Complete Real Analysis Mastery Examination

A Final Proof Practicum

This final set brings several recurring themes together: compactness turns boundedness into subsequential convergence, continuity lets us pass information to limits, and uniform estimates control whole families of values at once. The problems below are designed to test how these tools fit together, rather than to introduce a new collection of techniques in isolation.

The central problem concerns a sequence of continuous functions on a compact domain. If the functions converge uniformly, do their minimum values converge? What can be said about points where those minima occur? We will prove precise answers, then examine examples involving pointwise convergence, integration, and series. Earlier results such as the Extreme Value Theorem, the Sequential Compactness of a Compact Metric Space, the Dominated Convergence Theorem, and the Weierstrass M-Test will be used by name.

Problem 1: Stability of Minimum Values

Theorem (Uniform Stability of Compact Minimum Problems): Let $K$ be a nonempty compact metric space. Suppose $f_n:K\to\mathbb R$ and $f:K\to\mathbb R$ are continuous, and $f_n\to f$ uniformly on $K$. Write $$ m_n=\min_{x\in K}f_n(x) \qquad\text{and}\qquad m=\min_{x\in K}f(x). $$ Then $m_n\to m$. Moreover, if $x_n\in K$ satisfies $f_n(x_n)=m_n$, every convergent subsequence of $(x_n)$ converges to a point where $f$ attains its minimum.

Proof: By the Extreme Value Theorem, each continuous function $f_n$ attains a minimum on the nonempty compact set $K$, and $f$ attains a minimum there as well. Define $$ \delta_n=\sup_{x\in K}|f_n(x)-f(x)|. $$ Uniform convergence gives $\delta_n\to0$. Choose $x_*\in K$ with $f(x_*)=m$. For every $x\in K$, $f_n(x)\geq f(x)-\delta_n\geq m-\delta_n$, so $m_n\geq m-\delta_n$. At $x_*$, $m_n\leq f_n(x_*)\leq f(x_*)+\delta_n=m+\delta_n$. Thus $$ |m_n-m|\leq\delta_n, $$ and consequently $m_n\to m$.

Now choose any $x_n\in K$ with $f_n(x_n)=m_n$. Since $|f_n(x)-f(x)|\leq\delta_n$ for every $x\in K$, we have $$ m\leq f(x_n)\leq f_n(x_n)+\delta_n=m_n+\delta_n\leq m+2\delta_n. $$ Let $(x_{n_j})$ be a convergent subsequence with limit $p\in K$. The last inequalities and $\delta_{n_j}\to0$ imply $f(x_{n_j})\to m$. Continuity of $f$ gives $f(p)=m$, as required. $\square$

The estimate $|m_n-m|\leq\delta_n$ is the key quantitative part: an error bound for the functions gives the same error bound for their minimum values. The statement about subsequential limits requires a separate step. Convergence of the numbers $m_n$ alone does not identify where minimizers go; the pointwise comparison between $f_n$ and $f$, followed by compactness and continuity, does that work.

Worked Example: A Perturbed Quadratic Minimum

Let $K=[-1,1]$, let $f(x)=x^2$, and for each positive integer $n$ define $$ f_n(x)=x^2+\frac{x}{n}. $$ For every $x\in K$, $|f_n(x)-f(x)|=|x|/n\leq1/n$, so $f_n\to f$ uniformly. Completing the square gives $$ f_n(x)=\left(x+\frac{1}{2n}\right)^2-\frac{1}{4n^2}. $$ The point $-1/(2n)$ belongs to $[-1,1]$, since $0<1/(2n)\leq1/2$. The square is nonnegative and vanishes at that point, so $m_n=-1/(4n^2)$. For example, substitution gives $f_n(-1/(2n))=1/(4n^2)-1/(2n^2)=-1/(4n^2)$. Meanwhile $f$ has minimum $m=0$ at $x=0$. Therefore $m_n\to0=m$, and the minimizers $-1/(2n)$ converge to the minimizer $0$ of $f$.

Corollary (Convergence When the Limiting Minimizer Is Unique): Under the hypotheses of the theorem, suppose $f$ has a unique minimizer $x_*$, and choose a minimizer $x_n$ of each $f_n$. Then $x_n\to x_*$.

Proof: Suppose $x_n$ does not converge to $x_*$. Then there is an $\varepsilon>0$ and a subsequence $(x_{n_j})$ such that the distance from $x_{n_j}$ to $x_*$ is at least $\varepsilon$ for every $j$. Compactness of $K$, by the Sequential Compactness of a Compact Metric Space, gives a convergent subsequence of $(x_{n_j})$, with limit $p\in K$. The theorem shows that $p$ minimizes $f$, so uniqueness implies $p=x_*$. But continuity of the distance function means the distances along this subsequence converge to the distance from $p$ to $x_*$, which is zero. This contradicts their being at least $\varepsilon$. Hence $x_n\to x_*$. $\square$

Problem 2: Why Uniform Convergence Matters

Worked Example: Pointwise Convergence Does Not Control Minima

For $n\geq2$ and $x\in[0,1]$, define $$ \phi_n(x)=\max\{0,1-n|x-1/n|\}, \qquad g_n(x)=1-\phi_n(x). $$ Each $g_n$ is continuous and nonnegative. At $x=1/n$, $\phi_n(1/n)=1$, so $g_n(1/n)=0$, and consequently $\min_{[0,1]}g_n=0$ for every $n$.

