The Continuity Strategy

Use a Strict Margin to Control Nearby Values

The Density Strategy used continuity to transfer information from approximating points to a limit point. A related proof technique starts with a different question: if a function satisfies a strict inequality at one point, what can be said about its values at nearby points? Continuity provides the answer when there is a positive gap to preserve.

For example, if \(f(a)>c\), the difference \(f(a)-c\) is a positive margin. Choose the allowed change in \(f\) to be smaller than that margin. Continuity then guarantees that nearby values remain above \(c\). This local argument is useful whenever a proof needs to show that a sign, nonzero value, or strict comparison persists near a point.

Definition (Continuity at a Point): Let \(E\) be a subset of a metric space \((X,d)\), let \(a\in E\), and let \(f:E\to\mathbb{R}\). The function \(f\) is continuous at \(a\) if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(x\in E\) and \(d(x,a)<\delta\) imply \(|f(x)-f(a)|<\varepsilon\).

The restriction \(x\in E\) matters: neighborhoods are taken relative to the domain. The central move is to choose \(\varepsilon\) from the margin in the conclusion, rather than treating it as an arbitrary tolerance.

Theorem (Local Persistence of a Strict Inequality): Let \(f:E\to\mathbb{R}\) be continuous at \(a\in E\). If \(f(a)>c\), then there is a \(\delta>0\) such that \(f(x)>c\) for every \(x\in E\) with \(d(x,a)<\delta\).

Proof. Set \(m=f(a)-c\), so \(m>0\). By continuity at \(a\), there is a \(\delta>0\) such that \(x\in E\) and \(d(x,a)<\delta\) imply \(|f(x)-f(a)|<m/2\). In particular,

f(x)>f(a)-\frac{m}{2} =c+m-\frac{m}{2} =c+\frac{m}{2} >c.

This proves the claimed inequality for every such \(x\). \(\square\)

The same reasoning applies to \(f(a)<c\), by using the margin \(c-f(a)>0\). It also applies to a strict comparison between two functions: if \(f\) and \(g\) are continuous at \(a\) and \(f(a)<g(a)\), then \(g-f\) is continuous at \(a\) and is positive there. Local persistence gives \(f(x)<g(x)\) for all domain points sufficiently close to \(a\).

Worked Example: Preserving Positivity Near a Point

Consider \(p(x)=2+x-x^2\), and ask where near \(0\) it remains greater than \(1\). At \(0\), \(p(0)=2\), so the margin above \(1\) is \(1\). For \(|x|<1/4\),

|p(x)-p(0)|=|x-x^2| \leq |x|+x^2 <\frac{1}{4}+\frac{1}{16} =\frac{5}{16} <1.

Consequently, \(p(x)>p(0)-1=1\) whenever \(|x|<1/4\). This calculation illustrates the margin method explicitly: the change in the function is kept smaller than the gap between its value at \(0\) and the threshold \(1\).

Turn a Failed Local Claim into a Sequence

There is another useful way to recognize this strategy. If no neighborhood preserves a strict inequality, then points violating it can be found arbitrarily close to the given point. Choose one such point within distance \(1/n\) for each positive integer \(n\). The resulting sequence converges to the point, so continuity forces its function values to converge to the value there. A fixed strict margin cannot survive that contradiction.

This sequence construction is a general proof technique: when a local conclusion fails at every scale, select a counterexample at each scale. The sequence records the failure while approaching the point where continuity applies.

Worked Example: A Strict Comparison of Two Functions

Let \(f(x)=x^2\) and \(g(x)=3x\). At \(x=1\), \(f(1)=1<3=g(1)\). The difference \(h(x)=g(x)-f(x)=3x-x^2\) is continuous and satisfies \(h(1)=2>0\). By local persistence, \(h(x)>0\), and therefore \(f(x)<g(x)\), throughout some neighborhood of \(1\).

One can verify a particular neighborhood directly. Write \(x=1+t\). Then

$$ h(x)=3(1+t)-(1+t)^2 =3+3t-(1+2t+t^2) =2+t-t^2. $$

If \(|x-1|=|t|<1/4\), then

|h(x)-2|=|t-t^2| \leq |t|+t^2 <\frac{1}{4}+\frac{1}{16} =\frac{5}{16}.

Thus \(h(x)>2-5/16=27/16>0\), so \(x^2<3x\) whenever \(|x-1|<1/4\). The general theorem explains why some neighborhood must work; the calculation supplies an explicit one.

From Local Control to a Uniform Bound

Local persistence gives a neighborhood around each point, but the neighborhood may depend on the point. On a compact set, continuity and compactness can sometimes turn these separate local facts into one global bound. In particular, a continuous function that never vanishes on a compact set cannot have values whose absolute values approach zero.

