Piecewise-Linear Perturbations with Steep Secants

Small Changes, Large Secant Slopes

A continuous function can change very little in height under a small perturbation and still acquire a large secant slope. The reason is that a secant slope divides a change in function value by the distance between two inputs: a fixed, small vertical change can produce a large quotient when the inputs are sufficiently close. Here we make this observation precise using piecewise-linear perturbations of functions on the unit interval.

Write \(C([0,1])\) for the set of continuous real-valued functions on \([0,1]\). For \(f,g\in C([0,1])\), the sup norm distance is \(\|g-f\|_\infty=\sup_{t\in[0,1]}|g(t)-f(t)|\). Thus \(\|g-f\|_\infty<\varepsilon\) means that the change in function value is less than \(\varepsilon\) at every input. If \(x<y\), the secant slope of \(g\) between \(x\) and \(y\) is \((g(y)-g(x))/(y-x)\). We will produce a perturbation \(u\) for which \(g=f+u\) is uniformly close to \(f\), while this quotient is greater than a specified number.

The key construction prescribes opposite values for \(u\) at two nearby points. The change in the numerator of the secant quotient is then twice the size of either prescribed value. Linear interpolation connects those values to zero at the endpoints of the domain without making the perturbation any larger.

Controlling the Size of the Interpolation

Proof. Define \(u\) by linear interpolation through the four data points \((0,0)\), \((x,-\eta)\), \((y,\eta)\), and \((1,0)\). Since \(0<x<y<1\), these points specify three intervals of positive length, and the interpolated function is continuous and piecewise linear. On each interval, every value of \(u\) lies between its endpoint values: a linear interpolation has the form \((1-s)v_0+sv_1\) for \(0\le s\le1\), which lies between \(v_0\) and \(v_1\). All four specified values belong to \([-\eta,\eta]\), so \(|u(t)|\le\eta\) everywhere. Since \(|u(x)|=\eta\), the supremum is attained and equals \(\eta\). \(\square\)

This proposition is the size control we need. The important quantity for the secant is not the largest value of \(u\) alone, but the difference \(u(y)-u(x)=2\eta\). The next result uses continuity of the original function to ensure that its own change between the chosen points does not cancel too much of this increase.

The Perturbation Theorem

Proof. Set \(\eta=\varepsilon/2\), so \(0<\eta<\varepsilon\). Choose \(c\in(a,b)\); this is possible because \(a<b\). By continuity of \(f\) at \(c\), there is \(\delta>0\) such that \(t\in[0,1]\) and \(|t-c|<\delta\) imply \(|f(t)-f(c)|<\eta/2\). Choose a number \(d>0\) satisfying \(d<\min\{b-c,\delta,\eta/(M+1)\}\). All three entries in this minimum are positive, so such a \(d\) exists. Set \(x=c\) and \(y=c+d\). Then \(a<x<y<b\), and \(|f(y)-f(x)|<\eta/2<\eta\), by the choice of \(\delta\).

Apply the proposition to \(x,y,\eta\) and let \(u\) be its piecewise-linear function. Since \(\|u\|_\infty=\eta=\varepsilon/2<\varepsilon\), the required norm bound holds. For \(g=f+u\), we have \(u(y)-u(x)=\eta-(-\eta)=2\eta\), and therefore \(g(y)-g(x)=f(y)-f(x)+2\eta>-\eta+2\eta=\eta\). Also \(0<y-x=d<\eta/(M+1)\). Dividing by the positive number \(d\) gives \(\dfrac{g(y)-g(x)}{y-x}>\dfrac{\eta}{d}>M+1>M\). This proves the assertion, including the strict inequality, for every \(\varepsilon>0\), every \(M\ge0\), and every nondegenerate interval \([a,b]\). \(\square\)

The proof has two independent choices. Continuity makes the change \(f(y)-f(x)\) small compared with the prescribed vertical change. The distance \(y-x\) is then chosen small enough that even the guaranteed positive change in \(g\) produces the desired quotient. The perturbation may affect the entire domain, but its sup norm stays controlled because its interpolated values never leave \([-\eta,\eta]\).

Worked Constructions

Why the Construction Matters—and a Common Error

The conclusion is local in its secant condition but global in its norm control: the two points can be chosen inside any prescribed nondegenerate interval, while the perturbation remains small everywhere on \([0,1]\). This gives a concrete way to find functions close to a given continuous function with a large secant on a specified interval. The result concerns existence of a secant, not differentiability at either endpoint or elsewhere.

Common pitfall. In the quadratic example, a wrong calculation might say that the perturbation changes the numerator by \(0.1\), giving a slope \((0.0101+0.1)/0.01=11.01\). This incorrectly counts only one endpoint change. The numerator uses the difference \(g(y)-g(x)\): the value at \(y\) increases by \(0.1\), and the value at \(x\) decreases by \(0.1\). The total increase is \(0.1-(-0.1)=0.2\), so the correct quotient is \((0.0101+0.2)/0.01=21.01\). The sup norm, by contrast, is \(0.1\), not \(0.2\): it measures the largest change at a single input, not the difference between changes at two inputs.