lvert x_{k,i}-x_i \rvert \leq\lVert x_k-x \rVert \leq\sum_{j=1}^n\lvert x_{k,j}-x_j \rvert.$ Let $\varepsilon>0$ . If $x_k\to x$ , the left inequality lets any $K$ that Definition 1.6 supplies for $\varepsilon$ serve the coordinate sequence $(x_{k,i})$ as well. Conversely, if every coordinate converges, choose for each $j$ an index $K_j$ beyond which $\lvert x_{k,j}-x_j \rvert<\varepsilon/n$ ; then $k\geq\max\{K_1,\ldots,K_n\}$ makes the right-hand sum, and hence $\lVert x_k-x \rVert$ , smaller than $\varepsilon$ .

Subsequences. The indices of a subsequence satisfy $k_j\geq j$ : this holds at $j=1$ because indices are natural numbers, and $k_{j+1}>k_j\geq j$ forces $k_{j+1}\geq j+1$ . Now let $\varepsilon>0$ and take the $K$ that Definition 1.6 supplies for $(x_k)$ . Every $j\geq K$ has $k_j\geq j\geq K$ , and therefore $\lVert x_{k_j}-x \rVert<\varepsilon$ . ◻

Limits also respect the arithmetic of $\mathbb{R}^{n}$ , one operation at a time.

Theorem 1.10 (Algebra of limits). Suppose $x_k\to x$ and $y_k\to y$ in $\mathbb{R}^{n}$ , and $a_k\to a$ and $b_k\to b$ in $\mathbb{R}$ . Then

$x_k+y_k\to x+y$ ,
$a_kx_k\to ax$ ,
$a_kb_k\to ab$ ,
$x_k\cdot y_k\to x\cdot y$ ,
$\lVert x_k \rVert\to\lVert x \rVert$ ,
$a_k/b_k\to a/b$ , provided $b\neq0$ and $b_k\neq0$ for every $k$ ,
$x\geq y$ , provided $x_k\geq y_k$ coordinatewise for every $k$ .

Its proof is left as an exercise; the claims follow from part (3) of Theorem 1.3, which gives $\lvert \lVert x_k \rVert-\lVert x \rVert \rvert\leq\lVert x_k-x \rVert$ , from Theorem 1.2, and from the coordinatewise criterion in part (3) of Theorem 1.9, together with the boundedness supplied by part (2).

2 Suprema and compactness

We first develop two ingredients for existence arguments: a way to describe the limiting upper value of a set, and a condition that makes sequences behave well inside a feasible set.

2.1 Suprema and maximizing sequences

Definition 2.1 (Bounds and extrema). Let $A\subseteq\mathbb{R}$ be nonempty.

A number $u$ is an upper bound of $A$ if $a\leq u\qquad\text{for every }a\in A,$ and $A$ is bounded above if it has an upper bound.
The supremum, or least upper bound, of $A$ , written $\sup A$ , is an upper bound $s$ of $A$ satisfying $s\leq u\qquad\text{for every upper bound }u\text{ of }A.$
A number $z$ is the maximum of $A$ , written $\max A$ , if $z\in A \qquad\text{and}\qquad a\leq z\ \text{ for every }a\in A.$

Lower bounds, bounded below, the infimum, or greatest lower bound, $\inf A$ , and the minimum $\min A$ are defined by reversing every inequality above.

If a maximum exists, it is automatically the supremum. Conversely, if the supremum exists and belongs to $A$ , it is the maximum. Thus $\max A\text{ exists} \quad\Longleftrightarrow\quad \sup A\text{ exists and }\sup A\in A,$ and then the two numbers are equal.

For example, the upper bounds of $(0,1)$ are precisely the numbers at least $1$ , so $\sup(0,1)=1, \qquad \inf(0,1)=0.$ Neither bound belongs to the interval, so it has neither a maximum nor a minimum. The interval $[0,1]$ has the same supremum and infimum, but it contains both; hence $\max[0,1]=1$ and $\min[0,1]=0$ .

Definition 2.1 does not guarantee that a supremum exists; it is possible for a set to have upper bounds but no least one. For instance, the set of rational numbers $q$ with $q^2<2$ is bounded above by $2$ , but it has no least upper bound in $\mathbb Q$ . When dealing with $\mathbb{R}$ , we make a fundamental assumption that every nonempty set that is bounded above has a supremum. This is known as the completeness property of the real numbers.

Axiom 2.2 (Completeness of $\mathbb{R}$ ). Every nonempty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$ .

