Exercise 5.4.8

Review the argument for the nondifferentiability of g ( x ) at nondyadic points. Does the argument still work if we replace g ( x ) with the summation n = 0 ( 1 2 n ) h n ( 3 n x ) ? Does the argument work for the function n = 0 ( 1 3 n ) h n ( 2 n x ) ?

Answers

The critical part to showing that g ( x ) is not differentiable at nondyadic points was showing that

h n + 1 ( y n + 1 ) h n + 1 ( x n + 1 ) y n + 1 x n + 1

does not converge to zero, preventing the limit defining the derivative to exist. For the case n = 0 ( 1 2 n ) h n ( 3 n x ) , the above term would diverge to infinity, since y n x n would decrease by a factor of 3 on each iteration while h n ( y n ) h n ( x n ) would only decrease by a factor of 2. For similar reasons, in the case of n = 0 ( 1 3 n ) h n ( 2 n x ) , the above term would converge to 0, and the argument is no longer valid.

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2022-01-27 00:00
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