APPROXIMATING CONTINUOUS FUNCTIONS: WEIERSTRASS, BERNSTEIN, AND RUNGE

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1 APPROXIMATING CONTINUOUS FUNCTIONS: WEIERSTRASS, BERNSTEIN, AND RUNGE WILLIE WAI-YEUNG WONG. Introduction This set of notes is meant to describe some aspects of polynomial approximations to continuous functions. It in particular concerns the apparent discord between the Weierstrass Approximation Theorem and Runge s Phenomenon. First let us recall the theorem due to Weierstrass: thm:weierstrass Theorem (Weierstrass Approximation, real-variable version). Let f : [0, ] R be continuous. Then there exists a sequence of polynomials p n that approximates f uniformly on [0,]. That is to say we have () lim sup f (x) p n (x) = 0. x [0,] In particular, the theorem establishes that the polynomials are dense in the set of continuous functions C([0,];R) in the topology of the uniform norm. Note also by re-scaling, the interval [0,] in Theorem thm:weierstrass can be replaced by by any compact interval [a,b]. On the other hand, Runge s example concerns the following function, () R(x) = + x, x [ 5,5]. By Weierstrass s theorem, there exists an approximating sequence of polynomials. Naïvely one may try to construct this sequence by polynomial interpolation. Definition. Let f (x) : [a,b] R be continuous. We denote by I n [f ] the nth order equidistant polynomial interpolation of f. That is to say, we denote by I n [f ] the unique nth order polynomial such that I n [f ](a + i(b a)/n) = f (a + i(b a)/n), i {0,,...,n}. More generally, for each n, let σ n be a subset of [a,b] consisting of n+ distinct points, two of which are a and b. We denote by I σ n [f ] the nth order polynomial interpolation of f relative to the points σ n. That is to say we let I σ n [f ] be the unique nth order polynomial such that I σ n [f ](x) = f (x), x σ n. Note that by definition, there exists a subsequence I n [R] which converges on every rational point in [ 5,5] (let n = + i= p i where p i is the ith smallest prime Version rev358 of :00: (Fri, 9 Nov 03).

2 W. W.-Y. WONG number). But as it turns out, for the function R(x), we not only have a lac of uniformity in the convergence of I n [R] to R, we have in fact that lim sup sup x [ 5,5] I n [R](x) R(x) = +. As we shall see below, the method of polynomial interpolation is not particularly good for constructing the approximation sequence. In turns out that a better approximation can be obtained explicitly using Bernstein polynomials (incidentally we can directly prove Theorem thm:weierstrass using these polynomials).. Weierstrass approximation through Bernstein polynomials We first recall the Bernstein polynomials. Definition 3. By B,n (t) we denote the th Bernstein polynomial of order n. They can be explicitly given by ( ) n (3) B,n (t) = t ( t) n. We note that they form a partition of unity: n B,n (t) = [t + ( t)] n =. Definition 4. Let f : [0,] R be continuous. Denote by B n [f ] the Bernsteinpolynomial approximation of f, which we define as B n [f ](x) = n f (/n)b,n (x). By construction B n [f ] is a polynomial of degree at most n. It is easily checed that since B,n (0) = δ,0 = B n,n (), that B n [f ](0) = f (0) and B n [f ]() = f (). We claim that eq:bernsteinconv (4) lim sup B n [f ](x) f (x) = 0 x [0,] which would imply Theorem. thm:weierstrass Proof of ( 4). eq:bernsteinconv Since B,n =, we have that B n [f ] is a weighted average. We will show that as n grows, the weight heavily favours the points where /n x. First we compute a few things about B,n (x). Observe that (5) n n B,n(x) = = n ( ) n x ( x) n n n ( ) n n x x ( x) (n ) ( ) n = n ( ) n = x x j ( x) n j j n = x B j,n (x) = x

