g(x) = P (y) Proof. This is true for n = 0. Assume by the inductive hypothesis that g (n) (0) = 0 for some n. Compute g (n) (h) g (n) (0)

Transcription

1 Mollifiers and Smooth Functions We say a function f from C is C (or simply smooth) if all its derivatives to every order exist at every point of. For f : C, we say f is C if all partial derivatives to every order exist and are continuous. Proposition. The function g(x) = { 0 if x 0 e x if x > 0 is C. This will follow from several lemmas. Note the only thing we need to prove is that g (n) (0) exists for all n. We will show g (n) (0) = 0 for all n. Lemma 2. For x > 0, n 0 the n th derivative g (n) (x) = P ( x )e x for P a polynomial. Proof. This is true for n = 0. Now inductively, if g (n) (x) = P ( x )e x, compute for x > 0 g (n+) (x) = P ( x ) ( x 2 ) e x + P ( x ) e x ( x 2 ). Now note that P ( x ) ( x 2 ) + P ( x ) ( x 2 ) is also a polynomial in x. Lemma 3. If P is a polynomial, then lim P ( x 0 + x )e x = 0. Proof. Mae the substitution y =, and note that x Lemma 4. For all n, g (n) (0) = 0. P (y) lim = 0. y e y Proof. This is true for n = 0. Assume by the inductive hypothesis that g (n) (0) = 0 for some n. Compute g (n+) (0) = lim g (n) (h) g (n) (0) h = lim g (n) (h) h by the inductive hypothesis. For h < 0, it is clear that g (n) (h) = 0, and thus that g (n) (h) lim = 0. h

2 For h > 0, use Lemma 2 to compute g (n) (h) lim + h = lim P ( + h h )e h = 0 by applying Lemma 3 for the polynomial P (y) = yp (y). Since the left and right limits are equal, we find g (n+) (0) = 0 and by induction this holds for all n. The Proposition is proved. Theorem. There is a function φ on which satisfies () φ(x) 0 for all x. (2) φ C ( ). (3) supp φ = B (0). (4) φ dm =. Proof. For g from Proposition above, let { φ(x) = c g( x 2 0 if x ) = c exp( ) if x < x 2 where c is defined so that φ dm =. Then φ is C since φ = c g h for h = x 2 = x 2 x 2. h is C since it s a polynomial. Then we can verify that all the various partial derivatives of φ = c g h are continuous by using the usual rules of differentiation (the chain rule in multiple variables, the product rule, etc.) to compute them. A mollifer is a family of functions φ δ based on φ given by φ δ (x) = δ φ( x ) for all δ > 0. Then it is easy to chec that δ Proposition 5. For all δ > 0, () φ δ (x) 0 for all x. (2) φ δ C ( ). (3) supp φ δ = B δ (0). (4) φ δ dm =. Proof. All of these properties are obvious except the last one. For the last one, we use the change of variables formula in multiple integrals to prove φ δ (x) dm(x) = δ φ( x) dm(x) = δ φ(y) dm(y) =, by using using the substitution y = x, which implies dm(y) = δ δ dm(x) for δ the Jacobian determinant. 2

