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Test Functions, Mollifiers and Convolution

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Test Functions, Mollifiers and Convolution Test functions, mollifiers and convolution Timo Rohner The material presented in this document was motivated by the desire to have a rigurous proof for a result that is a very useful tool and makes its appearance in mathematical literature frequently. One such example is the notion of weak derivatives. In particular the result that will be developed throughout this document is used to show that weak derivatives are unique, should they exist. The result that this document sets out to prove can be stated as follows. n 1 R Theorem 1. Given an open set Ω ⊂ R and a function f 2 Lloc(Ω) we have that if f'dx = 0 for all Ω 1 ' 2 C0 (Ω), then f = 0 almost everywhere in Ω. Proving this theorem rigurously requires quite a bit of work. This document attempts to introduce and develop everything needed for a complete proof. We first start out by introducing a relatively simple function and highlighting some of its basic properties. Lemma 2. ( − 1 e x for x > 0; f(x) = 0 for x ≤ 0; is of C1(R) regularity. Proof. Step 1. We first show that for x > 0 the k-th derivative of f is of the following form. (k) 1 f (x) = P ( x )f(x); 1 1 where P ( x ) is a polynomial in x . 1 (0) (1) 1 − x 1 For k = 0 and k = 1 this holds trivially, seeing as f (x) = f(x) and f (x) = x2 e = x2 f(x). By hypothesis of induction, we suppose that for some k = 1; 2; ::: f does indeed take the above form. Computing the k + 1-th derivative of f yields k+1 1 0 1 1 0 f (x) = − x2 P ( x )f(x) + P ( x )f (x) 1 1 0 1 = x2 P ( x ) − P ( x ) f(x) 1 = Q( x )f(x): 1 1 1 0 1 1 Since Q( x ) = x2 P ( x ) − P ( x ) is a polynomial in x we move on to the next step. 1 1 Step 2. We now show that for any polynomial in x , i.e. P ( x ), we have 1 lim P ( x )f(x) = 0: x!0+ Via series definition of the exponential, we can establish the following upper bound for f. 0 ≤ f(x) ≤ xnn!; for any n = 0; 1; 2; :::. 1 1 Given any polynomial in x P ( x ), any n bigger than the degree of P can be used to show that 1 lim P ( x )f(x) = 0. x!0+ Step 3. As a final step we show that f (k)(0) = 0 for all k = 0; 1; 2; :::. This is done via induction. For k = 0 by definition this holds as f (0)(0) = f(0) = 0. Suppose now that f (k)(0) = 0 for some k ≥ 0. (k) (k) (k) By the definition of the derivative, we know that f (k+1)(0) = lim f (t)−f (0) = lim f (t) . In the case t!0 t t!0 t f (k)(t) of t < 0 we trivially have the desired result of lim t = 0. For t > 0 it suffices to remark that t!0− f (k)(t) 1 1 1 1 1 lim t = lim P ( t ) t f(t). By virtue of P ( t ) t being a polynomial in t , the result established in t!0+ t!0+ f (k)(t) Step 2 allows us to directly conclude that lim t = 0, which concludes the proof. t!0+ 1 Lemma 3. There exists a function φ : Rn ! R, for which the following holds true. i) φ(x) ≥ 0 for all x 2 Rn, ii) φ 2 C1(Rn), iii) supp φ = B1(0), iv) R φ dx = 1. n R Proof. We define φ(x) by making use of the function f introduced in Lemma2. 2 φ(x) := cnf(1 − kxk ); n 2 where cn is a positive constant dependent only on n, meaning the dimension of R . kxk is the standard euclidian norm in Rn. To show the first property it suffices to remark that f as defined in Lemma2 is a positive function over R and since cn is a positive constant we conclude that φ ≥ 0. Proving C1 regularity requires a bit more work. φ is a compound function of the polynomial 1−kxk2 and the function f, for which we established C1 regularity in Lemma2. It suffices to use standard rules of differentiation, i.e. applying both the chain- and product-rule in multiple variables, to show the continuity of each partial derivative of φ. 2 The third property is trivial to prove. We know that 8x2 = B1(0) 1 − kxk < 0 and therefore f(x) = 0, which implies x 62 supp φ. Seeing as we have not made use of the constant cn in any way, we can set define cn such that the last property holds. Definition 4. A mollifier is a smooth function ' : Rn ! R, i.e. ' 2 C1(Rn), if the following conditions hold. i) ' is of compact support, ii) R '(x)dx = 1, n R −n x iii) lim '"(x) = lim " '( ) = δ(x), "!0 "!0 " where δ(x) is the Dirac delta function and the limit taken in the schwartz space S(Rn). n 1 n Remark 1. i) The Schwartz space S(R ) contains the space of all test functions Cc (R ), 1 n ii) Cc (R ) is also referred to as the space of bump functions. Definition 5. A mollifier ' is a i) positive mollifier, if ' ≥ 0 in Rn, 1 ii) symmetric mollifier, if '(x) = (jxj), with 2 C (R+). Lemma 6. The function φ, as defined in Lemma3, is a symmetric and positive mollifier. Additionally for any " > 0 Z φ"(x)dx = 1: n R 2 Proof. φ being a mollifier follows directly from the definition of mollifiers and the properties established R x for the function φ in Lemma3. To show that φ"(x)dx = 1 we use a change of variable y = " in the n R equality R φ(x)dx = 1, which follows from Lemma3. n R Remark 2. The function φ as defined in Lemma3 is considered the standard mollifier. We now introduce the notion of convolution. We denote the convolution of two functions f and ' by f ∗ ' and define it as follows. Z Z (f ∗ ')(x) := f(x − y)'(y)dy = f(y)'(x − y)dy: n n R R 1 n n Lemma 7. If f 2 Lloc(R ) and 2 Cc(R ), then f ∗ is continuous. n n Proof. We prove this claim by showing that for any convergent sequence fxmg ⊂ R , xm ! x 2 R the following holds. lim (f ∗ )(xm) = (f ∗ )(x): m!1 1 Before we attempt to prove this, please note that f 2 Lloc allows f to be redefined arbitrarily on any set of measure 0 without changing f ∗ . Therefore we may assume that f is defined everywhere. Since xm converges to x, there exists R > 0 such that fxmg ⊂ BR(x). Let r > 0 such that supp ⊂ Br(0). By continuity of , we have for all y f(y) (xm − y) ! f(y) (x − y): On top of that for any y such that ky−xk > R+r we have ky−xmk ≥ ky−xk−kx−xmk > r+R−R = r, which means that y 62 supp and therefore (xm − y) = 0. This allows us to establish the following bound jf(y) (xm − y)j ≤ sup(j j)jf(y)jX (y) Br+R(x) By the dominated convergence theorem we can integrate, since the function on the righthand side is integrable. Z lim (f ∗ )(xm) = lim f(y) (xm − y)dy m!1 m!1 n R Z = lim f(y) (xm − y)dy m!1 n R Z = f(y) (x − y)dy n R = (f ∗ )(x): 1 n 1 n Lemma 8. Given a function f 2 Lloc(R ) and a function 2 Cc (R ), the following holds @(f ∗ ) @ (x) = (f ∗ )(x); @xi @xi for any i = 1; 2; :::. 3 Proof. For some i we compute the following @(f ∗ ) 1 (x) = lim t (f ∗ )(x + t · ei) − (f ∗ )(x) @xi t!0 Z Z 1 = lim f(y) (x + t · ei)dy − f(y) (x − y)dy t!0 t n n R R Z = lim f(y) (x+t·ei)− (x−y) dy: t!0 t n R @ (x+t·ei)− (x−y) The mean value theorem gives us s(t) such that js(t)j ≤ jtj and (x−y+s(t)·ei) = . @xi t Therefore we have Z @ (f∗ ) (x) = lim f(y) (x+t·ei)− (x−y) dy @xi t!0 t n R Z @ = lim f(y) (x − y + s(t) · ei)dy t!0 @xi n R @ = lim f ∗ x + s(t) · ei t!0 @xi = (f ∗ @ )(x); @xi with the last equality being a consequence of Lemma7. 1 n 1 n 1 n Theorem 9. If f 2 Lloc(R ) and 2 Cc (R ), then (f ∗ ) 2 C (R ). @ (f∗ ) @ @ n Proof. Using Lemma8 we compute (x) = (f ∗ )(x). Since is Cc( ) we can apply Lemma7 @xi @xi @xi R and therefore we have that @ (f∗ ) is continuous. @xi Higher-order derivatives are continuous as well, which follows from repeating the above procedure. 1 n Lemma 10. For any function f 2 Lloc(R ), we have supp(f ∗ φ") ⊂ suppf + B(0) == fx + y j x 2 supp(f); y 2 B"(0)g; where φ is the standard mollifier. Proof. Z Z f"(x) = f(x − y)φ"(y)dy = f(x − y)φ"(y)dy: n R B"(0) Suppose that for some x, we have f"(x) = 0.
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