3 APPROXIMATING CONTINUOUS FUNCTIONS 3 eq:variance (6) So in particular (/n x)b,n(x) = 0. Next we see that n ( ) n ( ) n x B,n(x) = x n + x ( x) n n = = x + x n n n ( = x + n n n x l=0 j n + n xb l,n (x) + n n ) ( ) n x j ( x) (n ) j j B j,n (x) = x + n n x(x + /(n )) = x( x) n Now we are in a position to conclude the proof. Fix an ɛ > 0. Using that f (x) is continuous on a compact interval [0,], we have that f (x) is uniformly continuous, and hence there exists a δ such that f (x) f (y) < ɛ whenever x y < δ. Uniform continuity also implies that f (x) is bounded: f (x) < M. We consider (7) B n [f ](x) f (x) n f (/n) f (x) B n, (x) = /n x <δ + /n x δ f (/n) f (x) B n,(x). For the first sum, by uniform continuity we have that f (/n) f (x) < ɛ, and since B,n = we have that (8) f (/n) f (x) B n, (x) < ɛ. /n x <δ For the second sum, we estimate eq:secondsum (9) f (/n) f (x) B n, (x) /n x δ /n x δ by ( 6). eq:variance So by choosing n > M, we have that δ ɛ B n [f ](x) f (x) ɛ M δ /n x B n, (x) M δ n independent of x, and so we obtained the uniform convergence. An after-note: the above proof can be interpreted using the probabilistic point of view as saying something about the binomial distribution. Observe that if we have a biased coin that lands with probability p heads and probability ( p) tails, the probability of receiving heads after n tosses is exactly B,n (p). So by the Law of Large Numbers, we expect that as n, for a fixed x, the functions B,n (x) will be all mostly close to zero except for the few points where /n x. Hence in the summation B n [f ](x) we expect mostly only contributions from f (/n) where /n x. By continuity this means that B n [f ](x) x. The argument given above maes this intuition rigorous. Notice that also that in the construction of B n [f ], the only points at which B n [f ] is guaranteed to be equal to f are the two endpoints 0 and.

4 4 W. W.-Y. WONG Notice also that in the proof above, the rate of convergence can be quantitatively estimated if we have Lipschitz control on f. That is, suppose we now that f (x) f (y) N x y for all x,y [0,], then we have that δ ɛ/n, and also that ( 9) eq:secondsum can be replaced by N 3 eq:secondsumprime (9 ) f (/n) f (x) B n, (x) ɛ /n x B n, (x) N 3 ɛ n. /n x δ So we have a rate of convergence of at least ɛ n /3. 3. Divergence of interpolation polynomials In this section we will discuss a bit the case of divergence, namely that of the Runge example. A good part of the material is taen from Epperson. The exact behaviour of the Runge example requires a dose of complex analysis. Suppose now that f (x) extends to f (z), a function that is complex analytic in a neighborhood of the subset [a,b] {0} of the complex plane. For the explicit function R(x) we tae R(z) = ( + z ) which is holomorphic away from ±i. Let σ n = {z 0,...,z n } be a collection of distinct points in the region where f (z) is holomorphic. We let I σ n [f ] be the polynomial interpolation of f based on the control points z j. Then we have that f (z) I σ n [f ](z) is holomorphic whenever f is, and vanishes at z 0,...,z n. This means that we can write where (0) w σ n (z) = g σ n (z) def = f (z) Iσ n [f ](z) w σ n (z) n (z z i ) i=0 such that gn σ (z) is holomorphic in the domain of holomorphicity of f (z). Now, let γ be a closed contour, and let z be a point belonging to the domain bounded by γ. Writing Gn σ (ζ) = gn σ (ζ)/(ζ z) and applying to it the residue theorem, we have that πi Gn σ (ζ) dz = gn σ (z) + Res(Gn σ,a ) γ where {a } enumerates the poles of f (z) contained inside the domain bounded by γ. We can rewrite as Gn σ (ζ) dz = g σ n (z) + πi wn σ (a γ ) (a z) Res(f,a ). So finally, after rearranging the terms we get eqn:errorestimate () f (z) In σ [f ](z) = wn σ (z) f (ζ) πi wn σ (ζ) ζ z dz wn σ (z) wn σ (a ) γ Res(f,a ) a z So we see that the ey to understanding the error behaviour as n is through understanding the function w σ n (z). Obviously, the limiting behaviour of w σ n (z) as n would depend on the sequence of sets σ n. James Epperson, On the Runge example, Amer. Math. Monthly 94 (987),

5 APPROXIMATING CONTINUOUS FUNCTIONS 5 Denote by δ σn the counting measure attached to σ n. Let us assume that n+ δ σ n ρ in measure. Then n + log wσ n (z) = n + log z ζ dδ σn (ζ) and hence for each fixed point z outside the support of ρ we have that lim n + log wσ n (z) = log z ζ dρ(ζ). Hence we can interpret the function ( ) eqn:winfty () w (z) σ def = lim w σ n (z) n+ = exp log z ζ dρ(ζ). By differentiating under the integral sign, which we can do as long as we stay away from the support of the measure ρ, we have that w (z) σ is real-valued and smooth. Now, assume that ρ has empty interior: that is to say that for every point y supp(ρ) and every open set V y, V \supp(ρ). Furthermore assume that the function w σ defined by ( ) eqn:winfty extends continuously to the support of ρ. Then since wn σ (z) n+ are continuous functions, we conclude that the point-wise convergence established in ( ) eqn:winfty implies the convergence also on the support of ρ. Now we can discuss the convergence and divergence of the polynomial interpolation. Henceforth z is a fixed point and γ is a closed curve not intersecting the support of ρ (which guarantees that w σ γ > 0), with the property that () z is in the domain bounded by γ; () w σ (z) < inf ζ γ w σ (ζ). Then passing to the limit we have that lim w σ ( n (z) wn σ (ζ) lim w σ ) n+ (z) inf ζ γ w (ζ) σ = 0. Hence ( ) eqn:errorestimate implies that eqn:simplifiederror (3) lim f (z) I σ n [f ](z) = lim wn σ (z) wn σ (a ) Res(f,a ) a z Now, if z is a value such that we can choose γ to satisfy the above properties, and such that for every pole a enclosed by γ the value w σ (z) < w σ (a ), then using the exact same argument as above, we can conclude that I σ n [f ](z) f (z). On the other hand, if for every γ admissible we must include a such that w σ (z) > w σ (a ), then generically the right hand side must blow-up. For the Runge example, the measure ρ coming from the limit of the equidistant interpolation is the Lebesgue measure on the interval [ 5, 5]. Evaluating the integral explicitly we have that at the two end-points w σ (±5) = exp(0ln0 0), while at the midpoint w σ (0) = exp(0ln5 0). For the singularities of R(z) at ±i, we compute 5 ln + x dx = 5ln6 0 + arctan