3 A function f is mollified by convolution with φ δ. Define f δ (x) = (f φ δ )(x) = f(x y)φ δ (y) dm(y) = f(y)φ δ (x y) dm(y), as long as the integrals converge. In this case, note that the integrals in the line above are equal by maing the substitution z = x y, which implies dm(z) = dm(y) (the multiplicative factor is the absolute value of the Jacobian determinant, which is ). The idea is that f δ (x) is a weighted, smoothed average all the values of f in the ball B δ (x) of radius δ and center x. To see this, we mae an analogous construction using the function α(x) = m(b (0)) χ B (0)(x). Then α satisfies the same properties as φ in Theorem except it is not smooth. Then if we define α δ (x) = δ α( x ), the convolution δ (f α δ )(x) = f(y)α δ (x y) dm(y) = m(b δ (x)) f(y) dm(y) B δ (x) is the average value of f over the ball B δ (x) of radius δ and center x. Since φ δ 0 and has integral over its support B δ (0), we can consider (f φ δ )(x) to be a weighted average of f over B δ (x). We say a complex function f on is locally L if χ K f L for every compact subset K of. It is an easy consequence of Hölder s inequality that every f L p ( ) is locally L for p. Compute for q the conjugate exponent of p: fχ K dm = f dm f L p (K) L q (K) f L p ( ) m(k) q < K Theorem 2. If f : C is locally L and ψ : C is C with compact support, then f ψ is C. In order to prove this theorem, we need a few lemmas: Lemma 6. If f L loc ( ) and ψ C c ( ), then f ψ is continuous. Proof. Let x n x be a convergent sequence. The lemma is proved if we can show (f ψ)(x n ) = f(y)ψ(x n y) dm(y) f(y)ψ(x y) dm(y) = (f ψ)(x). Note that the definition of f ψ is unchanged if f is redefined on a set of measure 0. Thus, without loss of generality, assume f is defined everywhere. 3

4 We may assume x n B (x). Choose r > 0 so that supp ψ B r (0) (since ψ has compact support). Since ψ is continuous, we have for all y, f(y)ψ(x n y) f(y)ψ(x y). Moreover, if C = sup ψ <, then for all x n, we have f(y)ψ(x n y) C f(y) χ Br+ (x) (y). (To show the last term is justified, note that if y > r +, then x n y y x n > (r + ) = r and so ψ(x n y) = 0.) By assumption, the function on the right-hand side is integrable. So the Dominated Convergence Theorem applies to show that lim (f ψ)(x n) = lim f(y)ψ(x n y) dm(y) n = lim f(y)ψ(x n y) dm(y) n = f(y)ψ(x y) dm(y) n = (f ψ)(x). Lemma 7. If f L loc ( ) and ψ C c ( ), then for all i =,...,, ψ (f ψ) = f xi x. i Proof. We may assume ψ is real, since we may otherwise consider the real and imaginary parts. Compute at x, for e i the i th coordinate vector (f ψ)(x) = lim x [(f ψ)(x + he i h i) (f ψ)(x)] [ = lim f(y)ψ(x + he i y) dm(y) h ] f(y)ψ(x y) dm(y) = lim f(y) h = lim f(y) ψ x (x y + c(h)e i) dm(y), i ( ψ(x + hei y) ψ(x y) 4 ) dm(y)

5 where we use the Mean Value Theorem to find a value c(h) so that c(h) h (here we use that ψ is real). Continue to compute at x x i (f ψ) = lim = ( f ψ x i ( f ψ x i ) (x) ) (x + c(h)e i ) 5 where we apply Lemma 6 to tae the last limit. Proof of Theorem 2. Apply Lemma 7 to compute the first partial derivative ψ (f ψ) = f xi x. i This function is continuous by Lemma 6 and since ψ C x i c ( ). The higher-order partial derivatives can be handled by induction. Lemma 8. If f is locally L and f δ = f φ δ is the standard mollifier, then Proof. If suppf δ suppf + B δ(0) = {x + y x suppf, y Bδ(0) }. 0 f δ (x) = f(x y)φ δ (y) dm(y) = B δ (0) f(x y)φ δ (y) dm(y), then f(x y) cannot be identically zero for all y B δ (0). So for some such y, x y supp f, and thus x = (x y) + y supp f + B δ (0). Theorem 3. If f : C is continuous, and φ δ is the standard mollifier, then f δ = f φ δ f uniformly on compact subsets of. Proof. Let K be compact. Then there is an r > 0 so that K B r (0). On the compact set B r+ (0), f is uniformly continuous. Choose ɛ > 0. There is an η > 0 so that if z, w B r+ (0), then z w < η implies f(z) f(w) < ɛ. We may additionally assume η <. Let δ (0, η).

