Mollifiers and its applications in L p (Ω) space

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1 Mollifiers and its alications in L () sace MA Shiqi Deartment of Mathematics, Hong Kong Batist University November 19, 2016 Abstract This note gives definition of mollifier and mollification. We illustrate the details of how to construct a smooth function that equals 1 in a comact subset and has comact suort, by using the mollifiers. We show that the mollification of a continuous function converges to this function uniformly in any comact subset. Then we show that C 0() is dense in L () when 1 < +, and our roof is also alicable for oen but unbounded. Finally, we rove that every f L () (1 < + ) can be aroximated by its mollification, but we should state these aroximation results carefully with resect to the domain where they are defined. An Aendix is given to show some details. Through out this note, unless otherwise indicated, we assume that 1 < + and R n be a oen, but not necessarily bounded set. 1 Mollifiers If R n is oen and ɛ > 0, we write: ɛ {x dist(x, ) > ɛ}. Definition 1 (Mollifiers [1] ). (1) Define η C () by { 1 Ce x η 2 1, if x < 1 0, otherwise. where C is selected such that R n η(x) dx 1. (2) For each ɛ > 0, set η ɛ (x) 1 ɛ n η(x ɛ ). We call η the standard mollifier. The functions η ɛ are C and su η ɛ B(0, ɛ). So η ɛ C 0 () 1. Then we can mollify functions by mollifiers. Definition 2 (mollification [1] ). If f : R n R is locally integrable, define its mollification f ɛ as f ɛ (x) η ɛ (x y)f(y) dy η ɛ (y)f(x y) dy, x ɛ, (1) where ɛ {x dist(x, ) > ɛ}. (2) We should note that the domain of f ɛ is ɛ. 1 C 0 () {f C() su f }. 1

2 2 Smooth function Unbounded oen set can always contain a bounded oen set, so we only consider bounded situation. Firstly, we resent a theorem, and then we give an illustrative exlanation. Theorem 1. Let oen set R n be bounded. Then for any comact subset K, f C 0 () such that f 1 in K and 0 f(x) 1, x. We illustrate the oint using a icture Figure 1 and Figure 2. Figure 1: mollifier not smooth somewhere In Figure 1, the blue line reresent function f that we are going to mollify. f is obviously comactly suort with resect to the interval (, ) (in the one-dimension examle we regard as to a real number, so as to K, 1, etc), and we only focus on the circumstance around. f is comactly suort but not smooth, so we shall mollify it to get the f ɛ that is: not only comactly suort, but also smooth. We reresent the suort of the mollifier η ɛ by the black ellise. Therefore, in order to guarantee that f ɛ (x) equals 1 when x < K, ɛ must satisfies: ɛ < 1 K. Note that the right-most oint of domain of f ɛ is ɛ, not. So in order to guarantee f ɛ reach 0 within its domain, ɛ must satisfies: ɛ < 1 2 ( 1). Now we choose ɛ < min{ 1 K, 1 2 ( 1)}, and we can get f ɛ as the read line in Figure 1. ab segment equals 1, bc segment is decreasing from 1 to 0, and f ɛ is smooth from in ac segment. However, bd segment may be not smooth at oint x 2. In Order to circumvent this roblem, we should shrinkage ɛ to be smaller. Let s see Figure 2. Searate 1 into three ieces: 1 2, 2 3 and 3. Let ɛ < 1 3 ( 1) we see what haens. ab, bc segments of f ɛ is the same. After we mollify f to oint 2, we can continue mollifying it u to 3 such that f ɛ 0 in segment cd. Now because f ɛ from a to d are all results of mollifier, so it is smooth from a to d, including at b and at c. In 3, we extend f ɛ to be 0 in 3, then f ɛ is of course smooth in segment de. Now, f ɛ is all smooth from a to e, only if we require ɛ to satisfy: ɛ < min{ 1 K, 1 3 ( 1)}. (3) As for higher-dimension situation, we just change 1 K to dist(k, ), etc. Fortunately, for any bounded oen set and a comact subset K (assuming the toology is Hausdorff toology), we can always find a oen set 1 that is re-comact in and satisfies K 1 2. And then we choose ɛ as in the inequality (3), then we can easily roof Theorem 1. 2 See Aendix for details. 2

