508F08Ex2solns

# Dy and seek a fixed point u of t clearly for all 0 a

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dy and seek a fixed point u of T . Clearly for all 0 < a 2 we have T : C ([0 , a ]) to C ([0 , a ]). We need only show that T is contracting for some a 2. But ( x ) - ( x ) = Z x 0 K ( x, y ) [ ϕ ( y ) - ψ ( y )] dy so | ( x ) - ( x ) | ≤ Ma k ϕ - ψ k . Pick a < 1 /M . Since the right side of the above inequality is independent of x [0 , a ], then with c = Ma , k - k ≤ c k ϕ - ψ k , so T is contracting. Thus the original integral equation has a unique solution. Remark. By modifying this reasoning we can even use a = 2. One approach to do this is to show that some power of T is contracting on [0 , 2].

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4 B–5. Let ϕ n ( t ) be a sequence of smooth real-valued functions with the properties ( a ) ϕ n ( t ) 0 , ( b ) ϕ n ( t ) = 0 for | t | ≥ 1 /n, ( c ) Z -∞ ϕ n ( t ) dt = 1 . Note: because of (b), this integral is only over - 1 /n t 1 /n . Assume f ( x ) is uniformly continuous for all x R and define f n ( x ) := Z -∞ f ( x - t ) ϕ n ( t ) dt. Show that f n ( x ) converges uniformly to f ( x ) for all x R . [ Suggestion: Use f ( x ) = f ( x ) R -∞ ϕ n ( t ) dt = R -∞ f ( x ) ϕ n ( t ) dt . Also, note explicitly where you use the uniform continuity of f ]. Remark: One can show that the approximations f n are also smooth. Thus, this proves that you can approximate a continuous function uniformly on any compact set by a smooth function. Solution Using the suggestion, f n ( x ) - f ( x ) = Z 1 /n - 1 /n [ f ( x - t ) - f ( x )] ϕ n ( t ) dt. Since f is uniformly continuous, given any > 0 there is a δ > 0 so that | f ( y ) - f ( x ) | < for any x , y that satisfy | y - x | < δ . Pick some N with 1 /N < δ . Then for any n N , if | t | ≤ 1 /n then | ( x - t ) - x | ≤ 1 /n < δ so | f ( x - t ) - f ( x ) | < for all x R . Consequently | f n ( x ) - f ( x ) | ≤ Z | t |≤ 1 /n ϕ n ( t ) dt = .
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