Sequence of functions?

Question

Let {f_n} be a sequence of continuous functions defined on [0,1] such that f_n(x) -> 0 as n -> inf. for all x in [0,1]. Show that for each positive number e, there is a subinterval [a,b] of [0,1] and an integer N such that |f_n(x)|N and x in [a,b].

Answer 1

Well, if we are allowed to have a=b, then this is trivial. Just take a=b=0, and use the pointwise convergence of f.

If, on the other hand, we must take a0. Define the following sets: let S_n = {x∈[0, 1]: ∃m>n s.t. |f_m(x)| > ε}. I claim that the S_n are all open sets, the intersection of all the S_n is the empty set, and finally that at least one of the S_n is not dense in (0, 1).

First, we prove that each of the S_n is open. Suppose x∈S_n, then ∃m>n s.t. |f_m(x)| > ε. Since f_m is continuous, ∃δ>0 s.t. |y-x| < δ ⇒ |f_m(y) - f_m(x)| < |f_m(x)| - ε. Suppose |y-x|<δ, then by the second form of the triangle inequality, |f_m(x)| - |f_m(y)| ≤ |f_m(y) - f_m(x)| < |f_m(x)| - ε, so |f_m(y)| > ε, thus y∈S_n. Since we can find such δ for any x∈S_n, it follows that all the S_n are open.

Next, we wish to show that the intersection of all of the S_n is empty. Let x∈[0, 1]. Then since f_n(x) → 0, we have that ∃n∈ℕ s.t. ∀m>N, |f_n(x)|<ε. But if x∈S_n, then ∃m>n s.t. |f_n(x)|>ε, a contradiction. Therefore, x∉S_n, and thus a fortiori is not in the intersection of all the S_n. Since this is true for arbitrary x∈[0, 1], it follows that the intersection of all the S_n is empty.

Finally, we note that at least one of the S_n is not dense in (0, 1), for if all the S_n were dense, then the empty set would be the intersection of countably many dense open sets in (0, 1), contradicting the Baire category theorem.

So now choose n so that S_n is not dense. Since S_n is not dense, there exists a point c∈(0, 1) and δ>0 such that |x-c|<δ ⇒ x∉S_n. Choose a and b in (0, 1) such that c-δ < a < c < b < c+δ. Then x∈[a, b] ⇒ |x-c| < δ ⇒ x∉S_n ⇒ ∀m>n, |f_m(x)| ≤ ε. So [a, b] and n are the required interval and integer.

Well, okay, technically we were supposed to have x∈[a, b] ⇒ ∀m>n, |f_m(x)| < ε (as opposed to |f_m(x)| ≤ ε). If it bothers you, go back and do the proof with ε/2 instead of ε and then at the end note that ε/2<ε.

Answer 2

Sounds like the definition of a continuous function; the old give me an epsilon and I'll give you a delta.