Start by Reading the Logical Shape
In the previous tutorial, a composition of disk automorphisms moved an arbitrary point to the origin, where the Schwarz lemma applied. That proof illustrates a general lesson: before trying to prove a statement, identify what its form asks you to do and which hypotheses are missing for a known result. A useful strategy is not a guess about how the proof will look; it is a plan whose steps match the logical obligations of the claim.
For example, a statement beginning “for every” asks you to fix an arbitrary object and prove the required property. A statement beginning “there exists” asks you to produce a suitable object and verify it meets the conditions. A statement of the form “if \(P\), then \(Q\)” can be proved by assuming \(P\) and deriving \(Q\), or by assuming \(P\) and the failure of \(Q\) and deriving a contradiction. These forms are not interchangeable without care: each creates a different proof obligation.
This principle is a diagnostic tool, not a substitute for proof. A plan such as “use compactness” is incomplete until you identify the set to which compactness applies and explain how a finite subcover yields the desired conclusion. Likewise, “use contradiction” is not an argument until the contradiction is explicitly derived.
Match Common Clues to Their Proof Obligations
| Clue in the statement | Likely strategy | What the proof must establish |
|---|---|---|
| “For every \(x\)” | Direct argument | Fix an arbitrary eligible \(x\), and derive the claim without adding assumptions. |
| “There exists \(x\)” | Construction or an existence theorem | Give a candidate and verify it, or show a theorem’s hypotheses apply. |
| “There exists a unique \(x\)” | Separate existence from uniqueness | Produce at least one candidate, then show any two candidates must agree. |
| “If \(P\), then \(Q\)” | Direct proof or contrapositive | Either derive \(Q\) from \(P\), or derive \(\neg P\) from \(\neg Q\). |
| An equality case in an inequality | Normalize and use rigidity | Move the equality to a setting where a known theorem describes when equality can occur. |
These are clues, not rigid rules. The shortest proof may use a different strategy from the one suggested by the statement’s first words. Still, the clues help reveal the shape of the work before details obscure it. In particular, an equality hypothesis often contains more information than an ordinary inequality hypothesis. If a theorem says a quantity is at most \(1\), knowing that it equals \(1\) may activate a rigidity conclusion.
Negating a Claim Before Looking for a Counterexample
A counterexample is not just an object for which the conclusion fails: it must also satisfy every hypothesis. Quantifier negation makes that requirement precise. The following elementary equivalence is useful whenever a claim says that every eligible object has a property.
Proof. For a fixed \(x\), the implication \(P(x)\Rightarrow Q(x)\) fails exactly when its assumption is true and its conclusion is false. Thus its negation is \(P(x)\land\neg Q(x)\). Negating the assertion that this implication holds for every \(x\) gives the assertion that it fails for at least one \(x\). Therefore the full negation is \(\exists x\,[P(x)\land\neg Q(x)]\), as claimed. \(\square\)
Worked Example: Finding a Genuine Counterexample
Consider the claim: for every real \(x\), if \(x^2=1\), then \(x=1\). Its hypothesis is \(P(x): x^2=1\), and its conclusion is \(Q(x): x=1\). The proposition shows exactly what a counterexample must satisfy: \(x^2=1\) and \(x\neq1\). Choose \(x=-1\). Then \((-1)^2=1\), so the hypothesis holds, while \(-1\neq1\), so the conclusion fails. This disproves the claim.
The example also shows why choosing a value merely because the conclusion fails is not enough. A value such as \(x=0\) has \(x\neq1\), but it does not satisfy \(x^2=1\), so it is not a counterexample to this implication.
The same check helps when a proposed theorem may be false. Negate the theorem first, including all of its quantifiers and assumptions. Then ask whether you can produce an object satisfying the entire negation. If you cannot, that does not prove the original theorem; it only means that this counterexample search has not succeeded.
Normalization as a Strategy for Equality Cases
In the Schwarz–Pick proof, the input point \(a\) and its image \(f(a)\) are moved to \(0\). This is normalization: transform a problem into a form where a known result applies, then translate the result back. For equality cases, the same procedure can uncover a stronger conclusion than the original inequality provides.
Recall from the previous tutorial that for \(a\in\mathbb{D}\), \(\phi_a(z)=(z-a)/(1-\overline{a}z)\) is a disk automorphism, and that the Schwarz lemma characterizes equality in its bound. The Schwarz–Pick proof used the normalized map \(F=\phi_{f(a)}\circ f\circ\phi_a^{-1}\). It satisfies \(F(0)=0\). If equality holds in the Schwarz–Pick inequality for two distinct points, then this normalized map attains equality in the Schwarz lemma at a nonzero point. That is a direct signal to use the Schwarz lemma’s equality case.
Proof. Set \(F=\phi_{f(a)}\circ f\circ\phi_a^{-1}\). As in the Schwarz–Pick proof, \(F\) is a holomorphic map from \(\mathbb{D}\) to itself and \(F(0)=0\). For the first condition, put \(w=\phi_a(z)\). Since \(\phi_a\) is one-to-one and \(z\neq a\), we have \(w\neq0\). The Schwarz–Pick equality at \(a,z\) becomes \(|F(w)|=|w|\). The equality case of the Schwarz lemma implies that \(F(w)=\lambda w\) for all \(w\in\mathbb{D}\), for some constant \(\lambda\) with \(|\lambda|=1\).
