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330 HW 05

# 330 HW 05 - MAT 330 Fourier Series and Partial Differential...

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MAT 330: Fourier Series and Partial Diﬀerential Equations Professor Sergiu Klainerman Problem Set 5 Rik Sengupta March 30, 2010 1. Stein and Shakarchi, p. 162, problem 4 Bump functions. Examples of compactly supported functions in S ( R ) are very handy in many applications in analysis. Some examples are: (a) Suppose a < b , and f is the function such that f ( x ) = 0 if x a or x b and f ( x ) = e - 1 / ( x - a ) e - 1 / ( b - x ) if a < x < b. Show that f is indeﬁnitely diﬀerentiable on R . (b) Prove that there exists an indeﬁnitely diﬀerentiable function F on R such that F ( x ) = 0 if x a , F ( x ) = 1 is x b , and F is strictly increasing on [ a,b ]. (c) Let δ > 0 be so small that a + δ < b - δ . Show that there exists and indeﬁnitely diﬀerentiable function g such that g is 0 if x a or x b , g is 1 on [ a + δ,b - δ ], and g is strictly monotonic on [ a,a + δ ] and [ b - δ,b ]. Solution. We will solve the parts one by one. (a) First of all, note that f is clearly inﬁnitely diﬀerentiable at all points other than a and b , from the fact that the product of two inﬁnitely diﬀerentiable functions is also inﬁnitely diﬀerentiable (since d ( uv ) = udv + vdu ). We will only consider the point x = a , because for x = b the argument is completely symmetric and therefore identical (a fact that is evident from the expression for f ). Clearly, the term e - 1 / ( b - x ) will not give us any problems, because for all small enough neighborhoods we will not face diﬀerentiability issues. So consider the term e - 1 / ( x - a ) . The following proof is presented in its generality, though we would have been justiﬁed in considering a = 0 to simplify the proof, because of translation invariance. The derivative of e - 1 / ( x - a ) is e - 1 / ( x - a ) ( x - a ) 2 , and also, approaching this point from the left will always give us 0 from the deﬁnition, and so we need only consider the case of approaching from the right. Therefore, the problem reduces to proving lim x a + e - 1 / ( x - a ) ( x - a ) 2 = 0 . We will use a small but neat trick to prove this fact. Since 1 3!

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330 HW 05 - MAT 330 Fourier Series and Partial Differential...

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