Dense Open Sets of Functions with Steep Secants

Steep Secants as Open Dense Conditions

A continuous function can be uniformly close to a given function while changing rapidly over a very short interval. That observation lets us study steep secants as conditions on an entire space of functions, rather than on a single graph. We will prove that, in any fixed rational subinterval of \([0,1]\), the condition of having a secant whose slope magnitude exceeds a prescribed positive integer is both open and dense in the sup-norm topology.

Let \(C([0,1])\) denote the set of continuous real-valued functions on \([0,1]\). We use the metric induced by the sup norm \(\|f\|_\infty=\sup_{t\in[0,1]}|f(t)|\), so the distance between \(f\) and \(g\) is \(\|f-g\|_\infty\). A rational subinterval will mean an interval \(J=(p,q)\), where \(p,q\in\mathbb{Q}\), \(0\le p<q\le1\). For distinct \(x,y\in[0,1]\), the secant slope of \(f\) between them is \((f(y)-f(x))/(y-x)\); we measure steepness by its absolute value.

The two parts of the proof use different facts. Openness follows because a secant that exceeds \(n\) by a positive margin continues to exceed \(n\) after sufficiently small uniform perturbations. Density follows by adding a bump of small height and sufficiently narrow support. Its height controls the distance moved in the sup norm, while its width controls the secant slope.

Why Each Steep-Secant Set Is Open

Proof. Let \(f\in U_{J,n}\). By definition, there are \(x,y\in J\) with \(x<y\) such that \(S=\left|(f(y)-f(x))/(y-x)\right|>n\). Write \(d=y-x>0\), and set \(\eta=d(S-n)/2>0\). Suppose \(g\in C([0,1])\) satisfies \(\|g-f\|_\infty<\eta\). The reverse triangle inequality gives $$ \left|\frac{g(y)-g(x)}{y-x}\right| \ge \left|\frac{f(y)-f(x)}{y-x}\right| -\frac{|g(y)-f(y)|+|g(x)-f(x)|}{d} >S-\frac{2\eta}{d}=n. $$ The same pair \(x,y\) therefore witnesses \(g\in U_{J,n}\). We have found an open sup-norm ball about each \(f\in U_{J,n}\) that is contained in \(U_{J,n}\), which proves openness. \(\square\)

The estimate depends on the separation \(d\) of the witnessing points. If the points are close together, a small uniform change can have a larger effect on that secant. This does not obstruct openness: for the particular witness, \(d>0\) and the margin \(S-n>0\), so the radius \(\eta\) is positive.

A Small Perturbation Creates a Steep Secant

For density, start with any \(f\in C([0,1])\), any positive tolerance \(\varepsilon\), any rational subinterval \(J=(p,q)\), and any \(n\in\mathbb{N}\). We need to find a function in \(U_{J,n}\) within sup-norm distance \(\varepsilon\) of \(f\). Choose a center \(c\) strictly inside \(J\), and add a triangular bump whose peak is at \(c\). By continuity of \(f\), the original function changes very little between \(c\) and a sufficiently nearby point to its right. The bump can then dominate that change, while its narrow width makes the resulting secant steep.

Proof. Fix \(f\in C([0,1])\) and \(\varepsilon>0\). Write \(J=(p,q)\), choose \(c=(p+q)/2\), and let \(\delta=\varepsilon/2>0\). Continuity of \(f\) at \(c\) gives a positive number \(\rho\) such that, whenever \(t\in[0,1]\) and \(|t-c|<\rho\), we have \(|f(t)-f(c)|<\delta/4\). Choose \(r>0\) small enough that \(r<\rho\), \(c+r<q\), \(c-r>p\), and \(nr<\delta/4\). In particular \(c,c+r\in J\).

Define $$ h(t)=\delta\max\left(1-\frac{|t-c|}{r},0\right),\qquad g(t)=f(t)+h(t)\quad (t\in[0,1]). $$ The function \(h\) is continuous, since it is formed from continuous functions by taking a maximum, and hence \(g\) is continuous. The quantity \(\max(1-|t-c|/r,0)\) lies between \(0\) and \(1\), and equals \(1\) at \(c\). Therefore \(\|h\|_\infty=\delta\), so \(\|g-f\|_\infty=\delta<\varepsilon\). Also \(h(c)=\delta\) and \(h(c+r)=0\). It follows that $$ g(c+r)-g(c)=f(c+r)-f(c)-\delta. $$ Since \(|f(c+r)-f(c)|<\delta/4\), the reverse triangle inequality yields \(\left|g(c+r)-g(c)\right|>3\delta/4\). The secant magnitude between \(c\) and \(c+r\) is consequently greater than \(3\delta/(4r)\). Our choice \(nr<\delta/4\) gives \(n<\delta/(4r)<3\delta/(4r)\). This secant is therefore greater than \(n\), and \(g\in U_{J,n}\). We have found such a \(g\) within \(\varepsilon\) of every \(f\), proving density. \(\square\)

Countably Many Conditions and a Common Pitfall

There are countably many rational subintervals \(J\) and countably many positive integers \(n\). Thus the sets \(U_{J,n}\) form a countable family of open dense subsets of \(C([0,1])\). The space \(C([0,1])\) with the sup-norm metric is complete, a standard result from the earlier material on the sup-norm space. The complete-metric Baire theorem says that in a complete metric space, every countable intersection of open dense sets is dense. Applying that theorem, \(\bigcap_{J,n}U_{J,n}\) is dense. Its members have secants of magnitude greater than each prescribed integer within every rational subinterval. This is the category-theoretic payoff of proving openness and density separately.

Common pitfall. If one writes \(h(t)=\delta(1-|t-c|/r)\) without the maximum with \(0\), the claim \(\|h\|_\infty=\delta\) can fail. For example, take \(c=0.5\), \(r=0.1\), and \(\delta=0.02\). At \(t=0\), the factor is \(1-|0-0.5|/0.1=1-5=-4\), so \(h(0)=-0.08\). Hence \(\|h\|_\infty\ge0.08>0.02\). The corrected definition \(h(t)=\delta\max(1-|t-c|/r,0)\) gives \(h(0)=0\), keeps all values between \(0\) and \(\delta\), and still has peak value \(\delta\) at \(c\). The truncation is essential both for the support and for the claimed perturbation bound.

The two proofs also illustrate why the strict inequality in the definition matters. A witnessing secant has a positive margin above \(n\), which provides a positive openness radius. For density, choosing \(r\) so that \(nr<\delta/4\) makes the constructed secant strictly exceed \(n\), rather than merely reach it.