pde7 - P DE LECTURE NOTES M ATH 237A-B 83 7 Test Functions...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
PDE LECTURE NOTES, MATH 237A-B 83 7. Test Functions and Partitions of Unity 7.1. Convolution and Young’s Inequalities. Letting δ x denote the “delta— function” at x, we wish to de fi ne a product ( ) on functions on R n such that δ x δ y = δ x + y . Now formally any function f on R n is of the form f = Z R n f ( x ) δ x dx so we should have f g = Z R n × R n f ( x ) g ( y ) δ x δ y dxdy = Z R n × R n f ( x ) g ( y ) δ x + y dxdy = Z R n × R n f ( x y ) g ( y ) δ x dxdy = Z R n ·Z R n f ( x y ) g ( y ) dy ¸ δ x dx which suggests we make the following de fi nition. De fi nition 7.1. Let f, g : R n C be measurable functions. We de fi ne f g ( x ) = Z R n f ( x y ) g ( y ) dy whenever the integral is de fi ned, i.e. either f ( x · ) g ( · ) L 1 ( R n , m ) or f ( x · ) g ( · ) 0 . Notice that the condition that f ( x · ) g ( · ) L 1 ( R n , m ) is equivalent to writing | f | | g | ( x ) < . Notation 7.2. Given a multi-index α Z n + , let | α | = α 1 + · · · + α n , x α := n Y j =1 x α j j , and α x = μ ∂x α := n Y j =1 μ ∂x j α j . Remark 7.3 (The Signi fi cance of Convolution) . Suppose that L = P | α | k a α α is a constant coe cient di ff erential operator and suppose that we can solve (uniquely) the equation Lu = g in the form u ( x ) = Kg ( x ) := Z R n k ( x, y ) g ( y ) dy where k ( x, y ) is an “integral kernel.” (This is a natural sort of assumption since, in view of the fundamental theorem of calculus, integration is the inverse operation to di ff erentiation.) Since τ z L = z for all z R n , (this is another way to characterize constant coe cient di ff erential operators) and L 1 = K we should have τ z K = z . Writing out this equation then says Z R n k ( x z, y ) g ( y ) dy = ( Kg ) ( x z ) = τ z Kg ( x ) = ( z g ) ( x ) = Z R n k ( x, y ) g ( y z ) dy = Z R n k ( x, y + z ) g ( y ) dy. Since g is arbitrary we conclude that k ( x z, y ) = k ( x, y + z ) . Taking y = 0 then gives k ( x, z ) = k ( x z, 0) =: ρ ( x z ) .
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
84 BRUCE K. DRIVER We thus fi nd that Kg = ρ g. Hence we expect the convolution operation to appear naturally when solving constant coe cient partial di ff erential equations. More about this point later. The following proposition is an easy consequence of Minkowski’s inequality for integrals. Proposition 7.4. Suppose q [1 , ] , f L 1 and g L q , then f g ( x ) exists for almost every x, f g L q and k f g k p k f k 1 k g k p . For z R n and f : R n C , let τ z f : R n C be de fi ned by τ z f ( x ) = f ( x z ) . Proposition 7.5. Suppose that p [1 , ) , then τ z : L p L p is an isometric isomorphism and for f L p , z R n τ z f L p is continuous. Proof. The assertion that τ z : L p L p is an isometric isomorphism follows from translation invariance of Lebesgue measure and the fact that τ z τ z = id. For the continuity assertion, observe that k τ z f τ y f k p = k τ y ( τ z f τ y f ) k p = k τ z y f f k p from which it follows that it is enough to show τ z f f in L p as z 0 R n .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern