A IMPROVEMENT OF LAX S PROOF OF THE IVT

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1 A IMPROVEMEN OF LAX S PROOF OF HE IV BEN KORDESH AND WILLIAM RICHER. Introduction Let Cube = [, ] n R n. Lax gave an short proof of a version of the Change of Variables heorem (CV), which aylor clarified using Stokes s theorem on rectangles [Lax99, ay2], and then deduced an Intermediate Value heorem (IV). We believe the best exposition of their amazing work is our proof a rectangular version of Lax s IV: heorem.. ake y in the interior of Cube. here is no continuous function φ: Cube R n {y} which is the identity on Cube. his result is close to π n (R n {}), a basic result of homotopy theory, and we believe we have the simplest proof of it here. Milnor s beautiful proof [Mil65] e.g. requires Sard s theorem, whose proof is much harder than our proofs here. heorem. follows easily from heorem.2. here is no continuous function φ: Cube R n {} which is the identity on Cube. We deduce this, as Lax does, via smoothing theory and a C result: heorem.3. Let O R n be an open set containing Cube. here is no C function φ: O R n {} which is the identity on an open set containing Cube. Lax s used mollification and convolution for his smoothing theory, but as we could not find an elementary reference for it, we will use a generalization of the Weierstrass Approximation heorem [Kor7]. heorem.3 follows from heorem 2. and Lemmas.5 and.4 below. Define a smooth function f : R n {} R n by f(z) = (/ z )z. Let θ be the smooth (n )-form on R n defined by θ z (v 2,..., v n ) = det(z, v 2,..., v n ). Consider the smooth (n )-form ω = f θ on R n {}. We need: Lemma.4. For all z R n {} and v 2,..., v n ω z (v 2,..., v n ) = (/ z ) n det(z, v 2,..., v n ). R n, we have

2 2 BEN KORDESH AND WILLIAM RICHER Lemma.5. he integral of our (n )-form ω over the boundary of Cube is nonzero. More precisely, () ω ι = 2n ( + x x 2 n) n/2 dx 2 dx n Cube [,] n hanks to Jared Wunsch for pointing out Lax s and aylor s papers and for much valuable encouragement. 2. Stokes s theorem on rectangles Stokes s theorem on rectangles is proved by Lang [Lan83, Ch. XXI 2]: heorem 2.. For a closed rectangle in R n and a C differential (n )-form ω on an open set O R n containing, we have (2) dω = ω ι. We give a leisurely proof of this here, partly because Lang does not state the result so precisely. First we make sense of ω ι in (2). A k-form α on a subset S R n is a function α: S Λ k (R n ) from S to the set of alternating k-tensors on R n, which has a standard basis of size ( n k). hus given another subset U R n and a function f : U S, we can take the composite α f : U Λ k (R n ) which is a k-form α f on U. his defines the (n )-form ω ι on. We now define the integral on the LHS of (2). ake a C differential n-form β defined on (an open set containing). We use the notation dx {,...,n} = dx dx n dx {,...,n} i = dx dx i dx n for all i {,..., n}. We can write β = h dx {,...,n} for a unique C function h: R. hen we define the differential form integral to be the Riemann integral β := h. We can define h directly as h(x) = β x (δ,..., δ n ), for all x, using the standard basis {δ i } of R n. Now we look at the RHS. ake a rectangle = n [a i, b i ] in R n and a differential (n )-form α on. For i {,..., n}, define the (n )-dimensional rectangle i = [a, b ] [a i, b i ] [a n, b n ],

