MT2002 Analysis

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(Exercises 9)

Exercises 8

  1. If P and Q are any two points in R2, prove that d1(P, Q) gte d2(P, Q) gte dinfinity(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 dinfinity.
    Prove also that d1(P, Q) lte sqrt2 d2(P, Q) and d2(P, Q) lte sqrt2 dinfinity(P, Q).
    Deduce that if a sequence (a1, a2, ...) in R2 is convergent in one of the metrics d1, d2, dinfinity then it is convergent in all of them.

    Solution to question 1

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


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

    Prove that the sequence (fn) 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 fn on the interval [0, 1].
    What is the limit of this maximum value as n rarrow infinity? (Use Exercises 6, question 7.)
    Deduce that (fn) does not converge to the 0-function in the norm dinfinity.

    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

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JOC September 2002