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Multivariable Analysis · Tutorial 815 of 1000

Multivariable Proof Workshop

Learn to organize compactness-based proofs, control near-maximizers, and quantify the effect of uniformly small changes to an objective.

Advanced 10 min read

What You'll Learn

  • Use sequential compactness and continuity to analyze sequences whose objective values approach a maximum
  • Prove that every cluster point of a maximizing sequence is a maximizer
  • Show that near-maximizers must approach the set of maximizers on a compact domain
  • Bound the change in optimal values when two objectives are uniformly close
  • Identify where compactness and continuity enter a proof, and recognize what can fail without them

A Workshop in Proof Architecture

The previous tutorial established that a continuous function on a nonempty compact subset of \(\mathbb{R}^n\) attains its maximum and minimum. This tutorial turns that existence result into a working proof method. We will follow sequences of points whose function values approach an optimum, and we will ask how much an optimum can change when the objective is altered slightly.

The recurring architecture is simple but powerful: choose points that capture the quantity of interest, use compactness to extract a convergent subsequence, and use continuity to identify the value at its limit. Each step has a distinct role. The choice of points translates a supremum or approximation into a sequence; compactness keeps a limit inside the domain; continuity connects the limiting function value to the limiting point.

We will use the Extreme Value Theorem in \(\mathbb{R}^n\) and the Compactness and Sequential Compactness in \(\mathbb{R}^n\) Theorem from earlier in this course. The aim is not to repeat those results, but to practice building new conclusions from them while keeping track of every hypothesis.

Maximizers and Sequences Approaching the Maximum

Definition (Set of Maximizers): Let \(K\subseteq\mathbb{R}^n\) be nonempty and compact, and let \(f:K\to\mathbb{R}\) be continuous. Write \(M=\max_{x\in K} f(x)\). The set of maximizers of \(f\) on \(K\) is \(A=\{x\in K:f(x)=M\}\).

The Extreme Value Theorem ensures that \(A\) is nonempty. A sequence \((x_j)\) in \(K\) is called a maximizing sequence if \(f(x_j)\to M\). Its points need not themselves be maximizers. The important question is what compactness and continuity force about the sequence’s possible limit points.

Theorem (Cluster Points of Maximizing Sequences): Let \(K\subseteq\mathbb{R}^n\) be nonempty and compact, and let \(f:K\to\mathbb{R}\) be continuous with maximum value \(M\). If \(x_j\in K\) and \(f(x_j)\to M\), then every convergent subsequence of \((x_j)\) has a limit in the maximizer set \(A=\{x\in K:f(x)=M\}\). In particular, every maximizing sequence has a subsequence converging to a maximizer.

Proof. Let \((x_{j_\ell})\) be a convergent subsequence, and write \(x_{j_\ell}\to a\). Since all its terms belong to \(K\), compactness and sequential compactness imply that its limit \(a\) belongs to \(K\). Continuity of \(f\) at \(a\) gives \(f(x_{j_\ell})\to f(a)\). On the other hand, \(f(x_j)\to M\), so the subsequence of function values also tends to \(M\). Uniqueness of limits in \(\mathbb{R}\) yields \(f(a)=M\). Therefore \(a\in A\).

For the final assertion, apply sequential compactness directly to the sequence \((x_j)\) in \(K\). It has a convergent subsequence; the argument just given shows that the subsequence’s limit is a maximizer. \(\square\)

The theorem also gives a useful stronger description: a maximizing sequence cannot stay a fixed positive distance away from all maximizers. To make that statement precise, define the distance from a point \(x\) to the nonempty set \(A\) by \(\operatorname{dist}(x,A)=\inf_{a\in A}\|x-a\|_2\). For every maximizing sequence, \(\operatorname{dist}(x_j,A)\to0\). Indeed, if this failed, there would be some \(\varepsilon>0\) and a subsequence with \(\operatorname{dist}(x_{j_\ell},A)\geq\varepsilon\) for every \(\ell\). Compactness would give a further subsequence converging to \(a\in A\), by the theorem. But then \(\operatorname{dist}(x_{j_\ell},A)\leq\|x_{j_\ell}-a\|_2\to0\), a contradiction.