3 RECANGLE IV 3 i.e. the product of the intervals of except [a i, b i ]. Given i {,..., n} and c {a i, b i }, we define the injective function I i,c : i by I i,c (x,..., x n ) = (x,..., x i, c, x i+,..., x n ), n so = I i,ai,( i ) I i,bi ( i ). Now we define the differential form integral over the boundary of a rectangle to be an alternating sum of Riemann integrals: Definition 2.2. For an (n )-form α = n Q i dx {,...,n} i on, α := ( ) i Q i I i,bi Q i I i,ai. i Note that in the definition above, Q i : R, and for all x, Q i (x) = α x (δ,..., δ i, δ i+,..., δ n ). Proof of heorem 2.. Write ω = n P i dx {,...,n} i, for C functions P i : O R. So dω = h dx {,...,n}, where h = n ( )i P i / x i, and dω = h = ( ) i P i ι I i,bi P i ι I i,ai, i by Fubini s theorem and the Fundamental heorem of Calculus. Since ω ι = n (P i ι) dx {,...,n} i, we obtain through Definition 2.2 dω = ω ι. 3. Proofs Proof of Lemma.4. By definition, ω z (v 2,..., v n ) = f θ z (v 2,..., v n ) = det(f(z), df z v 2,..., df z v n ). We ll repeatedly use the skew n-linear properties of det. We know ω z (v 2,..., v n ) = (/ z ) det(z, df z v 2,..., df z v n ). o calculate this, first take z R n {} and v R n, and write v = az + π(v) where a R and π is the orthogonal projection onto z. Clearly df z z =. We ll show that df z v = df z π(v) = π(v). First write π(v) = bw for b R and w R n with w = v. hen define the curve γ(t) = cos(t)z+sin(t)w so γ() = z and γ () = w. Since γ(t) = z for all t, we have (f γ)(t) = (/ z )γ(t). So df z w = (f γ) () = (/ z )w, and hence by linearity, df z π(v) = (/ z )π(v). Hence ω z (v 2,..., v n ) = (/ z n ) det(z, π(v 2 ),..., π(v n )) = (/ z n ) det(z, v 2,..., v n ),

4 4 BEN KORDESH AND WILLIAM RICHER since v i = a i z + π(v i ) and since the determinant doesn t change when we add multiples of the first column to other columns. Before proving Lemma.5, we discuss our (n )-form ω on R n {}. We write ω in standard form as above ω = n P i dx {,...,n} i, where P i (x) = ( ) i x i / x n, since by Lemma.4, ω x (δ,..., δ i, δ i+,..., δ n ) = (/ x n ) det(x, δ,..., δ i, δ i+,..., δ n ). hus (P i ι I i,c )(x) = ( ) i c(c 2 + x 2 ) n/2, for c {, }, so (3) (P i ι I i, )(x) (P i ι I i, )(x) = 2( ) i ( + x 2 ) n/2. Simple calculus proves that P i / x i (x) = ( ) i ( x 2 nx 2 i )/ x n+2, for all x R n {}. hus dω = ( ) i ( P i / x i ) dx {,...,n} = h dx {,...,n}, where h(x) = n ( x 2 nx 2 i )/ x n+2 =. hus dω =. Proof of Lemma.5. We re integrating over the boundary of the rectangle Cube = [, ] n. So for all i {,..., n}, i = [, ] n, a i = and b i =. By Definition 2.2 applied to α = ω ι and (3), ω ι = ( ) [,] i P i ι I i, P i ι I i, n Cube = ( ) [,] i 2( ) i ( + x 2 ) n/2 dx dx n n = 2n ( + x 2 ) n/2 dx dx n. [,] n Proof of heorem.3. Assume for the sake of contradiction there is a C function φ: O R n {}, for some open set O R n containing Cube, which is the identity on an open set containing Cube. Define the (n )-form α = φ ω on O. hen dα = dφ ω = φ dω =, since dω =. By definition, α x (v 2,..., v n ) = ω φ(x) (dφ x (v 2 ),..., dφ x (v n )). hen for all x Cube, φ(x) = x and dφ x is the identity, and hence α x = ω x. hus α ι = ω ι, for the inclusion map ι: Cube R n {}. By Stokes s theorem on rectangles applied to α and Cube we have = dα = ω ι Cube Cube by Lemma.5, and we have a contradiction.

