a) Suppose that a < b and n ? N is even. If f is continuous on [a, b] and b a f (x)xndx = 0, prove that f (x) = 0 for at least one x ? [a, b]. b) Show that part a) might not be true if n is odd. c) Prove that part a) does hold for odd n when a ?0. Solution f is uniformly continuous if, for every e > 0, there exists d > 0 such that for all x, y in [a,b] such that |x-y| < d, then |f(x)-f(y)| < e. The hint suggests you try to prove the contrapositive: if f is not uniformly continuous, then f is not continuous. So suppose f is not uniformly continuous. Then there exists e > 0 such that, for all d > 0, there exist x, y in [a,b] such that |x-y| < d and |f(x)-f(y)| ? e. Let d = 1/n, n a natural number. Then there exist x_n, y_n in [a,b] such that |x_n-y_n| < 1/n and |f(x_n)-f(y_n)| ? e. Since [a,b] is compact, we know that x_n has a convergent subsequence. (Or, using less advanced concepts, any sequence of reals contains a monotonic subsequence; since our sequence is in [a,b], it\'s bounded above and below, so the monotone sequence theorem says that the subsequence must converge.) Let\'s replace x_n by this subsequence, and let\'s replace y_n by the the subsequence obtained by taking the same indices as the subsequence of x_n. Say x_n converges to p. Then it\'s easy to see that y_n also converges to p. (Show this.) Now if f were continuous, then since x_n and y_n converge to p, we would have f(x_n) and f(y_n) converge to f(p). But f(x_n) and f(y_n) cannot converge to the same point, since |f(x_n)- f(y_n)| ? e and so, if they both do converge to some values a and b, we must have |a-b| ? e. (Make this more explicit.) So we cannot have that both f(x_n) and f(y_n) converge to f(p), and so f is not continuous..

1. a) Suppose that a < b and n ? N is even. If f is continuous on [a, b] and b
a f (x)xndx = 0, prove that f (x) = 0 for at least one x ? [a, b].
b) Show that part a) might not be true if n is odd.
c) Prove that part a) does hold for odd n when a ?0.
Solution
f is uniformly continuous if, for every e > 0, there exists d > 0 such that for all x, y in [a,b] such
that |x-y| < d, then |f(x)-f(y)| < e. The hint suggests you try to prove the contrapositive: if f is not
uniformly continuous, then f is not continuous.
So suppose f is not uniformly continuous. Then there exists e > 0 such that, for all d > 0, there
exist x, y in [a,b] such that |x-y| < d and |f(x)-f(y)| ? e. Let d = 1/n, n a natural number. Then there
exist x_n, y_n in [a,b] such that |x_n-y_n| < 1/n and |f(x_n)-f(y_n)| ? e.
Since [a,b] is compact, we know that x_n has a convergent subsequence. (Or, using less
advanced concepts, any sequence of reals contains a monotonic subsequence; since our sequence
is in [a,b], it's bounded above and below, so the monotone sequence theorem says that the
subsequence must converge.) Let's replace x_n by this subsequence, and let's replace y_n by the
the subsequence obtained by taking the same indices as the subsequence of x_n. Say x_n
converges to p. Then it's easy to see that y_n also converges to p. (Show this.)
Now if f were continuous, then since x_n and y_n converge to p, we would have f(x_n) and
f(y_n) converge to f(p). But f(x_n) and f(y_n) cannot converge to the same point, since |f(x_n)-
f(y_n)| ? e and so, if they both do converge to some values a and b, we must have |a-b| ? e. (Make
this more explicit.) So we cannot have that both f(x_n) and f(y_n) converge to f(p), and so f is
not continuous.