18.100B Problem Set 9

Due in class Wednesday, May 8. You may discuss the problems with other students, but you should write solutions entirely on your own.

1. Find a sequence {f_{n}} of non-negative continuous functions on [0, 1] such that

lim_{n→∞} f_{n}(x) = 0, all x ∈ [0, 1],

and also

∫_{0}^{1} f_{n}(x)dx = 1.

(So for this sequence, you can’t interchange integration and the limit.)

2. Suppose {f_{n}} is a sequence of functions on [0, 1] with the following properties:

a) f_{n} is continuous on [0, 1];

b) f_{n} is differentiable on (0, 1);

c) there is a constant M (independent of n) so that |f′_{n}| ≤ M; and

d) f_{n} converges pointwise on [0, 1] to a limit function f. Prove that the sequence must converge uniformly (so that the limit function

is necessarily continuous. (Hint: by the Mean Value Theorem, the functions f_{n} cannot change by more than hM on an interval of length h. So break [0, 1] into a finite number of intervals each of length no more than (say) ϵ/M. By choosing n large enough, you can make f_{n} close to f at all the endpoints of your intervals (since there are only finitely many of those). Now deduce that f_{n} has to be close to f everywhere.)

3. (This problem is a version of the one in the text, page 169, number 21. The point of it is to show the need for one of the hypotheses in Theorem 7.33; but you don’t need to know about that theorem to do the problem. For a hint, see the problem in the text.) Let

A = functions f(t) = ∑_{n=0}^{N} c_{n}e^{int}, with N a non-negative integer and c_{n} ∈ ℂ.

This is an algebra of continuous complex-valued functions on [0, 2π], all satisfying f(0) = f(2π). Prove that there is a continuous complex-valued function g on [0, 2π] such that g(0) = g(2π), but nevertheless

sup_{x∈[0,2π]} |g(x) − f(x)| ≥ 1, f ∈ A.

4. Text, page 170, number 24.