5 RECANGLE IV 5 4. Continuous IV We will call the identity function id. We have been writing x for the Euclidean norm of x R n, and will write E r (a) for the closed r ball around a in the Euclidean norm. We will write x for the max norm, and B r (a) for the closed r ball around a in the max norm. Note that x x and x n x for all x R n. hus E r (a) B r (a) and B r (a) E nr(a), for all a R n and r >. Note Cube = B (), and Cube = {x R n : x = } is the unit sphere in the max norm. Proof of heorem.2. Let φ satisfy the assumptions of the heorem. Since φ is continuous and B () is compact, there exists ɛ > such that φ(x) ɛ for all x B (). Since φ is the identity on B (), we may extend φ to R n by defining a continuous function { φ(x) if x, α(x) = x if x. Define β : R n R n {} by β(x) = (/k)α(kx), for a positive integer k 3 n. Clearly β also satisfies the assumptions of the theorem, and equals id outside of B /k (). Let γ = β id. By [Kor7, hm. 4.3], there exists a C function η : R n R n such that γ(x) η(x) < ɛ/2 for all x B (). By [Lee, Lemma. 2.4], there exists a C (bump) function P : R n [, ] such that P (x) = if x /3, P (x) if /3 x 2/3, P (x) = if 2/3 x. Let ζ = P η + id. If x /3, then β(x) ζ(x) = γ(x) P (x)η(x) = γ(x) η(x) < ɛ/2. Note B k () E /3 () and β = id outside of B k (). If /3 x 2/3, β(x) ζ(x) = P (x)η(x) = P (x) η(x) η(x) = γ(x) η(x) < ɛ/2. If 2/3 x, then β(x) ζ(x) = γ(x) P (x)η(x) =. hus β ζ < ɛ/2. By the triangle inequality, ζ > ɛ/2 on B (). Since ζ = P η + id and E 2/3 () B 2/3 (), we know ζ = id outside of B 2/3 (), and thus ζ = id on an open neighborhood of B (). Since ζ is C, by heorem.3 we have a contradiction. Now we use a common technique, the homotopy extension property:

6 6 BEN KORDESH AND WILLIAM RICHER Proof of heorem.. Assume, for a contradiction, that φ: B () R n {y} satisfies the assumptions. Define h: B () [, ] R n {} by h(x, t) = x ( t)y. hen define g : B () R n {} by { φ(2x) y if x /2, g(x) = By heorem.2, g does not exist. h( x, 2 x ) if x /2. x 5. some Lemma.5 calculations For n = 2, the () integral is easy: /( + y2 ) dy = 2 tan () = π/2. hus ω ι = 4π/2 = 2π. [,] 2 For n = 3, we calculate the () integral. First, /( + x 2 + y 2 ) 3/2 dxdy = 4 Now ( + x 2 + y 2 ) dx = x 3/2 ( + y 2 ) + y 2 + x 2 But as we learned from Mathematica, ( ) ( + y 2 ) y dy = tan 2 + y2 2 + y 2 hus /( + x 2 + y 2 ) 3/2 dxdy. = ( + y 2 ) 2 + y 2 = tan ( 3 ) = π/6. [,] 3 ω ι = (2 3) 4 π/6 = 4π. hese are well-known integration on manifolds results. We would like to calculate the () integral for all n to be the volume of S n. References [Kor7] B. Kordesh, A Proof of the Weierstrass Approximation heorem for R k Using Binomial Distributions, 6 pages, available at northwestern.edu/~richter/bernsteindistribution.pdf, 27. [Lan83] S. Lang, Undergraduate Analysis, Undergraduate exts in Math., Springer, New York, 983. [Lax99] P. Lax, Change of Variables in Multiple Integrals, Amer. Math. Monthly 6 (999), 497 5, available from [Lee] J. Lee, Introduction to Smooth Manifolds, Springer, New York, 2. [Mil65] J. Milnor, opology from the Differential Viewpoint, Univ. Va. Press, Charlottesville, 965. [ay2] M. aylor, Differential Forms and the Change of Variable, Journal of Math. Analysis and Applications 268 (22),