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- If
*P*and*Q*are any two points in**R**^{2}, prove that*d*_{1}(*P*,*Q*) ≥*d*_{2}(*P*,*Q*) ≥*d*_{∞}(*P*,*Q*).

Hence prove that if a sequence (*a*_{1},*a*_{2}, ...) in**R**^{2}is convergent in the metric*d*_{1}then it is also convergent in*d*_{2}and*d*_{∞}.

Prove also that*d*_{1}(*P*,*Q*) ≤ √2*d*_{2}(*P*,*Q*) and*d*_{2}(*P*,*Q*) ≤ √2*d*_{∞}(*P*,*Q*).

Deduce that if a sequence (*a*_{1},*a*_{2}, ...) in**R**^{2}is convergent in one of the metrics*d*_{1},*d*_{2},*d*_{∞}then it is convergent in all of them. - Define the function
*f*_{n}(*x*) =*nx*^{n}(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*d*_{1}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*_{∞}.

- 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.]

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