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

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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 ) .

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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 .
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