3 Figure 2: mollifier smooth everywhere Last but not the least, the coefficient 1 3 in inequality (3) is just for simlicity. According to the analysis above, in fact any ositive coefficient that least than 1 2 is OK. 3 Aroximate continuous function on comact subset Only if is oen, we can aroximate a continuous function using its mollified version on any comact subset. In fact, the aroximation is uniformly convergence. Theorem 2. Suose R n is oen and g C(), then for any comact subset K, g ɛ g (ɛ 0). Proof. Suose K is a comact subset of. Because g is continuous, so δ > 0, ɛ x > 0, corresonding to x K, such that f(y) f(x) < δ, y B(x, ɛ x ). The family of sets {B(x, ɛ x )} x K is a oen cover of K. K is comact. Therefore there are {B(x i, ɛ xi )} N i1 such that it is a oen cover of K. Denote ɛ min{ɛ x1,, ɛ xn, 1 2 dist(k, )} 3. Now, x K, according to (1), we have 4 : f ɛ (x) f(x) η ɛ (x)[f(x y) f(x)] dx η ɛ (x) f(x y) f(x) dx δ η ɛ (x) dx δ. And of course it holds for all ɛ : 0 < ɛ < ɛ. Therefore f ɛ f in K. 4 C 0 () is dense in L () In this section, we roof that C 0 () is dense in L (), rovided is oen, no matter whether is bounded or not. Theorem 3 (Lusin s Theorem [2] ). Suose E R n is measurable and has finite measure, let function f : E R be measurable and finite almost everywhere. Then for any δ > 0, there exists closed subset F δ E, such that m(e F δ ) < δ and f restricted to F δ is continuous. Theorem 4 (Tietze s Theorem [3] ). Suose F R n is closed, function f : F R restricted to F is continuous. Then there exists function g : R n R, such that g is continuous on R n and f(x) g(x), x F. Furthermore, g can be chosen such that su x R n g(x) su x F f(x). From Theorem 3 and Theorem 4, we can deduce the following lemma. 3 dist(k, ) is strictly larger than zero. See the Aendix for details 4 Note that let x K 2ɛ is to ensure x y ɛ in the following integration. 3

4 Lemma 5. Suose E R n is measurable and has finite measure, let function f : E R be measurable and finite almost everywhere. Then for any δ > 0, there exists closed subset F δ E, and function g : R n R, such that g is continuous on R n and f(x) g(x), x F δ. Furthermore, g can be chosen such that su x R n g(x) su x F f(x). Now we resent the main theorem of this section. Theorem 6. Let be any oen subset (not necessarily bounded) of R n ( can also be R n ), then C 0 () is dense in L () for 1 < +. Proof. Let f L () and we extend f as f(x) 0, x /. Then 0 m( f > m) 1 dx 1 M f dx 1 R M f L () 0 (M + ). n m( f >m) Therefore lim M + m( f > M) 0. lim M + { f >M} f dx 0. So, According to the absolute continuity of Lebesgue integral 5, we have ɛ > 0, M ɛ > 0 s.t. M >M ɛ, { f >M} f dx < ɛ. (4) Denote M, if f(x) > M {f} M (x) M, if f(x) < M f(x), otherwise. (5) Then according to (4), f {f} M dx f {f} M dx f {f} M dx R n { f >M} f dx < ɛ. (6) { f >M} {f} R n M dx f dx < +. K R n ɛ > 0 s.t. K > K ɛ, we have: {f} M dx {f} M dx {f} M R n E K R dx < ɛ, (7) n E K where E K {x R n x < K}. (8) According to Lemma 5, δ > 0, there exists closed subset F δ E K and g C(R n ) such that m(e k F δ ) < δ, {f} M (x) g(x) x F δ and su x R n g(x) M. Let s choose δ be less than ɛ/(2m) : m(e K F δ ) < δ < ɛ/(2m). (9) According to Theorem 1, η > 0, w C0 (E K ) such that w 1 in EK η and 0 w(x) 1. Let { g(x)w(x), if x E K, g(x) 0, if x R n E K. Then g C 0 () (in R n, bounded closed set is comact). Notice that E K is a ball in R n, it has finite volumn, so we can find small enough η such that 5 lim A 0 A f(x) dx 0, rovided f is integrable in E and A E. m(e K E K η ) < ɛ/(2m). (10) 4

