Prove the result on a dense subset and then prove that the set where the result holds is closed

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Example 1: Weierstrass approximation

The Weierstrass approximation theorem states that any continuous function on a closed, bounded interval can be uniformly approximated arbitrarily well by a polynomial. There are different ways of establishing this, but a proof due to Lebesgue of this theorem makes use of the technique being discussed. In this setting, the technique tells us to try to find a particular case of a continuous function that we can approximate arbitrarily well, and uniformly, by a polynomial, and then to try to use this to deduce the full result. An obvious choice of a continuous function that can be approximated well by a polynomial is the function $f$ given by $f(x) = x$ or, indeed, any polynomial. However, this does not help us reduce the problem at all; we cannot carry through the second part of the technique. We therefore need to think harder about what would make a suitable choice for the 'special case' $f$ . We want to be able to

establish the theorem for $f$ and
establish the general case using the theorem for $f$ .

One can thus go about trying to find $f$ in two ways; one can either look at which $f$ one can establish the theorem for, or one can look at which $f$ make it easier to establish the full result. If one follows the second route, then it seems reasonable to try to uniformly approximate an arbitrary continuous function on a closed, bounded interval by straight-line segments. It is fairly easy to see that one can do this, and our task is therefore reduced to trying to approximate piecewise linear functions on the closed, bounded interval; if one can prove the theorem for these special functions, then one can establish it for all functions by an $\epsilon/2$ argument, this being the transitivity part of the technique. Now, what is the problem with approximating piecewise linear functions by polynomials? Well, it is the sharp 'bends'; other than these, we just have the linear polynomial $f(x) = cx$ to approximate. This quickly leads one to consider approximating the simplest example of a piecewise linear function with a sharp bend: the absolute value function $f$ with $f(x) = |x|$ on the interval $[-1,1]$ . Once one has thought of this as the special case for which to establish the result, one arrives at Lebesgue's proof of the Weierstrass approximation theorem. To complete the proof the theorem, we need to show that piecewise linear functions can be approximated well by combinations of the absolute value function (and possibly polynomials) and that we can approximate the absolute value function well using polynomials. The former property follows quickly after one thinks to use the function $x + |x|$ and its translates and multiples, and the latter follows quickly if one uses the idea that $|x| = \sqrt{x^2}$ and that one can approximate $\sqrt{1 - u}$ well by its Taylor expansion.

General discussion

One could regard the above argument as having two stages. The first, which fits naturally into the framework of using generators and closure properties, is a proof that if $|x|$ can be approximated, then so can every piecewise linear function. The second is the deduction of the result from the fact that piecewise linear functions are dense in the uniform norm, and the set of all functions that can be uniformly approximated is closed in the uniform norm.

Example 2: Equidistribution

This section is in need of attention. This example was written up in a bit of a rush and could use some clarification.

A sequence $(a_j)$ of real numbers between 0 and 1 is said to be equidistributed if one eventually finds the 'expected' proportion of the numbers lying in any subinterval $[a,b]$ of $[0,1]$ ; that is, if

$\frac{|\{a_1, a_2, \dots, a_n\} \cap [a,b]|}{n} \to b-a$

as $n \to \infty$ for any intervals $[a,b] \subseteq [0,1].$ Let us say that the sequence $(a_j)$ is uniformly distributed in $[0,1]$ if for any integrable function $[0,1] \to \C$ one has that the averages of $f$ over the initial points $a_1, \dots, a_n$ gives a fair approximation to the integral of $f$ over the whole interval $[0,1]$ :

$\frac{1}{n} \sum_{j=1}^n f(a_j) \to \int_0^1 f(x) dx$

as $n \to \infty$ . We wish to establish that these two definitions are actually equivalent; a sequence is equidistributed if and only if it is uniformly distributed.

One may use the idea in this article to establish that equidistribution implies uniform distribution, though in this application one makes use of a key tool about equidistribution known as Weyl's criterion. This gives a test for equidistribution in terms of trigonometric sums: it says that the sequence $(a_j)$ is equidistributed if and only if

$\frac{1}{n}\sum_{j=1}^n e^{2\pi i a_j r} \to 0$

as $n \to \infty$ for any non-zero integer $r$ . Given this criterion for equidistribution, one does not have to look hard for the special cases for which one needs to show that equidistribution gives uniform distribution: since $\int e^{2\pi i r x} = 0$ for $r$ non-zero, Weyl's criterion establishes uniform distribution for the linear exponentials $x \mapsto e^{2\pi i r x}.$ We would therefore be done if we could approximate an arbitrary integrable function $f$ by linear exponentials. This should sound quite similar to Example 1, and, indeed, there is an approximation theorem similar to that of Weierstrass approximation that works with trigonometric polynomials $\sum_{r=-k}^k f_r e^{2\pi i r x}$ instead of the usual polynomials. Both of these approximation theorems are special cases of the Stone–Weierstrass theorem; the trigonometric polynomial version also follows directly from Fejér's theorem. Using this, the equivalence of equidistribution and uniform distribution follows quickly. We could add some more detail here, perhaps. Also, it should be noted that the terms equidistribution and uniform distribution do not always have the specific definitions given to them in this example.