Real Analysis Comprehensive Proof Set III

From Compactness to Global Minimization

Compactness has appeared throughout this course as a way to turn local or sequential information into a global conclusion. In this proof set, we use it to answer an optimization question: when must a continuous function attain its smallest value, even if its domain is unbounded? The key is to show that values near the infimum can occur only in a bounded region. Coercivity supplies that control, and closedness makes the resulting bounded region compact.

The argument combines familiar tools rather than relying on a new calculus formula. We use the Heine–Borel Theorem to obtain compactness, the Extreme Value Theorem to attain a minimum on a compact set, and Bolzano–Weierstrass to study minimizing sequences. The main result and its sequential companion make precise how these ingredients fit together.

Coercivity and Attainment of a Minimum

Definition: Let $E\subseteq\mathbb R$. A function $f:E\to\mathbb R$ is coercive on $E$ if, for every $M\in\mathbb R$, there exists $R>0$ such that whenever $x\in E$ and $|x|\geq R$, we have $f(x)\geq M$. Informally, the function values become arbitrarily large as points in the domain go far from the origin.

The quantifiers matter. The radius $R$ may depend on $M$, and the condition concerns only points that belong to $E$. If $E$ is bounded, the condition can hold vacuously: choose $R$ larger than every $|x|$ in $E$. This is one reason coercivity alone is not enough for a minimum-attainment theorem.

Theorem (Global Minimum for a Continuous Coercive Function): Let $E$ be a nonempty closed subset of $\mathbb R$. If $f:E\to\mathbb R$ is continuous and coercive on $E$, then there exists $x_*\in E$ such that $f(x_*)\leq f(x)$ for every $x\in E$.

Proof: Choose $x_0\in E$ and set $c=f(x_0)$. By coercivity, there is an $R>0$ such that for every $x\in E$ with $|x|\geq R$, $f(x)\geq c+1$. Consider the sublevel set $$ K=\{x\in E:f(x)\leq c\}. $$ It is nonempty because $x_0\in K$. It is bounded: if $x\in K$, then $|x|<R$, since $|x|\geq R$ would imply $f(x)\geq c+1>c$. It is closed in $\mathbb R$, because $E$ is closed and $f$ is continuous on $E$: if $x_n\in K$ and $x_n\to x$, then $x\in E$ and continuity gives $f(x)=\lim_n f(x_n)\leq c$, so $x\in K$. By the Heine–Borel Theorem, $K$ is compact.

The restriction of $f$ to $K$ is continuous, so the Extreme Value Theorem gives a point $x_*\in K$ at which $f$ attains its minimum on $K$. In particular, $f(x_*)\leq f(x_0)=c$. If $x\in E\setminus K$, then $f(x)>c\geq f(x_*)$. If $x\in K$, the choice of $x_*$ gives $f(x_*)\leq f(x)$. These two cases cover every $x\in E$, proving that $x_*$ is a global minimizer. $\square$

Notice that the proof does not need to show in advance that $f$ is bounded below on all of $E$. It first confines every point whose value is at most $f(x_0)$ to a compact set, then finds the minimum there. Points outside that set have values strictly larger than $f(x_0)$, so they cannot improve on the minimum found inside.

Worked Examples

Worked Example: A Minimum on an Unbounded Closed Domain

Let $E=[0,\infty)$ and define $f(x)=x^2-6x+11$. Completing the square gives $$ f(x)=(x-3)^2+2. $$ Since a square is nonnegative, $f(x)\geq2$ for every $x\in E$. At $x=3$, which belongs to $E$, we have $f(3)=(3-3)^2+2=2$. Thus the global minimum is $2$, attained at $x=3$.

The theorem applies as well. The domain $E$ is nonempty and closed. The function is continuous, and $f(x)=(x-3)^2+2\to\infty$ as $x\to\infty$ within $E$, so it is coercive. The square-completion calculation identifies the minimizer directly; the theorem guarantees attainment even when a direct formula is not available.

Worked Example: Two Global Minimizers

On $E=\mathbb R$, consider $g(x)=x^4-4x^2+1$. Rewrite it as $$ g(x)=(x^2-2)^2-3. $$ Because $(x^2-2)^2\geq0$, we have $g(x)\geq-3$. Equality holds exactly when $x^2-2=0$, which means $x=\sqrt{2}$ or $x=-\sqrt{2}$. Substitution verifies both values: $$ g(\sqrt{2})=(2-2)^2-3=-3,\qquad g(-\sqrt{2})=(2-2)^2-3=-3. $$ The global minimum is therefore $-3$, attained at two distinct points.

The function is continuous. It is also coercive: as $|x|\to\infty$, $x^2\to\infty$, and the expression $(x^2-2)^2-3$ tends to infinity. The theorem guarantees at least one minimizer, while the calculation shows that uniqueness need not follow from continuity and coercivity.

Worked Example: Closedness Cannot Be Omitted

Let $E=(0,1)$ and $f(x)=x$. This function is continuous on $E$, but it has no global minimum there: for every $x\in E$, the point $x/2$ also lies in $E$ and satisfies $f(x/2)=x/2<x=f(x)$. The infimum is $0$, but $0\notin E$, so it is not attained.