5 Therefore, {f} M g dx {f} M g dx E K E K {f} M g dx + E K η (E K E K η ) {f} M g dx {f} M g dx + E K η (E K E K η ) (2M) dx {f} M g dx + (2M) m(e K E K η ) E K {f} M g dx + (2M) m(e K E K η ) E K F δ (2M) m(e K F δ ) + (2M) m(e K E K η ) (refer to (9) and (10)) < 2 ɛ. (11) Now, we have the following control: f g dx 2 (f {f} M ) + ({f} M g) dx f {f} M dx + 2 {f} M g dx {f} M g dx R ( n (refer to (6)) 2 ɛ + 2 ) 2 ɛ + 2 {f} M g dx + {f} M g dx E K R n E ( ) K (refer to (11)) < 2 ɛ ɛ + {f} M dx R n E K (refer to (7)) < 2 ɛ + 2 (2 ɛ + ɛ) 4 2 ɛ So far, we know there exists {g n } + n1 C 0() such that f g n L () 0 (n + ). This means C 0 () is dense in L (). 5 f ɛ f in L () In this section, we show that any f L () can be aroximate by its mollified version f ɛ. To make the conclusion, we need Theorem 2 and Theorem 6. With the hel of some g C 0 () which is close enough to f L (), we can roof the following theorem. Theorem 7. Suose R n is oen, f L () and 1 < +, (1) if R n, then f ɛ is well-defined in R n for all ɛ > 0, and f ɛ f (ɛ 0) in R n, w.r.t the L norm; (2) if R n, then f ɛ is well-defined in ɛ, and a > 0, f ɛ f (ɛ 0) in a ɛ, w.r.t the L norm; or, (3) if R n, we do zero extend to f ɛ outside ɛ, then f ɛ is well-defined in, and f ɛ f (ɛ 0) in, w.r.t the L norm. Proof. Fixed f L (), for any δ > 0, according to Theorem 6, there exists g C 0 () such that g f, < δ/3. Secondly, Let su g be the comact subset K in Theorem 2, then according to Theorem 2, g ɛ g in su g, when ɛ be small enough. This means g ɛ g, < δ/3, when ɛ be small enough. According to Definition 2, the domain of g ɛ is ɛ {x dist(x, ) > ɛ}. However, this g has comact suort, therefore we can safely do zero extend to g ɛ in \ ɛ. Thus, g ɛ is well-defined within the whole. As to f ɛ, if is not R n, we do zero extend to f ɛ outside ɛ. Now f ɛ is also well-defined within the whole. 5

6 Thirdly, after zero extension, f ɛ g ɛ, is controlled by f g, : f ɛ g ɛ, (f g) ɛ (x) dx η ɛ (y)(f g)(x y) dy dx ( η ɛ (y) (f g)(x y) dy) dx ( ( ) 1 ( q 1) (η 1 q q ɛ (y)) q dy 1 η ɛ (y) (f g)(x y) dx dy η ɛ (y) η ɛ (y) f g,. (f g)(x y) dx dy (f g)(x) dx dy η ɛ (y) f g, dy (η 1 ɛ (y)) (f g)(x y) dy So f ɛ g ɛ, f g,, if we do the zero extension. Or, following the same deduction, we have f ɛ g ɛ,ɛ f g,, if we do not do the zero extension. Now, f ɛ f, f ɛ g ɛ, + g ɛ g, + g f, < δ. This means f ɛ f (ɛ 0) in, if we do the zero extend. Or, a > 0, f ɛ f (ɛ 0) in a, if we do not do the zero extension. ) 1 dx 6 Aendix We can always find a oen subset 1 that is re-comact in and satisfies K 1. This is due to the following reason. Because K, K is closed and is oen, dist(k, ) is greater than 0. If not so, there are {x n } + n1 K and {y n } + n1 s.t. x n y n 0 (n + ). K is comact there is convergent subsequence {x nj } + j1 such that x nj x, where x K. Then x y nj 0 (j + ) x. Boundary is always closed, therefore x. However, is oen so and x /. This contradicts with x K. Therefore, dist(k, ) > 0. Denote δ dist(k, ) > 0, and let 1 {x dist(x, ) δ/2}. Then obviously 1 is comact in, and contains K. References [1] Lawrence C. Evans. Partial Differential Equations: Second Edition. AMS, [2] htts://en.wikiedia.org/wiki/lusin%27s theorem. [3] htts://en.wikiedia.org/wiki/tietze extension theorem. 6