Verified answer

a) Let T be the period of the function f(x).

To prove f(x) is bounded, it suffices to prove it is bounded over [0, T], since the function repeats itself over all other intervals. But this is immediate, since all continuous real-valued functions over a closed interval [0, T] are bounded.

A short proof that it is uniformly continuous is to just note that it is uniformly continuous over the interval [-1, T+1], and so by periodicity it must be uniformly continuous everywhere. *The longer proof is provided below if you are interested.

b) The function is a composition of continuous functions, and hence is continuous. To show it is not uniformly continuous, just take a derivative of f(x):

f'(x) = 2xcos(x^2).

Evaluate this at times sqrt{2*pi*k} for large k. Its derivative goes to infinity at these points as k goes to infinity. So, for example, fix epsilon = 0.01. We can find arbitrarily small values delta_k such that |f(sqrt{2*pi*k}) - f(sqrt{2*pi*k} + delta_k)| > 0.01.

c) *In more detail for part 1:

Longer proof that the function in part 1 is uniformly continuous:

Consider the function over the closed interval [-1, T+1] (extending by 1 unit to the left and right of the period of the function). Since it is continuous, it is uniformly continuous over this closed interval. Fix epsilon, and define delta(epsilon) as the value such that if x, y are in the interval [-1, T+1] and |x-y| <= delta(epsilon), then |f(x)-f(y)|<= epsilon (such a delta(epsilon)) exists by uniform continuity over that interval).

Define DELTA(epsilon) = min[delta(epsilon), 1]. We now show that this value DELTA(epsilon) works for all real numbers x, y, not just those in the interval [-1, T+1].

Fix epsilon>0. We now show that if x, y are real numbers and |x-y| <= DELTA(epsilon), then |f(x)-f(y)|<= epsilon. Note that since |x-y|<= DELTA(epsilon), they are within a distance of 1 from each other.

Now define k as the integer such that x-kT is in the interval [0, T]. Then clearly x-kT and y-kT are in the interval [-1, T+1], and they are within delta(epsilon) of each other. Thus:

|f(x) - f(y)|

= |f(x-kT) - f(y-kT)|

<= epsilon

Thus, f(x) is uniformly continuous over the entire real number line (not just over [-1, T+1]).

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