Posted by: holdenlee | April 26, 2010

B7s: April 17, 2010


B7s is the math problem-solving club at MIT. We meet every Saturday 8-10PM at Random Hall to work on a hard problem, such as one from the Miklos Schweitzer competition.

Here’s the problem from Sat. 4/17.

[MS 1964/9] Let E be the set of functions from [0,1] to \mathbb{R}. Prove that there does not exist a topology on E such that f_n converges to f on the topology if and only if f_n converges to f a.e. (almost everywhere).

Explanation: Recall that a sequence x_n converges to x in a topology X if for every open set N containing x, all but finitely many points of x_n are in N. A sequence of functions f_n converges to f almost everywhere if it converges pointwise except on a set of measure 0. (Measure \mu is like the size of a set, a generalization of the length of an interval. For this problem, we only need to know measure is countably subadditive- \mu(\bigcup_{i\in \mathbb{N}}A_i)\leq \sum_{i\in \mathbb{N}} \mu(A_i)– and the measure of an interval is the length of the interval.)

Solution: It seems like the measure of the set where two functions f,g differ could act like a “distance” between f and g… so let \Delta(f,g)=\mu\{x|f(x)\neq g(x)\}. Inspired by this idea, we formulate a condition on a set N that we hope will be satisfied by open sets.

Condition: If f\in N then there exists \epsilon so that N contains all functions differing from f on a set of measure \epsilon, i.e. if \Delta(f,g)<\epsilon then g\in N.

We show that if “f_n converges to f almost everywhere implies f_n converges to f in the topology” is true, then every open set N satisfies the condition. Then we show that if “f_n converges to f in the topology implies f_n converges to f almost everywhere” is true, then some open set N does not satisfy the condition. Thus both statements cannot simultaneously hold, and the problem will be solved.

For the first part, let N be an open set. Suppose by way of contradiction the condition fails for f\in N. Then for all i\in \mathbb{N} we can choose f_i\not\in N so that \Delta(f,f_i)<2^{-i}. By choosing f_i \not\in N, we ensure that f_i does not converge to f in the topology. We claim that f_i converges to f almost everywhere. Indeed, for every \epsilon>0 we can find n so that \sum_{i=n}^{\infty} 2^{-i}<\epsilon. Then the measure of the set of x such that f_n(x) does not converge a.e. to f(x) is at most the measure of the set of x such that f_i(x)\neq f(x) for some i\geq n, which is

\displaystyle \mu\left[\bigcup_{i=n}^{\infty} \{x|f(x)\neq f_i(x)\} \right] \leq \sum_{i=n}^{\infty}\mu( \{x|f(x)\neq f_i(x)\} )<\sum_{i=n}^{\infty} 2^{-i}<\epsilon.

Hence f_n converges to f on a set of measure 1-\epsilon for all \epsilon>0; it must converge to f on a set of measure 1. In summary, f_n converges to f almost everywhere, but does not converge to f in the topology, violating the hypothesis.

For the second part, consider the sequence f_1,f_2,f_3,\ldots defined by

\displaystyle \chi_{[0,1]},\chi_{[0,\frac{1}{2}]},\chi_{[\frac{1}{2},1]},\chi_{[0,\frac{1}{4}]},\chi_{[\frac{1}{4},\frac{1}{2}]},\chi_{[\frac{1}{2},\frac{3}{4}]},\chi_{[\frac{3}{4},1]},\ldots,

where \chi_A denotes the characteristic function of the set A– it equals 1 on A and 0 elsewhere. If the condition holds for all open sets, then any open set containing f\equiv 0 must contain all functions g such that \Delta(f,g)<\epsilon, for some \epsilon>0. Since \Delta(f,f_n) converges to 0 (the intervals are getting shorter), f_n converges to the zero function in the topology. But every point x\in [0,1] is in an infinite number of intervals of the form [\frac{k}{2^n},\frac{k+1}{2^n}]; k,n\in \mathbb{N}_0, as well as absent from an infinite number of those intervals, so f_n(x) does not converge pointwise to f(x). Thus f_n converges to f nowhere! This contradicts the hypothesis.

Here’s the problem Sat. 4/24.

Let X be a closed subset of \mathbb{R}^n. Prove that there exists a function f:\mathbb{R}^n\to\mathbb{R} that is zero on X, positive outside X, and whose partial derivatives all exist.

(I wasn’t able to attend the meeting, and I haven’t solved this yet.)

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