condition. The latter statement seems intuitively clear but its proof is rather technical. We

will restrict ourselves here to cones with angle θ ∈ (0, π/5], which is actually no restriction

at all.

Lemma 3.12 Let C = C(x0 , ξ, θ, r ) be a cone with angle θ ∈ (0, π/5] and radius r > 0.

De¬ne

r

z = x0 + ξ.

1 + sin θ

Then we have, for every x ∈ C,

r

x’z ¤ .

2

1 + sin θ

32 Local polynomial reproduction

Proof Without restriction we can assume that x0 = 0. Because 0 < θ ¤ π/5 we have

2 cos θ ≥ 1 + sin θ or 2 cos θ /(1 + sin θ ) ≥ 1. This means that

x’z =x +z ’ 2x T z

2 2 2

2 2 2

r2 r

=x + ’2 ξT x

2

(1 + sin θ) 1 + sin θ

2 2

r cos θ

r2

¤ x 2+ ’2 x2

(1 + sin θ) 1 + sin θ

2 2

r2

¤ x 2 ’r x 2 +

(1 + sin θ )2

r2

= x 2( x 2 ’ r ) +

(1 + sin θ )2

r2

¤

(1 + sin θ )2

for every x ∈ C.

Proposition 3.13 Suppose that C = C(x0 , ξ, θ, r ) is a cone with angle θ ∈ (0, π/5] and

radius r > 0. Then C satis¬es a cone condition with angle θ = θ and radius

3 sin θ

r= r.

4(1 + sin θ)

Proof Without restriction we set again x0 = 0. Let

sin θ

r0 := r,

1 + sin θ

and de¬ne

r

z := (r ’ r0 )ξ = ξ.

1 + sin θ

From Lemma 3.7 we see that the ball B(z, r0 ) is contained in the cone C. From Lemma 3.10

we know that we can ¬nd for every point x ∈ B(z, r0 ) a cone with angle π/3 > θ and radius

r0 > (3/4)r0 = r . This means that we can ¬nd a cone for every point inside this ball, and it

remains to show the cone property for points inside the global cone but outside this ball.

Thus, we ¬x a point x ∈ C with x ’ z 2 ≥ r0 . Then we de¬ne the direction of the small

cone to be ζ = (z ’ x)/ z ’ x 2 and we have to show that every point y = x + »0 · with

»0 ∈ [0, r ], · 2 = 1, and · T ζ ≥ cos θ lies in C. De¬ne

1/2

» := z ’ x cos θ + r0 + z ’ x 2 (cos2 θ ’ 1) .

2

2 2

Then » is well de¬ned, because Lemma 3.12 gives z ’ x ¤ r0 / sin θ and this means that

2

r0 + z ’ x 2 (cos2 θ ’ 1) = r0 ’ z ’ x 2 sin θ

2 2 2 2

2

2

r0

≥ r0 ’ sin2 θ = 0.

2

sin θ

2

3.3 Existence for regions with cone condition 33

Furthermore, restricting θ to (0, π/5] gives » ≥ z ’ x 2 cos θ ≥ r0 cos θ ≥ 3r0 /4 = r .

This means that if the point x + »· is contained in C then the convexity of C shows that

the point x + »0 · is also in C, which gives the cone property. This is done by an easy

manipulation:

z ’ (x + »·) = z’x + »2 ’ 2»· T (z ’ x)

2 2

2 2

= z’x + »2 ’ 2» z ’ x 2 · T ζ

2

2

¤ z’x + »2 ’ 2» z ’ x cos θ

2

2

2

= (» ’ cos θ z ’ x 2 )2 ’ z ’ x cos2 θ + z ’ x

2 2

= r0 .

2

Hence x + »· ∈ B(z, r0 ) ⊆ C.

Now we are able to formulate and prove our local version of Corollary 3.9. Again let us

point out that all the constants are in explicit form.

Theorem 3.14 Suppose that ⊆ Rd is compact and satis¬es an interior cone condi-

tion with angle θ ∈ (0, π/2) and radius r > 0. Fix m ∈ N. Then there exist constants

h 0 , C1 , C2 > 0 depending only on m, θ, r such that for every X = {x1 , . . . , x N } ⊆ with

h X, ¤ h 0 and every x ∈ we can ¬nd real numbers u j (x), 1 ¤ j ¤ N , with

N

j=1 u j (x) p(x j ) = p(x) for all p ∈ πm (R ),

d

(1)

N

j=1 |u j (x)| ¤ C 1 ,

(2)

(3) u j (x) = 0 provided that x ’ x j 2 > C2 h X, .

Proof Without restriction we may assume θ ¤ π/5. We de¬ne the constants as

16(1 + sin θ)2 m 2 r

C1 = 2, C2 = , h0 = .

3 sin2 θ C2

Let h = h X, . Since C2 h ¤ r the region also satis¬es an interior cone condition with

angle θ and radius C2 h. Hence, for every x ∈ we can ¬nd a cone C(x) := C(x, ξ, θ, C2 h),

that is completely contained in . By Proposition 3.13, the cone C(x) itself satis¬es a cone

condition with angle θ and radius

3 sin θ

r= C2 h.

4(1 + sin θ )

Moreover, by the de¬nition of C2 ,

sin θ

h=r ,

4(1 + sin θ )m 2

so that we can apply Corollary 3.9 to the cone C(x) and Y = X © C(x) with h = h X, .

Hence, we ¬nd numbers u j (x), for every j with x j ∈ Y , such that

u j (x) p(x j ) = p(x)

x j ∈Y

34 Local polynomial reproduction

for all p ∈ πm (Rd ) and

|u j (x)| ¤ 2.

x j ∈Y