Note the role of both hypotheses of the axiom. First, the set $\mathbb{R}$ is nonempty but not bounded above, so it has no supremum. Second, the empty set $\emptyset$ is bounded above by every real number, but it has no least upper bound.

2.2 Closed, bounded, and compact sets

Our existence argument will turn a sequence of feasible points into a candidate solution. For this to work, some subsequence must converge and its limit must remain in the feasible set. Compactness guarantees both.

Definition 2.3 (Closed, bounded, and compact sets). A set $D\subseteq\mathbb{R}^{n}$ is

closed if $x_k\in D$ and $x_k\to x$ imply $x\in D$ ;
bounded if there is $M>0$ such that $\lVert x \rVert\leq M$ for every $x\in D$ ;
compact if every sequence $(x_k)$ in $D$ has a subsequence converging to some point of $D$ .

There are two failures to remember: $x_k=k$ runs off to infinity in $\mathbb{R}$ , while $x_k=1/k$ converges to a point missing from $(0,1)$ . Compactness rules out both.

In Euclidean space these conditions are related by a fundamental result that we use without proof.

Theorem 2.4 (Heine–Borel). A set $D\subseteq\mathbb{R}^{n}$ is compact if and only if it is closed and bounded.

Thus every closed interval $[a,b]$ is compact, whereas $(0,1)$ and $\mathbb{R}$ are not. The empty set is compact, so compactness alone does not supply a feasible point.

3 Continuity and existence

3.1 Continuity

So far, a limit has described the behavior of a sequence. We will also need to describe the behavior of a function as its input approaches a point.

Definition 3.1 (Limit of a function). Let $D\subseteq\mathbb{R}^{n}$ , let $f:D\to\mathbb{R}^m$ , let $a$ be a limit point of $D$ , and let $L\in\mathbb{R}^m$ . We write $\lim_{x\to a}f(x)=L$ if, for every sequence $(x_k)$ in $D\setminus\{a\}$ such that $x_k\to a$ , we have $f(x_k)\to L$ .

In Euclidean space this sequential definition is equivalent to the familiar $\varepsilon$ – $\delta$ condition: for every $\varepsilon>0$ there is $\delta>0$ such that $x\in D,\quad 0<\lVert x-a \rVert<\delta \quad\Longrightarrow\quad \lVert f(x)-L \rVert<\varepsilon.$

The restriction $x\neq a$ is deliberate: a function limit depends on values near $a$ , not on the value at $a$ . The point $a$ need not be in the domain. For example, if $q(h)=h^2/\lvert h \rvert=\lvert h \rvert$ for $h\neq0$ , then Theorem 1.10 gives $\lim_{h\to0}q(h)=0$ , even though $q(0)$ is not defined.

For functions of one real variable, suppose $a$ is approached by points of $D$ from the indicated side. The notation $\lim_{x\downarrow a}f(x)=L \qquad\text{and}\qquad \lim_{x\uparrow a}f(x)=L$ denotes the right-hand limit and left-hand limit, respectively. The first applies Definition 3.1 to $D\cap(a,\infty)$ ; the second applies it to $D\cap(-\infty,a)$ . If the two-sided limit exists, then every one-sided limit that is defined exists and equals it.

Definition 3.2 (Continuity). Let $D\subseteq\mathbb{R}^{n}$ , let $f:D\to\mathbb{R}^m$ , and let $x\in D$ . The function $f$ is continuous at $x$ if, whenever $x_k\in D$ and $x_k\to x$ , we have $f(x_k)\to f(x).$ The function is continuous on $D$ , or simply continuous, if it is continuous at every point of $D$ .

At a limit point $x\in D$ , Definition 3.2 is equivalent to $\lim_{y\to x}f(y)=f(x).$ Equivalently, for every $\varepsilon>0$ there is $\delta>0$ such that, for every $y\in D$ , $\lVert y-x \rVert<\delta \quad\Longrightarrow\quad \lVert f(y)-f(x) \rVert<\varepsilon.$ We will continue to use the sequential formulation because it fits our existence argument. In Lecture 4, the same function-limit notation will define derivatives: a difference quotient is evaluated only for $h\neq0$ , while $h$ approaches $0$ .

Here are the rules that let us build continuous functions out of simpler ones.

Theorem 3.3 (Operations preserving continuity). Let $D\subseteq\mathbb{R}^{n}$ .

If $f,g:D\to\mathbb{R}$ are continuous and $a\in\mathbb{R}$ , then $f+g$ , $af$ , and $fg$ are continuous on $D$ , and the quotient $f/g$ is continuous on the subset of $D$ on which $g\neq 0$ .
A composition of continuous functions is continuous.

