Start with the Hypotheses and Aim at the Conclusion
In “Turning Definitions Into Proofs,” the central task was to expand definitions and carry out the obligations they specify. A direct proof uses that same discipline to prove an implication: assume its hypotheses, then derive its conclusion. The work lies in finding a valid chain of steps from what is given to what must be shown.
For a statement of the form “if \(P\), then \(Q\),” a direct proof does not begin by assuming \(Q\), and it does not assume the statement is false. It begins by supposing \(P\) holds and then establishes \(Q\). If the claim is universal, such as “for every \(x\in A\), if \(P(x)\), then \(Q(x)\),” first fix an arbitrary \(x\in A\). The proof must use no special property of that \(x\) beyond membership in \(A\) and the stated hypothesis \(P(x)\).
A direct proof is not a requirement to move immediately from the hypotheses to the final line. Often the efficient route is to expand a definition, obtain an estimate, and use that estimate to meet the conclusion’s definition. You may introduce an intermediate claim, provided you establish it and make clear how it advances the proof.
If the claim is universal, choose an arbitrary object in the domain. Do not impose extra conditions on it.
State the hypotheses for this object. Expand definitions when they specify useful bounds or witnesses.
Look for a definition, estimate, identity, or earlier theorem that connects the assumptions to the target.
Justify each inequality or implication, then state explicitly that the required conclusion follows.
The strategy depends on the shape of the claim, but the word “direct” does not mean “obvious.” An estimate may require care about signs, domains, or the order in which bounds are chosen. The following examples show how to make those choices visible.
Worked Examples: Building a Direct Argument
Worked Example: Deriving an Inequality from an Interval Assumption
Suppose \(x\) is real and \(0\leq x\leq 3\). We prove directly that \(x^2\leq 3x\). Since \(x\geq0\) and \(3-x\geq0\), the product of these two nonnegative numbers is nonnegative:
Expanding gives \(3x-x^2\geq0\), which is equivalent to \(x^2\leq3x\). The proof uses both parts of the interval assumption: \(x\geq0\) and \(x\leq3\). Without checking the signs of both factors, the product inequality would not be justified. Here the conclusion follows directly from the hypotheses by constructing a nonnegative expression that rearranges to the desired inequality.
A common direct-proof move is to use the definition of the property in the conclusion. If the conclusion says that a function is bounded, for example, its definition asks for one bound that works for every point in the domain. The next theorem illustrates how to choose that bound from the bounds in the hypotheses.
Proof. Since \(f\) and \(g\) are bounded, there are positive real numbers \(M\) and \(N\) such that \(|f(x)|\leq M\) and \(|g(x)|\leq N\) for every \(x\in E\). Fix an arbitrary \(x\in E\). The triangle inequality gives
The same number \(M+N\) works for every \(x\in E\), so \(f+g\) is bounded. For the product, multiplicativity of absolute value gives
Again, the bound \(MN\) is independent of \(x\), so \(fg\) is bounded. Thus both conclusions follow from the hypotheses. \(\square\)
Worked Example: Applying the Bounded-Function Theorem
Define \(f,g:\mathbb{R}\to\mathbb{R}\) by \(f(x)=1/(1+x^2)\) and \(g(x)=x/(1+x^2)\). Since \(x^2\geq0\), we have \(1+x^2\geq1\), so \(0<f(x)\leq1\). Also, \(1+x^2\geq2|x|\), because
The denominator \(1+x^2\) is positive, so this inequality gives \(|g(x)|\leq1/2\) for every real \(x\). The Bounded-Function Theorem therefore implies that \(f+g\) and \(fg\) are bounded. In this case the estimates can be made explicit:
Each bound holds for every \(x\in\mathbb{R}\), not just for a selected point. The example also shows how a direct proof may proceed in stages: first establish the hypotheses of a theorem, then use the theorem or its estimates to obtain the desired conclusion.