Under the definition above, $f$ is coercive on $E$ in the vacuous sense: for any $M\in\mathbb R$, choose $R=2$. There is no $x\in(0,1)$ with $|x|\geq2$, so the required implication holds. The domain is not closed, however, and the theorem does not apply. This example shows why the closedness assumption cannot be dropped when coercivity is defined relative to the domain.

Worked Example: Coercivity Cannot Be Omitted

Define $h(x)=e^{-x^2}$ on $E=\mathbb R$. The function is continuous and positive, but it has no global minimum. For every $x$, $h(x)>0$, while $h(x)\to0$ as $|x|\to\infty$. Thus its infimum is $0$ and no point attains it.

The failure of coercivity can be checked directly. Take $M=1/2$. For every $R>0$, choose $x$ with $|x|\geq R$ and $|x|>\sqrt{\log 2}$. Then $x^2>\log 2$, so $h(x)=e^{-x^2}<1/2$. No radius makes all sufficiently distant values at least $1/2$. Here the domain is closed and the function is continuous, but without coercivity the compact-sublevel argument cannot confine low values to a bounded region.

What Minimizing Sequences Reveal

A minimizing sequence is a sequence of points whose function values approach the infimum. It need not itself converge, and it need not consist of minimizers. Under the hypotheses of the global minimum theorem, however, it cannot escape to infinity. This makes it possible to extract a convergent subsequence and identify its limit.

Theorem (Subsequential Limits of Minimizing Sequences): Let $E\subseteq\mathbb R$ be nonempty and closed, and let $f:E\to\mathbb R$ be continuous and coercive. Let $m=\min_{x\in E}f(x)$. If $f(x_n)\to m$ for a sequence $(x_n)$ in $E$, then $(x_n)$ is bounded, and every convergent subsequence of $(x_n)$ converges to a point $p\in E$ with $f(p)=m$. In particular, the sequence has a subsequence converging to a global minimizer.

Proof: Apply coercivity with $M=m+1$. There is an $R>0$ such that $x\in E$ and $|x|\geq R$ imply $f(x)\geq m+1$. Since $f(x_n)\to m$, there is an $N$ such that $f(x_n)<m+1$ for every $n\geq N$. Therefore $|x_n|<R$ for every $n\geq N$. The tail is bounded, and adding the finitely many terms $x_1,\ldots,x_{N-1}$ shows that the whole sequence is bounded.

Now let a subsequence $(x_{n_j})$ converge to $p$. Every term belongs to $E$, and $E$ is closed, so $p\in E$. Continuity gives $$ f(p)=\lim_{j\to\infty}f(x_{n_j})=m, $$ because a subsequence of a convergent real sequence has the same limit. Hence $p$ is a global minimizer. Finally, Bolzano–Weierstrass gives a convergent subsequence of the bounded sequence $(x_n)$, and the argument just given shows that its limit is a minimizer. $\square$

The theorem also explains what can and cannot be concluded about the whole sequence. If the minimizer is unique, every convergent subsequence of a minimizing sequence must converge to that same point; the theorem by itself does not assert that the original sequence converges. If there are several minimizers, different subsequences may converge to different ones.

Corollary (Uniqueness Under Strict Convexity): Suppose $E$ is an interval and $f:E\to\mathbb R$ is strictly convex, meaning that for distinct $x,y\in E$ and $0<t<1$, $$ f(tx+(1-t)y)<t f(x)+(1-t)f(y). $$ If $f$ has a global minimum on $E$, then that minimizer is unique. Indeed, if distinct points $x$ and $y$ both had minimum value $m$, then their strict convex combination $tx+(1-t)y$ would belong to $E$ and would satisfy $f(tx+(1-t)y)<tm+(1-t)m=m$, contradicting minimality.

How to Use the Argument

When a problem asks whether a minimum exists on an unbounded domain, separate the tasks. First identify a nonempty closed domain and verify continuity. Then establish coercivity, or find another reason that the relevant low-value sublevel set is bounded. Choose one point $x_0$ to set a reference value, restrict attention to points with values no greater than $f(x_0)$, and use compactness and the Extreme Value Theorem there.

For a minimizing-sequence proof, first use coercivity to bound the sequence’s tail; finitely many initial terms cannot affect boundedness. Then apply Bolzano–Weierstrass, use closedness to keep the subsequential limit in the domain, and use continuity to pass the function values to the limit. Keeping these steps distinct prevents a common error: a bounded sequence in a nonclosed set can converge outside the domain, while a sequence of points with nearly minimal values can escape to infinity when coercivity is absent.

Check Your Understanding

Use the hypotheses and proof structure to decide which step supports each conclusion.

In the global minimum proof, why is the sublevel set $K=\{x\in E:f(x)\leq f(x_0)\}$ bounded?
Where are closedness of $E$ and continuity of $f$ both used to show that $K$ is closed?
Why does the function $f(x)=x$ on $(0,1)$ meet the relative coercivity condition but still fail to attain its infimum?
For a minimizing sequence, how does coercivity rule out an unbounded tail?
Why must every convergent subsequence of a minimizing sequence have a minimizing limit?
Can a strictly convex function have two distinct global minimizers on an interval? Explain using the strict convexity inequality.