Its proof is left as an exercise. Part (1) follows from Definition 3.2 applied to an arbitrary sequence $x_k\to x$ , together with Theorem 1.10; part (2) follows from Definition 3.2 applied twice.

For example, constant functions and the identity function $f(x)=x$ are continuous on $\mathbb{R}$ by Definition 3.2. Part (1) then makes every polynomial continuous on $\mathbb{R}$ , and every rational function continuous off the zeros of its denominator, such as $(x^2+1)/(x-2)$ on $\mathbb{R}\setminus\{2\}$ . Taking as known that $\exp$ is continuous on $\mathbb{R}$ and $\log$ on $(0,\infty)$ , part (2) gives combinations: $e^{-x^2}$ is continuous on $\mathbb{R}$ , and so is $\log(1+x^2)$ , since $1+x^2$ stays positive.

3.2 The Weierstrass theorem

We are now ready to show existence. First, let us formalize what a solution to an optimization problem is.

Definition 3.4 (Maximizers and minimizers). Let $D\subseteq\mathbb{R}^{n}$ be nonempty and let $f:D\to\mathbb{R}$ . The set of maximizers of $f$ on $D$ is $\mathop{\mathrm{arg\,max}}_{x\in D}f(x) =\{x^*\in D:f(x^*)\geq f(x)\text{ for every }x\in D\}.$ The set $\mathop{\mathrm{arg\,min}}_{x\in D}f(x)$ of minimizers is defined by reversing the inequality.

The main existence result is the Weierstrass theorem, which we can now state and prove.

Theorem 3.5 (Weierstrass). Let $D\subseteq\mathbb{R}^{n}$ be nonempty and compact, and let $f:D\to\mathbb{R}$ be continuous. Then $f$ attains a maximum and a minimum on $D$ ; equivalently, $\mathop{\mathrm{arg\,max}}_{x\in D}f(x)\neq\emptyset, \qquad \mathop{\mathrm{arg\,min}}_{x\in D}f(x)\neq\emptyset.$

Proof (optional). We first prove that $f$ has a maximizer; the minimizer will follow at the end.

Step 1: show that the attainable values $f(D)$ are bounded above. Suppose they were not. Then for each $k$ there would be $y_k\in D$ with $f(y_k)>k$ . By Definition 2.3, some subsequence satisfies $y_{k_j}\to y\in D$ , and Definition 3.2 gives $f(y_{k_j})\to f(y)$ . Thus $(f(y_{k_j}))$ is bounded by part (2) of Theorem 1.9. But $k_j\geq j$ , so $f(y_{k_j})>k_j\geq j$ for every $j$ , making the same sequence unbounded. This contradiction proves that $f(D)$ is bounded above.

Step 2: construct feasible points whose values approach the best possible value. The set $f(D)$ is nonempty because $D$ is nonempty. By Step 1 it is also bounded above, so completeness gives the real number $S=\sup f(D)$ . For each $k$ , the smaller number $S-1/k$ cannot be an upper bound of $f(D)$ . Hence there is $x_k\in D$ with $S-\frac1k<f(x_k)\leq S, \qquad\text{that is,}\qquad \lvert f(x_k)-S \rvert<\frac1k.$ Since $1/k\to0$ by part (1) of Example 1.7, this proves $f(x_k)\to S$ . The sequence $(x_k)$ is called a maximizing sequence.

Step 3: obtain a feasible candidate for the maximizer. Compactness of $D$ gives a subsequence $x_{k_j}$ and a point $x^*\in D$ such that $x_{k_j}\to x^*$ . The membership $x^*\in D$ is important: the candidate remains feasible.

Step 4: prove that the candidate is a maximizer. Continuity gives $f(x_{k_j})\to f(x^*)$ . On the other hand, because $f(x_k)\to S$ , its subsequence also satisfies $f(x_{k_j})\to S$ by part (4) of Theorem 1.9. Limits are unique, so $f(x^*)=S$ . Since $S$ is an upper bound of $f(D)$ , we have $f(x)\leq S=f(x^*)$ for every $x\in D$ . Therefore $x^*$ is a maximizer in the sense of Definition 3.4.

Step 5: obtain a minimizer. The function $-f$ is continuous by part (1) of Theorem 3.3. Applying Steps 1–4 to $-f$ gives a point that maximizes $-f$ , which is exactly a point that minimizes $f$ . ◻

Note that failure of any one of the three hypotheses — nonemptiness, compactness, or continuity — can lead to a function that does not attain its maximum or minimum.