6 6 W. W.-Y. WONG Observing that 3 0ln0 > 5ln6 + arctan5 9 > 0ln5 6 we see that in the segment x [ 5,5] we have both convergent and divergent points. In fact, we see from the above analysis the following stronger form of Runge s phenomenon: Theorem 5. Let I = [a,b] be an interval. Let σ n a sequence of finite subsets of n + distinct points, two of which are a,b. Assume that n+ δ σ n converges in measure to ρ, a measure on [a,b] that is absolutely continuous with respect to the Lebesgue measure. If we assume further that the associated eqn:infsup (4) inf w σ < sup w σ, I I then we can find an appropriate real analytic function f defined on I such that I σ n [f ] converges to f pointwise on some non-empty subset of I, while diverges on some nonempty subset of I. Setch of proof. By continuity there exists a point z with Iz > 0 such that w (z) σ between the supremum and infimum of w (z) σ on I. Then the function f (ζ) = is complex analytic in a neighborhood of I and taes real values on I. (ζ z)(ζ z) A natural next question to as is: when is ( eqn:infsup 4) voided? This would require dρ = ρ (x) dx where ρ (x) is Lebesgue integrable, such that (5) b a log ξ x ρ (ξ) dξ is constant in x [a,b]. As it turns out, one of the answers is given by Chebyshev nodes. When the limiting density ρ (x) / x (taing, for simplicity, a = b = ), a result of classical potential theory gives that w (z) σ log z + z where the complex function z is defined (branch selection) so that lim z z z =. With this branch chosen, we have that for x R and x, x = x i, and hence the denominator x + x = is constant along the interval. For completeness, we give the computation for the Chebyshev nodes. Taing the transformation x = cosθ, we have that π dx log z x = log z cosθ dθ. x π This condition is not strictly necessary, but this helps avoid cases where the integral defining w σ is divergent.

7 APPROXIMATING CONTINUOUS FUNCTIONS 7 We write cosθ = (eiθ + e iθ ). This prompts us to use the so-called Jouowsi transformation z = (ζ + ζ ) which is defined for every z [,]. This is because then we simplify z cosθ = ζ + ζ e iθ e iθ = ζ e iθ ζ e iθ. We can solve for ζ in terms of z to get ζ = z ± z. We choose the plus sign and the definition of the square root as above, which implies that ζ. The integral then becomes π ζ e iθ ζ e iθ log dθ. π By the properties of the real valued logarithm it suffices to consider the integral π π log ζ e iθ dθ which is physically the potential associated to a uniform distribution of charge on the unit circle. As is well-nown that for ζ this potential is proportional to log ζ. Putting this all together gives us the expression for w σ (z), and taing the limit we get the desired conclusion. 4. Some final remars The discussion above with regards to the Chebyshev nodes show only that they can be used to get good polynomial interpolation of real analytic functions. In practice it turns out that Chebyshev nodes are good for all absolutely continuous functions. On the other hand, once we go to the continuous case there are two interestingly opposing theorems. First, by Weierstrass approximation we now that we have uniform convergence to any continuous function by some sequence of polynomials. As in turns out due to the Chebyshev alternation theorem, we can find one such sequence that intersects the original function in sufficiently many points, showing that this sequence can be constructed by interpolation. However, the nodes here exists a fortiori, and are different for each continuous function. On the other hand, from an argument using the converse of the uniform boundedness principle, one can in fact show that for every set of subsets σ n there exists a continuous function for which the sequence of interpolating polynomials diverges. The Wiipedia article on polynomial interpolation 3 contains many references in its section on convergence properties. École Polytechnique Fédérale de Lausanne, Switzerland address: willie.wong@epfl.ch 3