Worked Example: A Maximizing Sequence on an Ellipse

Let \(K=\{(x,y)\in\mathbb{R}^2:x^2+4y^2\leq1\}\) and \(f(x,y)=2x-y\). The set \(K\) is closed and bounded, so it is compact by the Heine–Borel Theorem. To find the maximum, set \(u=x\) and \(v=2y\). Then \(u^2+v^2\leq1\) and \(f=2u-v/2\). By the Cauchy–Schwarz inequality,

$$ 2u-\frac{v}{2} \leq \sqrt{2^2+\left(-\frac12\right)^2}\sqrt{u^2+v^2} \leq \frac{\sqrt{17}}{2}. $$

Equality in the upper bound requires \((u,v)\) to point in the direction \((2,-1/2)\) and have length \(1\). Thus \((u,v)=(4/\sqrt{17},-1/\sqrt{17})\), which corresponds to \(a=(4/\sqrt{17},-1/(2\sqrt{17}))\). This point lies on the ellipse because

$$ \left(\frac{4}{\sqrt{17}}\right)^2 +4\left(-\frac{1}{2\sqrt{17}}\right)^2 =\frac{16}{17}+\frac{1}{17}=1, $$

and its function value is \(8/\sqrt{17}+1/(2\sqrt{17})=\sqrt{17}/2\). For integers \(j\geq2\), set \(x_j=(1-1/j)a\). Since \(0<1-1/j<1\), each \(x_j\) lies in \(K\), and linearity gives \(f(x_j)=(1-1/j)\sqrt{17}/2\to\sqrt{17}/2\). This is a maximizing sequence. It converges to \(a\), which is indeed a maximizer.

Worked Example: Many Maximizers on an Annulus

Consider \(K=\{(x,y)\in\mathbb{R}^2:1\leq x^2+y^2\leq4\}\) and \(f(x,y)=-(x^2+y^2-1)^2\). The annulus is closed and bounded, hence compact, and \(f\) is continuous. Since a square is nonnegative, \(f(x,y)\leq0\). Every point on the inner circle \(x^2+y^2=1\) has value \(0\), so \(M=0\) and the entire inner circle is the maximizer set.

For \(j\geq1\), define \(x_j=((-1)^j(1+1/j),0)\). These points belong to \(K\), since \(1\leq(1+1/j)^2\leq4\). Their values satisfy

$$ f(x_j) =-\left((1+1/j)^2-1\right)^2 =-\left(\frac{2}{j}+\frac{1}{j^2}\right)^2 \longrightarrow0=M. $$

The even-indexed terms converge to \((1,0)\), and the odd-indexed terms converge to \((-1,0)\). Both limits lie on the inner circle and are maximizers, exactly as the Cluster Points of Maximizing Sequences Theorem predicts. A maximizing sequence need not converge as a whole; compactness guarantees convergent subsequences, not convergence of the original sequence.

Stability of the Optimal Value

In applications, an objective may be approximated by a simpler function, or its values may contain small measurement or computational errors. A useful proof question is then: if two objectives are uniformly close on the same compact domain, how different can their maximum values be? Uniform closeness means that a single error bound works at every point of the domain.

Theorem (Stability of Maximum Values): Let \(K\subseteq\mathbb{R}^n\) be nonempty and compact, and let \(f,g:K\to\mathbb{R}\) be continuous. If \(|f(x)-g(x)|\leq\eta\) for every \(x\in K\), where \(\eta\geq0\), then $$ \left|\max_{x\in K}f(x)-\max_{x\in K}g(x)\right|\leq\eta. $$ The same bound holds for the absolute difference between the minimum values.

Proof. Let \(M_f=\max_{x\in K}f(x)\) and \(M_g=\max_{x\in K}g(x)\). By the Extreme Value Theorem, choose \(a,b\in K\) with \(f(a)=M_f\) and \(g(b)=M_g\). At \(a\), the pointwise bound gives \(f(a)\leq g(a)+\eta\), and \(g(a)\leq M_g\). Therefore \(M_f\leq M_g+\eta\). At \(b\), we have \(g(b)\leq f(b)+\eta\), and \(f(b)\leq M_f\), so \(M_g\leq M_f+\eta\). Together these inequalities imply \(|M_f-M_g|\leq\eta\).

