Density Proof Strategies

From an Error Goal to a Construction

The counterexamples in the previous tutorial showed how a shared constraint can prevent approximation. A density proof takes the opposite approach: for an arbitrary target and an arbitrary positive error tolerance, it constructs a candidate whose error is smaller than that tolerance everywhere on the domain. On a compact interval, a useful way to do this is to sample the target on a fine mesh and connect the sampled values by straight line segments.

This method turns continuity into a global error estimate. Uniform continuity says that values of the target at nearby points are close. A mesh ensures that every point is near the sample points used to define its interpolant. The key proof step is to show that the interpolated value is a weighted average of nearby sampled values, so it cannot drift far from the target value at that point.

Uniform Mesh Interpolation

Let \(a<b\), let \(f\in C[a,b]\), and choose a positive integer \(N\). Divide the interval into \(N\) equal subintervals with nodes

Define \(I_Nf\) to agree with \(f\) at each node and to be linear on every subinterval \([x_i,x_{i+1}]\). More explicitly, if \(x=x_i+th\) for \(0\leq t\leq1\), then

The coefficients \(1-t\) and \(t\) are nonnegative and sum to one. Thus the interpolated value lies between the two sampled values. To measure how close those values must be, define the modulus of continuity of \(f\) by

The supremum is finite because \(f\) is bounded on the compact interval. Uniform continuity of \(f\) means precisely that \(\omega_f(\delta)\) tends to zero as \(\delta\) tends to zero from above. Indeed, for any \(\varepsilon>0\), uniform continuity supplies a \(\delta_0>0\) such that \(|f(u)-f(v)|<\varepsilon\) whenever \(|u-v|<\delta_0\). If \(0\leq\delta<\delta_0\), every pair in the supremum defining \(\omega_f(\delta)\) satisfies this inequality, so \(\omega_f(\delta)\leq\varepsilon\). The converse follows directly from the definition of the supremum.

Proof. Fix \(x\in[a,b]\). If \(x\) is a mesh node, then \(I_Nf(x)=f(x)\), so the error is zero. Otherwise, \(x\) lies in some subinterval \([x_i,x_{i+1}]\) and can be written \(x=x_i+th\) with \(0<t<1\). By the definition of the interpolant and the triangle inequality,

Both distances \(|x-x_i|\) and \(|x-x_{i+1}|\) are at most \(h\). Therefore each difference of function values on the right is at most \(\omega_f(h)\). Since \(1-t\) and \(t\) sum to one, the right-hand side is at most \(\omega_f(h)\). The same bound holds at the mesh nodes, so it holds for every \(x\in[a,b]\). Taking the supremum over \(x\) proves the estimate. \(\square\)

The Density Proof

Proof. Let \(f\in C[a,b]\) and let \(\varepsilon>0\). By the Heine–Cantor Theorem, \(f\) is uniformly continuous on \([a,b]\). Choose \(\eta>0\) such that \(|f(u)-f(v)|<\varepsilon\) whenever \(u,v\in[a,b]\) and \(|u-v|<\eta\). Choose a positive integer \(N\) large enough that \(h=(b-a)/N<\eta\). The interpolation estimate gives

In fact, the continuous function \(|f-I_Nf|\) attains its maximum on the compact interval \([a,b]\) at some point \(x^*\). If \(x^*\) is a mesh node, the error is zero; otherwise, both endpoint differences in the interpolation estimate are less than \(\varepsilon\), since their distances from \(x^*\) are at most \(h<\eta\), and their nonnegative weights sum to one. Thus \(\|f-I_Nf\|_{\infty,[a,b]}<\varepsilon\). The function \(I_Nf\) is continuous and piecewise linear by construction. Thus, for every target \(f\) and every positive tolerance, there is a continuous piecewise linear function within that tolerance. This is exactly density in the supremum norm. \(\square\)

The proof follows a reusable pattern: choose a mesh fine enough to control variation of the target, define a candidate using only nearby data, and prove that the construction preserves the error bound throughout each mesh cell. The uniform interpolation estimate makes the central step explicit; the density theorem then follows by choosing the mesh width sufficiently small.

Worked Examples: Choosing and Using a Mesh

Proof. On a cell, write \(x=x_i+th\). The endpoint distances are \(th\) and \((1-t)h\). Applying the Lipschitz inequality to each term in the interpolation estimate gives an error at most \((1-t)Lth+tL(1-t)h=2Lt(1-t)h\). Since \(2t(1-t)\leq1/2\), this is at most \(Lh/2\). At nodes the error is zero, so the bound holds throughout the interval. \(\square\)

What the Strategy Does—and Does Not—Guarantee

A density proof must control the error at every point, not just at the mesh nodes. Matching the target at finitely many points is not by itself a uniform approximation argument. The interpolation estimate supplies the missing step: it relates the value between nodes to nearby target values and bounds the error across the whole interval.

The compact interval matters because continuity there implies uniform continuity, by the Heine–Cantor Theorem. On a noncompact domain, continuity alone need not provide one mesh scale that controls variation everywhere. The argument also proves density of piecewise linear functions, not density of polynomials; the latter was established earlier by polynomial approximation theorems. Different candidate families require different constructions, even when the overall proof architecture is similar.

For a prescribed error tolerance, the practical choices are clear. First obtain a scale on which the target varies little. Then choose a mesh whose cell width is smaller than that scale. If a Lipschitz constant or another quantitative regularity estimate is available, use it to select the mesh directly. If the target has special structure, as in the corner example, place nodes strategically and check whether the construction is exact.