convolution

# convolution - Convolution Let f(x and g(x be continuous...

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Convolution Let f ( x ) and g(x) be continuous real-valued functions for x R and assume that f or g is zero outside some bounded set (this assumption can be relaxed a bit). Deﬁne the convolution ( f * g )( x ) : = Z - f ( x - y ) g ( y ) dy (1) Since f or g is zero outside a bounded set, this integral is only over a bounded interval. One simple yet useful observation is f * g = g * f . (2) To prove this make the change of variable t = x - y in the integral (1). One small observation. If f * g ( x ) = R b a f ( y ) g ( x - y ) dy , then only the values of g on the interval [ x - b , x - a ] are used. Thus if x [ a , b ] , then the convolution only involves the values of g on [ a - b , b - a ] . By a simple rescaling, we will often use b - a = 1. S MOOTHNESS OF f * g . Theorem 1 If f C 1 ( R ) then f * g C 1 ( R ) . Better yet, if f C k ( R ) and g C ` ( R ) , then f * g C k + ` ( R ) . PROOF This is clearer if we write h ( x ) = f * g ( x ) . Then h ( x ) - h ( x 0 ) x - x 0 = Z - f ( x - y ) - f ( x 0 - y ) x - x 0 g ( x ) dx . (3) We will be done if we can show that f ( x - y ) - f ( x 0 - y ) / ( x - x 0 ) converges uniformly to f 0 ( x 0 - y ) . To do this we use the integral form of the mean value theorem: f ( x - y ) - f ( x 0 - y ) = Z 1 0 d f ( x 0 - y + t ( x - x 0 )) dt dt = ± Z 1 0 f 0 ( x 0 - y + t ( x - x 0 )) dt ² ( x - x 0 ) . Then f ( x - y ) - f ( x 0 - y ) x - x 0 - f 0 ( x 0 - y ) = Z 1 0 [ f 0 ( x 0 - y + t ( x - x 0 )) - f 0 ( x 0 - y )] dt (4) Since f 0 is assumed continuous and is zero outside of a bounded set, it is uniformly continuous. Thus, given any ε > 0 there is a δ > 0 so that if | x - x 0 | < δ then | f 0 ( z + t ( x - x 0 )) - f 0 ( z ) | < ε for all values of z . In our case z = x 0 - y . Thus the left side of (4) tends to zero uniformly for all choices of x 0 and y

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convolution - Convolution Let f(x and g(x be continuous...

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