Theorem (A Continuous Nonvanishing Function Is Bounded Away from Zero): Let \(K\) be a nonempty compact metric space, and let \(f:K\to\mathbb{R}\) be continuous. If \(f(x)\neq0\) for every \(x\in K\), then there is an \(m>0\) such that \(|f(x)|\geq m\) for every \(x\in K\).

Proof. Suppose no such \(m\) exists. Since \(|f(x)|>0\) at every point, the failure of a positive lower bound means that for each positive integer \(n\), there is an \(x_n\in K\) such that \(|f(x_n)|<1/n\). By the Compactness and Sequential Compactness in Metric Spaces theorem, the sequence \((x_n)\) has a convergent subsequence \((x_{n_k})\) with limit \(x\in K\). Continuity at \(x\) gives \(f(x_{n_k})\to f(x)\). Since \(|f(x_{n_k})|<1/n_k\) and \(n_k\to\infty\), it follows that \(|f(x_{n_k})|\to0\). The Absolute-Value Limit Theorem therefore gives \(|f(x)|=0\), so \(f(x)=0\). This contradicts the assumption that \(f\) never vanishes on \(K\). Hence a positive lower bound \(m\) exists. \(\square\)

The proof combines two ingredients with different jobs. Compactness supplies a convergent subsequence of points, while continuity transfers the limiting behavior of the function values to the value at the limit point. The contradiction arises because the limit would be a zero of \(f\).

Worked Example: Bounding a Denominator Away from Zero

Consider \(f(x)=x^2+x+1\) on \(K=[-2,2]\). This interval is a nonempty compact subset of the real line, and \(f\) is continuous. Completing the square gives

f(x)=x^2+x+1 =\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \geq\frac{3}{4}.

In particular, \(f(x)\neq0\) on \(K\). The theorem guarantees a uniform positive lower bound; this calculation identifies one explicitly, namely \(m=3/4\). It follows that division by \(f(x)\) is safe throughout the interval, and

\left|\frac{1}{f(x)}\right| =\frac{1}{|f(x)|} \leq\frac{4}{3} \qquad (x\in[-2,2]).

The theorem is useful even when completing the square or finding an explicit minimum is difficult. It provides the existence of a suitable denominator bound without requiring its exact value.

Check the Hypotheses Before Using the Strategy

A positive value at one point gives a local conclusion, not a conclusion over the whole domain. Likewise, pointwise nonvanishing on a noncompact set does not by itself give a uniform positive lower bound. The compactness hypothesis in the theorem is what rules out sequences of points whose function values approach zero without ever reaching zero.

Worked Example: Why Compactness Matters

Define \(f(x)=x\) on \(E=(0,1]\). This function is continuous and positive at every point of \(E\), but it is not bounded away from zero. Indeed, \(x_n=1/n\) belongs to \(E\), and \(f(x_n)=1/n\to0\). For any proposed \(m>0\), choose an integer \(n>1/m\); then \(x_n\in E\) and \(f(x_n)=1/n<m\). Thus no positive number is a lower bound for all the values.

There is no contradiction with the theorem: \(E\) is not compact. The example shows why pointwise positivity and continuity alone are insufficient for a uniform lower bound over an entire domain.

When applying the Continuity Strategy, first identify the precise conclusion. If it is a local strict inequality, locate the positive margin at the point and choose a smaller continuity tolerance. If it is a uniform bound on a whole set, check whether compactness can convert the local information into a global conclusion. These are related uses of continuity, but they require different hypotheses and deliver different kinds of control.

Find the margin.
For a strict inequality at a point, calculate the positive difference between the value and the threshold.

Choose the tolerance.
Use continuity with a function-value tolerance smaller than that difference.

Pass to nearby points.
Combine the continuity estimate with the margin to obtain the desired local inequality.

For a global bound, use compactness.
If values approach a forbidden limit, select a sequence and use a convergent subsequence to contradict continuity or nonvanishing.

The essential habit is to match the strength of the conclusion to the available information. Continuity controls values near a point; compactness can make pointwise control uniform across a compact set. Keeping those roles distinct prevents a local estimate from being used as though it were global.

Check Your Understanding

Use the margin argument and the compactness theorem to answer these questions.

If \(f(a)>c\), what positive number provides the margin for proving \(f(x)>c\) near \(a\)?
Why is the continuity tolerance chosen smaller than the margin in the proof of local persistence?
How does local persistence prove a strict comparison \(f(x)<g(x)\) near a point where \(f(a)<g(a)\)?
Where does compactness enter the proof that a continuous nonvanishing function has a positive lower bound on a compact set?
Why does the function \(f(x)=x\) on \((0,1]\) not contradict the nonvanishing theorem?

What You'll Learn