For each fixed $x\in[0,1]$, $g_n(x)\to1$. At $x=0$, $n|0-1/n|=1$, so $\phi_n(0)=0$ for every $n$. If $x>0$, then for all sufficiently large $n$, $2/n<x$, which gives $|x-1/n|=x-1/n>1/n$, and hence $\phi_n(x)=0$. Thus the pointwise limit is the constant function $g(x)=1$, whose minimum is $1$. Yet the minima of the $g_n$ are all $0$.

The convergence is not uniform: at $x=1/n$, $|g_n(1/n)-1|=1$. The example shows why the stability theorem requires uniform convergence; pointwise convergence does not prevent a low value from moving around the domain.

This distinction is useful well beyond minimization. Pointwise convergence controls the values at each fixed point eventually, but it does not supply one error bound that works simultaneously at every point. Uniform convergence supplies precisely that shared bound. When a conclusion concerns a supremum, an infimum, or all points of a domain at once, checking whether convergence is uniform is often an essential part of the proof.

Problem 3: Integration and Convergence

Worked Example: Pointwise Convergence with a Dominating Function

On $[0,1]$, set $h_n(x)=x^n$. For each $x\in[0,1)$, $x^n\to0$, while $h_n(1)=1$ for every $n$. Thus the pointwise limit is zero except at the single point $1$. Since a singleton has Lebesgue measure zero, $h_n\to0$ almost everywhere. Also $0\leq h_n(x)\leq1$ for every $x$, and the constant function $1$ is integrable on $[0,1]$. The Dominated Convergence Theorem therefore gives $$ \lim_{n\to\infty}\int_0^1x^n\,dx=0. $$ Direct calculation checks the conclusion: $\int_0^1x^n\,dx=1/(n+1)\to0$.

This sequence does not converge uniformly to zero, since $\sup_{x\in[0,1]}|h_n(x)|=1$ for every $n$. The example contrasts two kinds of control: uniform convergence controls errors at every point, while the Dominated Convergence Theorem uses almost-everywhere convergence together with an integrable bound to control integrals.

Problem 4: Uniform Series and Termwise Integration

Worked Example: A Geometric Series of Functions

For $n\geq1$, define $u_n(x)=x^n/3^n$ on $[0,1]$. Since $0\leq x\leq1$, we have $|u_n(x)|\leq1/3^n$, and the numerical series $\sum_{n=1}^{\infty}1/3^n$ converges. The Weierstrass M-Test shows that $\sum_{n=1}^{\infty}u_n$ converges uniformly on $[0,1]$. Its sum is $$ u(x)=\sum_{n=1}^{\infty}\left(\frac{x}{3}\right)^n=\frac{x}{3-x}. $$ The denominator is nonzero on $[0,1]$. Uniform convergence on this closed interval permits passage to the integral, using the earlier theorem on convergence of integrals under uniform convergence: $$ \int_0^1u(x)\,dx =\sum_{n=1}^{\infty}\int_0^1\frac{x^n}{3^n}\,dx =\sum_{n=1}^{\infty}\frac{1}{3^n(n+1)}. $$ The sum function also gives an explicit value. Since $x/(3-x)=3/(3-x)-1$, $$ \int_0^1\frac{x}{3-x}\,dx =3\log(3/2)-1. $$ These expressions agree because the integral can be computed either from the uniformly convergent series or from its sum function.

Putting the Proofs Together

The minimum-stability argument illustrates a useful structure for a comprehensive proof. First identify the exact uniform error, here $\delta_n$. Next compare the quantities of interest using that error, rather than relying only on convergence language. Then use compactness to extract a convergent subsequence when points, rather than just values, must be studied. Finally, use continuity to identify the limit. Each step has a distinct role.

The other examples show that similar-looking convergence hypotheses are not interchangeable. The moving-dip functions converge pointwise but their minima do not converge to the minimum of their pointwise limit. The functions $x^n$ also converge pointwise rather than uniformly, but an integrable dominating function allows the Dominated Convergence Theorem to control their integrals. For a series of functions, the M-Test supplies uniform convergence, which then supports both continuity of the sum and passage to the integral.

A reliable final check in any proof is to match each conclusion to the hypothesis that justifies it. Compactness gives convergent subsequences, not uniform estimates. Continuity passes limits through a function, but does not itself bound errors uniformly. Uniform convergence controls all points at once, while domination controls integrals under the measure-theoretic hypotheses. Keeping these roles separate is one of the most effective ways to avoid gaps in advanced analysis proofs.

Check Your Understanding

For each question, identify the relevant estimate or hypothesis before deciding what follows.

In the minimum-stability theorem, how does the bound $\delta_n=\sup_{x\in K}|f_n(x)-f(x)|$ give both inequalities needed to prove $|m_n-m|\leq\delta_n$?
Why does compactness matter when showing that every convergent subsequence of minimizers has a minimizing limit?
In the moving-dip example, why does pointwise convergence fail to imply convergence of minimum values?
Why can the Dominated Convergence Theorem apply to $x^n$ on $[0,1]$ even though the convergence is not uniform?
Which estimate allows the Weierstrass M-Test to establish uniform convergence of $\sum_{n=1}^{\infty}x^n/3^n$ on $[0,1]$?