Use the Conclusion’s Definition to Choose an Estimate
For a claim about a Cauchy sequence, the definition itself specifies the target: for every \(\varepsilon>0\), find a stage after which every pair of terms is within \(\varepsilon\) of one another. If a function controls distances by a fixed constant factor, that definition suggests how to transfer the Cauchy estimate.
Proof. Choose a Lipschitz constant \(L\geq0\) for \(F\). Let \(\varepsilon>0\). If \(L=0\), then \(|F(x)-F(y)|\leq0\) for all \(x,y\in A\), so \(F(x)=F(y)\) for all such \(x,y\). The sequence \((F(x_n))\) is then constant and hence Cauchy.
Now suppose \(L>0\). Since \((x_n)\) is Cauchy, apply its definition with the positive tolerance \(\varepsilon/L\). There exists \(N\) such that, whenever \(m,n\geq N\),
For those \(m,n\), the Lipschitz inequality gives
Thus for every \(\varepsilon>0\) there is an \(N\) such that all \(m,n\geq N\) satisfy the Cauchy estimate for \((F(x_n))\). Therefore \((F(x_n))\) is Cauchy. \(\square\)
The proof’s central choice is the input tolerance \(\varepsilon/L\). It is selected so the Lipschitz estimate turns it into the required output tolerance \(\varepsilon\). The case \(L=0\) must be treated separately because division by \(L\) would be invalid.
Worked Example: The Square Function on a Bounded Interval
Fix \(R>0\), and consider \(F(x)=x^2\) on the interval \([-R,R]\). For any \(x,y\in[-R,R]\), factor the difference of squares and apply the triangle inequality:
The last inequality holds because \(|x|\leq R\) and \(|y|\leq R\). Hence \(F\) is Lipschitz on \([-R,R]\) with constant \(2R\). If a Cauchy sequence \((x_n)\) has every term in \([-R,R]\), the theorem shows that \((x_n^2)\) is Cauchy.
The interval restriction matters: the estimate uses a uniform bound \(R\) for both inputs. The same calculation does not supply a single Lipschitz constant for all real \(x,y\), because \(|x|+|y|\) is not bounded over \(\mathbb{R}\). A direct proof must retain the domain assumptions at each step.
Check the Logic Before You Finish
A correct direct proof has a clear route from assumptions to conclusion, and each step is justified on the domain under discussion. Several common errors can break that route:
- Assuming the conclusion. The conclusion is what the proof must establish, not an additional hypothesis.
- Using a bound that depends on the arbitrary point. For boundedness, the same bound must work throughout the domain.
- Dropping a hypothesis that controls a sign or size. For example, \(x(3-x)\geq0\) above depends on both \(x\geq0\) and \(x\leq3\).
- Choosing a tolerance without checking the resulting estimate. When scaling by a Lipschitz constant, the case \(L=0\) needs separate treatment.
- Proving a nearby claim instead of the stated one. The final estimate should match the exact definition in the conclusion.
A useful audit is to identify the first line that uses each hypothesis, and the last line that establishes the requested conclusion. If a hypothesis is unused, check whether it is unnecessary or whether the argument has missed a needed step. If a line depends on a domain restriction, state that restriction where it is used. These checks keep a direct proof both efficient and complete.
Check Your Understanding
For each question, identify the assumptions, the proof obligation, or the estimate that makes the direct argument work.
- To prove a statement of the form “for every \(x\in A\), if \(P(x)\), then \(Q(x)\),” what should be fixed at the start, and what should not be assumed?
- In the proof that sums of bounded functions are bounded, why must \(M+N\) be independent of \(x\)?
- Why does \(1+x^2\geq2|x|\) hold for every real \(x\)?
- In the Lipschitz preservation theorem, why is the input tolerance chosen as \(\varepsilon/L\) when \(L>0\)?
- Which estimate shows that the square function is Lipschitz on \([-R,R]\), and where is the interval assumption used?