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.)