7. Comments on Homework 3
≪ 6. Differentiating Power Series, Taylor Series | Table of Contents | 8. Convolutions ≫Homework 3 seemed to be significantly more difficult than the previous two assignments, judging by the scores alone. There are two problems I wanted to highlight here. First was problem 3, which was using the bounded version of Tietze’s extension theorem to prove its unbounded version. There was a very common mistake that signalled a misunderstanding of the statement of the bounded version of the theorem.
The second was problem 6, which involved generalising Stone-Weierstrass to non-compact domains. The most common mistake here was simply misunderstanding what “uniform convergence on compact subsets” meant, and I was asked to write up a solution for this problem.
Let’s start with problem 3, which states:
Problem 3.
Let \(F\) be a closed subset of a metric space \(X\), and let \(f: F\to \mathbb{R}\) be continuous. Show that there exists a continuous function \(h:X\to \mathbb{R}\) such that \(\left. h\right\rvert_{F}=f\).
For this, you were allowed to use the bounded version of Tietze’s extension theorem, which states:
Theorem 4. Tietze's extension theorem
Suppose \(F\) is a closed subset of a metric space \(X\) and \(f:F\to \mathbb{R}\) is a bounded continuous function. Then, there exists a continuous function \(h:X\to \mathbb{R}\) such that \(\left. h\right\rvert_{F}=f\) and \[\sup _{x\in X} \left\lvert h(x) \right\rvert = \sup _{x\in F} \left\lvert f(x) \right\rvert.\]
So, to solve problem 3, one needs to convert the problem into one about a bounded function, then somehow go backwards and “undo” this conversion. For this, we consider the function \(\arctan \left( f(x) \right) : F\to \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)\). There is a continuous extension of this \(h_0: X\to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]\) such that \(\left. h_0\right\rvert_{F}=\arctan\circ f\).
At this step, it is very tempting to take \(\tan\circ h_0\) and call it a day. However, when appealing to to Tietze's theorem, although we can select \(h_0\) to obey \[\sup _{x\in X}\left\lvert h_0(x) \right\rvert=\pi/2,\] we cannot a priori avoid the scenario where this supremum is attained. In other words, it is possible for \(h_0(x)=\pm \frac{\pi}{2}\)! Here, we need Urysohn’s lemma to help us out.
Lemma 5. Urysohn's lemma
Let \(X\) be a metric space. Let \(A, B\subseteq X\) be disjoint closed subsets. Then there exists a continuous function \(f:X\to \left[ 0,1 \right]\) such that \(f(a)=0\) for all \(a\in A\) and \(f(b)=1\) for all \(b\in B\).
Applying this to this problem, we let \(Z = h _{0}^{-1}\left( \left\lbrace \pm \frac{\pi}{2} \right\rbrace \right)\). By continuity, \(Z\) is closed. Moreover, since \(\arctan\left( f(x) \right)\in \left( -\frac{\pi}{2}, \frac{\pi }{2} \right)\) for all \(x\in F\), \(Z\) and \(F\) are disjoint. Hence, we apply Urysohn's lemma to construct a function \(\chi \) such that \(\chi = 0\) on \(Z\) and \(\chi = 1\) on \(F\).
Now consider the function \(\chi(x) \cdot h_0(x)\). This still lands in the interval \(\left[ - \frac{\pi}{2}, \frac{\pi }{2} \right]\); however, it cannot take the values \(\pm \frac{\pi }{2}\), for \(\chi \) is zero when \(h_0 = \pm \frac{\pi }{2}\). Moreover, when \(x\in F\), we have \(\chi (x) = \arctan \left( f(x) \right)\). We have effectively “zeroed out” the bad behaviour of our original extension.
Finally, taking \(h(x) =\tan \left( \chi (x) \cdot h_0(x) \right)\) gives a continuous extension of \(f\) defined on all of \(X\). \(\square\)
Problem 6.
Let \(\mathcal{A}\) be a subalgebra of \(C\left( \mathbb{R}^n, \mathbb{R} \right)\) that contains the constants and separates points. Show that for any \(f\in C\left( \mathbb{R}^n, \mathbb{R} \right)\), there exists a sequence \(\left\lbrace f_k \right\rbrace\subseteq \mathcal{A}\) such that \(f_k\to f\) uniformly on compact subsets of \(\mathbb{R}^n\).
I have to pat myself on the back for lining up the numbering like this.
As remarked at the beginning, the main difficulty here is getting past the “uniformly on compact subsets”. This means that for any compact \(K\subseteq \mathbb{R}^n\), the restrictions \(\left. f_k\right\rvert_{K}\to \left. f\right\rvert_{K}\) uniformly. Importantly, \(f_k\) is not allowed to depend on this \(K\).
If it were allowed to depend on \(K\), then Stone-Weierstrass seals the deal. However, we need to be a bit more delicate. We also can’t directly apply Stone-Weierstrass, for \(\mathbb{R}^n\) is not compact.
Instead, we need to use a diagonal-style argument. Define for all positive integers \(N\) the sets \[B_N = \left\lbrace x\in \mathbb{R}^n : \left\lVert x \right\rVert \leq N \right\rbrace.\] In words, these are the balls of radius \(N\). They are closed and bounded, hence compact. One can check that for any \(N\), the sets \[\mathcal{A}_N = \left\lbrace \left. g\right\rvert_{B_N} : g\in \mathcal{A} \right\rbrace\] is a subalgebra of \(C\left( B_N, \mathbb{R} \right)\) that contains all constants and separates points. Now that we are in a compact setting, we can apply Stone-Weierstrass.
For every \(N\), there exists some \(f_N \in \mathcal{A}\) such that \[\sup _{x\in B_N} : \left\lvert f_N(x) - f(x) \right\rvert < \frac{1}{N}.\] Note that this is using the density of \(\mathcal{A}_N\) in \(C\left( B_N, \mathbb{R} \right)\). We now claim that the sequence \(f_N\to f\) uniformly on compact subsets.
Let \(K\subseteq \mathbb{R}^n\) be compact; by Heine-Borel, \(K\) is bounded, hence there is some \(M\) such that \(K\subseteq B_M\). Then, for all \(N \geq M\), we have \(K\subseteq B_M\subseteq B_N\), so \[\sup _{x\in K}\left\lvert f_N(x) - f(x) \right\rvert \leq \sup _{x\in B_N} \left\lvert f_N(x) - f(x) \right\rvert < \frac{1}{N}.\] It follows that \(f_N\to f\) uniformly on \(K\). \(K\) was arbitrary, so we are done. \(\square\)