Example 3.6 (One failed hypothesis at a time). None of the following problems has a maximizer, and in each one exactly one hypothesis of Theorem 3.5 fails.

In $\mathbb{R}^2$ , let $f(x)=x_1+x_2$ on $D=\{x\in\mathbb{R}^2:x_1\geq0,\ x_2\geq0,\ x_1+x_2\leq1,\ x_1x_2\geq1\}.$ The feasible set $D$ is empty, so $f$ has no maximizer.
$D=(0,1)$ and $f(x)=x$ . The domain is bounded but not closed, since $x_k=1/(k+1)$ lies in $D$ and converges to $0\notin D$ , so again it is not compact. Here $\sup f(D)=1$ , while $f(x)<1$ for every $x\in D$ .
$D=[0,1]$ , which is compact, and $f(x)= \begin{cases} x, & 0\leq x<1,\\ 0, & x=1. \end{cases}$ What fails is continuity, and only at $1$ : the sequence $x_k=1-1/k$ lies in $D$ and converges to $1$ , while $f(x_k)=1-1/k\to1\neq0=f(1)$ , which Definition 3.2 forbids. Again $\sup f(D)=1$ is not attained.

That said, a maximum might exist even if one of the hypotheses of Theorem 3.5 fails. In that case, one must prove existence by other means, such as by finding a candidate and verifying that it is indeed a maximizer.

3.3 Intermediate values and fixed points

While we need linear algebra to meaningfully study functions in $\mathbb{R}^{n}$ , we can already prove some useful results about continuous functions in $\mathbb{R}$ . The intermediate value theorem says that a continuous function on an interval takes on every value between its values at the endpoints.

Theorem 3.7 (Intermediate value theorem). Let $f:[a,b]\to\mathbb{R}$ be continuous. If $\min\{f(a),f(b)\}\leq c\leq\max\{f(a),f(b)\},$ then there is $x\in[a,b]$ such that $f(x)=c$ .

Proof (optional). Assume first that $f(a)\leq c\leq f(b)$ , and let $A=\{x\in[a,b]:f(x)\leq c\}.$ Then $A$ contains $a$ and is bounded above by $b$ , so $s=\sup A$ exists by Axiom 2.2 and satisfies $a\leq s\leq b$ . We show that $f(s)=c$ .

$f(s)\leq c$ . For each $k$ , the number $s-1/k$ is smaller than $s$ and therefore not an upper bound of $A$ , so we may pick $a_k\in A$ with $s-\frac1k<a_k\leq s.$ Then $a_k\to s$ , since $1/k\to0$ by part (1) of Example 1.7, and Definition 3.2 gives $f(a_k)\to f(s)$ . Each $a_k$ lies in $A$ , so $f(a_k)\leq c$ for every $k$ , and part (7) of Theorem 1.10 carries the inequality to the limit: $f(s)\leq c$ .

$f(s)\geq c$ . If $s=b$ , this is the hypothesis $f(b)\geq c$ . If $s<b$ , no point of $(s,b]$ lies in $A$ , so $f(x)>c$ on $(s,b]$ . The points $x_k=s+1/k$ lie in $(s,b]$ as soon as $1/k\leq b-s$ , and $x_k\to s$ , so Definition 3.2 gives $f(x_k)\to f(s)$ . Part (7) of Theorem 1.10 again carries $f(x_k)>c$ to the limit, giving $f(s)\geq c$ .

This proves $f(s)=c$ . In the remaining case $f(b)\leq c\leq f(a)$ , apply what we have just proved to $-f$ , continuous by part (1) of Theorem 3.3, and to the value $-c$ : this produces a point $x$ with $-f(x)=-c$ , that is, $f(x)=c$ . ◻

From that, we obtain our first fixed point theorem in $\mathbb{R}$ for free.

Theorem 3.8 (Brouwer in $\mathbb{R}$ ). Let $a\leq b$ . Every continuous function $f:[a,b]\to[a,b]$ has a fixed point: there is $x^*\in[a,b]$ such that $f(x^*)=x^*$ .

Proof. Let $g(x)=f(x)-x$ , continuous on $[a,b]$ by part (1) of Theorem 3.3. Since $f$ takes its values in $[a,b]$ , we have $f(a)\geq a$ and $f(b)\leq b$ , so $g(a)=f(a)-a\geq0, \qquad g(b)=f(b)-b\leq0.$ Hence $g(b)\leq0\leq g(a)$ , and Theorem 3.7 applied to $g$ with $c=0$ gives $x^*\in[a,b]$ with $g(x^*)=0$ , which means $f(x^*)=x^*$ . ◻