Course MT2002 Analysis

Solution 8


  1. Look at the picture:

    If a point Q is as shown, it is clear that d2(P, Q) lte r since "The diamond is inside the circle" and so d2(p, Q) lte d1(p, Q).
    Similarly, "The circle is inside the square and so dinfinity(P, Q) lte d2(P, Q).



    Alternatively, look at the picture:

    Then d2(P, Q) = c gte a + b = d1(P, Q).
    Also both alte c and b lte c implies dinfinity(P, Q) = max{a, b} lte c.


    Without a picture, one can argue
    (a + b)2 = a2 + b2 + 2ab gte a2 + b2 implies a + b gte sqrt(a2 + b2) implies d1gte d2.
    Also, if a gte b then max{a, b} = a and since a2+ b2gte a2we have sqrt(a2 + b2) gte a and so d2 gte dinfinity.

    If (an) rarrow alpha in d1, then given epsilon > 0 thereexists N such that n > N implies d1(an, alpha) < epsilon implies (since d1gte d2and dinfinity) that the sequence converges in d2 and dinfinity also.


    One can draw the pictures :

    to show that d1lte sqrt2 d2 and d2lte sqrt2 dinfinity.


    Alternatively, sqrt2 sqrt(a2 + b2) gte a + b iff 2(a2 + b2) gte a2 + 2ab + b2 iff a2 + 2ab + b2 gte 0 which is true since (a - b)2 gte 0. Hence sqrt2 d2 gte d1.

    If (an) rarrow alpha in dinfinity, then given epsilon > 0 thereexists N such that n > N implies dinfinity(an, alpha) < epsilon implies d2(an, alpha) < epsilon/sqrt2 and d2(an, alpha) < epsilon/2 and so the sequence converges in d1 and d2 also.

    Similarly, if convergence takes place in d2 it also takes place in d1. Combining this with the earlier result shows that sequences which converge in one of these metrics converge in them all.

  2. d1(fn, 0) = int01 nxn- nxn+1dx = [n/(n + 1) xn+1 - n/(n + 2) xn+2 eval01 = n/[(n + 1)(n + 2)] rarrow 0 as n rarrow infinity.
    Thus (fn) rarrow the 0-function.
    To find the maximum, solve d/dxfn(x)= 0 to get:
    n2xn-1 - n(n+1)xn = 0 implies x = n/(n + 1) or 0.
    The maximum is at n/(n + 1) and is thus [n/(n + 1)]n = [1 - 1/(n + !)]n+1 rarrow e-1as n rarrow infinity (by Exercises 6 Question 7).
    Thus dinfinity(fn, 0) does not converge to 0 in R as n rarrow infinity and so (fn) does not converge to the 0-function in dinfinity.

  3. (a)  The sequence (0.49, 0.499, 0.4999, ... ) rarrow 0.5 and applying f gives (0.9, 0.9, 0.99, 0.99, ... ) rarrow 1.0 while f(0.5) = 0. Hence f is not continuous at 0.5. In fact, a similar proof shows that f is not continuous at any terminating decimal -- though it is continuous at all other points.

    (b)  Apply g to the same sequence to give (0.50111... , 0.500111... , 0.5000111... , ... ) which converges to 0.5. However, g(0.5) = 0.6111... which is different and so g is discontinuous at 0.5.

    (c)  This map is actually the map h(x) = 1 - x as you can see most easily by writing 1 as 0.9999... . Hence it is continuous everywhere.