For the derivative condition, differentiating the normalized composition at \(0\), as in the previous tutorial, gives $$ F'(0)=f'(a)\frac{1-|a|^2}{1-|f(a)|^2}. $$ The assumed equality therefore gives \(|F'(0)|=1\). The derivative equality case of the Schwarz lemma again implies \(F(w)=\lambda w\) for all \(w\in\mathbb{D}\), with \(|\lambda|=1\).
In either case, \(\phi_{f(a)}\) and \(\phi_a\) are disk automorphisms, and \(w\mapsto\lambda w\) is also a disk automorphism. Since \(F=\phi_{f(a)}\circ f\circ\phi_a^{-1}\), we can solve for \(f\): $$ f=\phi_{f(a)}^{-1}\circ(\lambda\,\mathrm{id})\circ\phi_a. $$ A composition of bijections from \(\mathbb{D}\) onto itself is again such a bijection. Hence \(f\) is a disk automorphism. \(\square\)
Worked Example: Recognizing Equality as a Rigidity Clue
Let \(f(z)=(z-1/3)/(1-z/3)=\phi_{1/3}(z)\), and choose \(a=0\), \(z=1/2\). These points are distinct and belong to \(\mathbb{D}\). We have \(f(0)=-1/3\) and \(f(1/2)=(1/2-1/3)/(1-1/6)=(1/6)/(5/6)=1/5\). The left side of Schwarz–Pick is
The right side is \(\left|(1/2-0)/(1-\overline{0}(1/2))\right|=1/2\), so equality holds. The theorem predicts that \(f\) is a disk automorphism, which is also clear from its displayed form as \(\phi_{1/3}\). The useful strategic point is that equality is not merely a successful numerical check: it signals that the map has a rigid global form.
Worked Example: Using Rigidity to Get a Strict Inequality
Take \(g(z)=z^2\), a holomorphic map from \(\mathbb{D}\) into itself. It is not a disk automorphism: \(g(1/2)=1/4=g(-1/2)\) although \(1/2\neq-1/2\), so it is not one-to-one. The equality-rigidity theorem therefore shows that Schwarz–Pick equality cannot hold for any distinct pair of points for this map. At \(a=0\) and \(z=1/2\), the two sides can be checked directly:
Here \(g(0)=0\) and \(g(1/2)=1/4\), so both denominators equal \(1\). The direct calculation verifies strictness for this pair; the rigidity result explains why equality is impossible for every distinct pair.
Direct Proofs and Their Obligations
A direct proof is often the best choice when the hypotheses already provide the quantities needed to estimate the conclusion. Continuity proofs are a clear example. To prove continuity at a point, the task is to start with an arbitrary positive error tolerance and find a positive input tolerance that guarantees the required output bound. Each choice must be justified by an estimate valid for every input in the chosen neighborhood.
Worked Example: Choosing a Tolerance in a Direct Proof
We verify directly that \(x\mapsto x^2\) is continuous at \(2\). Let \(\varepsilon>0\), and choose \(\delta=\min(1,\varepsilon/5)\). If \(|x-2|<\delta\), then \(\delta\leq1\) gives \(|x-2|<1\), so \(1<x<3\) and therefore \(|x+2|<5\). Factoring the difference yields
Thus every \(\varepsilon>0\) has a suitable \(\delta>0\), which proves continuity at \(2\). The strategy is direct because the factorization exposes a term controlled by the hypothesis \(|x-2|<\delta\), while the choice \(\delta\leq1\) bounds the remaining factor.
A common error in this kind of proof is to choose a tolerance that depends on the input \(x\). The definition requires one \(\delta\), chosen from \(\varepsilon\) and the fixed point, that works for every \(x\) within that distance. Another error is to state a useful bound without showing where it comes from. Here \(|x+2|<5\) follows specifically from \(|x-2|<1\); without that restriction, the factor is not bounded by \(5\).
Check the Strategy Before You Finish
Before polishing a proof, compare its steps with the claim’s exact obligations. A proof of existence must actually supply or secure a witness. A uniqueness argument must start with arbitrary candidates satisfying the defining property. A contradiction proof must negate the conclusion correctly. A normalization argument must verify that the transformed objects meet the theorem’s hypotheses and that the conclusion translates back.
Write down the quantifiers, assumptions, and exact conclusion.
Ask which hypothesis of a relevant theorem is missing, or which estimate the conclusion requires.
Construct, compare, estimate, normalize, or assume the negation of the conclusion.
Check domains, hypotheses, quantifiers, and whether the final statement is exactly the one requested.
The most reliable sign that a strategy is working is not that its name sounds appropriate, but that each step reduces a specific proof obligation. In the equality-rigidity theorem, normalization made a known equality case available; in the continuity example, a factorization made a concrete tolerance possible. Recognizing that match is the central skill: choose a method because its conclusion fits the problem’s structure, then verify the fit in detail.
Check Your Understanding
Use the strategy clues and examples in this tutorial to answer the following questions.
- What two conditions must an object satisfy to be a counterexample to a universal implication?
- When a statement asserts existence and uniqueness, what are the two separate tasks in its proof?
- In the equality-rigidity theorem, why does equality at distinct points produce a nonzero point where the Schwarz lemma has equality?
- Why does the function \(z^2\) fail to be a disk automorphism?
- In the continuity proof for \(x^2\) at \(2\), what role does the condition \(\delta\leq1\) play?