Exercises 8

1. If P and Q are any two points in R², prove that d₁(P, Q) d₂(P, Q) d(P, Q).
   Hence prove that if a sequence (a₁, a₂, ...) in R² is convergent in the metric d₁ then it is also convergent in d₂ and d.
   Prove also that d₁(P, Q) 2 d₂(P, Q) and d₂(P, Q) 2 d(P, Q).
   Deduce that if a sequence (a₁, a₂, ...) in R² is convergent in one of the metrics d₁, d₂, d then it is convergent in all of them.

   Solution to question 1
   
   

2. Define the function f_n(x) = nxⁿ(1 - x).
   

   
   The graphs of f_n for n = 2, 4, 6 are shown on the right.
   
   

   Prove that the sequence (f_n) converges to the 0-function in the metric d₁ on the space C[0, 1] of continuous functions on [0, 1].
   Use the usual method for finding the turning point of a differentiable function to find the maximum value of f_n on the interval [0, 1].
   What is the limit of this maximum value as n ? (Use Exercises 6, question 7.)
   Deduce that (f_n) does not converge to the 0-function in the norm d.
   
   

   Solution to question 2

3. Which of the following real-valued functions on the open interval (0, 1) are continuous?

   a) Define a function f on a real-number x by taking the decimal expansion of x (terminating in infinitely many 0's rather than infinitely many 9's if it is an exact decimal) and discarding the first, third, fifth and so on, decimal places.
   So, for example, f( 0.1234) = 0.24, f(0.1415926536...) = 0.45256... .

   b) Define a function g on a real-number x by taking the decimal expansion of x and replacing 0's by 1's, replacing 1's by 2's and so on except that 9's are replaced by 0's.
   So, for example, g(0.1298) = 0.23091111... (since the infinitely many 0's at the end all get replaced), g(0.1415926536...) = 0.1526037647... .

   c) Define a function h on a real-number x by taking the decimal expansion of x and replacing 0's by 9's, replacing 1's by 8's , 2's by 7's, 3's by 6's, 4's by 5's and vice-versa.
   So, for example, h(0.1298) = 0.9701999999... = 0.9702 (since the infinitely many 0's at the end all get replaced), h(0.1415926536...) = 0.9594073463... .

   [Hint: Observe that the sequence (0.49, 0.499, 0.4999, ...) converges to 0.5 and use the sequential definition of convergence.]

   Solution to question 3