6 For x K B r (0), compute f δ (x) f(x) = f(x y)φ δ (y) dm(y) f(x) = f(x y)φ δ (y) dm(y) f(x)φ δ (y) dm(y) f(x y) f(x) φ δ (y) dm(y) = f(x y) f(x) φ δ (y) dm(y) < B δ (0) B δ (0) ɛ φ δ (y) dm(y) = ɛ The last line is justified since x B r (0) B r+ (0) and x y B r (0)+ B δ (0) B r+ (0). Since this estimate applies to all x in the compact set K, we have that f δ f uniformly on K. Corollary 9. If f C c ( ), then f δ f uniformly on. Theorem 4. For p <, C c ( ) is dense in L p ( ). Proof. Let f L p ( ) and ɛ > 0. Then Theorem 3.4 of udin implies there is a g C c ( ) so that f g p < ɛ. 2 Now let g δ = g φ δ for φ the standard mollifier. Then Corollary 9 implies g δ g uniformly as δ 0. Assume supp g B r (0) for some r > 0. Then supp g δ B r (0) + B δ (0) = B r+δ (0). Compute g δ g p p = g δ g p dm = g δ g p dm sup g δ g p m(b r+δ (0)). B r+δ (0) Now sup g δ g 0 since g δ g uniformly. Thus g δ g p 0 and there is a δ > 0 so that g δ g p < ɛ. 2 Therefore, f g δ p f g p + g g δ p < ɛ. g δ Cc ( ) since supp g δ supp g + B δ (0) is compact. g δ is C by Theorem 2. A similar but more precise result is Theorem 5. For any p < and f L p ( ), then f φ δ f p 0 as δ 0, where φ is any nonnegative measurable function on with total integral one. The proof is essentially contained in Theorems 9.5, 9.9 and 9.0 in udin. Proposition 0. Let f L p ( ) for p <, and let y. Let f y (x) = f(x y). Then the map y f y is a continuous map from to L p ( ). 6

7 Proof. Let ɛ > 0. By udin, Theorem 3.4, choose a continuous function g with compact support so that f g p < ɛ. Let > 0 be so that supp g B (0). Since g is continuous with compact support, it is uniformly continuous. Thus there is a δ (0, ) so that s t < δ implies g(s) g(t) < m(b 0 (2)) p ɛ, for m Lebesgue measure on. Then compute for s t < δ g s g t p p = g(x s) g(x t) p dx < m(b 2 (0)) ɛ p m(b +δ (s)) < ɛ p. Whenever s t < δ, f s f t p f s g s p + g s g t p + g t f t p = (f g) s p + g s g t p + (g f) t p = f g p + g s g t p + g f p < 3ɛ. Here we have used the change of variables z = x s for α = f g to compute α s p p = α(x s) dx = α(z) dz = α p p. Proof of Theorem 5. We need to prove that lim f φ δ f p = 0. δ 0 As in the proof of Theorem 3 above, we have (f φ δ )(x) f(x) f(x y) f(x) φ δ (y) dm(y). Since φ δ is a positive function with integral one, we may apply Jensen s Inequality for the convex function t t p to find ( ) p (f φ δ )(x) f(x) p f(x y) f(x) φ δ (y) dm(y) f(x y) f(x) p φ δ (y) dm(y). 7

8 Now we may integrate this inequality over in x and use Fubini s Theorem to find f φ δ f p p f(x y) f(x) p φ δ (y) dm(y) dm(x) = f(x y) f(x) p dm(x) φ δ (y) dm(y) = f y f p p φ δ (y) dm(y). Let g(y) = f y f p p. Proposition 0 above shows g is a continuous function, and it is clear that g(0) = 0. Moreover, g is bounded since g(y) = f y f p p ( f y p + f p ) p = (2 f p ) p. We continue computing to find f φ δ f p p g(y)φ δ (y) dm(y) = g(y)δ φ(yδ ) dm(y), = g(δs)φ(s) dm(s) for the change of variables s = yδ. As δ 0, g(δs)φ(s) g(0)φ(s) = 0 pointwise on. Moreover, the integrand g(δs)φ(s) g φ(s) for all δ. Since g is bounded and φ is integrable, the Dominated Convergence Theorem applies to show that lim g(δs)φ(s) dm(s) = 0. δ 0 This implies f φ δ f p 0 as δ 0. 8