For minima, apply the maximum statement to the continuous functions \(-f\) and \(-g\). Their maximum values are \(-\min_K f\) and \(-\min_K g\). Hence

$$ \left|\min_{x\in K}f(x)-\min_{x\in K}g(x)\right| =\left|\max_{x\in K}(-f(x))-\max_{x\in K}(-g(x))\right| \leq\eta. $$

This proof uses actual maximizing points, whose existence comes from compactness and continuity. The pointwise error bound is then evaluated at those points. Notice that the result controls the optimal values; it does not claim that the locations of the maximizers are close. Maximizers can move substantially, especially when the original objective has several nearly optimal points. \(\square\)

Worked Example: Perturbing an Objective on an Interval

On \(K=[-1,1]\), let \(f(x)=-(x-1/2)^2\) and \(g(x)=f(x)+x/10\). Both functions are continuous, and \(|g(x)-f(x)|=|x|/10\leq1/10\) throughout \(K\). The Stability of Maximum Values Theorem therefore gives a maximum-value difference of at most \(1/10\).

For \(f\), the maximum is \(0\), attained at \(x=1/2\). For \(g\), expanding and completing the square gives

$$ g(x)=-(x-1/2)^2+\frac{x}{10} =-x^2+\frac{11}{10}x-\frac14 =-\left(x-\frac{11}{20}\right)^2+\frac{21}{400}. $$

The point \(11/20\) belongs to \([-1,1]\), so the maximum of \(g\) is \(21/400\). The difference between the maximum values is \(21/400\), and \(21/400\leq40/400=1/10\), as the theorem requires. The maximizing points differ by \(1/20\) in this example, but the theorem’s conclusion concerns the values, not that distance.

Where the Proof Can Fail

Compactness is not decorative in these arguments. It prevents sequences of nearly optimal points from escaping the domain without a convergent subsequence. Continuity is equally essential: it is what allows a limit of points to determine the limiting function value. When either link is missing, the conclusions may fail.

Worked Example: A Maximizing Sequence That Escapes

On the noncompact domain \(D=\mathbb{R}\), consider \(f(x)=1-1/(1+x^2)\). For every real \(x\), \(f(x)<1\), while \(f(j)=1-1/(1+j^2)\to1\) as positive integers \(j\) tend to infinity. Thus the supremum is \(1\), but no point attains it. The sequence \(x_j=j\) approaches the supremum in function value while having no convergent subsequence in \(\mathbb{R}\). This is precisely the compactness step that the cluster-point proof needs and that this domain does not provide.

A reliable multivariable proof should make its dependencies visible. Before extracting a subsequence, identify the compact set containing the sequence. Before passing a function value to a limit, verify continuity at the limit point. Before asserting attainment, check that the domain is nonempty and that the Extreme Value Theorem applies. These checks are especially important when the domain is a constraint set: closedness relative to an open domain does not automatically mean closedness in \(\mathbb{R}^n\).

1
Translate the claim into points or values.
For a supremum, choose points with values approaching it; for a perturbation estimate, compare the functions at a maximizing point.
2
Check the hypotheses before using compactness.
Confirm that the sequence lies in a compact set, or that the Extreme Value Theorem applies to the functions under discussion.
3
Pass to a limit with the right justification.
Use sequential compactness for points and continuity for function values; do not interchange these roles.
4
State exactly what follows.
A convergent subsequence may have an optimal limit even when the full sequence does not converge; close optimal values do not necessarily imply close maximizers.

These techniques apply beyond optimization. In many multivariable arguments, a contradiction produces a sequence, compactness extracts a convergent subsequence, and continuity or differentiability identifies the limit. The proof becomes easier to audit when each of those moves is stated explicitly and the conclusion is no stronger than the argument establishes.

Check Your Understanding

Use the hypotheses and proof strategies developed in this workshop to answer the following questions.

  1. Why must the limit of a convergent subsequence of points in a maximizing sequence belong to the domain \(K\)?
  2. What does the Cluster Points of Maximizing Sequences Theorem say if the original maximizing sequence does not converge?
  3. In the stability theorem, where is the uniform bound \(|f(x)-g(x)|\leq\eta\) used to compare the maximum values?
  4. Does a small uniform difference between two objectives guarantee that their maximizing points are close? Explain what the theorem actually controls.
  5. In the example on \(\mathbb{R}\), which compactness step fails for the sequence \(x_j=j\)?